Forces, Rotation & Equilibrium — Print · Physics & Environmental Science, Grade 12

Newton's Three Laws of Motion

Learning goals

• I can state Newton's three laws of motion and explain inertia as an object's resistance to changes in motion.
• I can apply Newton's second law (F = ma) to solve for net force, mass, or acceleration.
• I can identify action-reaction force pairs and explain why the two forces act on different objects.

Readings for this unit

• The force on the box is not the force on the floor

Newton’s laws describe a powerful model for ordinary macroscopic motion in inertial frames. A force is an interaction that contributes to acceleration through the net-force account. Their range is not literally every physical situation: relativistic speeds, quantum-scale evidence and rotating frames need additional models. Within their useful range, identify the object, interactions, mass and coordinate directions to predict how its velocity changes.

Newton's first law, the law of inertia, states that an object keeps doing whatever it is already doing — sitting still or gliding at constant velocity in a straight line — unless a nonzero net force acts on it. Inertia is simply an object's built-in resistance to any change in its motion, and its mass is the direct measure of that resistance. This law overturned a 2,000-year-old belief, going back to Aristotle, that motion needs a continuous push to keep going. The truth is the reverse: a rolling ball slows only because friction and air resistance push back on it. Remove those forces, as in deep space, and the ball glides on forever with no engine at all.

Newton's second law turns that idea into an equation: ΣF = ma, the net force on an object equals its mass times its acceleration. Read carefully, it says two intuitive things at once. First, for a given mass, a larger net force produces a larger acceleration — push harder and it speeds up faster. Second, for a given net force, a larger mass produces a smaller acceleration — that is inertia again. The Greek letter Σ (sigma) means 'sum of,' a crucial reminder that acceleration depends on the *net* force, the single vector left after every push and pull is combined. Force is measured in newtons (N), where one newton accelerates one kilogram at one meter per second squared.

Because it is a vector equation, the second law also tells you the *direction* of the acceleration: an object always accelerates in the same direction as the net force acting on it. Rearranged, the same relationship lets you solve for whatever is unknown — acceleration a = ΣF ÷ m, or mass m = ΣF ÷ a. This one compact formula is the workhorse of the entire unit; nearly every problem you meet, from a sliding crate to an orbiting satellite, comes down to finding the net force and dividing by the mass.

Newton’s third law describes equal-magnitude opposite-direction interaction forces acting on two different bodies. They do not cancel in a free-body diagram of one body. A swimmer pushes water backward and receives a forward force from it. In walking, the foot can push the ground backward while static friction from the ground acts forward on the foot; vertical support is a separate component. Identify the receiving body for every arrow before summing forces.

This third law explains how a rocket accelerates through the vacuum of space, where there is nothing outside to push against. The rocket hurls hot exhaust gas downward, and by the third law that gas pushes the rocket upward with equal force — the action-reaction pair is entirely between the rocket and its own exhaust. From walking to swimming to spaceflight, the three laws work as a team: the first tells you an object's motion only changes when forces are unbalanced, the second tells you exactly how much it changes, and the third reminds you that forces always come in pairs. Learn to read these three ideas together and you can predict the motion of almost anything.

Model check: Action/reaction forces act on different objects, so they do not cancel in one object’s diagram. a=Fnet/m=2 m/s² along the net force.

Newton's Second Law

ΣF = ma

ΣF is the net force in newtons (N); m is mass in kg; a is acceleration in m/s². Rearranges to a = ΣF ÷ m or m = ΣF ÷ a.

The Newton

1 N = 1 kg·m/s²

one newton is the force that accelerates a 1 kg mass at 1 m/s².

Worked examples

Avoid these mistakes

• Action/reaction forces act on different objects, so they do not cancel in one object’s diagram.

Words to know

inertia

The tendency of an object to resist any change in its state of motion; measured by its mass.

Newton's first law

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a nonzero net force.

Newton's second law

The net force on an object equals its mass times its acceleration: ΣF = ma.

Newton's third law

For every force one object exerts on a second, the second exerts an equal and opposite force back on the first.

newton (N)

The SI unit of force; 1 N is the force that accelerates 1 kg at 1 m/s², so 1 N = 1 kg·m/s².

