Engineering Mechanics: Statics

Statics is the study of bodies that are in equilibrium: at rest, or moving with constant velocity, so that the net force and net turning effect on them are zero. Every bridge, building, crane, and bolted joint that holds still is a statics problem. If you can find the forces inside a structure while it stands still, you can decide whether each member is strong enough. That is the entire practical purpose of this course.

Key idea: Statics studies bodies whose acceleration is zero, so both the resultant force and the resultant moment acting on them must be zero.

Why a whole course about things that do not move

It is fair to ask why a body that does nothing deserves a semester of study. The answer is that "nothing is happening" on the outside hides a great deal happening inside. A highway girder that has not moved in forty years is carrying tension along its bottom flange, compression along its top flange, and shear near its supports every second of those forty years. Statics is the tool that converts the visible loads - traffic, wind, snow, the structure's own weight - into the invisible internal forces the material must survive.

That conversion is also the gateway to everything that follows. Mechanics of materials takes the internal forces statics finds and turns them into stresses and deflections. Structural analysis extends the same equilibrium equations to systems too complicated for statics alone. Machine design uses statics to size pins, bolts, and bearings. Even fluid mechanics borrows the equilibrium idea to get the pressure in a tank. Get statics right and every later course has a foundation; get it wrong and every later number inherits the error.

Key idea: Equilibrium is not a claim that nothing is happening; it is the condition that lets you compute the internal forces that are happening.

What a force is

A force is a push or a pull. It is a vector, so describing it fully takes three pieces of information: its magnitude (how hard, measured in newtons), its direction (which way it points, including its line of action), and its point of application (where it acts on the body). Leave out any one of these and the force is not defined.

Forces come in two flavors that you must keep separate. A contact force requires touching: a cable pulling, a floor pushing, a pin bearing on a hole, friction dragging along a surface. A body force acts at a distance throughout the whole volume of the object, and in statics that means weight, the pull of gravity. Every free-body diagram you draw is a list of contact forces at the boundary plus the one body force at the center of gravity.

One further property matters for rigid bodies: the principle of transmissibility. A force may be slid anywhere along its own line of action without changing its effect on a rigid body. Pushing a filing cabinet from behind or pulling it from the front along the same line produces identical reactions at the floor. This principle is what lets us pick the most convenient point on a line of action when computing moments later. It governs the external effect on a rigid body only, not the internal forces and not a deformable body.

Key idea: A force is fully specified by magnitude, direction, and point of application, and on a rigid body it may be slid freely along its line of action.

The SI unit of force is the newton (N), where one newton is the force that gives a one-kilogram mass an acceleration of one meter per second squared. On Earth, an object of mass m is pulled down by a weight W = m g, where g = 9.81 m/s squared. A 10 kg toolbox therefore weighs about 10 x 9.81 = 98.1 N.

Mass, weight, and units

Mass and weight are not the same quantity, and confusing them is the most common unit error in a first statics course. Mass, in kilograms, measures how much matter an object contains and how strongly it resists being accelerated; it is the same on Earth, on the Moon, and in orbit. Weight, in newtons, is a force: the gravitational pull on that mass. The 10 kg toolbox has a mass of 10 kg everywhere, but it weighs 98.1 N on Earth and only about 16 N on the Moon.

Because loads in real structures are large, statics uses the kilonewton (kN) constantly: 1 kN = 1000 N, so the toolbox weighs 0.0981 kN while a small car weighs roughly 12 kN. Moments are reported in newton-meters (N*m) or kilonewton-meters (kN*m). In the United States you will also meet the pound-force (lbf), the kip (1 kip = 1000 lbf), and the slug. One pound-force equals 4.448 N, and one foot equals 0.3048 m exactly. The safest habit is to convert everything into a single system before you start, then forget about units until the final answer.

The value g = 9.81 m/s squared is a rounded form of the standard acceleration of gravity, fixed by convention at 9.80665 m/s squared, and the local variation from pole to equator is far below the uncertainty in any real load.

Key idea: Mass in kilograms is a property of the object; weight W = m g in newtons is a force, and the two must never be added or compared directly.

Newton's laws behind statics

Statics rests on Newton's three laws. The first law says a body stays at rest unless a net force acts on it, which is exactly the condition of equilibrium. The second law, F = m a, tells us that if the acceleration is zero, the resultant force must be zero. The third law says every action has an equal and opposite reaction. That last law is what lets us cut a structure apart in our minds: at any imaginary cut, the two pieces push on each other with equal and opposite forces, and that is how internal forces appear.

It pays to be precise about the third law, because it is routinely misapplied. Action and reaction always act on different bodies. The Earth pulls the toolbox down with 98.1 N; the toolbox pulls the Earth up with 98.1 N. Those two forces never appear on the same free-body diagram and therefore never cancel one another. What does appear on the toolbox's diagram is its weight, 98.1 N down, and the table's normal force, 98.1 N up - and those two are an equilibrium balance, not an action-reaction pair. Telling the two situations apart is a skill worth practising from the first day.

Key idea: Action and reaction act on different bodies and never cancel; the forces that cancel on a free-body diagram are the ones acting on the same body.

The two equilibrium ideas

Because acceleration is zero in statics, two conditions must hold for any body:

• The forces must balance, so the vector sum of all forces is zero. We write this as sum Fx = 0 and sum Fy = 0 (and sum Fz = 0 in three dimensions).
• The turning effects must balance, so the sum of all moments about any point is zero, written sum M = 0. Moments arrive in Module 3.

A body treated as a single point, with no size, is a particle; for a particle only the force condition applies, because a point cannot rotate. A body with size, where forces act at different places, is a rigid body, and both conditions apply. We will master particles first, then rigid bodies.

Idealizations

Real objects bend and stretch, but in statics we usually treat them as rigid: their shape does not change under load. We also treat ropes and cables as able to pull but never push, and we ignore their weight unless told otherwise. These idealizations are not laziness; they are the deliberate simplifications that make a hard problem solvable, and they are accurate enough for the loads structures normally carry.

Three more idealizations appear constantly. A concentrated or point load acts at a single point even though the real contact is spread over an area. A smooth surface is one where friction is neglected, so the contact force is purely perpendicular. A weightless member is one whose own weight is small beside the loads it carries, usually true for a bolt or cable and usually false for a long concrete beam.

Key idea: Rigid bodies, weightless cables, smooth surfaces, and point loads are deliberate models chosen because they are accurate enough for the job.

Worked example: weight and a single supporting cable

Given: a transformer of mass m = 250 kg hangs at rest from a single vertical cable attached to a crane hook. Use g = 9.81 m/s squared.
Find: (a) the weight of the transformer in newtons and kilonewtons, and (b) the tension in the cable.

Free-body diagram in words: isolate the transformer. Cut the cable just above it and remove the ground and the air. Exactly two forces remain: the weight W pointing straight down at the center of gravity, and the cable tension T pointing straight up along the cable, away from the body.

Solution (a):
W = m g = 250 x 9.81 = 2452.5 N
W = 2452.5 / 1000 = 2.45 kN.
Solution (b): take upward as positive and apply the force condition.
sum F_y = 0: T - W = 0
T = W = 2452.5 N = 2.45 kN.

Sanity check: with only two forces on the body, equilibrium demands they be equal, opposite, and collinear, which is exactly what came out. The number is plausible too: 2.45 kN is about the weight of two motorcycles, well inside the rating of any crane cable. Note that the answer did not depend on the length of the cable or on how quickly the crane lifted it beforehand - only on the fact that the transformer is now at rest.

Worked example: two stacked crates and the third law

Given: crate A has mass 40 kg and rests on top of crate B, which has mass 60 kg and rests on the floor. Nothing else touches them. Use g = 9.81 m/s squared.
Find: the contact force between the two crates and the contact force between crate B and the floor.

Free-body diagram of crate A: its weight W_A acting down at its center of gravity, and the normal force N_AB from crate B pushing up on its bottom face. Nothing else touches A.
W_A = 40 x 9.81 = 392.4 N
sum F_y = 0: N_AB - 392.4 = 0, so N_AB = 392.4 N.

Free-body diagram of crate B: its own weight W_B down, the floor's normal force N_floor up, and - by Newton's third law - crate A pushing down on B with 392.4 N. That downward push is the reaction to the 392.4 N that B pushed up on A; it acts on a different body, so it belongs on B's diagram and not on A's.
W_B = 60 x 9.81 = 588.6 N
sum F_y = 0: N_floor - 588.6 - 392.4 = 0
N_floor = 981.0 N.

Sanity check: treat the two crates as a single combined body. Its mass is 40 + 60 = 100 kg, so its weight is 100 x 9.81 = 981.0 N, and the floor must push up with exactly that much. The two routes agree, which confirms both the arithmetic and the direction of the third-law pair. Notice that the 392.4 N contact force vanished when the bodies were combined: internal forces always cancel in pairs, which is precisely why isolating the right body is the whole game.

Common wrong turns

• Writing W = m without multiplying by g. A mass of 40 kg is not a force of 40 N. Every weight in a statics problem is a mass times 9.81, or is already given in newtons.
• Putting an action-reaction pair on one free-body diagram. The pair acts on two different bodies. If both appear on one sketch, the diagram is wrong and the equation collapses to 0 = 0.
• Mixing units mid-problem. Millimeters with newtons and meters with kilonewtons in the same moment equation is the fastest route to being wrong by a factor of 1000. Convert first.
• Assuming equilibrium means no forces. Equilibrium means the forces sum to zero. A truss member can carry 200 kN and still be in perfect equilibrium.
• Forgetting that a cable cannot push. If the algebra hands back a negative cable tension, the model is wrong, not merely the sign - that cable would have gone slack.

Recap

• Statics analyzes bodies with zero acceleration, so the resultant force and the resultant moment are both zero.
• A force is a vector needing magnitude, direction, and point of application, and on a rigid body it slides freely along its line of action.
• Mass is measured in kilograms; weight is the force W = m g in newtons, so 1 kg weighs 9.81 N on Earth.
• Newton's first and second laws give the equilibrium conditions; the third law lets us expose internal forces by cutting.
• Action-reaction pairs act on different bodies and never appear together on one free-body diagram.
• Rigid bodies, weightless cables, smooth surfaces, and point loads are deliberate idealizations that keep problems solvable.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Introduction to statics. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Newton's laws of motion. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Units. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. OpenStax. (2016). 5.4 Mass and weight. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2016). 5.5 Newton's third law. In University Physics Volume 1. Rice University. openstax.org
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Newton's first law. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. National Institute of Standards and Technology. (n.d.). SI units. NIST Physical Measurement Laboratory. physics.nist.gov

Some quantities are fully described by a size alone: mass, time, length, and speed are scalars, a single number with a unit. Forces are different. A force also has a direction, so it is a vector. Pushing a door with 50 N to the left is not the same as pushing with 50 N upward, even though the magnitude is the same. Handling vectors correctly is the central skill of statics, and the most reliable method is to work with rectangular components.

Key idea: A scalar is a single number with a unit; a vector carries a direction as well, and every force in statics is a vector.

What goes wrong if you ignore direction

Two workers each pull on a crate with 400 N. What is the total pull? The question cannot be answered until you say which way each pulls. Pulling the same way gives 800 N. Pulling directly against each other gives zero. Pulling at right angles gives sqrt(400 squared + 400 squared) = 566 N. The three answers are wildly different, and only a vector treatment tells them apart.

This is why statics almost never adds force magnitudes directly. Instead it converts every force into a pair of signed numbers - a horizontal part and a vertical part - and adds those. Signed components are what turn a geometry problem into arithmetic, and they are the reason the equilibrium equations of the next module are simple algebra rather than trigonometry puzzles.

Resolving a force into components

Any force in a plane can be split into a horizontal piece and a vertical piece that, added together, reproduce it. If a force F points at an angle theta measured counterclockwise from the positive x-axis, its components are:

• Fx = F cos(theta) (the horizontal part)
• Fy = F sin(theta) (the vertical part)

These come straight from the right triangle the force makes with the axes. The force is the hypotenuse; the components are the two legs.

Key idea: With theta measured counterclockwise from the positive x-axis, Fx = F cos(theta) and Fy = F sin(theta), and the correct signs appear automatically.

Which angle, measured from where

Those two formulas are true only for that one convention: theta measured counterclockwise from the positive x-axis. Real problems rarely hand you the angle that way. A cable may be described as 20 degrees from the vertical, a roof brace as 35 degrees to the rafter, a slope as one unit up for every eight along. Before touching a calculator, redraw the angle onto the axes, or decide by inspection which component is the adjacent leg and which is the opposite leg.

The safe habit is to reason with the triangle instead of memorizing the formula. The component next to the angle is the adjacent leg and uses cosine. The component across from the angle is the opposite leg and uses sine. So if a force is 20 degrees from the vertical, the vertical component is F cos(20) and the horizontal component is F sin(20) - exactly the reverse of the default formula. Getting this backwards is the most common error in the first month of statics.

One calculator detail matters as much as the trigonometry: check that the machine is in degree mode. In radian mode cos(30) returns 0.1543 instead of 0.8660, and every later number inherits that error with no warning.

Key idea: Cosine goes with the component beside the angle and sine with the component across from it, no matter which axis the angle is measured from.

Going back the other way

Given the components, the magnitude is the hypotenuse by the Pythagorean theorem, and the angle comes from the tangent:

• F = sqrt(Fx squared + Fy squared)
• theta = arctan(Fy / Fx)

Always check the signs of Fx and Fy to place the vector in the correct quadrant; a calculator's arctangent alone cannot tell left from right.

Getting the quadrant right

An arctangent key only ever returns an angle between -90 and +90 degrees, so it can place a vector in the first or fourth quadrant and nowhere else. The fix is a two-step routine. First compute the reference angle from the magnitudes, alpha = arctan(|Fy| / |Fx|). Then use the signs to place it:

• Fx positive, Fy positive: first quadrant, theta = alpha.
• Fx negative, Fy positive: second quadrant, theta = 180 - alpha.
• Fx negative, Fy negative: third quadrant, theta = 180 + alpha.
• Fx positive, Fy negative: fourth quadrant, theta = 360 - alpha.

In practice many engineers skip the compass angle and report the answer in words: "241 N acting down and to the left, 35 degrees below the horizontal" cannot be misread. Whatever you report, sketch the components first, because a picture makes the quadrant obvious and the rule unnecessary.

Key idea: Arctangent cannot distinguish opposite directions, so fix the quadrant from the signs of Fx and Fy or from a sketch.

Slope triangles instead of angles

Structural drawings often give a direction as a slope rather than an angle: a brace that rises 5 units for every 12 it runs, or a cable described by the familiar 3-4-5 triangle. You never need to convert that to an angle. If the slope triangle has horizontal leg a, vertical leg b, and hypotenuse c = sqrt(a squared + b squared), then

• Fx = F (a / c)
• Fy = F (b / c)

The fractions a/c and b/c are simply cosine and sine written without ever naming the angle. Working straight from the slope removes two rounding steps and is what most practising engineers do.

Unit vectors in the plane

The same force can be written in Cartesian vector form with the unit vectors i (one unit along +x) and j (one unit along +y): F = Fx i + Fy j. A force with a 173.2 N horizontal part and a 100.0 N vertical part becomes F = 173.2 i + 100.0 j newtons. Nothing new is being claimed, but the notation extends cleanly to three dimensions and makes vector addition look like ordinary algebra: add the i-terms, then add the j-terms.

Worked example: resolving a cable force

Given: a cable pulls on a bracket with a force of 200 N directed 30 degrees above the horizontal. Find: its horizontal and vertical components.
Solution: Apply the component formulas.
Fx = 200 cos(30) = 200 x 0.8660 = 173.2 N
Fy = 200 sin(30) = 200 x 0.5000 = 100.0 N
So the cable pulls the bracket 173.2 N to the right and 100.0 N upward. As a check, sqrt(173.2 squared + 100.0 squared) = sqrt(30000 + 10000) = sqrt(40000) = 200 N, which recovers the original magnitude.

Why components win

Once every force is written as an x-part and a y-part, adding forces becomes simple arithmetic: add all the x-parts, add all the y-parts. There is no need to draw scaled diagrams or use the law of cosines. This is the workhorse method you will use in every equilibrium problem from here on.

Worked example: a force given by a slope

Given: a diagonal brace runs from a base plate up to a beam, rising 5 m for every 12 m of horizontal run. It carries an axial force of 260 N directed up and to the right along the brace.
Find: the horizontal and vertical components of that force.

Solution: build the slope triangle first.
c = sqrt(12 squared + 5 squared) = sqrt(144 + 25) = sqrt(169) = 13
Fx = 260 x (12/13) = 260 x 0.9231 = 240.0 N (to the right)
Fy = 260 x (5/13) = 260 x 0.3846 = 100.0 N (upward)

Sanity check: rebuild the magnitude. sqrt(240 squared + 100 squared) = sqrt(57600 + 10000) = sqrt(67600) = 260 N, which recovers the original force exactly. As a second check, the angle is theta = arctan(5/12) = 22.62 degrees, and 260 cos(22.62) = 260 x 0.9231 = 240.0 N, agreeing with the slope route. Note that the larger component belongs to the longer leg of the slope triangle, which is the quick plausibility test to run before trusting any component calculation.

Worked example: components back to magnitude and direction

Given: a bracket carries a force with components Fx = -60 N and Fy = -80 N.
Find: the magnitude of the force and the angle it makes measured counterclockwise from the positive x-axis.

Solution: the magnitude comes from Pythagoras.
F = sqrt((-60) squared + (-80) squared) = sqrt(3600 + 6400) = sqrt(10000) = 100 N
Now the reference angle from the magnitudes only:
alpha = arctan(80 / 60) = arctan(1.3333) = 53.13 degrees
Both components are negative, so the force points down and to the left, into the third quadrant:
theta = 180 + 53.13 = 233.13 degrees.

Sanity check: run the components forward again. Fx = 100 cos(233.13) = 100 x (-0.600) = -60 N and Fy = 100 sin(233.13) = 100 x (-0.800) = -80 N, matching the given data. Watch what a careless calculator would have done: keying arctan(-80 / -60) gives arctan(1.3333) = +53.13 degrees, which points up and to the right, the exact opposite of the true force. The two minus signs cancelled inside the division and the calculator lost the quadrant, which is why the signs must be handled separately.

Common wrong turns

• Using sine where cosine belongs. When the angle is measured from the vertical, the vertical component takes the cosine. Draw the triangle and identify the adjacent leg every time.
• Leaving the calculator in radian mode. A single wrong mode setting corrupts every component in the problem while looking perfectly reasonable.
• Dropping the signs. Components are signed quantities. A force pointing left has a negative Fx; treating it as positive flips the whole resultant.
• Trusting arctangent for the quadrant. Compute the reference angle from magnitudes, then place it using the signs of the components.
• Adding magnitudes. A 300 N force and a 400 N force do not combine to 700 N unless they happen to be parallel and in the same direction.
• Rounding too early. Keeping only two figures in the middle of a chain of trigonometry can shift a final reaction by several percent. Round once, at the end.

Recap

• Scalars have magnitude alone; vectors such as force also have direction and must be added componentwise.
• With theta from the positive x-axis, Fx = F cos(theta) and Fy = F sin(theta); from any other reference, use the adjacent-cosine and opposite-sine rule.
• Reverse the process with F = sqrt(Fx squared + Fy squared) and a reference angle arctan(|Fy| / |Fx|) placed by the signs.
• A slope triangle gives components directly as F(a/c) and F(b/c), with no angle ever computed.
• Cartesian form F = Fx i + Fy j makes vector addition into simple term-by-term arithmetic.
• Always rebuild the magnitude from the components as a check before moving on.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Vectors. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). 2D coordinate systems. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Right triangle trigonometry. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. OpenStax. (2016). 2.1 Scalars and vectors. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2016). 2.2 Coordinate systems and components of a vector. In University Physics Volume 1. Rice University. openstax.org
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Vectors. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Baker, D. W., & Haynes, W. (2021). Forces and other vectors. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

When several forces act at the same point, they can be replaced by one single force, the resultant, that has exactly the same effect. Finding the resultant is the heart of combining forces, and the component method makes it routine.

Key idea: The resultant of a set of concurrent forces is the single force with the same net effect, found by adding the components: Rx = sum Fx and Ry = sum Fy.

Why replacing forces with one force is legitimate

Forces that share a common point are called concurrent. Because they all act at the same place, none of them can turn the body about that point, so the only thing that matters is how they push. Add the pushes as vectors and you have said everything there is to say about the system. That is why a concurrent system always reduces to exactly one force acting through the same point.

Forces that are not concurrent do not reduce so simply. Two equal and opposite forces on different lines of action have a resultant of zero yet still spin the body, which is a preview of couples in Module 3. For now every system in this lesson is concurrent, so a single resultant tells the whole story.

The old way: parallelogram and triangle rules

Before components became the standard, two forces were added geometrically. The parallelogram rule draws the two forces from a common point, completes the parallelogram, and takes the diagonal as the resultant. The triangle rule places the tail of the second force at the head of the first, and the closing side from the first tail to the second head is the resultant. Extend this to many forces and you get the force polygon: if the polygon closes on itself, the resultant is zero and the particle is in equilibrium.

These pictures are still worth drawing because they build intuition, and they can be solved numerically with the law of cosines, R squared = A squared + B squared - 2 A B cos(angle between the sides). But for three or more forces the geometry becomes a maze, while the component method stays the same length no matter how many forces there are. That scalability is why every modern text, and every piece of analysis software, works in components.

Key idea: The parallelogram, triangle, and polygon rules are geometric pictures of the same addition that components perform arithmetically, and a closed force polygon means equilibrium.

The component method for resultants

To add any number of coplanar forces:

1. Resolve each force into its x and y components, keeping signs (right and up are positive; left and down are negative).
2. Add all the x-components to get the resultant's x-part: Rx = sum Fx.
3. Add all the y-components to get the resultant's y-part: Ry = sum Fy.
4. Combine: R = sqrt(Rx squared + Ry squared) and theta = arctan(Ry / Rx).

The single most useful discipline here is to lay the work out as a table, one row per force and one column each for the magnitude, the angle, Fx, and Fy. Sum the last two columns and you are done. A tidy table catches dropped signs and forgotten forces long before the algebra does, and it is the format every reviewer expects to see.

Remember that the final step needs the quadrant treatment from the previous lesson. Compute the reference angle from the magnitudes of Rx and Ry, then place it using their signs. Reporting a resultant that points the wrong way is worse than reporting no answer at all, because the number looks right.

Worked example: two forces on a hook

Given: a hook is pulled by two cables. Force F1 = 300 N at 20 degrees above the positive x-axis, and force F2 = 200 N at 120 degrees from the positive x-axis.

