Physics I: Classical Mechanics

Physics is a quantitative science, so every measurement needs a unit. A number without a unit is meaningless: "5" could be 5 meters, 5 seconds, or 5 kilograms, and each describes a completely different physical situation. The international standard is the SI system (Systeme International), built on a small set of base units. In mechanics we use just three: the meter (m) for length, the kilogram (kg) for mass, and the second (s) for time. Every other mechanical unit is a combination of these, called a derived unit. Speed is meters per second (m/s), and force, as you will see, is measured in newtons, where one newton equals one kg times m/s squared. Because the whole of mechanics is assembled from just these three building blocks, keeping careful track of them is the single most reliable habit a physicist can have.

The SI base units and prefixes

Real problems span an enormous range of sizes, from the diameter of an atom to the distance to a galaxy. Rather than write out long strings of zeros, SI uses prefixes that multiply a unit by a power of ten. A kilometer (km) is 10^3 meters; a millimeter (mm) is 10^-3 meters; a microsecond is 10^-6 seconds. Learning the common prefixes - kilo, centi, milli, micro, mega, giga - lets you move fluidly between scales. The trick is that a prefix is just a shorthand for a numerical factor, so "3.0 km" and "3000 m" are literally the same quantity written two ways.

Dimensional analysis

Each quantity has a dimension: length [L], mass [M], or time [T]. The dimension is more fundamental than the unit - length is length whether you measure it in meters, feet, or light-years. A correct equation must have matching dimensions on both sides, because you cannot add a length to a time any more than you can add apples to hours. Checking dimensions is a fast way to catch mistakes before you ever plug in a number. Consider the kinematics equation x = v t. On the right, velocity [L/T] times time [T] gives [L], which matches the position on the left. The equation is dimensionally consistent. Now test a wrong guess, x = v t squared: the right side has dimension [L/T][T squared] = [L T], which is not a length, so the equation cannot be correct no matter what numbers you use. Dimensional analysis will not catch a missing factor of one half, but it reliably flags any error that changes the kind of quantity.

Significant figures

A measured value carries only as much precision as the instrument allows. The significant figures are the digits that are known reliably plus one estimated digit. A ruler marked in millimeters lets you report a length to the nearest tenth of a millimeter by estimating between the marks, but no further. When you multiply or divide, the answer keeps the same number of significant figures as the least precise input, because the result can be no more certain than its shakiest ingredient. When you add or subtract, you instead line up the decimal places and keep the answer only to the least precise decimal position. If a car travels 100.0 m (4 significant figures) in 8.2 s (2 significant figures), the speed is 100.0 / 8.2 = 12.195..., which we report as 12 m/s - two significant figures, matching the time. Reporting "12.195 m/s" would be dishonest: it claims a precision the data does not support.

Worked example: unit conversion

Given: a speed of 90 km/h. Find: the speed in m/s.
Solution: Multiply by conversion factors chosen so the unwanted units cancel. A conversion factor is just a ratio equal to one, such as (1000 m / 1 km), so multiplying by it never changes the physical quantity, only its units. There are 1000 m in a km and 3600 s in an hour.
90 km/h x (1000 m / 1 km) x (1 h / 3600 s) = 90 x 1000 / 3600 = 25 m/s.
Notice how "km" cancels against "km" and "h" cancels against "h", leaving m/s. So 90 km/h equals 25 m/s. A useful rule of thumb falls out of this: to convert km/h to m/s, divide by 3.6.

Why it matters

Getting units and precision right is not busywork. In real engineering and science, a units error can crash a spacecraft: the 1999 loss of a NASA Mars orbiter traced back to one team using pound-force and another using newtons. Building the habit now, on simple problems, is what makes the harder problems trustworthy later. Every quantity you compute in this course should carry a unit, and every final answer should carry the right number of significant figures. Treat these as part of the answer, not decoration.

Some quantities need only a size to be fully described. Mass, time, temperature, and speed are scalars - a single number with a unit. Other quantities also need a direction. Displacement, velocity, acceleration, and force are vectors: they have both a magnitude (size) and a direction. Walking 3 km is a distance (scalar); walking 3 km east is a displacement (vector). This distinction runs through all of mechanics, because the direction of a force or a velocity changes everything about what happens next. Two forces of equal size can add up, cancel out, or produce something in between, entirely depending on their directions.

Representing vectors

A vector is drawn as an arrow: its length shows the magnitude and the way it points shows the direction. We write a vector's magnitude with plain symbols, so a force of magnitude F might point at some angle. Two vectors are equal if they have the same length and direction, regardless of where they are drawn on the page - sliding an arrow around without rotating or resizing it does not change the vector it represents. This freedom is what lets us line vectors up head to tail when we add them.

Components

The most reliable way to work with vectors is to break each one into perpendicular pieces called components along the x and y axes. This turns a hard geometry problem into simple arithmetic. A vector of magnitude V pointing at an angle theta above the x-axis has components Vx = V cos(theta) and Vy = V sin(theta). These come straight from the definitions of sine and cosine in the right triangle the vector forms with the axes. To reverse the process, the magnitude is V = sqrt(Vx squared + Vy squared) by the Pythagorean theorem, and the direction is theta = arctan(Vy / Vx). Components can be positive or negative: a vector pointing down and to the left has both Vx and Vy negative, which automatically encodes its direction.

Adding vectors

To add two vectors, add their x-components to get the total x, and add their y-components to get the total y. The two sums are the components of the resultant, the single vector that has the same effect as the originals combined. This component method works for any number of vectors and any directions, which is why physicists prefer it over drawing and measuring. To subtract a vector, add its negative (the same arrow reversed), which flips the signs of both components. The graphical picture that matches this arithmetic is the head-to-tail rule: place the tail of the second arrow at the head of the first, and the resultant runs from the very first tail to the very last head.

Worked example: two displacements

Given: you walk 6.0 m east, then 8.0 m north. Find: your total displacement (magnitude and direction).
Solution: East is the +x direction and north is +y. The components of the total are x = 6.0 m and y = 8.0 m. The magnitude is sqrt(6.0 squared + 8.0 squared) = sqrt(36 + 64) = sqrt(100) = 10.0 m. The direction is arctan(8.0 / 6.0) = arctan(1.33) = 53 degrees north of east. So your displacement is 10.0 m at 53 degrees north of east. Notice this straight-line displacement (10 m) is shorter than the total distance walked (14 m), because displacement measures only the net change in position, not the path length.