From Aristotle to the Principia: a 2,000-year correction

For nearly two millennia, the ancient Greek philosopher Aristotle taught that objects have a 'natural' state of rest and that a continuous force is required to keep anything moving — an idea that matches everyday experience, since carts stop when you stop pushing. Galileo Galilei challenged this in the early 1600 s by imagining perfectly smooth surfaces, reasoning that a ball would roll forever if friction could be removed. In 1687 Isaac Newton published his Principia Mathematica, one of the most influential books in the history of science, which gathered these insights into three crisp laws. The first law's radical claim — that constant velocity, not rest, is the natural state when no net force acts — overturned common sense and launched modern physics.

Does motion require a force? The stubborn misconception

The single most common error in introductory physics is believing that a moving object must have a force pushing it in its direction of travel. It feels obvious: stop pedaling and your bike coasts to a halt. But the bike slows because of friction and air resistance pushing backward, not because it 'ran out of force.' A hockey puck on near-frictionless ice glides at nearly constant speed with no forward force at all. Whenever you see something moving steadily in a straight line, the correct conclusion is that the net force on it is zero — the forces are balanced, exactly as they are for an object sitting still.

Momentum and impulse: the deeper form of the second law

Newton originally framed his second law in terms of momentum, p = mv — a kind of 'quantity of motion.' In this form, force is the rate at which momentum changes: F = Δp/Δt, which rearranges to the impulse relationship F·Δt = Δp. This explains why airbags and crumple zones save lives: they stretch out the time Δt over which a crash brings you to a stop, so the same change in momentum happens with a much smaller force on your body. The familiar ΣF = ma is just this deeper law written for the common case where mass stays constant, which is why momentum ideas underlie collisions, rockets, and recoil.

Rockets in a vacuum: the third law at full throttle

A persistent myth says rockets work by 'pushing against the air,' which would make spaceflight impossible. In truth, a rocket obeys Newton's third law: it throws exhaust gas out the back at high speed, and the gas pushes the rocket forward with an equal and opposite force. The Saturn V that carried astronauts to the Moon weighed roughly 2.9 million kilograms fully fueled and produced about 34 million newtons of thrust at liftoff — enough to overcome its enormous weight. Because the action–reaction pair is entirely between the rocket and its own expelled gas, a rocket accelerates just as well in the vacuum of space as in the atmosphere.

In the real world

Every rocket launch is Newton's third law at full power: engines fling exhaust gas downward, and the gas pushes the rocket up — which is why rockets work even in the vacuum of space.

Exit check

A3 kg cart has 6N net force. Find acceleration.

Free-Body Diagrams & Net Force

Learning goals

• I can draw a free-body diagram that shows every force acting on a single object as a labeled vector.
• I can compute the net force from several forces, including forces at right angles, and use ΣF = ma to find the resulting acceleration.
• I can recognize when an object is in equilibrium because its forces sum to zero.

Readings for this unit

• The force on the box is not the force on the floor

Real objects rarely feel just one force. A crate being dragged across a warehouse floor feels its own weight pulling down, the floor pushing up, a rope pulling forward, and friction dragging back — four forces at once. To predict how it moves, you cannot juggle all of that in your head; you need a system. That system is the free-body diagram, the single most-used tool in all of mechanics, and it is the bridge between a messy real scene and the clean equation ΣF = ma.

To draw one, you mentally cut the object free from its surroundings and represent it as a simple dot or box. Then you draw an arrow for every force *acting on that object* — and only those. Each arrow starts at the object, points the way the force pushes or pulls, and is drawn longer for a stronger force. The usual cast of characters is small and worth memorizing: weight (gravity, always straight down), the normal force (a surface pushing perpendicular to itself), tension (a rope or cable pulling along its length), an applied force (a direct push or pull), and friction (always opposing sliding). Naming the forces correctly is more than half the battle.

Once every force is on the diagram, you find the net force by adding the arrows as vectors — quantities with both size and direction. Forces along the same line are easy: add them if they point the same way, subtract them if they oppose. The net force is what actually goes into Newton's second law, which in its full form reads ΣF = ma. A common and costly mistake is to plug a single force into that equation; only the *sum* of all forces determines the acceleration, and the object always accelerates in the direction of that net force.

Because forces are vectors, it usually pays to split the problem into directions and handle each independently. Physicists resolve forces into perpendicular components, typically horizontal (x) and vertical (y), then apply Newton's second law separately to each: ΣFₓ = maₓ and ΣF_y = ma_y. For a crate sliding along level ground, the vertical forces (weight down, normal up) cancel, so all the acceleration comes from the horizontal forces. Breaking a two-dimensional problem into two one-dimensional problems is the standard professional move, and it keeps the arithmetic honest.