Find: the resultant force on the hook.
Solution: Resolve each force.
F1x = 300 cos(20) = 281.9 N, F1y = 300 sin(20) = 102.6 N
F2x = 200 cos(120) = -100.0 N, F2y = 200 sin(120) = 173.2 N
Add the components:
Rx = 281.9 + (-100.0) = 181.9 N
Ry = 102.6 + 173.2 = 275.8 N
Combine:
R = sqrt(181.9 squared + 275.8 squared) = sqrt(33088 + 76066) = sqrt(109154) = 330.4 N
theta = arctan(275.8 / 181.9) = arctan(1.516) = 56.6 degrees above the positive x-axis.
So the two cables together pull with a resultant of 330 N at 56.6 degrees. Both Rx and Ry are positive, so the resultant lies in the first quadrant, which matches the angle.

The equilibrium connection

If a particle is in equilibrium, the resultant of all forces on it must be zero. That means Rx = 0 and Ry = 0 at the same time. In our worked example the resultant is 330 N, not zero, so the hook is not in equilibrium; something else (the hook's support) must supply an equal and opposite 330 N force to hold it. This is the bridge from adding forces to solving for unknowns, which is the next module.

One more idea helps: the single force that would restore equilibrium is called the equilibrant. It is equal in magnitude to the resultant but points in the exact opposite direction. Here the equilibrant would be 330 N at 56.6 + 180 = 236.6 degrees.

Key idea: A particle is in equilibrium exactly when Rx = 0 and Ry = 0; the equilibrant is the single force equal and opposite to a non-zero resultant.

Worked example: three forces on a bracket

Given: a gusset plate is pulled at one point by three cables. F1 = 500 N acts horizontally to the right. F2 = 400 N acts up and to the left along a 3-4-5 slope, 4 units left for every 3 units up. F3 = 300 N acts straight down.
Find: the magnitude and direction of the resultant, and the equilibrant that would hold the plate still.

Solution: resolve each force, keeping signs. For F2 the hypotenuse of the slope triangle is sqrt(4 squared + 3 squared) = 5, so its fractions are 4/5 horizontal and 3/5 vertical.
F1x = +500 N, F1y = 0
F2x = -400 x (4/5) = -320 N, F2y = +400 x (3/5) = +240 N
F3x = 0, F3y = -300 N

Sum the columns:
Rx = 500 - 320 + 0 = 180 N
Ry = 0 + 240 - 300 = -60 N
Combine:
R = sqrt(180 squared + 60 squared) = sqrt(32400 + 3600) = sqrt(36000) = 189.7 N
alpha = arctan(60 / 180) = arctan(0.3333) = 18.4 degrees
Rx is positive and Ry is negative, so the resultant lies in the fourth quadrant: 189.7 N directed 18.4 degrees below the positive x-axis, or 341.6 degrees measured counterclockwise.

Sanity check: project the answer back onto the axes. 189.7 cos(341.6) = 189.7 x 0.9487 = 180.0 N and 189.7 sin(341.6) = 189.7 x (-0.3162) = -60.0 N, which reproduces the sums exactly. The direction is also plausible by inspection: the 500 N pull to the right dominates the 320 N pull to the left, so the resultant should lean right, and the 300 N downward pull beats the 240 N upward pull, so it should also lean slightly down. It does.

The equilibrant is therefore 189.7 N at 341.6 - 180 = 161.6 degrees, that is, 189.7 N directed up and to the left at 18.4 degrees above the negative x-axis. Adding it to the three cable forces would make both column sums zero, which is precisely the equilibrium condition of the next module.

Patterns worth spotting before you calculate

A few configurations let you shortcut the arithmetic or check it instantly.

• Collinear forces. If all the forces lie on one line, the resultant is just the signed sum of magnitudes. No trigonometry is needed.
• Perpendicular pairs. Two forces at 90 degrees combine by Pythagoras alone, and the resultant angle is arctan of the ratio.
• Equal and opposite pairs. These contribute nothing to the resultant. Cancel them on sight rather than carrying them through the table.
• Symmetry. Forces placed symmetrically about an axis have components perpendicular to that axis that cancel, leaving only the along-axis sum. This is the trick that made the three-cable problem in Module 2 a one-line solve.

A final magnitude check that costs nothing: the resultant of any set of forces can never exceed the sum of their magnitudes, and can never be smaller than the largest magnitude minus the sum of all the others. In the bracket example the sum of magnitudes is 1200 N and the answer was 189.7 N, comfortably inside the bound.

Common wrong turns

• Adding magnitudes instead of components. The three cables above total 1200 N of pull yet produce a resultant of only 190 N, because most of it cancels.
• Dropping a sign in the table. A force pointing left or down carries a minus sign into the sum. One missed minus and the resultant can end up in the wrong half of the plane.
• Taking arctan(Ry / Rx) straight off the calculator. When both components are negative the minus signs cancel inside the fraction and the reported angle is 180 degrees wrong.
• Confusing resultant with equilibrant. They have the same magnitude but opposite directions. The resultant is what the forces do; the equilibrant is what you would have to add to stop them.
• Assuming a zero resultant means no force anywhere. It means the pushes cancel. The members can still be heavily loaded, and if the forces are not concurrent the body may still rotate.

Recap

• Concurrent forces reduce to a single resultant acting through their common point.
• Resolve, sum the x-components and the y-components separately, then combine with Pythagoras and a quadrant-corrected arctangent.
• The parallelogram, triangle, and polygon rules are the geometric version of the same operation; a closed polygon means equilibrium.
• A particle is in equilibrium exactly when Rx = 0 and Ry = 0.
• The equilibrant equals the resultant in magnitude and points 180 degrees away from it.
• Check every answer by projecting the resultant back onto the axes and by testing it against the sum-of-magnitudes bound.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Vector addition. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Oblique triangle trigonometry. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Forces. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Vector addition. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Concurrent forces. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. OpenStax. (2016). 2.3 Algebra of vectors. In University Physics Volume 1. Rice University. openstax.org
7. Ulm, F.-J., & Buehler, M. (2007). 1.050 Engineering mechanics I [Course materials]. MIT OpenCourseWare. ocw.mit.edu

Real structures live in three dimensions, and many forces do not lie in a single plane. The component idea extends directly: a 3D force has parts along three perpendicular axes, x, y, and z, written with the unit vectors i, j, and k that point one unit along each axis. A force is then F = Fx i + Fy j + Fz k, and its magnitude is F = sqrt(Fx squared + Fy squared + Fz squared).

Key idea: A three-dimensional force is written F = Fx i + Fy j + Fz k, and everything you learned in the plane carries over one axis at a time.

Setting up a right-handed frame

Before any 3D calculation, fix the axes and keep them fixed. Statics uses a right-handed coordinate system: point the fingers of your right hand along +x, curl them toward +y, and your thumb points along +z. Equivalently, i crossed into j gives k. Choosing a left-handed frame by accident reverses every cross product and therefore every moment direction later in the course, so it is worth checking with your hand the first few times.

Beyond handedness, the placement of the origin is free. Put it where the geometry is simplest, usually at a support or at the point where several members meet, because that choice turns many coordinates into zeros and removes arithmetic. A tower problem is far easier with the origin at the tower base than at an arbitrary corner of the site.

Key idea: Always use a right-handed frame and place the origin where it produces the most zeros in the coordinates.

Direction cosines

A 3D direction can be given by the three angles alpha, beta, and gamma that the force makes with the x, y, and z axes. Their cosines, called direction cosines, are the fractions of the force along each axis:

• Fx = F cos(alpha), Fy = F cos(beta), Fz = F cos(gamma)

The three direction cosines are not independent; they always satisfy cos squared(alpha) + cos squared(beta) + cos squared(gamma) = 1. This identity is a fast way to check a 3D direction or to find a missing third angle.

Two features of these angles surprise students. First, each coordinate direction angle runs from 0 to 180 degrees, not 0 to 90, because a force may point along the negative side of an axis; a direction cosine of -0.5 simply means the angle is 120 degrees. Second, the three angles are never all small. Since their cosines must square to one, a vector cannot make a 30 degree angle with all three axes at once - that would give 3 x 0.75 = 2.25, which is impossible.

The identity is also a repair tool. If a problem hands you two of the three angles, the third follows: cos squared(gamma) = 1 - cos squared(alpha) - cos squared(beta). For example, with alpha = 60 degrees and beta = 45 degrees, we get cos squared(gamma) = 1 - 0.25 - 0.50 = 0.25, so cos(gamma) = plus or minus 0.5 and gamma = 60 degrees or 120 degrees. Two answers appear because the identity cannot tell whether the force leans toward +z or -z; a sketch or a stated sign settles it.

Key idea: Direction angles range over 0 to 180 degrees and always satisfy cos squared(alpha) + cos squared(beta) + cos squared(gamma) = 1, which supplies a missing angle up to a sign.

Angles measured in two steps

Many 3D forces are described not by three direction angles but by a pair: a tilt from the vertical axis and a bearing in the horizontal plane. This is how surveyors, crane operators, and antenna drawings usually specify a direction. Resolve such a force in two stages. If phi is the angle from the z-axis and theta is the angle in the xy-plane measured from the x-axis, then

• Fz = F cos(phi)
• F_horizontal = F sin(phi), the projection onto the xy-plane
• Fx = F_horizontal cos(theta) and Fy = F_horizontal sin(theta)

The first step splits the force into vertical and horizontal parts; the second step is an ordinary 2D resolution of the horizontal part. Nothing new is required, only the discipline of doing it in that order.

Force along a line between two points

Most often you know a force acts from point A toward point B along a cable or bar, and you know the coordinates of both. Then:

1. Form the position vector from A to B: d = (xB - xA) i + (yB - yA) j + (zB - zA) k.
2. Find its length |d| = sqrt(dx squared + dy squared + dz squared).
3. Divide to get the unit vector u = d / |d|, which carries the direction only.
4. Multiply the unit vector by the magnitude: F = F u. The components of u are exactly the direction cosines.

Worked example: a cable force in 3D

Given: a cable runs from the anchor A at the origin (0, 0, 0) to a point B at (3, 4, 12) meters and carries a tension of 650 N. Find: the force's rectangular components and its direction cosines.
Solution: The position vector is d = 3 i + 4 j + 12 k, with length
|d| = sqrt(3 squared + 4 squared + 12 squared) = sqrt(9 + 16 + 144) = sqrt(169) = 13 m.
The unit vector is u = (3/13) i + (4/13) j + (12/13) k.

Multiply by 650 N:
Fx = 650 x 3/13 = 150 N
Fy = 650 x 4/13 = 200 N
Fz = 650 x 12/13 = 600 N
So F = 150 i + 200 j + 600 k newtons. The direction cosines are cos(alpha) = 3/13 = 0.231, cos(beta) = 4/13 = 0.308, and cos(gamma) = 12/13 = 0.923. Check: 0.231 squared + 0.308 squared + 0.923 squared = 0.053 + 0.095 + 0.852 = 1.000. The identity holds, confirming the direction.

This position-vector method is the standard tool for 3D equilibrium. Whenever a force acts along a known line, build its unit vector from the endpoints, and the components follow at once.

Key idea: For a force along a known line, the unit vector u = d / |d| built from the endpoint coordinates carries the entire direction, and F = F u gives all three components in one step.

Worked example: a guy wire on a tower

Given: a radio mast stands on the z-axis with its top at B(0, 0, 12) m. A guy wire runs from B down to a ground anchor at A(6, -4, 0) m and carries a tension of 1300 N. All coordinates are in meters.
Find: the components of the force the wire exerts on the top of the mast, and the three coordinate direction angles of that force.

Direction first. The wire pulls the mast top toward the anchor, so the force on B points from B toward A.
d = A - B = (6 - 0) i + (-4 - 0) j + (0 - 12) k = 6 i - 4 j - 12 k
|d| = sqrt(6 squared + 4 squared + 12 squared) = sqrt(36 + 16 + 144) = sqrt(196) = 14 m
u = (6/14) i - (4/14) j - (12/14) k = 0.4286 i - 0.2857 j - 0.8571 k

Components. Multiply the unit vector by the 1300 N magnitude. Note 1300 / 14 = 92.857, which makes the arithmetic clean.
Fx = 92.857 x 6 = 557.1 N
Fy = 92.857 x (-4) = -371.4 N
Fz = 92.857 x (-12) = -1114.3 N
So F = 557.1 i - 371.4 j - 1114.3 k newtons.

Direction angles. Take the arccosine of each direction cosine.
alpha = arccos(0.4286) = 64.6 degrees
beta = arccos(-0.2857) = 106.6 degrees
gamma = arccos(-0.8571) = 149.0 degrees

Sanity check: the direction cosines must square to one. 0.4286 squared + 0.2857 squared + 0.8571 squared = 0.1837 + 0.0816 + 0.7347 = 1.0000, exactly as required, because (36 + 16 + 144) / 196 = 1. The magnitude check works too: sqrt(557.1 squared + 371.4 squared + 1114.3 squared) = sqrt(310400 + 137940 + 1241660) = sqrt(1690000) = 1300 N. Physically the answers are sensible: the wire pulls the mast top mostly downward, which is exactly what a guy wire is for, and the two obtuse angles reflect the fact that the force points along -y and -z.

What the components are used for

In the next module these three numbers go straight into sum Fx = 0, sum Fy = 0, and sum Fz = 0. Notice how little geometry is left to worry about once the unit vectors are built: the hard part of any 3D equilibrium problem is the bookkeeping of coordinates, not the physics. Set up a small table with a row for each force and columns for d, |d|, u, and the three components, and a problem that looks intimidating becomes a routine fill-in exercise.

One more habit pays off. Because |d| is a length, it is always positive, but the components of d carry signs that encode the direction. If a force pulls the other way along the same line, do not recompute anything - just flip the sign of the whole unit vector. The wire above, viewed as a force on the ground anchor rather than the mast, is -557.1 i + 371.4 j + 1114.3 k newtons, the same 1300 N reversed.

Common wrong turns

• Subtracting the coordinates the wrong way round. The position vector runs from the tail to the head: d = (head) - (tail). Reversing it reverses the force.
• Forgetting to divide by the length. Multiplying the raw position vector by the force magnitude produces a vector with a magnitude of F |d|, off by a factor of the length. The unit vector step is not optional.
• Assuming direction angles are all acute. Any component that points along a negative axis gives a direction cosine that is negative and an angle above 90 degrees.
• Adding direction cosines instead of squaring them. The identity is a sum of squares equal to one, not a sum equal to one.
• Choosing a left-handed frame. It will pass every force calculation in this lesson and then reverse every moment in Module 3. Check the frame with your right hand once, at the start.

Recap

• A 3D force is F = Fx i + Fy j + Fz k, with magnitude sqrt(Fx squared + Fy squared + Fz squared).
• Direction cosines are the components of the unit vector; the angles run from 0 to 180 degrees.
• The identity cos squared(alpha) + cos squared(beta) + cos squared(gamma) = 1 both checks a direction and supplies a missing angle up to a sign.
• For a force along a line, form d from the endpoints, divide by |d| to get u, then multiply by the magnitude.
• A tilt-and-bearing description resolves in two stages: vertical first, then the horizontal projection.
• Use a right-handed frame throughout and place the origin to create the most zeros.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). 3D coordinate systems. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Unit vectors. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Forces and other vectors. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). Notation. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. OpenStax. (2016). 2.4 Products of vectors. In University Physics Volume 1. Rice University. openstax.org
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Vectors. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Massachusetts Institute of Technology. (2005). 16.01 Unified engineering I, II, III, IV [Course materials]. MIT OpenCourseWare. ocw.mit.edu

The single most important skill in statics is drawing a good free-body diagram, or FBD. A free-body diagram is a sketch of one body, imagined cut free from everything touching it, showing every external force that acts on it and nothing else. Almost every mistake in statics traces back to a wrong or incomplete free-body diagram, so it is worth doing slowly and carefully.

Key idea: A free-body diagram shows one chosen body, separated from everything else, with every external force that acts on it and nothing else at all.

Why the diagram does the hard work

The equilibrium equations are trivial: three sums set to zero. All the engineering judgement in a statics problem is spent deciding which body to isolate and which forces act on it. Choose the wrong body and the unknown you want never appears in the equations. Miss a force and the equations are simply false. The free-body diagram is where the problem is actually solved; the algebra afterwards is bookkeeping.

The choice of body is genuinely free. You may isolate a whole bridge, one truss, one member, half a member, a pin, or a bolt. Each choice exposes a different set of forces. A useful rule is to isolate the body on which the force you want becomes external. Want the force in a truss member? Cut through it, so what was internal to the truss becomes external to the piece you keep. Want a pin reaction? Isolate the member the pin holds.

Key idea: Pick the body so that the unknown you are hunting becomes an external force on it - that single choice turns most statics problems from hard to routine.

How to draw one

1. Isolate the body. Draw it alone, mentally removing the ground, cables, walls, and neighbors that touch it.
2. Replace every contact you removed with the force it exerted. A rope becomes a tension pulling away from the body along the rope. A surface becomes a normal force pushing perpendicular to the surface. A pin becomes a reaction force.
3. Add the weight. If the body has mass and gravity matters, draw its weight W = m g straight down, acting at the center of gravity.
4. Label every force with a symbol and, where known, an angle. Choose x and y axes.

Two habits make step 2 reliable. First, walk the boundary of the body with your finger and ask at every point of contact, "what touches here, and which way can it push or pull?" Anything that touches contributes a force; anything that does not, does not. Second, when a direction is unknown, draw it in a chosen positive sense and let the algebra correct you. A negative answer means the arrow was backwards, and the magnitude is still right.

For step 4, choose axes to suit the geometry rather than the page. On an inclined surface, tilting the axes so that x runs up the slope and y runs perpendicular to it usually splits the problem into two independent one-unknown equations. Aligning axes with the largest number of forces is almost always the fastest route.

Reading the common supports

• A cable or rope can only pull, never push. Its force points away from the body, along the cable. This is called tension.
• A smooth surface or contact pushes with a normal force, perpendicular to the surface, and can only push.
• A spring pushes or pulls with a force Fs = k x, where k is its stiffness and x is its stretch or compression from the natural length.
• A frictionless pulley changes the direction of a cable without changing its tension, so the same value appears on both sides of the wheel. The pulley pin then carries the vector sum of the two cable pulls.
• A pin or hinge can push or pull in any direction in the plane but cannot resist turning, so it contributes two unknown components and no moment.
• A fixed or built-in support resists sliding in both directions and rotation, so it contributes two force components plus a reaction couple.

Key idea: Each support type contributes exactly the reactions it needs to prevent the motions it blocks - one force for a roller or cable, two for a pin, two plus a moment for a fixed end.

What must never appear on a free-body diagram

An incomplete diagram is a common failure, but an overfull one is just as damaging. Four things do not belong:

• Forces the body exerts on other things. Only forces exerted on the isolated body appear. The crate's push on the floor lives on the floor's diagram, not the crate's.
• Internal forces. If two members are kept together in the isolated body, the forces between them are internal and cancel. They appear only if you cut between them.
• A resultant drawn alongside its own components. Show one or the other. Drawing both double-counts the force.
• Motion arrows. Velocity, acceleration, and "the direction it wants to go" are not forces. In statics the only hint of intended motion appears as the direction friction opposes, which is handled separately.

Worked idea: the ring at a knot

Imagine a lamp hanging from a small ring, and two cables run from the ring up to the ceiling. To analyze the ring, isolate just the ring. Three forces act on it: the pull of cable 1 (along cable 1, away from the ring), the pull of cable 2 (along cable 2, away from the ring), and the downward pull of the lamp's weight transmitted through the short cable below. The ceiling and the lamp itself are gone from the picture; only their forces remain. That clean three-force picture is exactly what the equilibrium equations need.

A tip that prevents errors: draw tensions pointing away from the body every time, because a flexible cable can never push. If the algebra later returns a negative number for a normal force or a strut, that is a signal your assumed direction was wrong, and you simply reverse it. For a cable, a negative tension usually means the cable would go slack and the model must be reconsidered.

Worked example: a crate on a smooth ramp

Given: a 250 N crate rests on a smooth ramp inclined at 30 degrees to the horizontal. A rope parallel to the ramp surface holds it from sliding down. Friction is negligible.
Find: draw the free-body diagram in words, then compute the rope tension and the normal force.

Free-body diagram in words: isolate the crate. Erase the ramp and the rope. Three forces remain. (1) The weight, 250 N, acting straight down at the center of gravity. (2) The normal force N, perpendicular to the ramp surface, pushing away from the ramp - never along it, because the surface is smooth. (3) The rope tension T, acting up the slope, away from the crate along the rope. Choose tilted axes: x up the slope, y perpendicular to the slope.

Solution: resolve the weight onto the tilted axes. The component down the slope is W sin(30) and the component pressing into the ramp is W cos(30).
sum F_x = 0: T - 250 sin(30) = 0
T = 250 x 0.5000 = 125.0 N
sum F_y = 0: N - 250 cos(30) = 0
N = 250 x 0.8660 = 216.5 N

Sanity check: T and N are perpendicular, and together they must exactly cancel the weight, so sqrt(125.0 squared + 216.5 squared) = sqrt(15625 + 46872) = sqrt(62497) = 250.0 N. That matches the weight, so the diagram and the algebra agree. The numbers are also sensible: on a shallow ramp most of the weight presses into the surface and only a little pulls along it, and at 30 degrees the split is exactly half and 0.866 of the weight.

Worked example: the pin of a frictionless pulley

Given: a cable runs horizontally from the left, passes over a small frictionless pulley mounted on a bracket, and then hangs vertically downward carrying a 300 N load.
Find: the force the bracket pin must exert on the pulley.

Free-body diagram in words: isolate the pulley wheel together with its pin hole. Cut both cable segments. Because the pulley is frictionless, the tension is the same throughout the cable, so both cut ends carry 300 N. The horizontal segment pulls the pulley to the left with 300 N; the vertical segment pulls it down with 300 N. The pin supplies unknown components Px and Py, drawn positive to the right and upward. The pulley is small, so all forces are treated as concurrent at its center and no moment equation is needed.

Solution:
sum F_x = 0: Px - 300 = 0, so Px = +300 N (to the right)
sum F_y = 0: Py - 300 = 0, so Py = +300 N (upward)
P = sqrt(300 squared + 300 squared) = sqrt(180000) = 424.3 N
theta = arctan(300 / 300) = 45 degrees above the positive x-axis.

Sanity check: the two cable pulls are 300 N left and 300 N down, whose resultant is 424.3 N pointing down and to the left at 225 degrees. The pin force came out as 424.3 N at 45 degrees, exactly opposite, as equilibrium demands. Notice the practical lesson: a pulley that turns a 300 N cable through a right angle loads its mounting with 424 N, about 41 percent more than the load itself. Students who assume the pin carries "just 300 N" undersize the bracket.