Worked example: adding two forces

Given: two forces act on a ring: F1 = 30 N pointing east, and F2 = 40 N pointing at 90 degrees to it, north. Find: the resultant force.
Solution: The components add directly: total Fx = 30 N, total Fy = 40 N. The magnitude is sqrt(30 squared + 40 squared) = sqrt(900 + 1600) = sqrt(2500) = 50 N, at arctan(40/30) = 53 degrees north of east. The single resultant of 50 N would have exactly the same effect as the two forces together, which is why we can always replace a set of forces with their vector sum.

Kinematics describes how things move without yet asking why - that "why" waits for Newton's laws in Module 3. In one dimension, position is measured along a line, with one direction chosen as positive. Displacement is the change in position, delta x = x_final - x_initial, and it can be negative if the object ends up behind where it started. Divide displacement by the time taken and you get average velocity, v = delta x / delta t. Velocity differs from speed in that it has a sign showing direction: a velocity of -5 m/s means moving in the negative direction at 5 m/s. Acceleration is the rate of change of velocity, a = delta v / delta t, measured in m/s squared. Crucially, acceleration is about changing velocity, so an object can be accelerating even while slowing down - deceleration is simply acceleration opposite to the motion.

Average versus instantaneous

Average velocity smooths over a whole time interval, but often we want the velocity at a single moment - the reading on a speedometer. This is the instantaneous velocity, the value the average approaches as the time interval shrinks toward zero. On a position-versus-time graph, average velocity is the slope of the line connecting two points, while instantaneous velocity is the slope of the tangent at one point. The same distinction applies to acceleration. For the constant-acceleration problems in this lesson, the average and instantaneous accelerations are equal, which is exactly what makes the equations below so clean.

The kinematic equations

When acceleration is constant, four equations connect the five quantities of motion: initial velocity v0, final velocity v, acceleration a, displacement x, and time t. Each equation leaves out exactly one variable, so the art of solving a problem is to identify what you know and what you want, then pick the equation that omits the quantity you neither know nor need.

Equation	Leaves out
v = v0 + a t	x
x = v0 t + (1/2) a t squared	v
v squared = v0 squared + 2 a x	t
x = (1/2)(v0 + v) t	a

A reliable recipe: (1) write down every known quantity with its sign, (2) identify the unknown, (3) choose the equation that contains your knowns and unknown but not the irrelevant variable, and (4) solve algebraically before substituting numbers. Keeping the algebra symbolic until the last step reduces arithmetic mistakes and makes the physics visible.

Worked example: an accelerating car

Given: a car starts from rest (v0 = 0) and reaches v = 27 m/s in t = 5.0 s. Find: its acceleration and the distance it covers.
Solution: For acceleration, use v = v0 + a t. Solving, a = (v - v0) / t = (27 - 0) / 5.0 = 5.4 m/s squared. For distance, use x = (1/2)(v0 + v) t = (1/2)(0 + 27)(5.0) = 67.5 m. So the car accelerates at 5.4 m/s squared and travels 67.5 m. As a check, the average velocity over the interval is (0 + 27)/2 = 13.5 m/s, and 13.5 x 5.0 = 67.5 m, which agrees.

Worked example: stopping distance

Given: a car moving at v0 = 20 m/s brakes with acceleration a = -5.0 m/s squared until it stops (v = 0). Find: the stopping distance.
Solution: Time is not asked for, so use v squared = v0 squared + 2 a x. Then 0 = 20 squared + 2(-5.0) x, so 0 = 400 - 10 x, giving x = 40 m. The car needs 40 m to stop. Notice the acceleration is negative because it opposes the motion. An important consequence hides in the v0 squared term: stopping distance grows with the square of speed, so a car at 40 m/s needs four times the distance of one at 20 m/s. This is the physics behind highway speed limits and following-distance rules.

Near Earth's surface, every object in free fall - moving under gravity alone, with air resistance neglected - accelerates downward at the same rate. This acceleration due to gravity is written g and equals about 9.8 m/s squared. A feather and a hammer dropped in a vacuum land together, a fact famously demonstrated by astronaut David Scott on the Moon in 1971. Mass does not affect the rate of fall; it only affects the gravitational force, as Module 4 explains. This is one of the most counterintuitive results in all of physics, because everyday experience - a leaf drifting while a stone plummets - seems to say the opposite. The difference is entirely air resistance, not gravity.

Setting up vertical motion

Free fall is just one-dimensional kinematics with a specific acceleration. Choose up as positive; then the acceleration is a = -g, pointing down, throughout the motion - on the way up, at the top, and on the way down alike. All four kinematic equations still apply, with the vertical position usually written as y. The key insight is that the same equations govern an object dropped, thrown down, or thrown up; only the initial velocity v0 changes. This means you do not learn new physics for free fall, only a new value for the acceleration.

Why mass does not matter

It is worth pausing on why a heavy object and a light one fall together. A heavier object is pulled harder by gravity, but it also has more inertia and is correspondingly harder to accelerate. These two effects cancel exactly, leaving the same acceleration for every mass. Galileo argued this centuries before Newton by a thought experiment: if heavy things fell faster, then tying a light object to a heavy one should slow the heavy one down (the light one lagging), yet the combination is heavier still and should fall faster - a contradiction that can only be resolved if all objects fall at the same rate.

Worked example: a dropped object

Given: a ball is dropped from rest (v0 = 0) from a height of 44.1 m. Take g = 9.8 m/s squared. Find: the time to reach the ground and the speed at impact.
Solution: Using y = (1/2) g t squared for the magnitude of the fall, 44.1 = (1/2)(9.8) t squared = 4.9 t squared, so t squared = 9.0 and t = 3.0 s. The impact speed is v = g t = 9.8 x 3.0 = 29.4 m/s. So the ball falls for 3.0 s and hits at 29.4 m/s. Note that the fall distance grows as time squared, so in the first second the ball falls only 4.9 m, but in the third second it covers over 24 m - it speeds up as it goes.

Worked example: thrown straight up

Given: a ball is thrown straight up at v0 = 19.6 m/s. Find: the time to reach the highest point and the maximum height.
Solution: At the highest point the velocity is momentarily zero. Using v = v0 - g t with v = 0: 0 = 19.6 - 9.8 t, so t = 2.0 s. For the height, v squared = v0 squared - 2 g y gives 0 = 19.6 squared - 2(9.8) y, so y = 384.16 / 19.6 = 19.6 m. The ball rises for 2.0 s to a height of 19.6 m. By symmetry it takes another 2.0 s to fall back, returning at the same 19.6 m/s it left with, but now directed downward.

The symmetry of flight and real air

The up-and-down symmetry - equal rise and fall times, equal launch and return speeds - is exact only when air resistance is ignored. A real thrown ball feels a drag force that grows with speed and always opposes motion, so it actually spends slightly longer falling than rising and returns a little slower than it left. When drag balances gravity entirely, an object stops accelerating and settles at a constant terminal velocity; this is why a skydiver reaches a steady speed and why raindrops do not arrive at lethal velocity despite falling from great heights. For the idealized problems in this course we neglect air, but it is good to know exactly which assumption we are making.