When two forces act at a right angle instead of along one line, you combine them with the Pythagorean theorem, exactly as in geometry. A 9 N pull east and a 12 N pull north do not add to 21 N; the true net force is √(9² + 12²) = √225 = 15 N, aimed diagonally between them. This is a beautiful moment where math and physics turn out to be the same tool: the resultant of two perpendicular forces is literally the hypotenuse of a right triangle whose legs are those forces.

A special and important case is equilibrium, when all the forces sum to exactly zero. A book resting on a table, a lamp hanging motionless from a cord, and an airplane cruising at a steady altitude and speed are all in equilibrium — not because no forces act, but because the forces perfectly cancel. By the first law, zero net force means the velocity never changes, so an object in equilibrium is either sitting still or moving at constant velocity. Reading a free-body diagram tells you instantly which case you are in: if the arrows balance, motion holds steady; if they don't, the leftover net force sets the acceleration through ΣF = ma.

Model check: Draw only forces exerted on the selected object; motion arrows are not forces. Net 12N right; a=12/4=3 m/s² right.

Net Force (same line)

ΣF = F₁ + F₂ + …

add forces pointing the same way; subtract forces pointing opposite ways, to get the net force along one direction.

Resultant of Perpendicular Forces

F net = √(Fₓ² + F_y²)

Fₓ and F_y are two forces (or force components) at right angles; their vector sum has this magnitude.

Worked examples

Avoid these mistakes

• Draw only forces exerted on the selected object; motion arrows are not forces.

Words to know

free-body diagram

A simplified sketch of a single object with an arrow drawn for every force acting on it, pointed in the correct direction and sized to show relative strength.

net force (ΣF)

The single overall force found by adding every force on an object as a vector, accounting for direction.

equilibrium

The condition in which the net force on an object is zero, so its velocity does not change.

normal force

The support force a surface pushes back with, always directed perpendicular to the surface.

vector

A quantity that has both magnitude and direction, such as force, velocity, or acceleration.

The free-body diagram: mechanics' most-used tool

Ask any physicist or engineer for the first step in a force problem and they will say the same thing: draw a free-body diagram. The technique is powerful precisely because it forces you to make three decisions explicit — which object you are analyzing, which forces act on it (and which do not), and which direction each one points. A frequent beginner mistake is to include forces the object exerts on other things; a free-body diagram shows only the forces acting on the chosen object itself. Done carefully, it converts a confusing real-world scene into a clean set of vectors ready for ΣF = ma.

Resolving forces into components

Because forces are vectors, the professional approach to any two-dimensional problem is to choose perpendicular axes (often x and y, sometimes tilted to match a ramp) and break every force into components along them. Then Newton's second law is applied separately in each direction: ΣFₓ = maₓ and ΣF_y = ma_y. A force F at angle θ splits into F cosθ and F sinθ, the same trigonometry used throughout geometry. This is why a rope pulled at an angle is less effective at dragging a sled than a horizontal pull — part of its force goes into lifting rather than pulling, and only the horizontal component contributes to forward motion.

Constant velocity is not the same as 'no forces'

It is tempting to think that if an object's engine is running, it must be accelerating — but an airplane cruising at a steady altitude and speed is in perfect equilibrium. Lift balances weight, and the engines' thrust balances air drag; the net force is zero even though four large forces are acting. The engines run constantly not to speed the plane up, but simply to cancel the drag trying to slow it down. The same logic explains terminal velocity: a skydiver stops accelerating once air resistance grows to equal their weight, after which they fall at a constant speed with zero net force.

Static equilibrium: why bridges and buildings stand

Structural engineering is, at heart, the art of keeping enormous objects in static equilibrium — every net force (and every net rotational effect, called torque) balanced to zero. Engineers draw free-body diagrams of individual joints, beams, and cables to verify that the forces of traffic, wind, snow, and a structure's own weight are exactly opposed by the supports and tension members holding it up. When a designer gets a single force balance wrong, the consequence is not a lost point on a test but a structure that can sag, crack, or collapse — which is why the humble free-body diagram is a genuine matter of public safety.

In the real world

Structural engineers draw free-body diagrams of every joint in a bridge to confirm the forces of traffic, wind, and the bridge's own weight all cancel — an unbalanced joint is a bridge that could fail.

Exit check

A4 kg block has 20N right and 8N left. Find acceleration.