Common wrong turns

• Leaving out the weight. Unless the problem says the body is light or weightless, gravity acts on it. Draw W = m g down at the center of gravity every time.
• Drawing a cable pushing. A rope, chain, or cable pulls only. Its arrow always points away from the body along the cable.
• Putting the normal force along the surface. A smooth contact gives a force perpendicular to the surface. If it can also act along the surface, that part is friction and must be labelled separately.
• Mixing two bodies on one diagram. One outline, one set of forces. Any force acting on a neighbour belongs on the neighbour's sketch.
• Assuming the pulley tension changes. On a frictionless pulley the tension is equal on both sides; the pulley changes direction only.
• Skipping the diagram because the problem "looks easy". The two-minute sketch is what catches the missing force that would otherwise cost an hour.

Recap

• A free-body diagram isolates one body and shows every external force on it and nothing else.
• Choose the body so that the unknown you want becomes an external force on it.
• Replace each removed contact by the force it exerted: tension along a cable, normal force perpendicular to a smooth surface, two components at a pin, two components plus a moment at a fixed support.
• Add the weight at the center of gravity, and label every force and angle.
• Never draw internal forces, reactions the body exerts on others, motion arrows, or a resultant beside its own components.
• Assume a direction for unknown forces and let a negative answer reverse the arrow for you.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Free-body diagrams. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Interactions between members. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Bodies and free body diagrams. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. OpenStax. (2016). 5.7 Drawing free-body diagrams. In University Physics Volume 1. Rice University. openstax.org
6. OpenStax. (2016). 5.6 Common forces. In University Physics Volume 1. Rice University. openstax.org
7. Baker, D. W., & Haynes, W. (2021). Equilibrium of particles. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

A particle in equilibrium obeys two scalar equations in a plane: the horizontal forces must balance and the vertical forces must balance.

• sum Fx = 0
• sum Fy = 0

Two equations can solve for at most two unknowns, which is usually two cable tensions or one tension and one normal force. The recipe is always the same: draw the free-body diagram, resolve each force into components, write the two sums, and solve.

Key idea: In the plane a particle gives exactly two independent equations, sum Fx = 0 and sum Fy = 0, so it can resolve at most two unknowns.

Counting unknowns before you start

The most valuable thirty seconds in any statics problem is spent counting. List the unknowns on the free-body diagram and compare with the number of equations available. Two equations and two unknowns means the problem is solvable as it stands. Two equations and three unknowns means you must either find another body to isolate or find another relation, such as a friction law or a geometric constraint. Two equations and one unknown means either the problem is over-determined, which usually signals a modelling mistake, or the second equation is a check you should not waste.

Counting first also stops the most demoralizing failure mode in the subject: grinding through a page of algebra on a problem that never had enough equations. If the count does not work, no amount of algebra will fix it.

Key idea: Count unknowns against equations before you compute anything; two-dimensional particle problems must reduce to two unknowns.

Two forces, three forces

Two special cases are worth recognizing on sight. If exactly two forces act on a particle in equilibrium, they must be equal in magnitude, opposite in direction, and along the same line. That is why a hanging weight puts exactly its own weight into a single vertical cable, and it is the seed of the two-force-member idea that makes truss analysis so quick.

If exactly three forces act on a particle, a shortcut called Lami's theorem applies. Writing the three forces as F1, F2, F3 and the angles between the other two as a1, a2, a3 respectively:

F1 / sin(a1) = F2 / sin(a2) = F3 / sin(a3)

It is the sine rule applied to the closed force triangle, and it solves a three-cable problem in one line with no components at all. It is not a replacement for the component method, which handles any number of forces, but it makes an excellent independent check on a three-force answer.

Worked example: a hanging sign held by two cables

Given: a sign of weight W = 150 N hangs from a ring at point A. Cable AC runs up and to the left to the ceiling, making an angle whose slope is 3 across and 4 up (so it makes 53.13 degrees with the horizontal, with cos = 0.6 and sin = 0.8). A second cable AB runs horizontally to the right and ties to a wall. Find: the tension in each cable.
Free-body diagram: at the ring A, three forces act. The weight 150 N pulls straight down. Cable AB pulls horizontally to the right with tension T_AB.

Cable AC pulls up and to the left with tension T_AC, whose components are T_AC cos(53.13) = 0.6 T_AC to the left and T_AC sin(53.13) = 0.8 T_AC upward.
Equilibrium:
sum Fy = 0: 0.8 T_AC - 150 = 0, so T_AC = 150 / 0.8 = 187.5 N.
sum Fx = 0: T_AB - 0.6 T_AC = 0, so T_AB = 0.6 x 187.5 = 112.5 N.
The tensions are T_AC = 187.5 N and T_AB = 112.5 N.

Both are positive, so both cables are indeed in tension, which makes physical sense for cables. As a check, the vertical component of AC (150 N) exactly supports the weight, and the horizontal wall cable simply keeps the ring from swinging in.

Choosing which equation to write first

A useful habit: write the equilibrium equation that contains only one unknown first, so you can solve it immediately without substitution. In the example, the y-equation contained only T_AC, so we solved it first, then used its value in the x-equation. Picking the order well turns a system of equations into a short chain of single-step solves.

When no equation contains a single unknown, you have a genuine simultaneous system. Solve it by substitution: take the simpler equation, express one unknown in terms of the other, and substitute into the second. Two unknowns in two linear equations always yields to that method in a few lines, and it is far less error-prone than trying to eliminate by adding scaled equations in your head.

Key idea: Write first the equation containing the fewest unknowns; if none has only one, substitute from the simpler equation into the other.

Worked example: block on a frictionless incline

Given: a 500 N block rests on a smooth incline tilted 25 degrees from horizontal, held from sliding by a rope running straight up the slope. Find: the rope tension and the normal force.
Solution: Tilt the axes so x points up the slope and y is perpendicular to the slope.

The weight's component down the slope is W sin(25) = 500 x 0.4226 = 211.3 N, and its component into the slope is W cos(25) = 500 x 0.9063 = 453.2 N.
sum Fx = 0 (up-slope positive): T - 211.3 = 0, so T = 211.3 N.
sum Fy = 0: N - 453.2 = 0, so N = 453.2 N.
The rope carries 211 N and the surface pushes back with a normal force of 453 N. Notice how choosing tilted axes made each equation contain just one unknown.

Worked example: a traffic light on two unequal cables

Given: a traffic light of weight 500 N hangs from a ring. Cable 1 runs up and to the left at 30 degrees above the horizontal; cable 2 runs up and to the right at 45 degrees above the horizontal.
Find: the tension in each cable.

Free-body diagram in words: isolate the ring. Three forces act on it. The weight of the light, 500 N, pulls straight down through the short vertical hanger. Cable 1 pulls up and to the left along its own line with tension T1. Cable 2 pulls up and to the right with tension T2. Both tensions point away from the ring, as cables always do. Two unknowns, two equations: the count works.

Equilibrium. Resolve with x to the right and y upward.
sum F_x = 0: -T1 cos(30) + T2 cos(45) = 0
-0.8660 T1 + 0.7071 T2 = 0, so T2 = 1.2247 T1.
sum F_y = 0: T1 sin(30) + T2 sin(45) - 500 = 0
Substitute T2 = 1.2247 T1:
0.5000 T1 + 0.7071 (1.2247 T1) = 500
0.5000 T1 + 0.8660 T1 = 500
1.3660 T1 = 500, so T1 = 366.0 N
T2 = 1.2247 x 366.0 = 448.3 N.

Sanity check by Lami's theorem. The three forces make angles of 105 degrees between the two cables, 120 degrees between cable 1 and the weight, and 135 degrees between cable 2 and the weight, which sum to 360 as they must. Lami then gives
T1 = 500 sin(135) / sin(105) = 500 x 0.7071 / 0.9659 = 366.0 N
T2 = 500 sin(120) / sin(105) = 500 x 0.8660 / 0.9659 = 448.3 N,
agreeing with the component solution to the last digit. A final plausibility test: the steeper cable, cable 2 at 45 degrees, carries the larger tension because it does more of the lifting, and both tensions exceed half the weight because each also has to fight the other's horizontal pull. That is a general pattern worth remembering - the flatter the cables, the larger the tensions, which is why a tightrope pulled nearly horizontal can snap under a modest load.

Reading the sign of an answer

Every unknown in this lesson came out positive, which confirmed the assumed directions. When one comes out negative, the interpretation depends on what the force is. A negative normal force means the surface would have to pull rather than push, which is impossible, so the body has actually lifted off and the model must change. A negative cable tension means the cable would have to push, so it has gone slack and carries nothing. A negative force in a rigid strut or link is perfectly acceptable: it simply means compression instead of tension. Always ask which kind of force you are looking at before deciding whether a negative answer is a result or a red flag.

Common wrong turns

• Splitting the weight into components as well as using it whole. Pick one representation. On tilted axes the weight becomes W sin(theta) and W cos(theta), and the original W then disappears from the equations.
• Using sine and cosine from the wrong reference. A cable at 30 degrees from the horizontal contributes T cos(30) horizontally; a cable at 30 degrees from the vertical contributes T sin(30) horizontally. Read the figure, not the number.
• Forgetting a sign on a leftward or downward component. The x-equation of the traffic light needed a minus on cable 1; without it both cables appear to pull the same way and no solution exists.
• Assuming symmetric geometry gives equal tensions. Equal angles give equal tensions, but 30 and 45 degrees do not, and the difference here is over 20 percent.
• Trying to find three unknowns from two equations. Count first. If the count fails, isolate a different body or bring in another physical law.

Recap

• Particle equilibrium in the plane is sum Fx = 0 and sum Fy = 0, two equations for at most two unknowns.
• Draw the free-body diagram, resolve, count unknowns, then write the equation with the fewest unknowns first.
• Two forces on a particle in equilibrium must be equal, opposite, and collinear.
• Three forces obey Lami's theorem, F1 / sin(a1) = F2 / sin(a2) = F3 / sin(a3), a one-line independent check.
• Tilting the axes to match a slope usually turns a simultaneous system into two single-unknown equations.
• A negative answer reverses the assumed arrow, but for a cable or a smooth contact it means the model itself needs revisiting.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). 2D particle equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Equilibrium of particles. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Equilibrium analysis for concurrent force systems. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Static equilibrium. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. OpenStax. (2016). 6.1 Solving problems with Newton's laws. In University Physics Volume 1. Rice University. openstax.org
6. OpenStax. (2016). 12.1 Conditions for static equilibrium. In University Physics Volume 1. Rice University. openstax.org
7. Ulm, F.-J., & Buehler, M. (2007). Lecture notes. In 1.050 Engineering mechanics I. MIT OpenCourseWare. ocw.mit.edu

In three dimensions a particle in equilibrium must balance forces along all three axes:

• sum Fx = 0, sum Fy = 0, sum Fz = 0

Three equations solve for up to three unknowns. The method mirrors 2D but uses the position-vector technique from Module 1 to get each cable's components.

Key idea: A particle in space gives three independent equations, so a 3D concurrent-force problem is solvable exactly when it has three or fewer unknowns.

What actually changes from two dimensions

Conceptually nothing changes: the resultant must still vanish. Practically, two things do. First, you can rarely read a component off a sketch, because a drawing on paper flattens the third dimension and hides the true angles. Second, three simultaneous equations are enough algebra that sloppy organization becomes fatal.

Both problems have the same cure: build unit vectors from coordinates and keep everything in a table. Never estimate a 3D angle from a picture. Get the coordinates of the two ends of every cable or bar, form d, divide by |d|, and read off the three fractions. Those fractions are exact, they carry their own signs, and they eliminate every guess about which trigonometric function to use.

Key idea: In space, replace trigonometry with coordinates: every direction comes from a unit vector built from endpoint positions.

The procedure

1. Isolate the joint or particle and list every force, including the weight.
2. For each force along a known line, build its unit vector from the endpoints and write the force as magnitude times unit vector.
3. Collect the x, y, and z components and set each sum to zero.
4. Solve the three equations for the unknowns.

Step 3 is where the table earns its keep. Give every force a row and give the columns headings d, |d|, ux, uy, uz, then Fx, Fy, Fz. Fill it in completely before writing a single equilibrium equation. Summing the last three columns then produces the three equations mechanically, and any missing force shows up as a blank row rather than as a wrong answer.

A second habit: look for an equation that involves only one unknown before solving the full system. In many real geometries one axis is untouched by most of the forces, and its equation collapses immediately. Solving that one first often reduces a three-by-three system to a pair of easy substitutions.

Worked example: a weight held by three symmetric cables

Given: a 600 N weight hangs from a ring at the origin A(0, 0, 0). Three identical cables run up to anchors evenly spaced around a circle of radius 3 m at a height of 4 m above the ring. By symmetry each cable makes the same angle, and the slant length of each cable is L = sqrt(3 squared + 4 squared) = 5 m. Find: the tension in each cable.
Solution: Because the three anchors are evenly spaced and equally high, the three tensions are equal by symmetry, call each one T.

The horizontal parts of the three cables cancel around the circle, leaving only the vertical parts to hold the weight. Each cable's vertical fraction is height / length = 4 / 5 = 0.8. The three vertical components must sum to the weight:
sum Fz = 0: 3 x T x (4/5) - 600 = 0
3 x 0.8 x T = 600
2.4 T = 600, so T = 250 N.
Each cable carries 250 N. As a sanity check, the total upward pull is 3 x 250 x 0.8 = 600 N, exactly balancing the weight, and the horizontal pulls cancel by the three-fold symmetry.

When there is no symmetry

If the anchors are not symmetric, you cannot assume equal tensions. Then you build a unit vector for each cable from its endpoints, write all three tensions as unknowns times their unit vectors, and solve the three component equations simultaneously. The algebra is longer, but the idea is unchanged: three balance equations, three unknowns. Symmetry, when present, is simply a shortcut that spares you the full system.

Symmetry can also be partial and still useful. In the jib crane above the geometry was symmetric about the xz-plane, so the y-equation alone proved the two cables equal before any other work was done. Look for a plane of symmetry that contains both the load and the structure; if it exists, the two mirrored members carry equal forces, and one unknown disappears at the cost of a single line of algebra.

When no symmetry exists at all, the honest approach is a three-by-three linear system. Write it as a tidy set of three equations in the three unknown magnitudes, solve by substitution or elimination, and then substitute the answers back into all three equations to confirm they close. That back-substitution takes a minute and is worth every second, because a 3D system offers a great many places for a sign to go astray.

Three-dimensional equilibrium is where careful bookkeeping pays off. Keep a small table of each force's three components, sum each column, and set each column to zero. The tidy layout is what keeps sign errors away.

Key idea: Organize a 3D solve as a component table, exploit any genuine symmetry, and always back-substitute to confirm all three columns sum to zero.

Worked example: a wall-mounted jib crane

Given: a horizontal boom runs from a wall bracket at O(0, 0, 0) straight out along the x-axis to its tip at A(4, 0, 0) m. Two cables run from the tip back to the wall, one to B(0, 2, 4) and the other to C(0, -2, 4) m. A load of 1500 N hangs vertically from the tip, that is, in the -z direction. The boom is a straight two-force member, so its force on the tip acts along the boom, in the +x direction if the boom is in compression.
Find: the tension in each cable and the force in the boom.

Free-body diagram in words: isolate the small joint at the tip A. Four forces meet there and all pass through the same point, so it is a particle problem. (1) The 1500 N load pulling straight down. (2) Cable AB pulling from A toward B. (3) Cable AC pulling from A toward C. (4) The boom pushing along +x with an unknown compressive force F_OA. Three unknowns, three equations: the count works.

Unit vectors.
d_AB = B - A = -4 i + 2 j + 4 k, |d_AB| = sqrt(16 + 4 + 16) = sqrt(36) = 6, so u_AB = -0.6667 i + 0.3333 j + 0.6667 k.
d_AC = C - A = -4 i - 2 j + 4 k, |d_AC| = sqrt(16 + 4 + 16) = 6, so u_AC = -0.6667 i - 0.3333 j + 0.6667 k.

Equilibrium, one axis at a time.
sum F_y = 0: 0.3333 T_AB - 0.3333 T_AC = 0, so T_AB = T_AC = T.
sum F_z = 0: 0.6667 T_AB + 0.6667 T_AC - 1500 = 0
1.3333 T = 1500, so T = 1125 N in each cable.
sum F_x = 0: F_OA - 0.6667 T_AB - 0.6667 T_AC = 0
F_OA = 0.6667 x (1125 + 1125) = 0.6667 x 2250 = 1500 N.

Sanity check: the vertical pulls are 0.6667 x 1125 = 750 N from each cable, and 750 + 750 = 1500 N, exactly balancing the load. The horizontal pulls toward the wall are also 750 N each, and the boom pushes back with 1500 N, so the x-column closes too. Physically the answers make sense: the boom is squeezed, the cables are stretched, and each cable carries more than half the load because it is inclined rather than vertical. Notice that the y-equation, involving nothing but the two cables, fell out first and cost one line - exactly the payoff the procedure promised. Note also that the boom force did not depend on the boom being long or short; only the direction of each member matters in a concurrent force system.

Reading the answers

The two cable tensions came out positive, confirming that cables can indeed carry these forces. The boom force came out positive under the assumption of compression, so the boom really is in compression, which is what a horizontal jib boom under a hanging load must be. Had it come out negative, the boom would be in tension - possible for a bar, impossible for a rope, and a signal that the geometry was misread.

A useful design observation hides in the numbers. Each cable carries 1125 N to support a 1500 N load, and the boom carries a full 1500 N of compression. Every member is loaded at or above the level of the load itself. That is normal in structures: geometry usually amplifies forces rather than reducing them, and the flatter the cables, the worse the amplification becomes.

Common wrong turns

• Guessing 3D angles from a flat drawing. A perspective sketch distorts every angle. Always work from coordinates.
• Forgetting a component because it looks small. Every non-zero fraction in a unit vector belongs in the sum, however modest.
• Assuming symmetry that is not there. Equal cable lengths do not by themselves guarantee equal tensions; the loads must be placed symmetrically too.
• Building the position vector from the wrong end. The force on the joint points from the joint toward the anchor, so d = (anchor) - (joint).
• Using the boom's length in place of its direction. In a concurrent system only directions matter, so a 4 m boom and a 40 m boom along the same line give the same answer.
• Skipping the column check. Adding each of the three component columns to zero at the end takes seconds and catches almost every arithmetic slip.

Recap

• A particle in space satisfies sum Fx = 0, sum Fy = 0, and sum Fz = 0: three equations, three possible unknowns.
• Get every direction from a unit vector built out of endpoint coordinates rather than from an angle read off a picture.
• Lay out a component table before writing any equation, then sum the three columns.
• Look for an axis that involves only one or two unknowns and solve that equation first.
• Symmetry, when genuinely present, replaces the full system with a single balance.
• Check the sign of every answer against what the member can physically do: cables pull only, struts and booms may push.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). 3D particle equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). 3D rigid body equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). 3D coordinate systems. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Equilibrium analysis for concurrent force systems. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. OpenStax. (2016). 12.2 Examples of static equilibrium. In University Physics Volume 1. Rice University. openstax.org
6. Massachusetts Institute of Technology. (2006). 2.001 Mechanics and materials I [Course materials]. MIT OpenCourseWare. ocw.mit.edu
7. Baker, D. W., & Haynes, W. (2021). Equilibrium of particles. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

A force can do more than push a body along; it can turn it. The moment of a force about a point measures that turning effect. Loosen a bolt with a wrench and you feel it directly: the same hand force turns the bolt more easily on a long wrench than a short one, because the moment depends on both the force and its distance.

Key idea: The moment of a force about a point measures its turning effect and equals the force times the perpendicular distance from the point to the force's line of action.

Why statics needs a second kind of equation

Everything up to this lesson used only sum F = 0, which is enough for a particle because a point cannot rotate. A real body has size, and forces applied at different places can balance perfectly while still spinning it. Push the left edge of a wheel forward and the right edge backward with equal force: the net force is zero and yet the wheel turns. The moment equation is what statics adds to rule that out.

This is also why the moment is the quantity engineers watch most closely. Overturning a retaining wall, tipping a crane, snapping a bolt in a flange, bending a beam - all are moment problems, and all can occur while the net force is comfortably balanced.

Force times perpendicular distance

The moment of a force F about a point O is
M_O = F x d,
where d is the perpendicular distance (the moment arm) from O to the line of action of the force. The unit is the newton-meter (N*m). A moment has a sense of rotation: counterclockwise is taken as positive, clockwise as negative, by convention. In three dimensions the sense is given by the right-hand rule: point the fingers of the right hand along the direction the force would rotate the body, and the thumb points along the moment axis.

Two features of this definition are worth dwelling on. First, the distance d is perpendicular to the line of action, not the distance to the point where the force happens to be applied. A force applied 2 m away along a bar but pointing straight at the pivot has a moment arm of zero and produces no turning at all. Second, because the moment uses the line of action, sliding the force along that line changes nothing - which is exactly the principle of transmissibility from Lesson 1 showing up again.

The sign convention is a choice, not a law. This course takes counterclockwise as positive, which matches the right-hand rule with z out of the page. Any consistent choice works, but it must be stated and then never changed inside a single problem.

Key idea: The moment arm is the perpendicular distance to the line of action, so a force pointed straight at the moment centre produces zero moment no matter how far away it is applied.

Varignon's theorem

Finding a true perpendicular distance in an awkward geometry is unpleasant. Varignon's theorem removes the need: the moment of a force about a point equals the sum of the moments of its components about that same point. Break the force into any convenient pieces, take each piece's moment separately, and add.

This is what makes the component formula below legitimate, and it also justifies a habit worth acquiring: slide the force along its line of action to a point where one of its components has an obvious moment arm, then take moments component by component. A force applied at the end of an L-shaped bracket is far easier to handle once its horizontal and vertical parts are treated separately.

Key idea: Varignon's theorem lets you replace one awkward moment with the sum of two easy ones, one per component.

The component method

Finding the true perpendicular distance can be awkward, so we often use components. Place O at the origin and let the force act at a point with position (x, y). Then
M_O = x Fy - y Fx.
This formula handles the geometry automatically. A positive result means counterclockwise; a negative result means clockwise.

Read the two terms physically and the formula stops being something to memorize. The term x Fy is a vertical force acting at a horizontal offset: push up on the right of O and the body turns counterclockwise, so a positive x with a positive Fy gives a positive moment. The term -y Fx is a horizontal force acting at a vertical offset: push right above O and the body turns clockwise, so a positive y with a positive Fx gives a negative moment. The minus sign is doing exactly that bookkeeping.

