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.

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. 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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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. As an independent check, take moments about B: Ay(8) - 600(6) - 400(2) - 300 = 512.5(8) - 3600 - 800 - 300 = 4100 - 4700 = 0. It balances, so the answer is confirmed. (The clockwise 300 N·m couple appears with the same sign about either point, because a couple's moment is the same everywhere.)

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

The truss we will solve

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.