When an object moves in two dimensions under gravity, the secret is a powerful idea called the independence of motion: the horizontal and vertical parts of the motion do not affect each other. Horizontally, gravity exerts no force, so velocity is constant. Vertically, the object is in free fall with acceleration g. This means a two-dimensional problem is really two one-dimensional problems running in parallel, linked only by the shared clock. A ball rolled off a table and a ball dropped from the same height hit the floor at the same instant, because their vertical motions are identical - the horizontal motion of the first ball changes where it lands but not when.

The strategy: split, solve, recombine

Every projectile problem follows the same three steps. First, resolve the initial velocity into horizontal and vertical components. Second, treat each direction with its own kinematics: horizontal uses x = vx t with no acceleration, while vertical uses the free-fall equations with a = -g. Third, use the time t as the bridge between the two, since both motions happen over the same interval. The moment you internalize that time is the only quantity the two directions share, projectile motion stops feeling like a special topic and becomes routine.

Horizontal launch

For an object launched horizontally at speed vx from height h, the initial vertical velocity is zero, so the vertical motion determines the fall time from h = (1/2) g t squared. The horizontal distance (the range) is then simply x = vx t. The path traced out is a parabola, curving downward more and more steeply as the growing vertical speed combines with the steady horizontal speed.

Worked example: off a cliff

Given: a ball leaves a 20 m high cliff horizontally at vx = 15 m/s. Take g = 9.8. Find: the time in the air and how far from the base it lands.
Solution: Fall time from 20 = (1/2)(9.8) t squared = 4.9 t squared gives t squared = 4.08, so t = 2.02 s. Horizontal range is x = 15 x 2.02 = 30.3 m. The ball is airborne for 2.02 s and lands 30.3 m out. The horizontal launch speed had no effect on the fall time - the ball would hit the ground at the same moment whether it left the cliff at 1 m/s or 100 m/s.

Angled launch

Launch at speed v0 and angle theta above the horizontal. Split the initial velocity: v0x = v0 cos(theta) stays constant, while v0y = v0 sin(theta) behaves exactly like a ball thrown straight up. Three useful results follow, valid when launch and landing heights are equal:

• Time of flight: t = 2 v0 sin(theta) / g
• Maximum height: H = (v0 sin(theta)) squared / (2 g)
• Range: R = v0 squared sin(2 theta) / g

Because of the sin(2 theta) term, the range is greatest at a launch angle of 45 degrees, where sin(2 theta) = sin 90 = 1. A neat consequence is that angles equally above and below 45 - such as 30 and 60, or 40 and 50 - give the same range, though the higher angle produces a taller, slower-arriving arc. This is why a punt in football and a low line-drive can travel the same distance by very different routes.

Worked example: a kicked ball

Given: a ball is kicked at v0 = 20 m/s at theta = 30 degrees. Take g = 9.8. Find: the maximum height and the range.
Solution: The vertical launch speed is v0 sin 30 = 20 x 0.5 = 10 m/s. Maximum height is H = 10 squared / (2 x 9.8) = 100 / 19.6 = 5.1 m. For range, sin(2 x 30) = sin 60 = 0.866, so R = 20 squared x 0.866 / 9.8 = 400 x 0.866 / 9.8 = 35.3 m. The ball rises 5.1 m and lands 35.3 m away. To confirm with the split method: the horizontal speed is 20 cos 30 = 17.3 m/s, time of flight is 2 x 10 / 9.8 = 2.04 s, and 17.3 x 2.04 = 35.3 m, which matches.

Kinematics tells us how objects move; dynamics tells us why. The foundation is three laws set out by Isaac Newton in his 1687 work the Principia, which together explain the motion of everything from a rolling ball to the planets. A force is a push or pull, a vector measured in newtons (N). What matters for motion is never a single force in isolation but the net force, the vector sum of all forces acting on an object. Getting comfortable with the idea of net force is the whole game: once you can total up the forces on an object, Newton's second law tells you exactly what it will do.

The first law: inertia

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net force. This tendency to resist changes in motion is called inertia, and mass is its quantitative measure - a loaded truck has far more inertia than a bicycle. The first law overturns the old Aristotelian intuition that a force is needed to keep something moving. In reality, moving objects slow down only because forces like friction and air resistance act on them; remove those forces, as on a frictionless surface or in the vacuum of space, and no force at all is needed to maintain constant velocity forever. A spacecraft coasting between planets with its engines off is the first law in its purest form.

The second law: F = ma

When a net force does act, it produces an acceleration in the same direction, given by the most important equation in mechanics: F_net = m a. The acceleration is directly proportional to the net force and inversely proportional to the mass. Double the force and you double the acceleration; double the mass and you halve it. This single equation is the bridge between the force concepts of this module and the kinematics of Module 2: find the net force, divide by mass to get acceleration, and then the kinematic equations tell you the motion. It is worth noting that the first law is really just the special case of the second when the net force is zero, giving zero acceleration and hence constant velocity.

Worked example: pushing a car

Given: a 1200 kg car experiences a net forward force of 3600 N. Find: its acceleration.
Solution: From F_net = m a, solve a = F_net / m = 3600 / 1200 = 3.0 m/s squared. The car accelerates at 3.0 m/s squared. If you then wanted the speed after 5 seconds starting from rest, you would hand this acceleration to kinematics: v = a t = 3.0 x 5 = 15 m/s. This handoff between dynamics and kinematics is the pattern behind most mechanics problems.

The third law: action and reaction

For every action there is an equal and opposite reaction. If object A pushes on object B with some force, then B pushes back on A with a force equal in magnitude and opposite in direction. The single most important thing to understand about the third law is that these two forces act on different objects, which is precisely why they do not cancel each other out - cancellation only happens between forces on the same object. When you walk, your foot pushes backward on the ground and the ground pushes forward on you; that forward reaction force is what propels you, and it acts on you, not on the ground. A rocket works the same way, pushing exhaust gas downward so the gas pushes the rocket upward, which is why rockets function even in the vacuum of space where there is nothing to "push against" but their own exhaust.

Why it matters

These three laws are not abstract - they underlie seatbelts (your inertia keeps you moving forward when the car stops suddenly), the recoil of a fired gun, the way a swimmer pushes water backward to move forward, and the design of every vehicle and structure. Master them and you have the conceptual core of all classical mechanics.