Note also that x and y are the coordinates of the point of application measured from the moment centre. If you take moments about a point other than the origin, subtract that point's coordinates first: M_A = (x - xA) Fy - (y - yA) Fx.

Worked example: a wrench

Given: a force of 100 N is applied at the end of a horizontal wrench, 0.20 m from the bolt O, and the force is directed 60 degrees above the horizontal. Find: the moment about O.
Solution, component method: the force acts at (0.20, 0) m with components Fx = 100 cos(60) = 50 N and Fy = 100 sin(60) = 86.6 N.

Then
M_O = x Fy - y Fx = (0.20)(86.6) - (0)(50) = 17.3 N*m, counterclockwise.
Check, perpendicular-distance method: only the vertical part of the force turns the bolt about O, because the horizontal part points straight at O and has zero moment arm. Its perpendicular distance is 0.20 m, so M_O = 86.6 x 0.20 = 17.3 N*m. The two methods agree.

The 3D moment: the cross product

In three dimensions the moment is the cross product M_O = r x F, where r is the position vector from O to any point on the line of action.

If r = (rx, ry, rz) and F = (Fx, Fy, Fz), the components are
Mx = ry Fz - rz Fy, My = rz Fx - rx Fz, Mz = rx Fy - ry Fx.
For example, a downward force F = (0, 0, -200) N acting at r = (2, 3, 0) m gives Mx = (3)(-200) - 0 = -600, My = 0 - (2)(-200) = 400, Mz = 0, so M_O = -600 i + 400 j N*m. Notice the 2D formula x Fy - y Fx is just the z-component of this cross product.

The cross product is most reliably evaluated as a determinant with i, j, k across the top row, the components of r in the second row, and the components of F in the third. Expanding along the top row reproduces the three formulas above. Two properties follow immediately and both are useful checks: the moment vector is always perpendicular to both r and F, and reversing the order to F x r flips its sign, so the order matters.

Key idea: In space the moment is the vector M_O = r x F, whose magnitude is F times the perpendicular distance and whose direction is the axis of rotation given by the right-hand rule.

Moment about a point versus about an axis

A bolt does not care about the whole moment vector; it cares about the part that twists it around its own axis. The moment about an axis is found by projecting the moment vector onto a unit vector u along that axis: M_axis = (r x F) dot u. The result is a single signed number rather than a vector. A door hinge, a shaft, and a lug nut are all axis problems, and the projection step is what separates the useful component from the part the bearing simply absorbs.

Worked example: force at a point on a bracket

Given: a 500 N force acts at point P(4, 3) m on a bracket and is directed 30 degrees below the positive x-axis.
Find: the moment of that force about the origin O, using the component formula, and verify it with Varignon's theorem.

Components first. The force points right and down, so Fy is negative.
Fx = 500 cos(30) = 500 x 0.8660 = 433.0 N
Fy = -500 sin(30) = -500 x 0.5000 = -250.0 N

Component formula.
M_O = x Fy - y Fx = (4)(-250.0) - (3)(433.0)
M_O = -1000 - 1299 = -2299 N*m,
that is, about 2300 N*m clockwise.

Check by Varignon. Take the two components separately. The 433.0 N horizontal component acts 3 m above O, and a rightward force above the centre turns the body clockwise, contributing -3 x 433.0 = -1299 N*m. The 250.0 N downward component acts 4 m to the right of O, and a downward force to the right turns the body clockwise too, contributing -4 x 250.0 = -1000 N*m. Their sum is -2299 N*m, matching exactly.

Sanity check: both components turn the bracket the same way, so the answer had to be larger in magnitude than either one alone, and it is. The equivalent perpendicular distance is d = 2299 / 500 = 4.60 m, which is plausible for a point sitting 5 m from the origin with the force not quite perpendicular to that radius. If the force had been exactly perpendicular to OP, the arm would have been the full 5 m and the moment 2500 N*m.

Worked example: a 3D moment by cross product

Given: a vertical downward force F = -600 k newtons is applied at the point A(3, 4, 0) m of a horizontal platform.
Find: the moment of this force about the origin, and its magnitude.

Solution: here r = 3 i + 4 j + 0 k and F = 0 i + 0 j - 600 k. Apply the three component formulas.
Mx = ry Fz - rz Fy = (4)(-600) - (0)(0) = -2400 N*m
My = rz Fx - rx Fz = (0)(0) - (3)(-600) = +1800 N*m
Mz = rx Fy - ry Fx = (3)(0) - (4)(0) = 0
So M_O = -2400 i + 1800 j newton-meters, with magnitude
|M_O| = sqrt(2400 squared + 1800 squared) = sqrt(5760000 + 3240000) = sqrt(9000000) = 3000 N*m.

Sanity check: the force is vertical and acts along a vertical line passing through (3, 4) in plan view, so the perpendicular distance from the origin to that line is sqrt(3 squared + 4 squared) = 5 m. The scalar moment must then be 600 x 5 = 3000 N*m, which matches the vector magnitude exactly. As a second check, M_O should be perpendicular to r: their dot product is (3)(-2400) + (4)(1800) + 0 = -7200 + 7200 = 0, as required. The moment vector lies in the horizontal plane, which is right - a downward force tips the platform about a horizontal axis, never about the vertical one.

Common wrong turns

• Using the distance to the point of application instead of the perpendicular distance. Only the perpendicular offset from the line of action counts. When in doubt, use components and let the formula handle it.
• Forgetting that a force through the moment centre contributes nothing. This is a feature, not an oversight: choosing the moment centre at an unknown reaction is how that unknown is eliminated.
• Losing the minus sign in x Fy - y Fx. Writing x Fy + y Fx makes horizontal forces turn the body the wrong way.
• Measuring coordinates from the origin when taking moments about another point. Subtract the moment centre's coordinates first.
• Reversing the cross product to F x r. The order is position first, force second; swapping it reverses every moment direction.
• Switching sign conventions mid-problem. Choose counterclockwise-positive at the start and keep it for every term, including applied couples.

Recap

• The moment of a force about a point is M_O = F d, with d the perpendicular distance to the line of action.
• Counterclockwise is taken as positive here, matching the right-hand rule with z out of the page.
• Varignon's theorem lets you replace one moment with the sum of the moments of the components.
• In the plane, M_O = x Fy - y Fx, with coordinates measured from the moment centre.
• In space, M_O = r x F, a vector perpendicular to both r and F.
• The moment about a particular axis is the projection (r x F) dot u onto that axis.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Moment of a force. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Scalar addition of moments. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Varignon's theorem. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). 3D moments. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Moment about a point. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. OpenStax. (2016). 10.6 Torque. In University Physics Volume 1. Rice University. openstax.org
7. Nave, R. (n.d.). Torque and equilibrium. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

Two forces that are equal in magnitude, opposite in direction, and separated by a distance form a couple. A couple produces pure rotation: its forces cancel, so it exerts no net push, yet it still turns the body. Think of the two hands turning a steering wheel in opposite directions.

Key idea: A couple is two equal, opposite, non-collinear forces; it exerts zero net force but a definite moment, so it produces pure rotation.

Couples in real hardware

Couples are not a textbook curiosity. A torque wrench applies one: your two hands, or the wrench body and the socket, push in opposite directions across the handle. An electric motor delivers torque to a shaft as a couple, which is why a motor rated at 40 N*m says nothing at all about any net force. A screwdriver, a steering wheel, a tap turning a thread, and the bending action on a welded joint are all couples. Whenever a device is specified by a torque rating rather than a force rating, the physical action behind it is a couple.

The moment of a couple

The moment of a couple is
M = F x d,
where F is the magnitude of one of the forces and d is the perpendicular distance between their two lines of action. A remarkable fact makes couples easy to work with: the moment of a couple is the same about every point in the plane. It is a free vector; only its magnitude and sense matter, not where you measure it. That is why you can slide a couple anywhere on a body without changing its effect.

The claim is easy to verify. Put the two forces of magnitude F on vertical lines at x = a and x = a + d, the first pointing up and the second down. Take moments about any point x = p. The first contributes (a - p)(F) and the second contributes (a + d - p)(-F). Adding gives F(a - p) - F(a + d - p) = -F d. Every appearance of p has cancelled, so the answer does not depend on where you measured from. That cancellation is the whole reason a couple is called a free vector.

The practical payoff is large. When a couple appears in an equilibrium equation, you write the same number no matter which point you chose as the moment centre. There is no moment arm to compute and no geometry to get wrong - a rare and welcome simplicity in this subject.

Key idea: A couple's moment is the same about every point, so it enters any moment equation as a single fixed number with no moment arm.

Worked example: a couple

Given: two forces of 50 N each act in opposite directions, their lines of action 0.4 m apart. Find: the moment of the couple.
Solution: M = F d = 50 x 0.4 = 20 N*m. If the pair tends to rotate the body counterclockwise, the couple is +20 N*m, and this value is the same no matter which point you take moments about.

Sanity check: confirm the free-vector property numerically. Put the two 50 N forces on vertical lines at x = 0 (upward) and x = 0.4 m (downward), and take moments about the origin: (0)(50) - (0.4)(50) = -20 N*m. Now take moments about x = 3 m instead: (0 - 3)(50) - (0.4 - 3)(50) = -150 + 130 = -20 N*m. Identical, as promised. Note too that the net force is 50 - 50 = 0, so the couple cannot translate the body at all.

Moving a force: the force-couple system

Often we want to slide a force from where it acts to a more convenient point, such as a support. You cannot simply move a force to a new line of action without changing its turning effect. But you can move it if you add a couple that makes up for the change.

Moving a force F from its point of application to a new point O requires adding a couple equal to the moment that the force had about O:
M_O = (moment of the original force about O).
The result is an equivalent force-couple system: the same force F now acting at O, plus a couple M_O. It has exactly the same effect on the rigid body as the original force.

Worked example: replace a force by a force-couple at O

Given: an 80 N downward force acts at a point 0.5 m to the right of O. Find: the equivalent force-couple system at O.
Solution: The force stays the same, 80 N downward, but now placed at O. The couple we must add equals the moment the force had about O:
M_O = x Fy - y Fx = (0.5)(-80) - 0 = -40 N*m, i.e. a 40 N*m clockwise couple.
So the equivalent system at O is an 80 N downward force plus a 40 N*m clockwise couple.

This move-and-add-a-couple idea is exactly how distributed loads and off-center forces get reduced to something the equilibrium equations can digest.

Sanity check on that example: the two systems must have the same net force and the same moment about every point. Net force: 80 N down in both. Moment about O: originally (0.5)(-80) = -40 N*m; in the replacement the force sits at O and contributes nothing, while the added couple supplies exactly -40 N*m. Moment about a second point, say x = 2 m: originally (0.5 - 2)(-80) = +120 N*m; in the replacement (0 - 2)(-80) + (-40) = 160 - 40 = +120 N*m. The two systems agree everywhere, which is what "equivalent" means.

Key idea: Any force can be moved to a new point provided you add a couple equal to the moment that force had about the new point.

Reducing a whole system of forces

The move-and-add-a-couple trick generalizes. Take any collection of forces and couples acting on a rigid body, pick a point O, and slide every force to O, adding the appropriate couple each time. What is left is a single resultant force R = sum F acting at O, plus a single resultant couple M_O = sum M_O. Every force system in two dimensions, however complicated, reduces to that pair of numbers.

Because the couple depends on the choice of O while the resultant force does not, you can often choose O to make the couple vanish. For a coplanar system with R not zero, there is always a line of action along which a single force R, with no couple at all, reproduces the whole system. Finding it is a one-step calculation: place the resultant at the distance d = M_O / R from O, on whichever side makes the sign come out right.

Three outcomes are possible, and reading which one you are in tells you a great deal about the loading:

• R not zero: the system reduces to a single force on a definite line of action.
• R = 0 but M_O not zero: the system is a pure couple, turning the body without pushing it.
• R = 0 and M_O = 0: the system is in equilibrium, which is the whole subject of the next lesson.

Worked example: reducing a beam loading to one force

Given: a horizontal beam lies along the x-axis with its left end at the origin O. It carries a 300 N downward force at x = 1 m, a 500 N downward force at x = 4 m, and an applied counterclockwise couple of 200 N*m.
Find: the equivalent force-couple system at O, and then the single force that would replace the entire loading, including where its line of action crosses the beam.

Resultant force. Only vertical forces act, and the couple contributes none.
R = -300 - 500 = -800 N, that is 800 N downward.

Resultant couple at O. Take counterclockwise as positive and use M = x Fy for each vertical force.
M_O = (1)(-300) + (4)(-500) + 200
M_O = -300 - 2000 + 200 = -2100 N*m, a 2100 N*m clockwise couple.
So the force-couple system at O is 800 N down plus 2100 N*m clockwise.

Single equivalent force. Move the 800 N downward force to a position x = d chosen so that its own moment about O reproduces -2100 N*m:
(d)(-800) = -2100
d = 2100 / 800 = 2.625 m.
A single 800 N downward force applied 2.625 m from the left end is statically equivalent to the entire original loading.

Sanity check: test the equivalence at a different point, the right-hand location x = 4 m. Original system: (1 - 4)(-300) + (4 - 4)(-500) + 200 = 900 + 0 + 200 = 1100 N*m. Replacement: (2.625 - 4)(-800) = (-1.375)(-800) = 1100 N*m. They match, so the reduction is correct. The location is also plausible: 2.625 m lies between the two loads and nearer the larger one, but pulled left by the counterclockwise couple. Without the couple the resultant would sit at 2300 / 800 = 2.875 m, so the 200 N*m couple shifts the line of action a quarter of a metre.

Common wrong turns

• Giving a couple a moment arm. A couple's moment is a fixed number. Multiplying it by a distance is a dimensional error that produces N*m squared.
• Believing a couple has a net force. The two forces cancel exactly, so a couple never appears in sum Fx = 0 or sum Fy = 0, only in sum M = 0.
• Sliding a force to a new line of action without adding the couple. That changes the moment about every point and silently produces a different problem.
• Using the wrong separation distance. In M = F d for a couple, d is the perpendicular distance between the two lines of action, not the distance between the points where the forces are drawn.
• Assuming the reduced single force can always be found. If the resultant force is zero, no single force can represent the system; it is a pure couple.
• Changing the sign convention between the force step and the couple step. One convention, applied to every term.

Recap

• A couple is a pair of equal, opposite, parallel forces on different lines of action, with moment M = F d.
• Its net force is zero and its moment is the same about every point, so it is a free vector.
• A force may be moved to any point provided a couple equal to its moment about that point is added.
• Any planar force system reduces to a resultant force at a chosen point plus a resultant couple.
• If the resultant force is not zero, the couple can be absorbed by shifting the force to the line d = M_O / R.
• Zero resultant force with a non-zero couple means pure rotation; both zero means equilibrium.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Couples. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Equivalent transformations. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Statically equivalent systems. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Couples. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Equivalent force couple system. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Statically equivalent systems. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Baker, D. W., & Haynes, W. (2021). Moments and static equivalence. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

A rigid body has size, so forces act at different places and can turn it. In a plane, equilibrium of a rigid body requires three conditions:

• sum Fx = 0
• sum Fy = 0
• sum M = 0 about any point

Three equations can solve for three unknowns, typically the reactions at the supports. A key strategy: take moments about a point where an unknown force acts, so that force drops out (its moment arm is zero), and the moment equation contains fewer unknowns, often just one.

Key idea: A rigid body in the plane gives three independent equations, so it can resolve exactly three unknown reaction components.

Alternative sets of three equations

The set sum Fx = 0, sum Fy = 0, sum M_A = 0 is the familiar one, but it is not the only valid choice. Two others are often more convenient:

• One force and two moments: sum Fx = 0, sum M_A = 0, sum M_B = 0, provided the line AB is not perpendicular to the x-axis.
• Three moments: sum M_A = 0, sum M_B = 0, sum M_C = 0, provided A, B, and C do not lie on a straight line.

These alternatives are useful because a moment equation taken at a support kills two unknowns at once. Choosing two well-placed moment centres can hand you two reactions directly, with no simultaneous algebra at all. The restrictions matter, though: three collinear moment points do not give three independent equations, and using them produces a system that looks solvable but is not.

Whichever set you use, remember that only three of them are independent. Writing a fourth equation is never wrong, but it can never produce new information - which makes it the perfect check, as the worked example below demonstrates.

Key idea: Any three independent equilibrium equations work; a fourth is always a check and never a new fact.

Distributed loads

A load spread along a length, like the weight of a floor or the pressure of wind, is a distributed load, measured in newtons per meter. For a uniform distributed load of intensity w over a length L, the total force is w L, and it acts at the centroid of the load, which for a uniform strip is at its midpoint. Replace the distributed load with this single equivalent force before writing equilibrium, exactly as the previous lesson taught.

Not every distributed load is uniform. Two more shapes cover most of what a first course needs:

• A triangular load rising from zero to a peak intensity w over a length L has a total of w L / 2, the area of the triangle, acting at the triangle's centroid: one third of the length from the tall end, two thirds from the zero end.
• A trapezoidal load is easiest handled by splitting it into a rectangle plus a triangle and treating each piece separately, exactly as with composite areas in Module 6.

In every case the rule is the same and worth stating once: the equivalent force equals the area under the loading diagram, and it acts through the centroid of that area. Units are a good check here. An intensity in N/m multiplied by a length in m gives newtons, so if your equivalent load comes out in N/m you have forgotten to multiply by the length.

Key idea: Replace a distributed load by a single force equal to the area under the load diagram, placed at that area's centroid.

Worked example: a simply supported beam

Given: a horizontal beam 8 m long rests on a pin at A (left end, x = 0) and a roller at B (right end, x = 8). It carries a 600 N downward point load at x = 2 m, a uniform distributed load of 100 N/m over the right half from x = 4 to x = 8, and an applied clockwise couple of 300 N*m. Find: the support reactions.
Free-body diagram: the pin at A supplies horizontal and vertical reactions Ax and Ay; the roller at B supplies a vertical reaction By.

The distributed load totals 100 x 4 = 400 N and acts at the middle of the right half, x = 6 m.
Solve the horizontal equation first: no horizontal loads act, so
sum Fx = 0: Ax = 0.
Take moments about A (counterclockwise positive) to eliminate Ax and Ay:
sum M_A = 0: By(8) - 600(2) - 400(6) - 300 = 0
8 By = 1200 + 2400 + 300 = 3900
By = 487.5 N.
Vertical balance gives the remaining reaction:
sum Fy = 0: Ay + By - 600 - 400 = 0
Ay = 1000 - 487.5 = 512.5 N.
The reactions are Ax = 0, Ay = 512.5 N, By = 487.5 N.

Independent check. Write the fourth equation, moments about B, keeping counterclockwise positive throughout. Measuring each position from B at x = 8:
sum M_B = (0 - 8)(Ay) + (2 - 8)(-600) + (6 - 8)(-400) - 300
sum M_B = -8(512.5) + 3600 + 800 - 300
sum M_B = -4100 + 4100 = 0.
It balances, so the reactions are confirmed. Note that the clockwise 300 N*m couple enters with the same -300 about either point, because a couple's moment is the same everywhere; the load terms, by contrast, change both magnitude and sign because their moment arms are measured from a different centre.

Worked example: triangular load with a horizontal thrust

Given: a beam of span 6 m rests on a pin at A (x = 0) and a roller at B (x = 6). It carries a triangular distributed load that is zero at A and rises linearly to 900 N/m at B. A horizontal force of 400 N also acts on the beam, pointing to the right along the beam axis.
Find: all three support reactions.

Free-body diagram in words: the beam alone. At A the pin supplies Ax and Ay, both drawn positive to the right and upward. At B the roller supplies a single vertical force By, drawn upward. The triangular load is replaced by its equivalent point force, and the 400 N horizontal force acts along the beam axis so it has no moment about A.

Replace the distributed load. The area of the load triangle is
W = (1/2)(900)(6) = 2700 N,
acting at the centroid of the triangle, two thirds of the span from the zero end:
x_bar = (2/3)(6) = 4.0 m from A.

Equilibrium. Take moments about A first, so Ax and Ay drop out.
sum M_A = 0: By(6) - 2700(4.0) = 0
6 By = 10800, so By = 1800 N upward.
sum F_y = 0: Ay + 1800 - 2700 = 0, so Ay = 900 N upward.
sum F_x = 0: Ax + 400 = 0, so Ax = -400 N.

The negative sign on Ax is not an error. It means the pin actually pushes the beam 400 N to the left, opposite the direction drawn, which is exactly what must happen to resist a 400 N thrust to the right. Reverse the arrow and report 400 N to the left.

Sanity check: take moments about B as an independent test. sum M_B = -6(900) + (4.0 - 6)(-2700) = -5400 + 5400 = 0. It closes. The split of the vertical load is also physically sensible: the load is heaviest near B, so B carries twice as much as A, 1800 N against 900 N, and the two together equal the 2700 N total. Had the load been uniform, the two reactions would have been equal at 1350 N each, so the triangular shape has shifted a full 450 N from A onto B.

Statical determinacy

Three equations solve three unknown reactions. If a body has exactly three unknown reaction components, it is statically determinate and solvable by statics alone. With more unknowns than equations it is statically indeterminate, and you need extra information about how the body deforms, which is a topic for a later course. With fewer, it is a mechanism and cannot stay in equilibrium under general loads.

Counting is necessary but not sufficient. A body can have exactly three reaction components and still be unable to stand up, a condition called improper constraint. Two arrangements cause it:

• Concurrent reactions. If the lines of action of all three reactions pass through a single point, no combination of them can produce a moment about that point, so any load with a moment there spins the body.
• Parallel reactions. If all three reactions are parallel, say all vertical, nothing resists a horizontal load and the body slides away.

The practical test is simple: after counting to three, ask whether the supports can resist a push in every direction and a twist. If any one of those three motions is unopposed, the arrangement fails regardless of the count.

Key idea: Three reaction components are necessary for determinacy but not sufficient for stability - reactions that are all concurrent or all parallel leave a motion unrestrained.

Common wrong turns

• Placing a distributed load's resultant at the wrong point. A uniform strip acts at its midpoint, a triangle at one third of the way from its tall end. Using the midpoint for a triangular load is a classic and expensive slip.
• Forgetting the applied couple in the moment equation. A couple contributes to sum M even though it never appears in the force equations.
• Changing the sign convention between two moment equations. Pick counterclockwise-positive once and apply it to every term in every equation, including the couple.
• Measuring moment arms from the wrong point. When taking moments about B, every position must be measured from B, not from the origin.
• Treating a negative reaction as a mistake. It means the assumed arrow was backwards. Reverse it, keep the magnitude, and carry the corrected sense forward.
• Assuming a roller can take horizontal load. It cannot, which is why the pin in the second example had to absorb the full 400 N thrust.
• Counting three reactions and declaring victory. Check for concurrency and parallelism as well.

Recap

• A rigid body in a plane requires sum Fx = 0, sum Fy = 0, and sum M = 0 about any point.
• Alternative valid sets use two or three moment equations, subject to the non-perpendicular and non-collinear restrictions.
• Take moments about a point where an unknown acts to eliminate it from that equation.
• Replace a distributed load by the area under its diagram, acting at the centroid of that area.
• Always write one extra equation as a check; it costs a line and catches most arithmetic errors.
• Three reaction unknowns make a body statically determinate, but only if the reactions are neither all concurrent nor all parallel.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). 2D rigid body equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Equations of equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Stability and determinacy. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). Distributed loads. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Equilibrium analysis for a rigid body. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Distributed forces. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Baker, D. W., & Haynes, W. (2021). Rigid body equilibrium. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

Structures are held in place by supports, and each type of support can supply only certain reactions. Knowing exactly what each support can and cannot do is what turns a real connection into a solvable free-body diagram. The rule of thumb: a support provides a reaction force in any direction it prevents motion, and a reaction moment if it prevents rotation.

Key idea: Every support supplies exactly one reaction for each motion it blocks - no more and no fewer.

Degrees of freedom: the rule behind the table

A rigid body floating free in a plane has three degrees of freedom: it can slide horizontally, slide vertically, and rotate. Holding it still means removing all three. Each reaction component removes exactly one degree of freedom, which is why three is the magic number in two dimensions and why the count of reactions and the count of equations are always compared.

Read any real connection through that lens and its idealization becomes obvious. Ask, one motion at a time: can the body slide left or right here? Can it slide up or down? Can it turn? Every "no" is a reaction. A roller says no to motion perpendicular to its surface and yes to everything else, so one reaction. A pin says no to both slides but yes to rotation, so two. A welded or grouted end says no to all three, so three.

Key idea: A planar rigid body has three degrees of freedom, and each reaction component removes exactly one of them.

The common 2D supports

Support	Prevents	Reactions supplied	Unknowns
Roller / smooth surface	Motion perpendicular to the surface	One force, perpendicular to the surface	1
Cable	Motion away along the cable	One force (tension) along the cable	1
Pin / hinge	Motion in any direction	Two force components (Ax and Ay)	2
Fixed / built-in	Motion in any direction and rotation	Two force components plus a moment	3
Link or two-force member	Motion along the link	One force, along the line joining its two pins	1
Collar on a smooth shaft	Motion perpendicular to the shaft, and rotation	One perpendicular force plus a moment	2

Two entries in the table deserve emphasis. A roller on an inclined surface still supplies just one force, but that force is perpendicular to the inclined surface, not vertical. Students who automatically draw roller reactions straight up get every inclined-support problem wrong. A link is even more restrictive than a cable: it can push as well as pull, but only along the straight line joining its two end pins, which is the two-force-member result you will meet again in Module 4.

Reading the table

A roller can push only perpendicular to its surface; it cannot resist a force along the surface, so it gives one unknown. A pin (or hinge) stops the body from sliding in any direction but lets it rotate freely, so it gives two force components and no moment. A fixed support, like a beam cemented into a wall, stops both sliding and turning, so it supplies two force components and a reaction couple, three unknowns in all. This is why a cantilever beam, fixed at one end and free at the other, is held by exactly three reaction components and is statically determinate.

Counting for determinacy

Add up the reaction unknowns and compare with the number of equilibrium equations, which is three for a single rigid body in 2D.

• A beam on a pin plus a roller has 2 + 1 = 3 unknowns, matching three equations, so it is statically determinate.
• A fixed cantilever has 3 unknowns, also determinate.
• A beam on two pins has 2 + 2 = 4 unknowns, one more than the equations, so it is statically indeterminate.
• A beam on a single roller has 1 unknown, two fewer than the equations, so it is a mechanism: it will slide and rotate away under a general load.

Indeterminate does not mean badly designed. Most real structures are deliberately indeterminate, because redundant supports give a load path if one connection fails and reduce deflections. It only means statics alone cannot find the reactions; you also need to know how the structure deforms, which is the subject of a mechanics-of-materials or structural-analysis course. Statics simply reports honestly that its three equations are not enough.

Key idea: Reactions fewer than three means a mechanism, exactly three means determinate, more than three means indeterminate and solvable only with deformation information.

Supports in three dimensions

In space a rigid body has six degrees of freedom - three slides and three rotations - so equilibrium supplies six equations and six reaction components are needed. The common 3D supports follow the same motion-blocking logic:

• A ball-and-socket joint blocks all three slides but no rotation: three force components.
• A journal bearing or smooth pin blocks two slides and two rotations, though in most textbook problems the couples are assumed negligible, leaving two force components.
• A fixed or built-in support blocks everything: three force components plus three couple components, six unknowns in all.
• A smooth surface or roller blocks one slide: one force, normal to the surface.

The counting rule is unchanged: compare the number of reaction components with six rather than three.

Worked example: cantilever reactions

Given: a cantilever beam of length 4 m is built into a wall at A (x = 0) and carries a single 500 N downward load at its free end (x = 4).

Find: the wall reactions.
Solution: the fixed support supplies Ax, Ay, and a reaction couple M_A.
sum Fx = 0: Ax = 0.
sum Fy = 0: Ay - 500 = 0, so Ay = 500 N.
sum M_A = 0: M_A - 500(4) = 0, so M_A = 2000 N*m (counterclockwise, holding the beam up).
The wall must push up with 500 N and supply a 2000 N*m reaction moment. That large moment is why a cantilever feels the load most severely right at the wall, a fact the next module on internal forces makes precise.

Sanity check: test the answer with a moment equation about the free end at x = 4. Measuring positions from there, sum M = (0 - 4)(Ay) + M_A = -4(500) + 2000 = 0. It closes. The signs are also physically right: the wall pushes up to carry the weight and twists counterclockwise to stop the beam from rotating down, which is exactly what you feel when you hold a heavy plank out at arm's length.

Worked example: a beam with two overhangs

Given: a beam runs from x = 0 to x = 8 m. It is supported by a pin at A(x = 2) and a roller at B(x = 6), so both ends overhang. It carries a 500 N downward point load at x = 0, a uniform load of 200 N/m from x = 2 to x = 6, and an 800 N downward point load at x = 8.
Find: the reactions at A and B.

Free-body diagram in words: the pin at A supplies Ax and Ay; the roller at B supplies only a vertical By. That is three unknowns against three equations, so the beam is statically determinate. The uniform load is replaced by 200 x 4 = 800 N acting at the midpoint of its strip, x = 4.

Equilibrium. Take moments about A, measuring every position from x = 2, counterclockwise positive.
500 N down at x = 0: (0 - 2)(-500) = +1000 N*m
800 N down at x = 4: (4 - 2)(-800) = -1600 N*m
800 N down at x = 8: (8 - 2)(-800) = -4800 N*m
By up at x = 6: (6 - 2)(By) = +4 By
sum M_A = 0: 1000 - 1600 - 4800 + 4 By = 0
4 By = 5400, so By = 1350 N upward.
sum F_y = 0: Ay + 1350 - 500 - 800 - 800 = 0, so Ay = 750 N upward.
sum F_x = 0: Ax = 0, since no horizontal loads act.

Sanity check: take moments about B, measuring from x = 6. (0 - 6)(-500) + (4 - 6)(-800) + (8 - 6)(-800) + (2 - 6)(750) = 3000 + 1600 - 1600 - 3000 = 0. It balances. The result is also intuitive: the 800 N load hanging out past B is far from A, so it loads B heavily, while the 500 N load overhanging on the left partly relieves A. Total vertical check: 750 + 1350 = 2100 N, matching 500 + 800 + 800 = 2100 N.

One design remark. Had the left overhang carried nothing, Ay would have been larger and By smaller. Overhangs are used deliberately in bridges and balconies for exactly that reason: a load beyond a support pushes the far reaction up and can even lift it, which is why uplift anchors appear on cantilevered structures.

Common wrong turns

• Drawing a roller reaction vertically on an inclined surface. A roller pushes perpendicular to the surface it rides on, whatever that surface's angle.
• Giving a pin a reaction moment. A pin lets the body rotate, so it supplies two forces and no couple. Only a fixed support has a reaction couple.
• Forgetting the reaction moment at a fixed support. Leaving M_A out of a cantilever gives two equations for a body that needs three and produces nonsense.
• Assuming a cable can push. A cable is one unknown pulling along its own line; if the algebra makes it negative, it has gone slack.
• Counting reactions but not checking arrangement. Three parallel or three concurrent reactions still leave the body free to move.
• Treating "indeterminate" as "wrong". Most real structures are indeterminate on purpose; it just means statics alone cannot finish the job.

Recap

• Each support supplies one reaction for each degree of freedom it removes.
• Roller or cable gives one unknown, pin gives two, fixed support gives three; a link gives one along its own line.
• A planar body has three degrees of freedom, so three reaction components make it statically determinate.
• Fewer than three means a mechanism; more than three means indeterminate and needs deformation information.
• In three dimensions the same logic applies with six degrees of freedom and six equations.
• Always confirm reactions with a second moment equation taken at a different point.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Degree of freedom. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Rigid body equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Stability and determinacy. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Engineering structures. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. OpenStax. (2016). 12.1 Conditions for static equilibrium. In University Physics Volume 1. Rice University. openstax.org
6. Buyukozturk, O. (2003). 1.051 Structural engineering design [Course materials]. MIT OpenCourseWare. ocw.mit.edu
7. Baker, D. W., & Haynes, W. (n.d.). Engineering Statics: Open and Interactive [PDF]. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org

A truss is a structure built from straight members joined at their ends to form a rigid framework, most often a pattern of triangles. Roofs, bridges, cranes, and transmission towers are trusses. Their great advantage is efficiency: each member carries load along its length only, so material is used where it counts.

Key idea: A truss is a pin-jointed framework of straight members loaded only at its joints, so every member carries pure tension or pure compression.

Why triangles

Build a square from four bars with pinned corners and push on one corner: it collapses into a parallelogram without any bar changing length. Build a triangle from three bars and it cannot deform at all unless a bar stretches or shortens. The triangle is the only polygon whose shape is fixed by its side lengths alone, and that single geometric fact is why trusses are triangulated.

A simple truss is built by starting with one triangle and adding two new members and one new joint at a time. Every truss made this way is rigid and, as the counting rule below shows, statically determinate. Roof trusses, the Warren and Pratt patterns in railway bridges, crane booms, and stage lighting rigs are all built on this principle, and so are the great nineteenth-century truss bridges that are now recognized as engineering landmarks.

Key idea: Triangles are the only rigid pin-jointed polygons, so triangulation is what makes a truss stiff without making it heavy.

The idealizing assumptions

Truss analysis rests on two assumptions that make the problem clean:

1. Members are connected by frictionless pins at their ends, so joints transmit force but not moment.
2. All loads and reactions are applied at the joints, not along the members.

Real trusses violate both assumptions. Members are usually bolted or welded to gusset plates, which do resist a little rotation, and every member carries its own weight along its length. Engineers still use the idealization because the errors it introduces, called secondary stresses, are typically only a few percent of the axial forces. The usual convention for self-weight is to lump half a member's weight at each of its two end joints, which restores the joint-loading assumption without ignoring the mass.

A third assumption hides in the geometry: the members' centrelines must all meet at a single point at each joint. Connections built with an offset introduce genuine bending, which is why detailers work hard to make the work lines intersect.

Key idea: Pin joints and joint-only loading are idealizations that are accurate to within a few percent, provided member centrelines actually meet at each joint.

Why members are two-force members

Under these assumptions each member has forces applied at only two points, its two end pins, and no load in between. A body with forces at just two points, in equilibrium, is a two-force member: the two forces must be equal, opposite, and directed along the line joining the two points. Therefore every truss member carries a force purely along its own length, either pulling its joints together or pushing them apart.

• If a member pulls its end joints toward each other, it is in tension. We report tension as positive.
• If a member pushes its end joints apart, it is in compression. We report compression as negative.

This is a large simplification. Instead of a force with an unknown magnitude and an unknown direction, each member has a single unknown, its magnitude, because the direction is fixed along the member. Half the difficulty disappears.

Key idea: A body loaded at only two points must carry a force along the line joining them, so each truss member has one unknown magnitude instead of an unknown magnitude and direction.

Worked example: the force a two-force member applies

Given: a truss member AB runs from joint A(0, 0) to joint B(3, 4) m and carries 250 N of tension.
Find: the force components the member applies to each of its two end joints.

Solution: first the direction.
d = B - A = 3 i + 4 j, |d| = sqrt(9 + 16) = 5 m, so u = 0.6 i + 0.8 j.
A member in tension pulls each joint toward the other joint. So at A the member pulls in the direction of +u, and at B it pulls in the direction of -u.
On joint A: 250 x (0.6 i + 0.8 j) = 150 i + 200 j newtons.
On joint B: 250 x (-0.6 i - 0.8 j) = -150 i - 200 j newtons.

Sanity check: the two forces are equal and opposite, as Newton's third law and member equilibrium both demand, and each has magnitude sqrt(150 squared + 200 squared) = sqrt(62500) = 250 N, matching the stated tension. Had the member been in compression instead, every sign would flip: it would push its joints apart, giving -150 i - 200 j at A and +150 i + 200 j at B. Nothing else about the calculation would change, which is exactly why one signed number per member is enough.

Zero-force members

Some truss members carry no load at all under a given loading. Spotting them before you calculate saves large amounts of work, and two rules find nearly all of them:

• Two-member rule. If exactly two non-collinear members meet at an unloaded joint, both carry zero force. Resolving perpendicular to one member gives F sin(angle) = 0 for the other, so it is zero; then the first is zero too.
• Three-member rule. If three members meet at an unloaded joint and two of them are collinear, the third member carries zero force. Resolving perpendicular to the collinear pair leaves only the odd member in the equation.

To see the second rule in numbers, take a joint where members CD and DE lie along the horizontal and member BD rises at 40 degrees, with no external load applied at D. Summing forces perpendicular to CD-DE, only BD has a component:
sum F_perp = F_BD sin(40) = 0, and since sin(40) = 0.643 is not zero, F_BD = 0.

Zero-force members are not useless. They brace long compression members against buckling, they carry load under a different loading case such as wind from the other side, and they hold the geometry during construction. A member that is zero-force today may be the critical member tomorrow, which is why they are never simply deleted from a design.

Key idea: Two non-collinear members at an unloaded joint are both zero-force, and the odd member out of three at an unloaded joint with two collinear is zero-force.

Determinacy of a truss

A planar truss with m members, r reaction components, and j joints is statically determinate when
m + r = 2 j.
The reason is that each joint gives two equilibrium equations (it is a particle, so sum Fx = 0 and sum Fy = 0), for 2 j equations total, and the unknowns are the m member forces plus the r reactions. When they match, the truss can be solved by statics. If m + r exceeds 2 j, the truss is indeterminate; if it is less, the truss is unstable.

As an example, a truss with 7 members, 3 reaction components, and 5 joints satisfies 7 + 3 = 10 = 2 x 5, so it is determinate, and we can find every member force. That very truss is the one we analyze in the next two lessons, first joint by joint, then with a single strategic cut.

Worked example: determinacy of three related trusses

Given: the base truss has joints A(0, 0), C(4, 0), E(8, 0), B(2, 3), and D(6, 3), with the seven members AB, AC, BC, BD, CD, CE, DE. It is supported by a pin at A and a roller at E.
Find: whether the truss is determinate, and how the answer changes if (a) an extra diagonal BE is added, or (b) member CD is removed.

Base truss. Count first: m = 7 members, r = 3 reaction components (2 at the pin, 1 at the roller), and j = 5 joints.
m + r = 7 + 3 = 10
2 j = 2 x 5 = 10
They match, so the truss is statically determinate and every member force can be found from statics alone.

Case (a): add member BE. Now m = 8, r = 3, j = 5.
m + r = 11 against 2 j = 10.
There is one more unknown than there are equations, so the truss is statically indeterminate to the first degree. Statics can find the reactions and some member forces but cannot distribute the load between BE and the members it parallels without knowing how much each stretches.

Case (b): remove member CD. Now m = 6, r = 3, j = 5.
m + r = 9 against 2 j = 10.
There is one equation more than there are unknowns, which means the structure cannot satisfy equilibrium under a general load. It is unstable, a mechanism: the panel that CD used to brace can now distort like a parallelogram.

Sanity check: the count agrees with the geometry. The base truss is a chain of triangles ABC, BCD, and CDE, exactly the simple-truss construction of starting from one triangle and adding two members with each new joint. Adding BE creates a redundant load path; deleting CD leaves the quadrilateral BCDE with no diagonal, and a pin-jointed quadrilateral collapses. Counting and looking at the picture should always agree, and if they disagree, the picture is right and the count has an error in it.

Common wrong turns

• Applying a load in the middle of a member and still calling it a truss. A member loaded along its length is not a two-force member; it bends, and the structure becomes a frame.
• Assuming a zero-force member can be removed. It carries nothing under this loading only, and it may be bracing a compression member against buckling.
• Reporting a member force without saying tension or compression. The magnitude alone is only half the answer, and the sign convention must be stated.
• Miscounting joints or members in m + r = 2 j. Count reaction components, not supports: a pin is two, a roller is one.
• Concluding "determinate" means "safe". The count says the forces are computable, not that the members are strong enough.
• Forgetting that compression members can buckle. A slender member is far weaker in compression than in tension, which is why the top chord of a bridge truss is usually the heaviest section.

Recap

• A truss is a pin-jointed framework of straight members with all loads applied at the joints.
• Triangles are the only rigid pin-jointed shapes, so trusses are triangulated.
• Every member is a two-force member, carrying force along its own line: tension positive, compression negative.
• Two non-collinear members at an unloaded joint are both zero-force; the odd member of three with two collinear is zero-force.
• A planar truss is determinate when m + r = 2 j, indeterminate when greater, and unstable when less.
• The pin-joint idealization introduces only small secondary stresses provided member centrelines meet at each joint.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Trusses. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Structures. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Interactions between members. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Two force members. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Baker, D. W., & Haynes, W. (2021). Equilibrium of structures. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org
6. American Society of Civil Engineers. (n.d.). Historic civil engineering landmarks. ASCE. asce.org
7. Wikipedia contributors. (n.d.). Truss. In Wikipedia. en.wikipedia.org

The method of joints finds every member force by treating each joint as a particle in equilibrium. Since each joint gives two equations, sum Fx = 0 and sum Fy = 0, you can solve a joint that has at most two unknown member forces, then march to the next.

Key idea: Every joint of a truss is a particle in equilibrium, giving two equations, so start at a joint with no more than two unknown members and work outward.

The sign convention that makes this painless

Assume every unknown member is in tension. On the free-body diagram of a joint, that means drawing each member force pulling away from the joint, along the member. Do this without exception, even when you are sure a member is squeezed.

The payoff is a single unambiguous rule for reading the answers: a positive result means tension, a negative result means compression, and the magnitude is correct either way. You never have to redraw a diagram or reason about direction after the fact. Mixing assumed directions from joint to joint is the fastest way to end up with a member that is in tension at one end and compression at the other, which is of course impossible.

One more consequence matters. Because each member force appears at both of its end joints, and each joint sees it as a pull away from itself, the same member force enters the two joints' equations with opposite geometric signs automatically. Newton's third law is built into the convention.

Key idea: Always assume tension and draw member forces pulling away from the joint; then positive means tension and negative means compression, every time.

Worked example: a five-joint symmetric truss

Given and find: consider a symmetric truss with five joints. The bottom joints are A(0, 0), C(4, 0), and E(8, 0). The top joints are B(2, 3) and D(6, 3). The seven members are AB, AC, BC, BD, CD, CE, and DE, all lengths in meters. The truss is supported by a pin at A and a roller at E (vertical reaction). A single downward load of 6 kN is applied at the middle bottom joint C.

Before calculating, scan for zero-force members using the rules from the previous lesson. Joint A carries a support reaction and joint E carries one too, so neither qualifies as an unloaded joint. Joints B and D each have three members meeting with no external load, but no two of those three are collinear, so the three-member rule does not apply at either. This particular truss has no zero-force members, which means every one of the seven forces must be computed. Running that check first takes fifteen seconds and often saves several joints of work on larger trusses.

Step 1: support reactions

Treat the whole truss as one rigid body. The load acts at the center, x = 4, so by symmetry (or by taking moments about A) each support carries half:
sum M_A = 0: Ey(8) - 6(4) = 0, so Ey = 3 kN (up).
sum Fy = 0: Ay + Ey - 6 = 0, so Ay = 3 kN (up).
sum Fx = 0: Ax = 0.

Step 2: joint A

At A the unknown members are AB (up and to the right toward B) and AC (horizontal toward C), plus the reaction 3 kN up. The member AB has length sqrt(2 squared + 3 squared) = sqrt(13) = 3.606 m, so its direction fractions are 2/3.606 horizontal and 3/3.606 vertical.
sum Fy = 0: F_AB (3/3.606) + 3 = 0, so F_AB = -3 x 3.606/3 = -3.61 kN.

The negative sign means AB is in compression, 3.61 kN.
sum Fx = 0: F_AB (2/3.606) + F_AC = 0, so F_AC = -(-3.61)(2/3.606) = +2.0 kN, meaning AC is in tension, 2.0 kN.

Step 3: joint C

At C the members are CA (toward A, we already know its force), CE (toward E), CB (up-left toward B), and CD (up-right toward D), with the 6 kN load pulling down. By the left-right symmetry of the truss and load, F_CB = F_CD and F_CE = F_CA = 2.0 kN.

The two diagonals share the vertical load:
sum Fy = 0: F_CB (3/3.606) + F_CD (3/3.606) - 6 = 0. With F_CB = F_CD, this gives 2 F_CB (3/3.606) = 6, so F_CB = 6 x 3.606 / 6 = 3.61 kN, in tension. So F_CD = 3.61 kN tension as well.

Step 4: joint B

At B the members are BA (known, 3.61 kN compression), BC (known, 3.61 kN tension), and BD (horizontal toward D). Balancing horizontally at B gives F_BD = -4.0 kN, that is 4.0 kN compression. By symmetry F_DE = F_AB = 3.61 kN compression.

Written out in full, joint B carries three member forces and no external load. Member BA points from B(2, 3) toward A(0, 0), giving the direction fractions (-2/3.606, -3/3.606). Member BC points from B toward C(4, 0), giving (2/3.606, -3/3.606). Member BD points horizontally toward D, giving (1, 0). Then
sum F_y = 0: F_BA (-3/3.606) + F_BC (-3/3.606) = 0, so F_BA = -F_BC, consistent with -3.61 and +3.61 already found.
sum F_x = 0: F_BA (-2/3.606) + F_BC (2/3.606) + F_BD = 0
(-3.606)(-0.5547) + (3.606)(0.5547) + F_BD = 0
2.0 + 2.0 + F_BD = 0, so F_BD = -4.0 kN, that is 4.0 kN of compression.