To apply Newton's second law, you first need a clear picture of every force acting on your object. A free-body diagram is that picture: the object is isolated and drawn as a dot or box, and every force acting on it is drawn as an arrow pointing in its direction of action. Forces the object exerts on other things do not appear - only forces acting on the chosen body. Getting this diagram right is more than half the battle in any dynamics problem; most mistakes trace back to a missing force, an extra force that should not be there, or a wrong direction.

The common forces

A handful of forces appear again and again in mechanics. Weight (W = m g) always points straight down toward Earth's center and acts on every object with mass. The normal force (N) is the support a surface pushes back with, always perpendicular to that surface. Tension (T) is the pull transmitted along a rope, string, or cable, directed along its length away from the object. Friction (covered next lesson) acts along a surface, opposing relative sliding. And there may be any applied push or pull from a hand, engine, or other agent. Learning to recognize which of these act in a given situation is a skill that comes quickly with practice.

The solution recipe

Once the diagram is drawn, the procedure is always the same: choose x and y axes, write Newton's second law separately for each direction (sum of Fx = m ax and sum of Fy = m ay), and solve. On a flat surface with no vertical acceleration, the vertical equation gives N - W = 0, so the normal force balances the weight and N = m g. If the object is in equilibrium - at rest or moving at constant velocity - then every acceleration is zero, so the forces in each direction must sum to zero. This equilibrium condition is the basis of statics, the engineering of bridges and buildings that must not move.

Worked example: a box on a frictionless incline

Given: a box slides down a frictionless ramp tilted at theta = 30 degrees. Take g = 9.8. Find: its acceleration down the ramp.
Solution: The clever move is to tilt the axes so x points down the slope and y points away from the surface, matching the geometry. Then only the weight needs resolving. The component of gravity along the slope is W sin(theta) = m g sin(theta), and the component pressing into the surface is W cos(theta) = m g cos(theta). Newton's second law along the slope gives m g sin(theta) = m a, so the mass cancels and a = g sin(theta) = 9.8 x sin 30 = 9.8 x 0.5 = 4.9 m/s squared. The box accelerates at 4.9 m/s squared down the ramp, regardless of its mass. This is why the steepness, not the weight, sets how fast a frictionless slide speeds up. Perpendicular to the slope there is no acceleration, so N = m g cos(theta), which is less than the full weight - a result that matters as soon as friction enters.

Worked example: an elevator

Given: a 70 kg person stands in an elevator that accelerates upward at 2.0 m/s squared. Take g = 9.8. Find: the normal force from the floor (the "apparent weight").
Solution: Two forces act on the person: the normal force N up and the weight W = mg = 686 N down. Newton's second law upward gives N - mg = m a, so N = m(g + a) = 70 x (9.8 + 2.0) = 70 x 11.8 = 826 N. The person feels 826 N pressing up, heavier than their true weight of 686 N - the familiar sensation of being pushed into the floor as an elevator starts up. Were the elevator accelerating downward, the normal force would be less than the weight, and in free fall (a = -g) it would be zero, which is the origin of apparent weightlessness.

Friction is the force that resists sliding between two surfaces in contact. It arises from microscopic roughness and molecular adhesion where the surfaces touch, and it always acts along the surface in the direction that opposes relative motion or the tendency toward motion. Friction is often treated as a nuisance, but it is also indispensable: without it you could not walk, drive, hold a pencil, or tie a knot. Learning to handle friction turns the idealized frictionless problems of the last lesson into realistic ones.

Two kinds of friction

There are two distinct regimes. Static friction acts when the surfaces are not yet sliding. It is a responsive force that adjusts itself to exactly match whatever is trying to cause motion, up to a maximum value - push a heavy crate gently and static friction pushes back just as hard, so nothing moves; push harder and it pushes back harder, until you exceed its limit and the crate breaks free. Kinetic friction acts once the surfaces are actually sliding, and it is generally a bit weaker than the maximum static friction and roughly constant regardless of sliding speed. This is why it takes a big initial shove to get something moving but less effort to keep it sliding.

The friction model

Both types depend on how hard the surfaces are pressed together, that is, on the normal force N. The kinetic friction force is f_k = mu_k N, where mu_k is the coefficient of kinetic friction, a dimensionless number set by the pair of materials in contact. The maximum static friction is f_s(max) = mu_s N, with its own coefficient mu_s. Below that maximum, the actual static friction is only as large as it needs to be to prevent sliding. These coefficients are typically between about 0.1 for slippery pairs (ice on steel) and 1.0 or more for grippy ones (rubber on dry concrete). A surprising feature of this model is that friction does not depend on the contact area - a brick slides with the same friction whether laid on its broad face or its narrow edge, because spreading the same weight over more area reduces the pressure proportionally.

Worked example: pushing a box on the floor

Given: a 10 kg box on a level floor is pushed horizontally with 40 N. The coefficient of kinetic friction is mu_k = 0.30. Take g = 9.8. Find: the friction force and the acceleration.
Solution: On a level floor with no vertical acceleration, the normal force equals the weight: N = m g = 10 x 9.8 = 98 N. The kinetic friction force is f_k = mu_k N = 0.30 x 98 = 29.4 N, opposing the push. The net horizontal force is 40 - 29.4 = 10.6 N, so the acceleration is a = 10.6 / 10 = 1.06 m/s squared. The box speeds up at about 1.06 m/s squared. Notice that if you pushed with only 25 N - less than the friction the surface can provide once moving - and the box were already at rest, static friction would simply match your push and the box would stay put.

Worked example: will it slide?

Given: a 15 kg crate sits on a floor with mu_s = 0.40. You push horizontally with 50 N. Take g = 9.8. Find: whether the crate moves.
Solution: First find the maximum static friction: N = mg = 147 N, so f_s(max) = mu_s N = 0.40 x 147 = 58.8 N. Your push of 50 N is less than this 58.8 N limit, so static friction rises to match your 50 N and the crate does not move. You would need to push beyond 58.8 N to break it loose. This kind of threshold check - compare the applied force to mu_s N - is how you decide in any problem whether an object is about to slide.

Why static friction matters

Static friction is what lets you walk without slipping and lets car tires grip the road. A car accelerates because static friction from the road pushes the tires forward; the tire surface in contact with the road is momentarily at rest relative to it, so it is static, not kinetic, friction at work. If you demand more force than the maximum (mu_s N), the tires spin and you skid. Because mu_s is usually larger than mu_k, a rolling, gripping tire has more traction than a skidding one, which is exactly the physics behind anti-lock braking systems that prevent the wheels from locking up so they keep gripping rather than sliding.