The full result

Member	Force (kN)	State
AB, DE	3.61	Compression
BC, CD	3.61	Tension
AC, CE	2.0	Tension
BD	4.0	Compression

Every force is found. Notice the top chord BD is in compression and the bottom chord (AC, CE) is in tension, the classic pattern for a simply supported truss carrying a downward load. The symmetry let us solve four joints quickly, but even without it the march from joint to joint would work.

Checking the answer at an unused joint

Joint E was never solved, so it is a free and completely independent check. At E(8, 0) two members meet, plus the 3 kN roller reaction. Member ED points from E toward D(6, 3), with direction fractions (-2/3.606, 3/3.606) = (-0.5547, 0.8321). Member EC points horizontally toward C, with fractions (-1, 0).
sum F_y = 0: F_ED (0.8321) + 3 = 0, so F_ED = -3.606 kN, compression.
sum F_x = 0: F_ED (-0.5547) + F_EC (-1) = 0
(-3.606)(-0.5547) - F_EC = 0
2.000 - F_EC = 0, so F_EC = +2.0 kN, tension.
Both match the table exactly. When a check joint closes to zero like this, the whole solution is almost certainly right, because an error anywhere upstream would have to conspire to cancel here.

Key idea: Solve every joint but one, then use the remaining joint as an independent check; if it balances, the analysis is sound.

Worked example: a two-member cantilever truss

Given: a small bracket truss is bolted to a wall at A(0, 0) and B(0, 3) m. A horizontal member runs from A out to the tip C(4, 0), and a diagonal runs from B down to C. A load of 5 kN hangs vertically from the tip at C.
Find: the force in each member, stating tension or compression.

Free-body diagram of joint C in words: isolate the pin at C. Three forces act on it. The 5 kN load pulls straight down. Member CA pulls toward A, that is horizontally in the -x direction, with assumed-tension force F_CA. Member CB pulls toward B, along B - C = (-4, 3) whose length is sqrt(16 + 9) = 5, so its direction fractions are (-0.8, 0.6). Two unknowns, two equations.

Equilibrium at C. Take the vertical equation first, since only CB has a vertical component.
sum F_y = 0: F_CB (0.6) - 5 = 0
F_CB = 5 / 0.6 = 8.33 kN, positive, so the diagonal is in tension.
sum F_x = 0: F_CA (-1) + F_CB (-0.8) = 0
-F_CA - (8.33)(0.8) = 0
-F_CA - 6.67 = 0, so F_CA = -6.67 kN, negative, meaning the horizontal member is in compression at 6.67 kN.

Sanity check: resolve the diagonal's tension into components. It pulls the joint 8.33 x 0.6 = 5.0 kN upward, exactly carrying the load, and 8.33 x 0.8 = 6.67 kN toward the wall. The horizontal strut must push back with 6.67 kN, which is precisely the compression found. The pattern also makes physical sense: the diagonal above the load acts like a cable holding the tip up, so it stretches, while the horizontal member is squeezed between the wall and the tip. Notice too that the 8.33 kN in the diagonal is well above the 5 kN load, because the diagonal is inclined; shallow diagonals carry much larger forces than the loads they support.

Common wrong turns

• Starting at a joint with three unknown members. Two equations cannot resolve three unknowns. Find the reactions first, then begin at a joint with two or fewer unknowns.
• Flipping the assumed direction partway through. Assume tension everywhere and let the signs do the work; a member cannot be in tension at one end and compression at the other.
• Using the member length instead of its direction fractions. The component is F times (delta x / L), not F times delta x.
• Forgetting to find the reactions first. Almost every truss requires the support reactions before any joint has only two unknowns.
• Reporting a negative member force as a negative tension. Convert it: report 4.0 kN compression, not -4.0 kN tension, on the final drawing.
• Assuming symmetry of geometry implies symmetry of forces. The loading must be symmetric too, which it happened to be here.

Recap

• Each joint is a particle giving sum Fx = 0 and sum Fy = 0, so start where at most two members are unknown.
• Find the support reactions from the whole truss before touching any joint.
• Assume every member is in tension, drawn pulling away from the joint; positive answers then mean tension and negative mean compression.
• Get direction fractions from the joint coordinates, not from measured angles.
• For the lesson truss, the top chord BD is in 4.0 kN compression and the bottom chord in 2.0 kN tension, the classic pattern for a simply supported truss.
• Leave one joint unsolved and use it as an independent check on the whole analysis.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Method of joints. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Equilibrium of structures. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). 2D particle equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Method of joints. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Engineering structures. In Mechanics Map. Engineering LibreTexts. eng.libretexts.org
6. Buyukozturk, O. (2003). 1.051 Structural engineering design [Course materials]. MIT OpenCourseWare. ocw.mit.edu
7. Hibbeler, R. C. (2016). Engineering mechanics: Statics (14th ed.). Pearson. find source ↗

The method of joints finds every force but can be slow if you only need one member deep inside a large truss. The method of sections goes straight to that member. You imagine slicing the truss into two parts with a cut that passes through the member you want, then apply rigid-body equilibrium to one part. Because the cut part is a rigid body, you get three equations, sum Fx = 0, sum Fy = 0, and sum M = 0, and a cut through three members can solve all three at once.

Key idea: Cutting a truss turns the internal member forces into external forces on the piece you keep, so rigid-body equilibrium can find them directly.

Why cutting is legitimate

The move rests on Newton's third law, exactly as in Lesson 1. Before the cut, the two halves of the truss press on each other through the members that span the imaginary line. After the cut, those interactions become external forces on each piece, equal and opposite across the divide. Nothing physical has changed; only the choice of free body has.

Once cut, the piece you keep is a rigid body, not a particle, because its forces no longer all pass through one point. That upgrade is the whole advantage of the method: a rigid body supplies three equations instead of two, and one of them is a moment equation whose centre you get to choose.

The strategy

1. Find the support reactions first, from the whole truss.
2. Cut the truss with a line that passes through the member you want (and, ideally, through no more than three unknown members total).
3. Keep one side. At each cut member, draw its force along the member, assumed in tension (pointing away from the piece).
4. Take moments about the point where two of the three cut members intersect, so those two drop out and the third is found alone.

Three practical points about step 2. The cut does not have to be a straight line; a zig-zag cut is perfectly valid as long as it completely separates the truss into two pieces. The cut must pass through the member you want, and ideally through no more than three members whose forces are still unknown - a cut through four unknowns leaves you one equation short. And you may keep either side, so keep the one with fewer forces on it, which is usually the side away from the messier support.

The moment centre in step 4 may lie outside the piece you kept, or even off the truss entirely. That is allowed: a moment can be taken about any point in the plane, real or imaginary, and choosing a distant intersection of two members is often exactly the right move.

Key idea: Cut through at most three unknown members, keep the simpler side, and choose a moment centre where two unwanted member forces intersect - even if that point is off the structure.

Worked example: one cut on the lesson truss

Given: the same truss as before, joints A(0,0), C(4,0), E(8,0), B(2,3), D(6,3), with reactions Ay = Ey = 3 kN and a 6 kN load at C. Find: the force in the bottom-chord member CE, using a section.
Cut: slice vertically between the two halves, passing through the top chord BD, the diagonal CD, and the bottom chord CE. Keep the left part, which contains joint A (with its 3 kN reaction) and joint C (with the 6 kN load).
Choose the moment center at D(6, 3): the cut members BD and CD both pass through D, so their moments about D are zero, leaving only CE.
Compute the moment of every force on the left part about D, taking counterclockwise as positive.

Using M = x Fy - y Fx with positions measured relative to D:
The reaction Ay = 3 kN up acts at A(0, 0), position relative to D is (-6, -3): moment = (-6)(3) - (-3)(0) = -18.
The load 6 kN down acts at C(4, 0), position relative to D is (-2, -3): moment = (-2)(-6) - (-3)(0) = +12.
Member CE is horizontal at the bottom (y = 0); assumed in tension it pulls the left piece toward E, in the +x direction, acting at C(4, 0), position relative to D is (-2, -3): moment = (-2)(0) - (-3)(F_CE) = +3 F_CE.
Sum of moments about D:
-18 + 12 + 3 F_CE = 0
3 F_CE = 6, so F_CE = +2.0 kN.
The positive sign confirms CE is in tension, 2.0 kN, exactly matching the method-of-joints answer.

We found it with a single equation, without solving any other member. Compare that with the method of joints, which needed joints A, C, and B - six equations - before CE could be reached.

Worked example: the top chord from the same cut

Given: the same section, keeping the same left-hand piece with Ay = 3 kN at A(0, 0) and the 6 kN load at C(4, 0). The cut passes through BD, CD, and CE.
Find: the force in the top chord BD.

Choose the moment centre at C(4, 0). Members CD and CE both pass through C, so both drop out and only BD survives. Measure every position relative to C and take counterclockwise as positive.
The reaction Ay = 3 kN upward acts at A(0, 0), whose position relative to C is (-4, 0): moment = (-4)(3) - (0)(0) = -12.
The 6 kN load acts at C itself, so its moment arm is zero and it contributes nothing.
Member BD is horizontal at y = 3; assumed in tension it pulls the left piece toward D in the +x direction, at a height of 3 above C: moment = -(3)(F_BD) = -3 F_BD.
sum M_C = 0: -12 - 3 F_BD = 0
F_BD = -4.0 kN, so the top chord is in 4.0 kN of compression.

Sanity check: this matches the method-of-joints result from the previous lesson exactly, obtained there only after three joints of work. The sign is also physically right - the top chord of a simply supported truss under downward load is squeezed while the bottom chord is stretched, which is why bridge designers give the top chord the heavier section.

Worked example: the diagonal from the vertical equation

Given: the same cut and the same left-hand piece.
Find: the force in the diagonal CD, and then verify the whole section with the horizontal equation.

Solution: CD is the only cut member with a vertical component, so the force equation in y isolates it immediately. Member CD runs from C(4, 0) to D(6, 3), a length of sqrt(4 + 9) = 3.606, giving direction fractions (0.5547, 0.8321). Assumed in tension, it pulls the left piece toward D.
sum F_y = 0: 3 - 6 + 0.8321 F_CD = 0
0.8321 F_CD = 3, so F_CD = 3.606 kN, positive and therefore tension.

Sanity check with the third equation. All three cut member forces are now known, so the horizontal equation must close on its own:
sum F_x = F_BD + 0.5547 F_CD + F_CE
= -4.0 + (0.5547)(3.606) + 2.0
= -4.0 + 2.0 + 2.0 = 0.
It closes exactly, confirming all three member forces at once. This is the great economy of the method: one cut, three equations, three members, and a built-in check. Notice as well that the diagonal's 3.606 kN of tension carries the 3 kN of unbalanced shear across the cut - the reaction of 3 kN up minus the 6 kN load leaves 3 kN of downward shear on the left piece, and only the diagonal can resist it. Diagonals in a truss are the shear-carrying members, exactly as the web of a beam is.

Why the moment center matters

The power of the method is choosing the moment center wisely. By taking moments about the intersection of the two members you do not want, you make their moment arms zero and isolate the one you do want. If instead you needed the top chord BD, you would take moments about C(4, 0), where CE and CD meet, and BD would fall out alone. This targeted approach is why engineers reach for sections when checking a critical member in a large truss.

Sections or joints: which to use

The two methods are complementary rather than competing, and experienced analysts switch between them inside a single problem.

• Use joints when you need every member force, when the truss is small, or when you are near a support where only two members meet.
• Use sections when you need one or a few specific members, especially deep inside a large truss, or when you want an independent check on a joints solution.
• Use both when a section gets you into the middle of a truss quickly and joints then finish off the surrounding members.

A useful habit for design work: cut the truss at the panel where the shear is largest, usually next to a support, because that is where the diagonals carry the most force. Then cut at midspan, where the bending is largest, because that is where the chords carry the most. Two well-chosen cuts often reveal the two critical members of an entire bridge.

Key idea: Joints give everything slowly; sections give a chosen member instantly. Real analyses use both.

Common wrong turns

• Cutting through four or more unknown members. Three equations cannot resolve four unknowns, so the cut must be redrawn.
• Forgetting to find the reactions first. Whichever side you keep will normally include a support, and its reaction must already be known.
• Drawing the cut member forces at the joints instead of at the cut. The force acts along the member's line of action, so its position along the member is irrelevant, but its line is not - it is the line that sets the moment arm.
• Choosing a moment centre that eliminates nothing. The point of the method is to pick the intersection of two unwanted members. Any other centre leaves all three unknowns in the equation.
• Mixing sign conventions between the two pieces. Keep counterclockwise positive and the assume-tension rule on whichever side you retain.
• Including forces from the discarded side. Once cut, only what acts on the piece you kept belongs on its diagram.

Recap

• A section cuts the truss into two rigid bodies, turning internal member forces into external ones.
• A rigid-body piece supplies three equations, so a cut through three unknown members is fully solvable.
• Find the reactions first, cut through the member you want, and keep the simpler side.
• Choose the moment centre at the intersection of the two members you do not want; the point may lie off the structure.
• For the lesson truss, one cut gave F_CE = +2.0 kN tension, F_BD = -4.0 kN compression, and F_CD = +3.61 kN tension, with the third equation closing as a check.
• Chords carry the bending across a truss and diagonals carry the shear, which is where to cut when hunting critical members.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Method of sections. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Scalar addition of moments. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). 2D rigid body equilibrium. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Method of sections. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Engineering structures. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Baker, D. W., & Haynes, W. (2021). Equilibrium of structures. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org
7. Beer, F. P., Johnston, E. R., Mazurek, D. F., & Cornwell, P. J. (2019). Vector mechanics for engineers: Statics and dynamics (12th ed.). McGraw-Hill Education. find source ↗

Not every structure is a truss. A frame is a stationary structure that contains at least one multi-force member, a member with forces at more than two points or with a load along its length. A machine is similar but is meant to move and transmit or modify forces, such as pliers, a bolt cutter, or a crane. In both, the members are not two-force members, so their internal forces do not simply run along the member. We must analyze the parts separately.

Key idea: Frames and machines contain multi-force members, so their internal forces do not run along the members and the structure must be taken apart to be solved.

Three families of structure

It helps to see where frames and machines sit. A truss is rigid and built entirely of two-force members. A frame is rigid, stays put, and contains at least one multi-force member; scaffolding, a bookshelf bracket, and a building's beam-and-column skeleton are frames. A machine contains moving parts and exists to change the size or direction of a force; pliers, bolt cutters, a car jack, a backhoe arm, and a bicycle brake are machines.

The analysis is the same for frames and machines. The difference is only in what you are asked for: a frame problem usually wants support reactions and pin forces so the connections can be sized, while a machine problem usually wants the output force for a given input, that is, its mechanical advantage.

The method: dismember

The key technique is to take the structure apart at its pins and draw a free-body diagram of each member. At every pin that connected two members, the two members exert equal and opposite forces on each other, by Newton's third law. You represent each pin force by two unknown components, and you must be consistent: if you draw the force from member 1 on member 2 pointing a certain way, the force from member 2 on member 1 points exactly opposite.

1. Analyze the whole structure first to find the external support reactions where possible.
2. Separate the structure into individual members.
3. Draw each member's free-body diagram, showing external loads and the unknown pin-force components, with third-law pairs opposite.
4. Write equilibrium for each member and solve.

Step 1 is worth doing even when it does not finish the job. Treating the whole assembly as one rigid body hides every internal pin force, so its three equations often give the support reactions immediately, and those reactions then feed the member-by-member analysis. If the whole-structure equations are not enough, nothing is lost - you simply proceed to dismember.

Key idea: Analyze the whole structure first to get the support reactions, then dismember to get the internal pin forces.

Spot the two-force members first

Before drawing a single equation, scan the frame for members that are pinned at exactly two points and carry no load in between. Each such member is a two-force member, so the force it transmits must act along the line joining its two pins. That converts two unknown components into one unknown magnitude, and it fixes a direction you would otherwise have to solve for.

The saving compounds. A frame of four members with pins everywhere might start with eight unknown pin components. Recognizing two of the members as two-force members can cut that to five, often turning an intimidating simultaneous system into a short sequence of one-line solves. Identifying two-force members is the single most valuable habit in frame analysis.

Key idea: Any member pinned at just two points with nothing acting in between is a two-force member, and its force lies along the line joining those pins.

Worked example: a simple frame

Given: an L-shaped frame has a vertical member from A(0, 0) up to B(0, 3) and a horizontal member from B(0, 3) to C(3, 3). The frame is pinned to the ground at A and supported by a roller at C that provides a vertical reaction. A downward load of 1200 N is applied at the midpoint of the horizontal member, at (1.5, 3).

Find: the support reactions.
Solution (whole structure):
sum M_A = 0: Cy(3) - 1200(1.5) = 0, so Cy = 600 N (up).
sum Fy = 0: Ay + Cy - 1200 = 0, so Ay = 600 N (up).
sum Fx = 0: Ax = 0.
So the pin at A pushes up with 600 N and the roller at C pushes up with 600 N. To go further and find the force in the pin at B, you would isolate one member, say the horizontal one BC, which carries the 1200 N load, the roller reaction 600 N at C, and the unknown pin force at B, then apply its three equilibrium equations. Let us finish that calculation.

Free-body diagram of member BC in words: isolate the horizontal member. Three things act on it. The 1200 N load pushes down at (1.5, 3). The roller at C pushes up with 600 N at (3, 3). And at B the vertical member pushes on it with unknown components Bx and By, drawn positive to the right and upward.

Equilibrium of BC.
sum F_y = 0: By + 600 - 1200 = 0, so By = 600 N upward.
sum F_x = 0: Bx = 0, since nothing else acts horizontally on BC.
Check with moments about B at (0, 3): -1200(1.5) + 600(3) = -1800 + 1800 = 0. It closes, so the pin force at B is 600 N vertical, with no horizontal component.

Sanity check by the two-force rule: member AB is pinned at exactly two points, A and B, and carries nothing in between, so it is a two-force member. Its force must therefore act along the line AB, which is vertical - and that is precisely why Bx and Ax both came out as zero. Recognizing this at the start would have given both results without a single equation. Member AB carries 600 N of compression, squeezed between the ground at A and the beam at B.

Worked example: a machine (pliers)

Given: a pair of pliers pivots at a central pin. You squeeze the handles with 100 N each, applied 0.15 m from the pivot, and the gripped object sits 0.05 m from the pivot on the jaw side.

Find: the gripping force on the object.
Solution: analyze one handle-and-jaw piece, taking moments about the pivot pin so the pin force drops out:
sum M_pivot = 0: F_jaw (0.05) - 100 (0.15) = 0
F_jaw = 100 x 0.15 / 0.05 = 300 N.
The pliers multiply your 100 N squeeze into a 300 N grip, a mechanical advantage of 3, set purely by the ratio of the lever arms. This is why machines are analyzed member by member: the useful output force lives inside the linkage, not along any single member.

The pin force, which is what actually fails. The handle-and-jaw piece is in equilibrium under three forces: the 100 N squeeze pushing down on the handle, the 300 N reaction from the gripped object pushing down on the jaw on the other side of the pivot, and the pin force. Summing vertically, with upward positive:
sum F_y = 0: P - 100 - 300 = 0
P = 400 N, pushing up on the lever.

Sanity check: the pivot pin carries 400 N, which is four times the force your hand applies and more than the 300 N grip it delivers. That is the general lesson of machine analysis - the pin at the fulcrum of a force multiplier carries the sum of input and output, so it is usually the part that shears off first. Checking the moment equation confirms the geometry: about the pivot, the hand contributes (0.15)(-100) = -15 N*m and the object contributes (-0.05)(-300) = +15 N*m, which cancel, while the pin force acts at the pivot itself and has no moment arm.

Checking a dismembered analysis

Frames generate many numbers and many chances to slip a sign, so build in checks.

• Third-law pairs. Every pin force must appear on two diagrams with opposite signs. Read them off both members and confirm.
• Unused equations. If you solved a member with two of its three equations, the third must close to zero on its own.
• Reassembly. Add up the external loads and reactions for the whole structure. They must satisfy equilibrium with all the internal pin forces cancelling in pairs.
• Plausibility. Ask whether the magnitudes make sense. A pin force much smaller than the applied load in a compact frame is usually a dropped term.

Key idea: Every internal pin force appears twice, with opposite signs; using that fact as a check catches most errors in a frame analysis.

Common wrong turns

• Drawing the same direction for both halves of a third-law pair. If member 1 pushes member 2 to the right, member 2 pushes member 1 to the left. Drawing both to the right makes the internal forces add instead of cancel.
• Treating a frame member as a two-force member when a load acts on it. A load anywhere between the pins destroys the two-force property, and the pin force is then no longer along the member.
• Forgetting that a pin at a joint of three members needs care. Decide explicitly whether the pin belongs to one member or is its own free body, and be consistent.
• Applying an external load to a pin and then to a member as well. Each load acts once, on whichever body the problem says it acts on.
• Skipping the whole-structure step. The support reactions are usually free from the assembled structure, and they simplify everything afterwards.
• Assuming the pivot pin of a machine carries only the input force. It typically carries input plus output, which is why fulcrum pins are sized generously.

Recap

• Frames are rigid structures with at least one multi-force member; machines transmit or modify forces and contain moving parts.
• Analyze the whole assembly first for support reactions, then dismember at the pins.
• At every pin, the two members exert equal and opposite forces on each other.
• Identify two-force members before writing equations: their force acts along the line joining their two pins.
• In the L-frame, both reactions were 600 N and the pin at B carried 600 N vertically, because AB is a two-force member.
• The pliers gave a mechanical advantage of 3, and the pivot pin carried 400 N, the sum of input and output.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Frames and machines. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Interactions between members. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Free-body diagrams. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Analysis of frames and machines. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Two force members. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Newton's third law. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. OpenStax. (2016). 12.2 Examples of static equilibrium. In University Physics Volume 1. Rice University. openstax.org

So far we have found the forces at the outside of a member. But whether a beam is strong enough depends on the forces inside it. When you cut a loaded beam at some section, the material on each side pushes and pulls on the other to hold the beam together. Those internal actions have three parts:

• Normal force (N): the internal push or pull along the beam's axis.
• Shear force (V): the internal force perpendicular to the axis, tending to slide one side past the other.
• Bending moment (M): the internal couple that tends to bend the beam.

To find them, cut the beam at the section of interest, keep one side, and apply equilibrium to that piece. The three unknowns N, V, and M are whatever values make the cut piece balance.