An object moving in a circle at constant speed is in uniform circular motion. The word "uniform" refers only to the speed; the velocity is emphatically not constant, because velocity is a vector and its direction is continually changing as the object rounds the circle. Since any change in velocity - even a change in direction alone - is an acceleration, a circling object is always accelerating, and that acceleration points toward the center of the circle. This center-pointing acceleration is called centripetal acceleration, from the Latin for "center-seeking." Grasping that constant speed can coexist with nonzero acceleration is the conceptual key to this entire topic.

The centripetal equations

For an object moving at speed v in a circle of radius r, the centripetal acceleration is a_c = v squared / r, directed toward the center. Two things about this formula are worth noticing: the acceleration grows with the square of the speed, so doubling the speed quadruples the acceleration, and it shrinks with a larger radius, so a gentle wide curve requires less acceleration than a tight one at the same speed. By Newton's second law, producing this acceleration requires a net inward force, the centripetal force, of magnitude F_c = m a_c = m v squared / r.

Centripetal force is a role, not a new force

It is vital to understand that centripetal force is not a new kind of force to add to your list alongside gravity, tension, and friction. It is the name for the role played by whichever real force happens to point toward the center in a given situation. For a car turning on a flat road, the centripetal force is the static friction between tires and road. For a ball whirled on a string, it is the tension. For the Moon circling Earth, it is gravity. For a rider on a loop-the-loop, it can be the normal force from the track. Identifying which real force is doing the job is the crucial first step in any circular-motion problem. Remove that force - cut the string, hit an icy patch - and the object does not fly outward; it flies off along a straight line tangent to the circle, exactly as Newton's first law predicts, because with no inward force there is nothing to bend its path.

The "centrifugal" feeling

When a car turns sharply you feel flung toward the outside of the curve, which tempts you to believe an outward force is acting. There is no such force on you. What you feel is your own inertia: your body tends to continue in a straight line, so the car door must push inward on you to force you around the curve, and you experience that inward push from the door as if you were being pressed outward against it. The outward "centrifugal force" is an illusion of being in an accelerating (rotating) reference frame, not a real force in the fixed frame we use for these problems.

Worked example: a car rounding a curve

Given: a car takes a curve of radius r = 45 m at v = 15 m/s. Find: the centripetal acceleration.
Solution: a_c = v squared / r = 15 squared / 45 = 225 / 45 = 5.0 m/s squared, directed toward the center of the curve. The static friction between the tires and road must supply the force that produces this acceleration. If the required friction ever exceeds what the tires can provide (mu_s N), the car slides outward off the curve - which is why you must slow down for a tight or icy bend.

Worked example: a ball on a string

Given: a 0.50 kg ball whirls on a string in a horizontal circle of radius 1.5 m at v = 6.0 m/s. Find: the tension in the string.
Solution: The centripetal acceleration is a_c = 6.0 squared / 1.5 = 36 / 1.5 = 24 m/s squared. The tension provides the centripetal force: F_c = m a_c = 0.50 x 24 = 12 N. The string must pull with 12 N toward the center to keep the ball circling. If you whirled the ball faster, the required tension would climb with the square of the speed, which is why a string eventually snaps if you spin something too quickly.

Newton's great insight was that the same force pulling an apple to the ground also holds the Moon in orbit - that terrestrial and celestial motion obey one law. His law of universal gravitation states that every pair of masses attracts each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers:

F = G m1 m2 / r squared

Here m1 and m2 are the two masses, r is the distance between their centers, and G is the gravitational constant, 6.674 x 10^-11 in SI units (newton meters squared per kilogram squared). The force is always attractive and acts along the line joining the two masses, with each mass pulling on the other equally in accordance with Newton's third law. Because G is such a tiny number, gravity between everyday objects is utterly negligible - two people standing side by side attract each other with a force far too small to feel. Gravity becomes significant only when at least one of the masses is astronomically large, like a planet or star.

The inverse-square law

The 1 / r squared dependence is the heart of the law and deserves careful attention. Double the distance and the force drops to a quarter; triple it and the force drops to a ninth; increase it tenfold and the force falls to a hundredth. This rapid falloff is why gravity weakens quickly with altitude yet, because 1/r squared never actually reaches zero, extends its reach across the entire universe, binding galaxies together over unimaginable distances. The inverse-square pattern is not arbitrary; it reflects the way influence spreads out over the surface of an expanding sphere, whose area grows as r squared.

Weight is gravitation

An object's weight is simply the gravitational force the Earth exerts on it - the two are the same phenomenon, not merely related. Setting m g = G M_Earth m / R_Earth squared and canceling the object's mass m shows that g = G M_Earth / R_Earth squared. Plugging in Earth's mass (5.97 x 10^24 kg) and radius (6.37 x 10^6 m) yields about 9.8 m/s squared, exactly the free-fall value used throughout Module 2. This is a beautiful moment of unification: the constant g you have been using since the second module is a direct consequence of Newton's gravitation applied to the specific mass and size of Earth. On the Moon, with far less mass, the same formula gives a surface gravity of only about 1.6 m/s squared, which is why astronauts bounded so easily across it.

Gravity as the centripetal force for orbits

Combining this lesson with the last one explains orbits completely. A satellite or moon travels in a near-circular path because gravity supplies exactly the centripetal force needed to bend its motion into a circle. Setting the gravitational force equal to the centripetal force, G M m / r squared = m v squared / r, and canceling the orbiting mass, gives the orbital speed v = sqrt(G M / r). Everything in orbit is, in truth, continuously falling toward Earth - it simply moves sideways fast enough that it keeps missing, curving around the planet instead of hitting it. Newton himself pictured this with a cannon on a mountaintop firing ever faster until the ball's fall matched Earth's curvature and it circled the globe.

Worked example: comparing gravity to weight

Given: a satellite orbits at an altitude equal to one Earth radius, so its distance from Earth's center is 2 R_Earth. Find: how its gravitational acceleration compares to surface gravity.
Solution: Because gravity follows 1 / r squared, doubling the distance divides the acceleration by 2 squared = 4. So the acceleration is 9.8 / 4 = 2.45 m/s squared, about one quarter of the surface value. The satellite still very much feels gravity - that inward pull is precisely the centripetal force keeping it in orbit. Astronauts aboard such a satellite float not because gravity is absent but because they and their spacecraft fall together, an idea explored in the assignment.

In everyday speech, work means effort. In physics, work has a precise and sometimes surprising meaning: it is done only when a force moves an object through a distance. Specifically, W = F d cos(theta), where F is the magnitude of the force, d is the magnitude of the displacement, and theta is the angle between the force and the displacement. Work is a scalar measured in joules (J), where one joule is one newton-meter. This definition captures the idea that a force accomplishes something mechanical only to the extent that it acts along the direction of motion.