Key idea: Cutting a beam exposes three internal actions - axial force N, shear V, and bending moment M - and each is whatever value makes the cut piece satisfy equilibrium.

Why internal forces decide the design

Support reactions tell you how hard the ground pushes; they say nothing about whether the beam survives. What breaks a beam is what happens inside it. The bending moment stretches the fibres on one face and compresses those on the other, which is what fractures concrete and yields steel. The shear force tries to slide adjacent slices past each other, which is what splits timber along the grain and buckles the thin web of a steel section. The axial force adds uniform tension or compression on top of both.

Every beam design therefore begins with two questions: where along the beam is M largest, and where is V largest? Answer those and you know where to put the material. That is precisely the information a shear and moment diagram delivers.

Sign conventions

The standard convention: shear is positive when it tends to rotate the cut element clockwise, and the bending moment is positive when it bends the beam concave up (a smile), putting the bottom fibers in tension. Consistent signs let you plot how V and M vary along the beam, the shear and moment diagrams that designers rely on.

Said concretely: on the left-hand piece of a cut, a positive shear V acts downward on the cut face, and a positive bending moment M acts counterclockwise on that face. On the right-hand piece the same positive values act upward and clockwise. Drawing both pieces once and labelling the positive senses is the fastest way to internalize the convention.

The reason for defining positive moment as "sagging" is practical. A sagging beam is stretched along the bottom, and reinforcement or the larger flange goes where the tension is. A designer reading a moment diagram wants to know instantly which face is in tension, and the sign convention answers that at a glance.

Key idea: Positive shear tends to rotate the cut element clockwise, and positive moment sags the beam, putting the bottom fibres in tension.

How load, shear, and moment are linked

The three quantities are not independent. For a beam carrying a distributed load of intensity w (taken positive downward):

• dV/dx = -w, so the slope of the shear diagram equals minus the load intensity.
• dM/dx = V, so the slope of the moment diagram equals the shear.

Read as areas rather than slopes, these become the fastest way to draw diagrams by hand: the change in shear between two points equals minus the area under the load diagram between them, and the change in moment equals the area under the shear diagram. Three consequences follow immediately and are worth memorizing.

• Where there is no load, the shear is constant and the moment varies linearly.
• Under a uniform load, the shear varies linearly and the moment is a parabola.
• The moment reaches a maximum or minimum where the shear passes through zero, because that is where dM/dx = 0.

A point load makes the shear diagram jump vertically by the size of the load, and an applied couple makes the moment diagram jump by the size of the couple. Both jumps are instantaneous, which is why real diagrams have steps in them.

Key idea: dV/dx = -w and dM/dx = V, so shear is the area under the load curve and moment is the area under the shear curve, with the peak moment where the shear crosses zero.

Worked example: simply supported beam, central load

Given: a simply supported beam of length 6 m carries a single 1200 N downward load at the center (x = 3 m). Find: the shear and bending moment at the midspan, and the maximum bending moment.
Reactions: by symmetry each support carries 1200 / 2 = 600 N up.
Cut just left of center and keep the left piece, which has only the 600 N reaction at its left end.
sum Fy = 0: 600 - V = 0, so the shear is V = 600 N (constant over the left half).
sum M at the cut = 0: M - 600 x 3 = 0, so M = 1800 N*m at the center.
The bending moment is largest right under the load: M_max = 1800 N*m.

For a central point load this is P L / 4 = 1200 x 6 / 4 = 1800 N*m, a formula worth remembering. The shear jumps from +600 N to -600 N as you pass the load, and the moment peaks where the shear crosses zero, a general and useful rule.

Worked example: cantilever with an end load

Given: a cantilever beam of length 4 m, fixed at the left, carries a 500 N downward load at its free right end. Find: the internal shear and bending moment as functions of position, and their values at the wall.
Cut at a distance x from the fixed end and keep the right piece, which carries only the 500 N end load.
sum Fy = 0: V - 500 = 0, so V = 500 N everywhere along the beam.
sum M at the cut = 0: M + 500 (4 - x) = 0, so M = -500 (4 - x).
At the free end (x = 4), M = 0, as it must be.

At the wall (x = 0), M = -500 x 4 = -2000 N*m, the largest magnitude. This confirms what the supports lesson foreshadowed: a cantilever is most heavily loaded, in bending, right at its fixed support, which is where such beams are made deepest.

Sanity check on the cantilever: the reaction moment found back in the supports lesson was M_A = 2000 N*m, and the internal bending moment at the wall came out as -2000 N*m. They are the same magnitude with opposite sign, exactly as they must be: the wall's reaction couple is what holds the internal bending in balance. The negative sign means the cantilever hogs rather than sags, so its top fibres are in tension - which is why reinforcement in a cantilevered concrete balcony runs along the top, not the bottom.

Worked example: simply supported beam with a uniform load

Given: a simply supported beam of span L = 8 m carries a uniform distributed load of w = 5 kN/m over its entire length.
Find: the shear and bending moment as functions of position, the location and value of the maximum bending moment, and the maximum shear.

Reactions. The total load is 5 x 8 = 40 kN, acting at midspan, so by symmetry each support carries half:
R_A = R_B = 40 / 2 = 20 kN upward.

Cut at a distance x from the left support and keep the left piece. That piece carries the 20 kN reaction at its left end and the portion of the distributed load lying on it, which totals 5x kN acting at x/2 from the left end.
sum F_y = 0: 20 - 5x - V = 0
V(x) = 20 - 5x kN.
sum M at the cut = 0: M - 20x + 5x (x/2) = 0
M(x) = 20x - 2.5 x squared kN*m.

Maximum moment. Set the shear to zero, since that is where the moment peaks:
20 - 5x = 0, so x = 4.0 m, exactly midspan as symmetry predicts.
M_max = 20(4) - 2.5(16) = 80 - 40 = 40 kN*m.
Maximum shear occurs at the supports, where V(0) = +20 kN and V(8) = 20 - 40 = -20 kN.

Sanity check: the standard formula for a simply supported beam under a uniform load is M_max = w L squared / 8 = 5 x 64 / 8 = 40 kN*m, matching exactly. The end conditions also check out: M(0) = 0 and M(8) = 160 - 160 = 0, as they must be at simple supports, which cannot resist a moment. Compare with the point-load case earlier in this lesson: a central point load of the same total 40 kN would give P L / 4 = 40 x 8 / 4 = 80 kN*m, twice as much. Spreading a load out halves the bending it causes, which is why concentrated loads are the enemy of long spans.

Reading the diagrams

The two functions above are exactly what a shear and moment diagram plots. The shear diagram is a straight line falling from +20 kN at the left support to -20 kN at the right, crossing zero at midspan. The moment diagram is a parabola starting at zero, rising to 40 kN*m at midspan, and returning to zero at the right support. Every rule from the previous section is visible: the uniform load gives a linear shear and a parabolic moment, and the moment peaks precisely where the shear crosses zero.

Two quick checks apply to every diagram you will ever draw. First, both the shear and the moment must return to zero at a free end, and the moment must be zero at any simple support or pin at the end of a beam. Second, the total area under the shear diagram over the whole beam must equal the change in moment from one end to the other, which for a simply supported beam is zero - the positive and negative halves of the shear diagram must cancel.

Common wrong turns

• Forgetting to find the reactions before cutting. The cut piece almost always contains a support, and its reaction must already be known.
• Using the whole distributed load instead of the part on the cut piece. Only the load actually sitting on the retained piece belongs on its diagram, and its resultant acts at the centroid of that portion.
• Placing the partial distributed load's resultant at the cut. For a piece of length x under a uniform load, the resultant acts at x/2, not at x.
• Mixing up shear and moment sign conventions between the two pieces. Cutting and keeping the right-hand piece must give the same V and M as keeping the left-hand piece; if it does not, a sign is wrong.
• Looking for the maximum moment at the largest load rather than where the shear crosses zero. They coincide for a single central load and generally do not otherwise.
• Ignoring the jump at a point load or applied couple. Shear steps by the point load; moment steps by the applied couple.

Recap

• Cutting a beam exposes an internal normal force N, shear force V, and bending moment M.
• Positive shear rotates the cut element clockwise; positive moment sags the beam and stretches its bottom fibres.
• dV/dx = -w and dM/dx = V, so shear is the area under the load curve and moment the area under the shear curve.
• The maximum bending moment occurs where the shear passes through zero.
• Simply supported with a central point load gives M_max = P L / 4; with a uniform load it gives M_max = w L squared / 8.
• A cantilever with an end load has its largest moment at the wall, equal in magnitude to the support's reaction couple.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Internal forces. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Sign conventions. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Shear and bending moment diagrams. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). Relations between loading, shear and moment. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Internal forces. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Shear and moment diagrams. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Baker, D. W., & Haynes, W. (2021). Internal loadings. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

The centroid of an area is its geometric center, the average position of all its points. It matters in statics for two reasons: a distributed load acts at the centroid of its loading diagram, and the bending strength of a beam is measured about the centroid of its cross-section. Learning to locate centroids quickly is therefore essential groundwork.

Key idea: The centroid is the area-weighted average position of all the points of a shape, and it is where a distributed effect can be treated as acting.

Centroid, centre of mass, centre of gravity

Three related terms get used loosely, and it is worth separating them once. The centroid is a purely geometric property: the average position of area, or of volume, or of length. The centre of mass is the average position of mass. The centre of gravity is the point where the total weight can be taken to act.

They coincide whenever the material is uniform and gravity is uniform, which covers essentially every problem in this course. They differ when a body is made of two materials of different density - a steel plate bolted to an aluminium one has its centroid at the geometric middle but its centre of mass shifted toward the steel. In statics, when we say "the weight acts at the centre of gravity", we are relying on that coincidence.

Key idea: For a uniform body in uniform gravity the centroid, centre of mass, and centre of gravity are the same point; they separate only when density varies.

Centroids of simple shapes

For symmetric shapes the centroid sits on the axis of symmetry, and often you can read it off:

• A rectangle of width b and height h has its centroid at its center, b/2 from a side and h/2 from the base.
• A triangle has its centroid one-third of the way up from the base, at h/3.
• A circle or any doubly symmetric shape has its centroid at its center.
• A semicircle of radius r has its centroid at 4 r / (3 pi), roughly 0.424 r, from the flat diameter along the axis of symmetry.
• A quarter circle of radius r has its centroid at 4 r / (3 pi) from each straight edge.

Symmetry is the most powerful shortcut available. If a shape has an axis of symmetry, the centroid lies on it; if it has two, the centroid is where they cross and no calculation is needed at all. Always look for symmetry before reaching for a formula, because it can turn a two-coordinate problem into a one-coordinate problem or into no problem at all.

Key idea: The centroid always lies on any axis of symmetry, so two axes of symmetry locate it exactly with no arithmetic.

The composite-area method

Most real cross-sections are combinations of simple shapes. To find the centroid of a composite area, treat each simple piece as if its whole area sits at its own centroid, then take the area-weighted average:

• x_bar = (sum Ai xi) / (sum Ai)
• y_bar = (sum Ai yi) / (sum Ai)

Here Ai is the area of each piece and (xi, yi) is its own centroid. It is the same idea as a weighted average of positions, with area playing the role of weight.

The reliable way to run the method is as a table with one row per piece and columns for Ai, xi, yi, Ai xi, and Ai yi. Sum the last three columns and divide. The products Ai xi are called first moments of area, and the whole method is really just the statement that the first moment of the composite equals the sum of the first moments of its pieces - the same additivity that Varignon's theorem gave us for force moments.

Choosing where to put the origin is free, and a good choice saves work. Placing it at a corner of the shape makes every coordinate positive, which removes sign errors, and placing it on an axis of symmetry makes one coordinate come out by inspection.

Key idea: Tabulate area and centroid for each piece, sum the first moments Ai xi and Ai yi, and divide by the total area.

Worked example: an L-shaped area

Given: an L-shape made of two rectangles. Rectangle 1 is 2 wide and 6 tall, occupying x from 0 to 2 and y from 0 to 6. Rectangle 2 is 4 wide and 2 tall, occupying x from 2 to 6 and y from 0 to 2 (all in centimeters).

Find: the centroid.
Solution: tabulate each piece.
Rectangle 1: area A1 = 2 x 6 = 12, centroid at (1, 3).
Rectangle 2: area A2 = 4 x 2 = 8, centroid at (4, 1).
Total area = 12 + 8 = 20.
x_bar = (12 x 1 + 8 x 4) / 20 = (12 + 32) / 20 = 44 / 20 = 2.2 cm.
y_bar = (12 x 3 + 8 x 1) / 20 = (36 + 8) / 20 = 44 / 20 = 2.2 cm.
The centroid is at (2.2, 2.2) cm. It lies inside the thicker corner of the L, as expected, since more area is concentrated near the bottom-left.

Handling a hole

If a shape has a hole, treat the hole as a piece of negative area and subtract it. For a 6 by 6 square (area 36, centroid at its center (3, 3)) with a circular hole of radius 1 (area pi x 1 squared = 3.14) at the center, the centroid stays at (3, 3) by symmetry, because the removed area is itself centered. In general, use x_bar = (A_solid x_solid - A_hole x_hole) / (A_solid - A_hole), letting the hole carry a minus sign in both the numerator and denominator.

Worked example: a plate with an off-centre hole

Given: a rectangular steel plate 100 mm wide and 80 mm tall, with its lower-left corner at the origin. A circular hole of diameter 30 mm is drilled with its centre at (70, 40) mm.
Find: the centroid of the remaining material.

Tabulate. Treat the hole as negative area.
Plate: A1 = 100 x 80 = 8000 mm squared, centroid at (50, 40).
Hole: A2 = -pi (15) squared = -706.9 mm squared, centroid at (70, 40).
Total area: 8000 - 706.9 = 7293.1 mm squared.

First moments.
sum Ai xi = 8000(50) - 706.9(70) = 400000 - 49483 = 350517
sum Ai yi = 8000(40) - 706.9(40) = 320000 - 28276 = 291724
Divide.
x_bar = 350517 / 7293.1 = 48.06 mm
y_bar = 291724 / 7293.1 = 40.00 mm

Sanity check: the whole shape is symmetric about the horizontal line y = 40, since the hole sits on that line, so y_bar had to come out as exactly 40 mm - and it did, which validates the arithmetic. The horizontal result is also plausible: material was removed from the right-hand side, so the centroid must shift left of the 50 mm midpoint, and it moved by about 1.9 mm. The shift is small because the hole is only about 9 percent of the plate's area. A quick estimate confirms it: the centroid moves roughly by (A_hole / A_total) x (distance from plate centroid to hole) = (706.9 / 7293.1) x 20 = 1.94 mm, matching the full calculation.

Worked example: locating the resultant of a trapezoidal load

Given: a beam carries a distributed load that varies linearly from 200 N/m at x = 0 to 600 N/m at x = 6 m.
Find: the magnitude of the equivalent single force and the position of its line of action.

Solution: the load diagram is a trapezoid, so split it into a rectangle and a triangle and treat them as composite areas.
Rectangle: intensity 200 N/m over 6 m, so A1 = 200 x 6 = 1200 N at its midpoint x1 = 3.0 m.
Triangle: peak intensity 600 - 200 = 400 N/m at the right end, so A2 = (1/2)(400)(6) = 1200 N at two thirds of the span from the zero end, x2 = 4.0 m.
Total = 1200 + 1200 = 2400 N
x_bar = (1200 x 3.0 + 1200 x 4.0) / 2400 = (3600 + 4800) / 2400 = 8400 / 2400 = 3.50 m

Sanity check: the closed-form centroid of a trapezoidal load is x_bar = (L/3)(2 w2 + w1) / (w2 + w1), which gives (6/3)(1200 + 200) / (600 + 200) = 2 x 1400 / 800 = 3.50 m, an exact match. The position also passes the common-sense test: the load is heavier on the right, so the resultant must sit right of midspan, and 3.50 m is indeed past 3.0 m but well short of the 4.0 m a purely triangular load would give. This is exactly the calculation that lets a trapezoidal load be dropped into the rigid-body equilibrium equations of Module 3.

Common wrong turns

• Averaging the piece centroids without weighting by area. A small piece far away moves the centroid far less than a large piece nearby; area is the weight.
• Forgetting to subtract the hole from the total area as well. The hole's negative area must appear in the denominator, not only in the numerator.
• Using the triangle's midpoint instead of its third-point. A triangle's centroid is one third of the height from its base, never halfway.
• Measuring each piece's centroid from its own corner instead of the common origin. All coordinates must be measured in one shared frame.
• Assuming the centroid lies inside the shape. For an L, a C, or a ring it often does not, which is perfectly acceptable.
• Confusing centroid with centre of mass in a two-material part. They separate as soon as density varies.

Recap

• The centroid is the area-weighted average position of a shape, and coincides with the centre of gravity for a uniform body.
• The centroid lies on every axis of symmetry, so two axes locate it with no arithmetic.
• Rectangle at its centre, triangle at h/3 from the base, semicircle at 4r/(3 pi) from the flat side.
• For composites, x_bar = (sum Ai xi) / (sum Ai) and likewise for y_bar, tabulated piece by piece.
• Holes are handled as negative areas in both the numerator and the denominator.
• The same method locates the resultant of any distributed load, since the load diagram is just another composite area.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Centroids. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Centroids using composite parts. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Weighted averages. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). Center of gravity. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Centroid, centre of mass, and mass moment of inertia in composite bodies. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. OpenStax. (2016). 9.6 Center of mass. In University Physics Volume 1. Rice University. openstax.org
7. Baker, D. W., & Haynes, W. (2021). Centroids and centers of gravity. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

The area moment of inertia, also called the second moment of area, measures how an area is distributed relative to an axis. It captures the resistance of a beam's cross-section to bending: the farther the material lies from the bending axis, the larger the moment of inertia and the stiffer the beam. This single geometric property is why I-beams put their material in the flanges, far from the center.

Key idea: The area moment of inertia I measures how far a shape's area is spread from an axis, and it controls how stiff a cross-section is in bending.

Why distance is squared

When a beam bends, its cross-section rotates slightly about the centroidal axis. A fibre at distance y from that axis stretches in proportion to y, so the force it develops is proportional to y, and the moment that force contributes is that force times its arm y again. Two factors of y give the y squared in the integral. That single squaring is why material far from the axis is disproportionately valuable, and why moving a flange twice as far out makes it four times as effective.

Definition and basic formulas

About a horizontal axis through the centroid, the moment of inertia is I = integral of y squared dA, summing each bit of area times the square of its distance from the axis. Two results you can use directly:

• A rectangle of width b and height h, about its own centroidal axis (horizontal), has I = b h cubed / 12.
• A circle of radius r, about a centroidal axis, has I = pi r to the fourth / 4.

The units are length to the fourth power, such as cm to the fourth. Notice the height enters cubed for a rectangle: doubling a beam's depth multiplies its bending stiffness by eight, which is why beams are built tall.

Three more results complete the toolkit:

• The same rectangle about its vertical centroidal axis has Iy = h b cubed / 12. Which axis you use matters enormously: a plank laid flat is far more flexible than the same plank stood on edge.
• A triangle of base b and height h, about the horizontal centroidal axis, has I = b h cubed / 36. About its base it is b h cubed / 12.
• The polar moment of inertia about an axis through the centroid perpendicular to the plane is J = Ix + Iy. It governs resistance to twisting rather than bending.

The radius of gyration k = sqrt(I / A) is a useful companion. It is the distance at which the entire area could be concentrated to give the same moment of inertia, and it is the parameter that decides how easily a compression member buckles. A slender column with a small k buckles readily; a hollow tube with the same area and a large k does not.

Key idea: I depends on which axis you take, on the cube of the depth, and it is summarized by the radius of gyration k = sqrt(I/A).

The parallel-axis theorem

Formulas like b h cubed / 12 give the moment of inertia about the shape's own centroid. To find it about a different, parallel axis a distance d away, use the parallel-axis theorem:
I = I_centroid + A d squared,
where A is the area and d is the distance between the two parallel axes. The added term A d squared is always positive, so an area's moment of inertia is smallest about its own centroid and grows as you move the axis away. This theorem is the engine that lets us build up complicated sections from simple ones.

One restriction is easy to overlook and expensive to get wrong: the starting value I_centroid must be taken about an axis through the shape's own centroid. You cannot apply the theorem twice from an arbitrary axis, and you cannot shift from a base axis to another base axis in one step. If you already have I about some non-centroidal axis, first come back to the centroid by subtracting A d squared, then go out to the new axis by adding the new A d squared.

The theorem also tells you something design-relevant at a glance. Because A d squared grows quadratically, the position of a piece of area usually matters far more than its own shape. In a typical I-beam, the two flanges contribute the great majority of the section's I almost entirely through their A d squared terms; their own b h cubed / 12 contributions are almost negligible.

Key idea: I = I_centroid + A d squared, where I_centroid must be about the piece's own centroidal axis, and the transfer term usually dominates.

Worked example: rectangle about its base

Given: a rectangle 3 wide and 6 tall (cm). Find: its moment of inertia about its centroidal axis and about its base.
Centroidal: I_c = b h cubed / 12 = 3 x 6 cubed / 12 = 3 x 216 / 12 = 54 cm to the fourth.
About the base: the base is d = h/2 = 3 cm below the centroid, and the area is A = 18.

By the parallel-axis theorem,
I_base = I_c + A d squared = 54 + 18 x 3 squared = 54 + 162 = 216 cm to the fourth.
As a check, the direct formula for a rectangle about its base is b h cubed / 3 = 3 x 216 / 3 = 216, which agrees.

Worked example: a T-section

Given: a T-section with a flange 8 wide and 2 tall on top (y from 6 to 8) and a web 2 wide and 6 tall below (y from 0 to 6), in centimeters. Find: the moment of inertia about the horizontal centroidal axis of the whole T.
Step 1, centroid: web area A_w = 12 at y = 3; flange area A_f = 16 at y = 7.

So
y_bar = (12 x 3 + 16 x 7) / 28 = (36 + 112) / 28 = 148 / 28 = 5.29 cm.
Step 2, each piece about the T centroid using I_own + A d squared, with d measured from each piece's own centroid to y_bar = 5.29:
Web: I_own = 2 x 6 cubed / 12 = 36; distance d = 5.29 - 3 = 2.29; contribution = 36 + 12 x 2.29 squared = 36 + 62.9 = 98.9.
Flange: I_own = 8 x 2 cubed / 12 = 5.33; distance d = 7 - 5.29 = 1.71; contribution = 5.33 + 16 x 1.71 squared = 5.33 + 46.8 = 52.1.
Total: I = 98.9 + 52.1 = 151 cm to the fourth.
The moment of inertia of the T about its centroid is about 151 cm to the fourth.