When work is zero, positive, or negative

The cos(theta) factor makes work behave in ways that clash with intuition. If you push in the direction of motion, theta = 0 and cos 0 = 1, so W = F d is maximal and positive. A force perpendicular to the motion (theta = 90 degrees) does zero work, because cos 90 = 0 - which is why carrying a heavy box across a level floor does no work on the box against gravity, since the upward carrying force is perpendicular to the horizontal motion. And a force opposing the motion (theta = 180 degrees, cos 180 = -1) does negative work, meaning it removes energy; friction is the classic example. Positive work speeds things up or stores energy; negative work slows them down or releases it.

Worked example: pushing a crate

Given: you push a crate with F = 50 N over d = 8.0 m in the direction of the push. Find: the work done.
Solution: W = F d cos(0) = 50 x 8.0 x 1 = 400 J. If instead you pushed at 60 degrees to the motion, W = 50 x 8.0 x cos 60 = 50 x 8.0 x 0.5 = 200 J - only half as much, because only the component of force along the motion, F cos(theta), does work. The perpendicular component merely presses the crate into or out of the floor without contributing to its forward progress.

Kinetic energy

Kinetic energy is the energy an object has by virtue of its motion, given by KE = (1/2) m v squared, also measured in joules. Two features stand out. First, kinetic energy is always positive, since it depends on speed squared and mass, both positive. Second, and more importantly, it grows with the square of speed, so doubling an object's speed quadruples its kinetic energy and tripling it multiplies the energy ninefold. This quadratic dependence is why a small increase in highway speed sharply raises both stopping distance and collision severity, and why a bullet, though light, carries devastating energy at high speed.

The work-energy theorem

Work and kinetic energy are tied together by one of the most useful results in mechanics, the work-energy theorem: the net work done on an object equals its change in kinetic energy, W_net = KE_final - KE_initial. This holds regardless of the details of the forces or the path. Its great value is that it lets you relate forces and speeds directly, bypassing acceleration and time entirely - if you know the net work done, you know the change in speed, and vice versa. Whenever a problem gives you forces and distances and asks about speeds (or the reverse), the work-energy theorem is often the fastest route to the answer.

Worked example: speeding up a cart

Given: a 1500 kg cart starts from rest and a net work of 30000 J is done on it. Find: its final speed.
Solution: By the work-energy theorem, W_net = (1/2) m v squared - 0, so 30000 = (1/2)(1500) v squared = 750 v squared. Then v squared = 40 and v = 6.3 m/s. The cart ends up moving at about 6.3 m/s. Notice we never needed the force, the acceleration, or the time - only the total work, which is the whole point of the energy method.

Power: the rate of doing work

Two engines can do the same total work but differ greatly in how fast they do it. Power is the rate at which work is done, P = W / t, measured in watts (W), where one watt is one joule per second. A powerful motor does a given amount of work in less time. Power also equals force times velocity (P = F v) for a force acting along the motion, which is why a car needs far more engine power to sustain high speed against air resistance than to cruise slowly. This is why engines and appliances are rated in watts or kilowatts.

Potential energy is stored energy that depends on an object's position or configuration rather than its motion. It represents work that has been done to place an object somewhere and that can be recovered later as motion. The most familiar form is gravitational potential energy, the energy an object has because of its height: PE = m g h, where h is the height above a chosen reference level. Lift an object and you do work against gravity that is stored as potential energy; release it and that stored energy is converted back into motion as the object falls. A second common form is elastic potential energy stored in a stretched or compressed spring: PE = (1/2) k x squared, where k is the spring's stiffness (its "spring constant") and x is the distance it is stretched or compressed from its natural length.

The reference level is your choice

An important subtlety: only changes in gravitational potential energy have physical meaning, so you are free to place the h = 0 reference wherever is convenient - the floor, a tabletop, the ground below a cliff. A book on a shelf has a different potential energy depending on whether you measure height from the desk or the floor, but the change in potential energy as it falls a given distance is the same either way. Choosing a sensible reference level, usually the lowest point in the problem, keeps the numbers simple.

Conservation of mechanical energy

The mechanical energy of a system is the sum of its kinetic and potential energy: E = KE + PE. Here is the powerful result: when only conservative forces like gravity and ideal spring forces act, and no energy is lost to friction, this total remains constant. This is the conservation of mechanical energy, written KE_initial + PE_initial = KE_final + PE_final. Energy is never created or destroyed; it merely transforms back and forth between kinetic and potential forms. A conservative force is one for which the work done depends only on the start and end points, not on the path taken - lift a book straight up or along a winding ramp to the same shelf and gravity does the same work either way. Friction is not conservative, because the energy it removes as heat depends on the path length and cannot be recovered.

Worked example: a falling object

Given: a ball is dropped from rest at height h = 10 m. Take g = 9.8. Find: its speed just before hitting the ground.
Solution: At the top, all the energy is potential: PE = m g h, and KE = 0 since it starts from rest. At the bottom, all of that energy has become kinetic: KE = (1/2) m v squared, and PE = 0 at the reference level. Setting initial energy equal to final energy, m g h = (1/2) m v squared. The mass cancels, giving v squared = 2 g h = 2 x 9.8 x 10 = 196, so v = 14 m/s. The ball lands at 14 m/s - the same answer kinematics gives, but reached with no time calculation at all.

Worked example: a roller coaster

Given: a coaster car starts from rest at the top of a 20 m frictionless hill. Take g = 9.8. Find: its speed at the bottom.
Solution: Again m g h = (1/2) m v squared, so v squared = 2 g h = 2 x 9.8 x 20 = 392 and v = 19.8 m/s. Notice two things: the mass never mattered, and neither did the shape of the track - only the total height dropped. A car descending a straight ramp, a gentle curve, or a wild corkscrew all arrive at the bottom with the same speed, provided friction is negligible. That is the elegance and the power of the energy method: it cuts straight through geometric complexity that would defeat a force-by-force analysis.

Where does the energy go with friction?

In real systems friction and air resistance drain mechanical energy, so a real roller coaster arrives a bit slower than the ideal calculation predicts. But energy is still conserved overall - the "lost" mechanical energy has simply been converted into thermal energy, warming the track, the wheels, and the air. This broader conservation of energy, counting heat and every other form, is one of the deepest and most universal laws in all of physics: the total energy of an isolated system is absolutely constant, no matter what transformations occur within it. Mechanical energy conservation is the special case that holds when no energy leaks into thermal or other forms.

Linear momentum measures how much motion an object carries, combining its mass and velocity into a single vector: p = m v. It points in the direction of the velocity and is measured in kilogram meters per second. Momentum captures something that neither mass nor velocity alone conveys: a slow-moving loaded truck and a fast-flying bullet can carry comparable momentum, because momentum weighs mass and speed together. As you will see in the next lesson, momentum is conserved in interactions even when kinetic energy is not, which makes it in some ways an even more fundamental quantity for analyzing collisions.

Momentum is a vector

Because momentum is a vector, direction matters and signs count. Two objects of equal mass moving toward each other at the same speed have momenta that are equal in size but opposite in sign, so their total momentum is zero even though each is moving. This vector nature is exactly what makes momentum so useful for collisions, where objects approach from different directions and the bookkeeping of pluses and minuses does the work of tracking who ends up going which way.

Impulse

To change an object's momentum you must apply a force over a period of time. The product of the force and the time interval over which it acts is called impulse: J = F t. The impulse-momentum theorem states that the impulse delivered to an object equals its change in momentum: F t = delta p = m v_final - m v_initial. This is really just Newton's second law rearranged, since F = m a = m (delta v / delta t) gives F delta t = m delta v. The theorem is the workhorse for any situation involving forces acting over short times, especially impacts, where the force is large and brief and hard to measure directly but the change in momentum is easy to compute.

Worked example: catching a baseball

Given: a 0.145 kg baseball arrives at 40 m/s and is brought to rest by a catcher in 0.020 s. Find: the average force on the ball.
Solution: The change in momentum is delta p = m v_final - m v_initial = 0.145 x 0 - 0.145 x 40 = -5.8 kg m/s. The average force is F = delta p / t = -5.8 / 0.020 = -290 N. The magnitude is 290 N, and the negative sign shows the force opposes the ball's motion. That is a substantial force - about the weight of a 30 kg mass - delivered in a fraction of a second, which is why catching a fast ball stings.

Why time matters for safety

The impulse-momentum theorem explains an enormous amount of safety engineering, and understanding it changes how you see the world. Rearrange it to F = delta p / t: for a fixed change in momentum - say, bringing a moving body to rest - the force is inversely proportional to the stopping time. Stretch the collision out over a longer time and the peak force drops proportionally. This single insight is the design principle behind airbags, which inflate to extend the time over which your head decelerates; automobile crumple zones, which crush progressively to lengthen the crash; padded dashboards and helmets; the netting under a trapeze artist; and the instinct to bend your knees when you land from a jump. Catching a raw egg by drawing your hands back as it arrives is the very same physics as a car's crumple zone, scaled down. In every case the change in momentum is fixed by the situation, and safety comes entirely from managing the time.

Worked example: comparing stopping times

Given: the same baseball as above, but a skilled catcher lets the glove move back so the ball stops in 0.10 s instead of 0.020 s. Find: the new average force.
Solution: The momentum change is unchanged at 5.8 kg m/s. Now F = 5.8 / 0.10 = 58 N, one-fifth the earlier 290 N. By extending the stopping time fivefold, the catcher cuts the force to a fifth - which is exactly why experienced catchers "give" with the ball rather than holding a rigid glove.

One of the deepest and most reliable principles in physics is the conservation of momentum: in the absence of a net external force, the total momentum of a system of objects stays constant, no matter how the objects interact within it. This follows directly from Newton's third law - when two objects push on each other, the forces are equal and opposite, so the momentum one gains the other loses, and the total is unchanged. For any interaction, then, total momentum before = total momentum after. This one statement handles collisions, explosions, and recoil with equal ease, and because it involves only the states before and after, it sidesteps the complicated and often unknowable forces that act during the interaction itself.

What "isolated" means

Momentum conservation requires an isolated system, meaning no net external force acts on the objects you are tracking. In a collision the forces the objects exert on each other are internal and cancel, but forces from outside - a sustained push, or friction with the ground - are external and can change the total momentum. In practice, collisions happen so quickly that external forces like friction have negligible time to act, so momentum is very nearly conserved during the brief instant of impact even when it is not conserved over longer times. Choosing the right system and the right instants is part of the skill.

Two kinds of collision

Momentum is conserved in every collision, but kinetic energy is a different story. In an elastic collision, kinetic energy is also conserved: the objects bounce apart with no energy lost to heat, sound, or permanent deformation. Collisions between hard objects like billiard balls or steel bearings come close to elastic, and collisions between atoms and subatomic particles can be perfectly elastic. In an inelastic collision, some kinetic energy is converted into other forms - heat, sound, and the energy of bending metal - so the total kinetic energy afterward is less than before, even though momentum is unchanged. The extreme case is a perfectly inelastic collision, in which the objects stick together and move off as a single combined mass; this loses the most kinetic energy consistent with conserving momentum. Real-world crashes are almost always inelastic, which is precisely why they are destructive: the "lost" kinetic energy is what crumples the cars.

Worked example: a perfectly inelastic collision

Given: a 2000 kg car moving at 20 m/s rear-ends a stationary 3000 kg truck, and they lock together. Find: their common velocity just after impact.
Solution: Conserve momentum. Before the collision, the total is p = 2000 x 20 + 3000 x 0 = 40000 kg m/s, all carried by the car. After, the two vehicles move together with combined mass 5000 kg at a common velocity v, so p = 5000 v. Setting before equal to after, 40000 = 5000 v, giving v = 8.0 m/s. The wreck moves off at 8.0 m/s in the car's original direction. As a check on the "inelastic" label, the kinetic energy before is (1/2)(2000)(20 squared) = 400000 J, while after it is (1/2)(5000)(8 squared) = 160000 J - well over half the kinetic energy has been converted to deformation and heat, exactly the energy that damages the vehicles.

Worked example: recoil

Given: a 60 kg skater at rest on frictionless ice throws a 2.0 kg ball horizontally at 10 m/s. Find: the skater's recoil speed.
Solution: The total momentum starts at zero, since nothing is moving, so it must remain zero after the throw. Therefore 0 = m_ball v_ball + m_skater v_skater = 2.0 x 10 + 60 x v_skater. Solving, 60 v_skater = -20, so v_skater = -0.33 m/s. The skater glides backward at about 0.33 m/s, in the direction opposite the ball. This is exactly the physics that propels a rocket (throwing exhaust one way to move the other), recoils a fired gun, and lets a swimmer push off a wall - all cases where a system with zero initial momentum splits into parts moving in opposite directions so the total stays zero.

Why momentum, not just energy

You might wonder why we need momentum at all when energy served so well in Module 5. The reason is that in an inelastic collision, kinetic energy is not conserved - an unknown amount vanishes into heat and deformation - so you cannot use energy conservation to find the final velocities. Momentum, however, is always conserved regardless of how much energy is lost, so it gives you a reliable equation every time. For elastic collisions you can use both conservation laws together, momentum and kinetic energy, which provides enough equations to solve for two unknown final velocities. Knowing which conservation laws apply to which type of collision is the key to choosing your approach.