Each piece needed its own centroidal inertia plus a parallel-axis shift, and summing gives the whole. This composite procedure is exactly how the section properties of real beams are computed.

Sanity check on the T-section: the answer must lie between two easy bounds. If the whole 28 square-centimetre area sat right on the centroidal axis, I would be tiny; if it all sat at the extreme fibre, 2.7 cm away, I would be about 28 x 2.7 squared = 204. The computed 151 sits sensibly inside that range. Note also which term dominates: the web's transfer term A d squared = 62.9 and the flange's 46.8 together account for 110 of the 151 total, so more than two thirds of the stiffness comes from where the material is rather than from the shape of the pieces themselves.

Worked example: a hollow box section

Given: a rectangular hollow section with outside dimensions 100 mm wide by 150 mm deep and a uniform wall thickness of 10 mm, so the inner void is 80 mm wide by 130 mm deep. Both rectangles share the same centroid.
Find: the moment of inertia about the horizontal centroidal axis, and compare it with the solid rectangle of the same outside dimensions.

Solution: because the void is concentric with the outside, no parallel-axis shift is needed. Simply subtract.
I_outer = b h cubed / 12 = (100)(150 cubed) / 12 = (100)(3375000) / 12 = 28125000 mm to the fourth
I_inner = (80)(130 cubed) / 12 = (80)(2197000) / 12 = 14646667 mm to the fourth
I = 28125000 - 14646667 = 13478333 mm to the fourth, or about 1.35 x 10^7 mm to the fourth.

Sanity check and the design lesson: the box keeps 13478333 / 28125000 = 0.479, roughly 48 percent of the solid section's stiffness. Its area, though, is only (100 x 150) - (80 x 130) = 15000 - 10400 = 4600 mm squared against 15000, which is about 31 percent. So the hollow section delivers 48 percent of the bending stiffness for 31 percent of the material and weight. That ratio is the whole argument for tubes, box girders, and I-beams: the material near the neutral axis contributes almost nothing to I, so removing it costs stiffness slowly and saves weight quickly.

Common wrong turns

• Applying the parallel-axis theorem from a non-centroidal starting axis. Always begin from the piece's own centroid, or return there first by subtracting.
• Forgetting to find the composite centroid before shifting. The distances d are measured from each piece's centroid to the centroid of the whole section, which must be located first.
• Subtracting A d squared instead of adding. The transfer term is always positive, because I is smallest about the centroidal axis.
• Using the wrong dimension as the depth. In b h cubed / 12, h is measured perpendicular to the axis you are taking the moment about. Swapping b and h can change the answer by an order of magnitude.
• Mixing units. Moments of inertia carry length to the fourth power, so a factor of 10 in the length becomes a factor of 10000 in I.
• Confusing area moment of inertia with mass moment of inertia. The area version has units of length to the fourth and governs bending; the mass version has units of mass times length squared and governs rotational dynamics.

Recap

• I = integral of y squared dA measures how far a shape's area lies from an axis; units are length to the fourth power.
• Rectangle: b h cubed / 12 about the centroidal axis and b h cubed / 3 about the base; circle: pi r to the fourth / 4; triangle: b h cubed / 36.
• The parallel-axis theorem is I = I_centroid + A d squared, starting always from the piece's own centroid.
• For composites, locate the overall centroid first, then sum each piece's I_own + A d squared.
• The transfer term usually dominates, which is why flanges and tube walls carry most of a section's stiffness.
• Removing material near the neutral axis costs little stiffness and saves much weight, the reason for I-beams and hollow sections.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Integral properties of shapes. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Moments of inertia of common shapes. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Parallel axis theorem. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Baker, D. W., & Haynes, W. (n.d.). Composite shapes. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Parallel axis theorem. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Area moments of inertia via composite parts. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Baker, D. W., & Haynes, W. (2021). Moments of inertia. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

Every real contact between surfaces resists sliding through friction. Friction is what lets us walk, grip, and brake, and it appears in almost every practical statics problem. The simplest and most useful model is dry friction, also called Coulomb friction, which describes unlubricated surfaces sliding on each other.

Key idea: Dry friction acts along the contact surface, opposing the tendency to slide, and it can grow only up to a maximum of mu_s N.

Why the contact area does not matter

The claim that friction is independent of the apparent contact area feels wrong the first time you meet it. A wide tyre surely grips better than a narrow one. The resolution is that the apparent area and the real area of contact are different things. No surface is truly flat; contact happens only at microscopic high spots, and the real contact area is a tiny fraction of the apparent one.

Press harder and those high spots flatten, so the real contact area grows in proportion to the normal force. Spread the same load over a larger apparent area and the pressure falls, the high spots flatten less, and the real contact area stays about the same. The two effects cancel, which is exactly why friction depends on N and not on how the load is spread. The Coulomb model captures that cancellation in one line.

Key idea: Friction tracks the real microscopic contact area, which is proportional to the normal force, so the apparent area drops out of the model.

The laws of dry friction

The friction force F acts along the surface, opposing the tendency to slide, and its size follows a few rules:

• As long as the body does not slide, friction takes exactly the value needed to keep equilibrium, up to a maximum. Below that maximum, friction is whatever balance requires, no more.
• The maximum static friction is F_max = mu_s N, where mu_s is the coefficient of static friction and N is the normal force pressing the surfaces together.
• Once sliding begins, the friction drops slightly to the kinetic value F_k = mu_k N, with mu_k the coefficient of kinetic friction, usually a little less than mu_s.
• Friction does not depend on the contact area, only on the normal force and the surface pair.

The moment when the body is just about to slip, with friction at its maximum mu_s N, is called impending motion. That is the critical case most problems ask about.

Three regimes, and the inequality that governs them

Almost every friction mistake comes from writing F = mu_s N when it does not apply. The correct statement is an inequality:

F is less than or equal to mu_s N

Three distinct situations follow, and identifying which one you are in is the first step of any friction problem.

• Static, below the limit. The body is at rest and the applied load is small. Friction takes exactly the value equilibrium requires, which is less than mu_s N. Find it from sum F = 0, never from the friction formula.
• Impending motion. The body is on the verge of slipping. Only now does friction equal its maximum, F = mu_s N, and the friction relation becomes a usable equation.
• Sliding. Motion has begun and friction drops to the kinetic value F_k = mu_k N, which acts opposite to the actual velocity rather than to a mere tendency.

A practical test settles the regime: compute the friction required for equilibrium and compare it with the maximum available, mu_s N. If required is less than available, the body holds and friction equals the required value. If required exceeds available, the body slides and friction equals mu_k N.

Key idea: Compare friction required against friction available; write F = mu_s N only at impending motion, never before.

The friction angle

Combine the normal force and the friction force into a single resultant contact force. At impending motion that resultant leans away from the surface normal by an angle

phi_s = arctan(mu_s),

called the angle of static friction. Rotate it about the normal and you sweep out the cone of friction. The geometric picture is compact and useful: if the total contact force needed lies inside the cone, the surface can supply it and the body stays put; if the required force lies outside, no amount of friction can hold and sliding begins.

The same angle appears as the angle of repose of a slope, which is why a pile of dry sand always settles to the same characteristic slope. Sand with mu_s = 0.6 forms cones at about arctan(0.6) = 31 degrees, no matter how much you pour.

Key idea: The friction angle phi_s = arctan(mu_s) defines a cone; contact forces inside it can be sustained, those outside cannot.

Typical coefficients

Values vary with roughness, cleanliness, and moisture, so handbooks give ranges rather than exact numbers. Rough guides for dry surfaces: steel on steel around 0.6, steel on ice near 0.03, rubber on dry concrete between 0.6 and 0.9, wood on wood around 0.4, and lubricated metals near 0.05. Kinetic coefficients typically run 10 to 30 percent below the static value, which is why a stuck object lurches when it finally breaks free.

Worked example: pushing a crate on the floor

Given: a crate weighing 200 N sits on a floor with mu_s = 0.40. You push horizontally. Find: the push needed to make it start sliding.
Solution: vertical balance gives the normal force N = W = 200 N (no vertical push). The maximum static friction is F_max = mu_s N = 0.40 x 200 = 80 N. The crate starts to move when your push just exceeds this, so the required push is 80 N at the point of impending motion. Any push below 80 N is fully balanced by an equal friction force and the crate stays put.

Worked example: will a block slide down an incline?

Given: a 400 N block rests on a 20 degree incline with mu_s = 0.30. Find: whether it stays or slides.
Solution: resolve along and perpendicular to the slope. The component of weight pulling the block down the slope is W sin(20) = 400 x 0.342 = 136.8 N.

The normal force is N = W cos(20) = 400 x 0.940 = 375.9 N, so the maximum friction available is F_max = mu_s N = 0.30 x 375.9 = 112.8 N.
Compare: the driving force down the slope is 136.8 N, but friction can supply only 112.8 N. Since 136.8 > 112.8, friction cannot hold the block, and it slides. Equivalently, the block slides whenever the slope angle exceeds the angle of repose arctan(mu_s) = arctan(0.30) = 16.7 degrees; here 20 degrees is steeper, confirming the slide.

That angle-of-repose shortcut is worth keeping: on an incline with a rope or nothing but friction, a block is on the verge of sliding exactly when tan(angle) = mu_s. Below that angle it holds; above it, it goes.

Worked example: friction below its maximum

Given: the same 200 N crate on a floor with mu_s = 0.40 and mu_k = 0.30. This time you push horizontally with only 50 N.
Find: the actual friction force acting on the crate, and whether it moves. Then repeat for a push of 90 N.

Free-body diagram in words: isolate the crate. Its weight, 200 N, acts down. The floor pushes up with a normal force N and sideways with a friction force F opposing the tendency to slide, so F points backward against the push. The 50 N push acts horizontally.

Case 1, a 50 N push.
sum F_y = 0: N - 200 = 0, so N = 200 N.
Friction available: F_max = mu_s N = 0.40 x 200 = 80 N.
Friction required for equilibrium: sum F_x = 0 gives F = 50 N.
Since 50 is less than 80, the crate stays put and the friction force is exactly 50 N - not 80 N. Writing F = mu_s N here would overstate the friction by 60 percent and would break the equilibrium equation.

Case 2, a 90 N push. The required friction is now 90 N but only 80 N is available, so the crate slides. Once moving, friction drops to the kinetic value:
F_k = mu_k N = 0.30 x 200 = 60 N.
The crate is no longer in equilibrium; the unbalanced force is 90 - 60 = 30 N, and it accelerates. Statics has done its job by identifying the threshold; what happens after is dynamics.

Sanity check: the two cases bracket the transition at 80 N, exactly the value computed in the earlier example, so all three results are consistent. Notice the discontinuity at the threshold: just below it the crate is held by 80 N of friction, and just above it friction collapses to 60 N. That sudden drop is why heavy furniture jerks forward the instant it starts to move, and why anti-lock brakes work hard to keep a tyre just below the slipping point.

Common wrong turns

• Writing F = mu_s N when the body is not on the verge of slipping. Below the limit, friction is whatever equilibrium requires, and it must be found from sum F = 0.
• Assuming N equals the weight. It does only when nothing else acts vertically. An inclined push, a rope at an angle, or a slope all change N, and friction changes with it.
• Using W instead of W cos(theta) for the normal force on a slope. Only the component perpendicular to the surface presses the surfaces together.
• Drawing friction in the direction of motion. It always opposes the sliding or the tendency to slide.
• Believing a wider contact patch gives more friction. In the Coulomb model only N and the surface pair matter.
• Mixing up static and kinetic coefficients. Use mu_s to decide whether motion starts and mu_k only once it has.

Recap

• Friction acts along the surface, opposing the tendency to slide, and obeys F less than or equal to mu_s N.
• Only at impending motion does F = mu_s N become an equation; below that, friction equals whatever equilibrium demands.
• Once sliding, friction falls to F_k = mu_k N, which is why stuck objects lurch when they break free.
• Friction is independent of apparent contact area because the real contact area scales with the normal force.
• The friction angle phi_s = arctan(mu_s) defines a cone of sustainable contact forces and equals the angle of repose of a slope.
• Always compare friction required against friction available before deciding which formula applies.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Dry friction. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Friction. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Flexible belts. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Dry friction. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. OpenStax. (2016). 6.2 Friction. In University Physics Volume 1. Rice University. openstax.org
6. Nave, R. (n.d.). Friction. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. Baker, D. W., & Haynes, W. (2021). Friction. In Engineering Statics: Open and Interactive. Engineering LibreTexts. eng.libretexts.org

With the friction laws in hand, we can tackle richer problems: pushing at an angle, driving a block up a slope, and deciding whether a body slides or tips over. Each uses the same equilibrium equations plus the friction relation F = mu_s N at impending motion.

Key idea: Every applied-friction problem is ordinary equilibrium plus one extra equation, F = mu_s N, written only at the instant motion impends.

The one habit that solves them all

Applied friction problems look varied but share a structure. First decide what motion is about to happen, because that fixes the direction of the friction force: friction always opposes the impending slide. Second, write the equilibrium equations. Third, substitute F = mu_s N and solve. The hardest step is the first, and it is the one most often skipped.

Sign errors mostly come from getting that direction wrong. Pushing a block up a slope and letting it slide down a slope give friction in opposite senses, which is why the same geometry produces two very different answers depending on which way motion impends.

Worked example: pushing a block at an angle

Given: a 250 N block rests on a floor with mu_s = 0.25. You push with a force P directed 20 degrees below the horizontal (pressing down and forward). Find: the push needed to start it sliding.
Solution: the downward slant of the push increases the normal force.

Vertical balance:
N = W + P sin(20).
At impending motion the horizontal push equals the maximum friction:
P cos(20) = mu_s N = mu_s (W + P sin(20)).
Expand and collect the P terms:
P cos(20) - mu_s P sin(20) = mu_s W
P (cos(20) - 0.25 sin(20)) = 0.25 x 250
P (0.9397 - 0.0855) = 62.5
P (0.8542) = 62.5, so P = 73.2 N.
You need about 73 N. Pushing downward is inefficient here, because it presses the block harder into the floor and raises the friction you must overcome. Pushing slightly upward would take less force.

Compare with pulling upward at the same angle. Now the vertical component relieves the normal force, so N = W - P sin(20) and the impending-motion condition becomes
P cos(20) = mu_s (W - P sin(20))
P (cos(20) + mu_s sin(20)) = mu_s W
P (0.9397 + 0.25 x 0.3420) = 62.5
P (1.0252) = 62.5, so P = 61.0 N.
Sanity check: 61.0 N against 73.2 N, a saving of about 17 percent, purely from tilting the force the other way. The number is also bracketed correctly: a purely horizontal push would need mu_s W = 62.5 N, which sits between the two angled cases, exactly as it should. This is why a hand truck's handle is pulled upward and why dragging a sled is easier with a rope that rises to your hand.

Worked example: force to move a block up an incline

Given: a 400 N block on a 20 degree incline with mu_s = 0.30, pushed by a force P directed up along the slope. Find: the P that starts it moving up.
Solution: to move up, friction acts down the slope (opposing the impending upward motion).

Along the slope,
P = W sin(20) + mu_s W cos(20).
Compute each term: W sin(20) = 400 x 0.342 = 136.8 N and mu_s W cos(20) = 0.30 x 400 x 0.940 = 112.8 N.
P = 136.8 + 112.8 = 249.6 N.
It takes about 250 N to push the block up the slope, the sum of the gravity component and the friction, both of which oppose upward motion.

The reverse case: holding it from sliding down. If instead the block is about to slide down, friction reverses and acts up the slope, helping you. The required holding force becomes
P = W sin(20) - mu_s W cos(20) = 136.8 - 112.8 = 24.0 N.
Sanity check: the answer is positive, which means friction alone cannot hold this block and a genuine 24 N push is needed. That agrees with the previous lesson, where a 20 degree slope with mu_s = 0.30 was found to exceed the angle of repose of 16.7 degrees. Had the result come out negative, it would have meant friction was more than sufficient and no force was required at all. The gap between 24.0 N to hold it and 249.6 N to move it up is enormous, and that range - over ten to one - is entirely due to friction switching direction.

Wedges

A wedge is a simple machine that converts a large horizontal driving force into an even larger, usually vertical, lifting force. Doorstops, splitting mauls, machine-levelling shims, and the tapered keys that lock a pulley onto a shaft are all wedges. The analysis is ordinary friction analysis applied to two or three contact surfaces at once: draw a free-body diagram of the wedge and another of the block it lifts, put a friction force on every contact opposing the impending relative sliding, and solve the two sets of equilibrium equations together.

The most useful single result concerns self-locking. A wedge stays in place after the driving force is removed if its taper angle is smaller than the friction angle of its surfaces:

theta less than phi_s, where phi_s = arctan(mu_s).

For example, a wedge with a 12 degree taper working against surfaces with mu_s = 0.25 has phi_s = arctan(0.25) = 14.0 degrees. Since 12 < 14.0, the wedge is self-locking and will hold its load without being held in. Increase the taper to 20 degrees and it would spit back out the moment the hammer stopped. This is exactly why levelling shims, wood screws, and machine tapers are made shallow: the shallower the angle, the more certainly the device holds itself.

Key idea: A wedge is self-locking when its taper angle is less than the friction angle arctan(mu_s), which is why practical wedges and screws use shallow angles.

Key idea: Tilting an applied force changes the normal force and therefore the friction, so a pull that lifts slightly always beats a push that presses down.

Sliding versus tipping

When a horizontal push is applied high on a block, the block may either slide or tip about its leading bottom edge, whichever happens at the smaller force. Two checks decide it:

• Slide check: sliding impends when the push reaches P_slide = mu_s N = mu_s W.
• Tip check: tipping impends when the push produces a moment about the leading bottom edge equal to the weight's restoring moment. Taking moments about that edge, tipping starts when P h = W (b/2), where h is the push height and b is the block's width, giving P_tip = W b / (2 h).

Whichever P is smaller occurs first. A tall, narrow block pushed high tends to tip; a short, wide block pushed low tends to slide. For example, a block with W = 500 N, width b = 0.6 m, mu_s = 0.4, pushed at height h = 0.9 m: sliding needs P_slide = 0.4 x 500 = 200 N, while tipping needs P_tip = 500 x 0.6 / (2 x 0.9) = 300 / 1.8 = 167 N. Since 167 N is less than 200 N, the block tips before it slides.

The critical height. There is one push height at which the two thresholds coincide. Setting P_slide = P_tip:
mu_s W = W b / (2 h)
h_crit = b / (2 mu_s).
For the block above, h_crit = 0.6 / (2 x 0.4) = 0.75 m. Push below 0.75 m and it slides first; push above and it tips first. The 0.9 m push height in the example is above that threshold, confirming the tipping answer. Notice the weight cancels out entirely - whether a body tips or slides depends only on its width, the push height, and the friction coefficient, never on how heavy it is.

Worked example: a filing cabinet

Given: a filing cabinet weighs 800 N, is 0.5 m wide and 1.6 m tall, and stands on a floor with mu_s = 0.35. Someone pushes horizontally at the very top, so h = 1.6 m.
Find: the push at which motion begins, whether the cabinet slides or tips, and the height at which the behaviour would change.

Free-body diagram in words: isolate the cabinet. Its weight, 800 N, acts down through the centre of gravity, taken at the geometric centre so it is b/2 = 0.25 m from either edge. The floor supplies a normal force N and a friction force F opposing the push. The applied push P acts horizontally at 1.6 m above the floor. At the point of tipping the normal force migrates entirely to the leading bottom edge, which is why moments are taken about that edge.

Sliding threshold.
N = W = 800 N, so P_slide = mu_s N = 0.35 x 800 = 280 N.
Tipping threshold. Take moments about the leading bottom edge, where the push moment must overcome the weight's restoring moment.
P h = W (b/2)
P_tip = 800 x 0.5 / (2 x 1.6) = 400 / 3.2 = 125 N.

Result and sanity check. Since 125 is less than 280, the cabinet tips first, at a push of only 125 N - well under half the force needed to slide it. The critical height is h_crit = b / (2 mu_s) = 0.5 / 0.7 = 0.714 m, and the push at 1.6 m is far above it, which confirms the tipping answer independently. Substituting back, P_tip at that critical height would be 400 / (2 x 0.714) = 280 N, exactly matching P_slide as it must at the crossover. The practical lesson is familiar: tall, narrow furniture tips easily when pushed high, which is why anti-tip brackets are shipped with bookcases and why a loaded filing cabinet should have only one drawer open at a time.

Common wrong turns

• Assuming N = W when the applied force is inclined. A downward-slanting push increases N, an upward-slanting pull decreases it, and friction follows.
• Keeping friction in the same direction for up-slope and down-slope cases. Friction always opposes the impending motion, so it reverses between the two cases.
• Taking tipping moments about the centre of the base. Tipping occurs about the leading bottom edge, where the normal force concentrates at the instant of rotation.
• Using the full height instead of the push height. The moment arm of P is the height at which it is applied, which may be well below the top of the body.
• Computing only one threshold. Both P_slide and P_tip must be found; the smaller one is what actually happens.
• Expecting weight to decide the outcome. It cancels from the slide-or-tip comparison entirely; geometry and mu_s decide it.
• Treating a wedge as frictionless. Friction is the entire reason a wedge stays put, and neglecting it makes every practical wedge look useless.

Recap

• Decide the direction of impending motion first; friction opposes it, and that fixes every sign.
• An inclined push changes the normal force: N = W + P sin(theta) pushing down, N = W - P sin(theta) pulling up.
• To move a block up a slope, P = W sin(theta) + mu_s W cos(theta); to keep it from sliding down, P = W sin(theta) - mu_s W cos(theta).
• A wedge is self-locking when its taper angle is less than the friction angle arctan(mu_s).
• Sliding impends at P_slide = mu_s W; tipping impends at P_tip = W b / (2 h); the smaller threshold wins.
• The crossover height is h_crit = b / (2 mu_s), and the weight cancels out of the comparison.

Sources

1. Baker, D. W., & Haynes, W. (n.d.). Wedges. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
2. Baker, D. W., & Haynes, W. (n.d.). Slipping vs. tipping. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
3. Baker, D. W., & Haynes, W. (n.d.). Friction. In Engineering Statics: Open and Interactive. Colorado State University and Massachusetts Maritime Academy. engineeringstatics.org
4. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Slipping vs. tipping. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
5. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Wedges. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
6. Moore, J., Miller, N., Fahrenholtz, S., Robertson, C., Fox, C., & Kirtley, N. (n.d.). Dry friction. In Mechanics Map. Pennsylvania State University. mechanicsmap.psu.edu
7. Meriam, J. L., Kraige, L. G., & Bolton, J. N. (2020). Engineering mechanics: Statics (9th ed.). Wiley. find source ↗