So far objects have moved along paths, but a huge fraction of the motion around us is rotation: wheels, gears, turbines, planets, hard drives, and spinning tops. Rotational motion has its own set of quantities, and the good news is that they parallel the linear ones you already know so closely that learning rotation is largely a matter of translation. Instead of position we track angular displacement theta, the angle an object has turned through, measured in radians. Instead of velocity we have angular velocity omega, the rate of spin in radians per second. Instead of acceleration we have angular acceleration alpha, the rate at which the spin speeds up or slows down, in radians per second squared.

Why radians

Angles could be measured in degrees, but physics uses radians because they connect rotation to distance so naturally. One radian is the angle that subtends an arc equal in length to the radius, and a full turn is 2 pi radians, about 6.28. The reason radians are the natural choice is the clean relationship they give: the arc length s traced by a point at radius r through an angle theta is simply s = r theta, with no awkward conversion factor. Every linear-angular link in this lesson depends on measuring angles in radians.

Linking linear and angular motion

A point on a rotating object at radius r from the axis moves along a circle, and its linear motion connects to the object's rotation through two key relations. Its linear (tangential) speed is v = r omega, and its linear tangential acceleration is a = r alpha. The farther a point sits from the axis, the faster it moves for the same rotation - which is why the rim of a spinning wheel moves faster than a point near its hub, why the outer horses on a carousel travel faster than the inner ones, and why a longer propeller blade has faster-moving tips. All points share the same angular velocity omega, but their linear speeds differ in proportion to their distance from the axis.

The rotational kinematic equations

Because the angular definitions parallel the linear ones exactly, the constant-angular-acceleration equations have precisely the same form as the linear kinematic equations, with each linear symbol replaced by its angular partner. Anything you learned to do in Module 2 you can now do for rotation by simple substitution.

Linear	Rotational
v = v0 + a t	omega = omega0 + alpha t
x = v0 t + (1/2) a t squared	theta = omega0 t + (1/2) alpha t squared
v squared = v0 squared + 2 a x	omega squared = omega0 squared + 2 alpha theta

Worked example: a spinning wheel

Given: a wheel starts from rest (omega0 = 0) and reaches omega = 12 rad/s in t = 4.0 s. Find: its angular acceleration and the number of revolutions it makes.
Solution: Using omega = omega0 + alpha t, solve alpha = (omega - omega0)/t = (12 - 0)/4.0 = 3.0 rad/s squared. The wheel's spin increases at 3.0 rad/s squared. For the angle turned, use theta = omega0 t + (1/2) alpha t squared = 0 + (1/2)(3.0)(4.0 squared) = 24 radians. Dividing by 2 pi gives 24 / 6.28 = 3.8 revolutions. Every equation from linear kinematics carries straight over, so once you master linear motion, rotation costs you almost nothing extra.

Worked example: a point on the rim

Given: the same wheel has radius 0.30 m. Find: the linear speed of a point on its rim at the moment omega = 12 rad/s.
Solution: v = r omega = 0.30 x 12 = 3.6 m/s. A point on the rim is moving at 3.6 m/s, while a point halfway to the center, at r = 0.15 m, moves at only 0.15 x 12 = 1.8 m/s - half as fast, because it is half as far from the axis.

Force causes linear acceleration; its rotational counterpart, torque, causes angular acceleration. Torque measures the turning effect of a force, and the crucial insight is that it depends not just on how hard you push but on where and in what direction. Its magnitude is tau = r F sin(theta), where r is the distance from the axis to the point where the force is applied, F is the force, and theta is the angle between the direction of r and the direction of the force. Torque is measured in newton-meters. The combination r sin(theta) is called the lever arm or moment arm - it is the perpendicular distance from the axis to the line along which the force acts.

Why placement matters

The dependence on r explains a great deal of everyday mechanics. A longer wrench loosens a stubborn bolt more easily because a larger r multiplies the same force into a larger torque. A door handle is placed far from the hinges, not near them, for the same reason: pushing far from the axis produces more turning effect. The angle matters too. Pushing perpendicular to the door (theta = 90 degrees, sin 90 = 1) gives the maximum torque, while pushing straight toward the hinge (theta = 0, sin 0 = 0) produces no torque at all, no matter how hard you shove - the door simply will not swing. This is why you instinctively push a door near its outer edge and at right angles to its face.

Worked example: a wrench

Given: you push with F = 20 N perpendicular to a wrench, r = 0.50 m from the bolt (theta = 90 degrees). Find: the torque.
Solution: tau = r F sin(90) = 0.50 x 20 x 1 = 10 N m. You apply 10 N m of torque. If you pushed at only 30 degrees to the wrench instead, the torque would drop to 0.50 x 20 x sin 30 = 0.50 x 20 x 0.5 = 5 N m, half as effective. And if you gripped a wrench twice as long, r = 1.0 m, the same 20 N perpendicular push would give 1.0 x 20 x 1 = 20 N m, twice the torque - the mechanical advantage of a long handle.

Rotational equilibrium

An object is in rotational equilibrium when the total torque on it is zero, so it does not start spinning or change a spin it already has. Torques that tend to cause clockwise rotation must exactly balance those that tend to cause counterclockwise rotation. This is the principle of the lever and the seesaw: a lighter person can balance a heavier one by sitting farther from the pivot, trading distance for weight so that the two torques match. Full mechanical equilibrium actually requires two conditions - the net force must be zero (so the object does not accelerate linearly) and the net torque must be zero (so it does not angularly accelerate) - and engineers apply both when designing any structure that must stay put.

Worked example: balancing a seesaw

Given: a child weighing 300 N sits 2.0 m to the left of the pivot. On the right, an adult weighs 600 N. Find: where the adult must sit to balance.
Solution: For balance, the counterclockwise torque from the child must equal the clockwise torque from the adult: 300 x 2.0 = 600 x d. So 600 = 600 d, giving d = 1.0 m. The adult must sit 1.0 m from the pivot, exactly half the child's distance because the adult weighs twice as much - the extra weight is offset by the shorter lever arm. This torque balance is the foundation of levers, balance scales, cranes, and countless machines that trade force for distance.

Levers as force multipliers

The same balance principle lets a small force overcome a large one. On a lever, if you push down on the long end, a short lever arm on the other side can lift a heavy load, because equal torques on both sides mean the side with the longer arm needs less force. A crowbar, a bottle opener, a wheelbarrow, and a pair of pliers all exploit this: the ratio of the two lever arms sets the mechanical advantage. Archimedes captured the idea when he reportedly said that given a long enough lever and a place to stand, he could move the Earth. Torque is the physics that makes that boast, in principle, true.