Friction & Common Forces

Learning goals

• I can distinguish static from kinetic friction and calculate a friction force using f = μN.
• I can identify the normal force, tension, and weight acting on an object.
• I can resolve weight into components on an inclined plane and predict the resulting motion.

Readings for this unit

• The force on the box is not the force on the floor

Friction at a dry contact resists relative slipping or its tendency. Microscopic contact interactions involve adhesion, deformation, and surface structure; roughness alone is not a universal predictor. In a simple model, sliding friction has magnitude approximately μkN, while static friction adjusts up to a limit μsN. Both models depend on specified surfaces and conditions. They should not be treated as one exact force formula for every contact.

Static friction adjusts to meet a no-slip requirement up to its modeled maximum μsN. Kinetic friction applies while surfaces slide and is often modeled as μkN. In many introductory examples μk is smaller than μs, but that relationship and constant coefficients are not universal for all materials and conditions. Decide whether contact is resting, impending slip, or sliding before selecting a force model.

To use f = μN you must first find the normal force, and this is where students most often stumble. The normal force is the perpendicular push a surface gives back, and it is *not* automatically equal to weight. On flat, level ground with nothing pressing down but gravity, it happens that N = mg — but tilt the surface, add a downward push, or pull upward on the object, and the normal force changes accordingly. Because friction depends on N, anything that changes the normal force also changes the friction. Getting the normal force right is the whole game.

Near Earth in a suitable uniform-field approximation, weight is gravitational force mg downward, with a commonly used g around 9.8 m/s² unless another value is supplied. Its value varies with location and is not a universal exact constant. Tension in an ideal taut rope pulls along the connector. A massive rope, nonideal pulley, or changing contact can require additional forces and constraints beyond the simplest weight-normal-tension-friction model.

Inclined planes are where all of this comes together, and they reveal a slick trick. When an object sits on a ramp tilted at angle θ, its weight still points straight down, but it is convenient to split that weight into two pieces: a component *along* the slope, mg sinθ, which tries to slide the object downhill, and a component *pressing into* the slope, mg cosθ, which the surface must support. The normal force therefore equals only mg cosθ, not the full weight — steeper ramps press less directly into the surface. This is why, on a frictionless ramp, an object accelerates downhill at exactly a = g sinθ, independent of its mass: the heavier object feels a bigger pull but has proportionally more inertia, and the two effects cancel.

For many dry contacts, a useful first approximation takes kinetic friction proportional to normal force and relatively insensitive to apparent contact area over a suitable range. Real contact mechanics, pressure, deformation, surface treatment, and speed can change that behavior. An area-independent model should be tested under stated conditions rather than justified by an exact cancellation of macroscopic area and pressure. Record variation and limitations when comparing different object orientations.

Model check: Static friction adjusts up toμsN; it is not always equal to that maximum. Static friction is 2N opposite the push, so net force is zero. It need not reach 5N.

Friction Force

f = μN

N is the normal force in newtons; μ is the coefficient of friction (μₛ for static, μₖ for kinetic, with μₛ > μₖ); f is the friction force.

Weight

W = mg

m is mass in kg; g = 9.8 m/s² near Earth's surface; W (weight) comes out in newtons.

Incline Components

F∥ = mg sinθ, N = mg cosθ

θ is the incline angle; F∥ is the weight component down the slope; N is the normal force. On a frictionless ramp, a = g sinθ.

Worked examples

Avoid these mistakes

• Static friction adjusts up toμsN; it is not always equal to that maximum.

Words to know

static friction

The friction force between two surfaces that are not yet sliding; it adjusts up to a maximum to prevent motion.

kinetic friction

The friction force between two surfaces that are already sliding past each other; f = μₖN.

coefficient of friction (μ)

A number describing how rough a pair of surfaces is; larger μ means more friction. Static μₛ is usually greater than kinetic μₖ.

normal force (N)

The support force a surface exerts perpendicular to itself; on level ground it equals the object's weight, but on an incline it is only mg cosθ.

tension

The pulling force carried along a rope, string, or cable, directed along its length.

What friction really is, at the atomic scale

Under a microscope, even a mirror-polished surface is a landscape of jagged peaks and valleys. When two surfaces touch, they actually meet only at a few tiny high points, where atoms come close enough to form weak bonds — a process sometimes called cold welding. Sliding requires continually breaking and reforming these microscopic contacts, and that resistance is what we experience as friction. This picture explains why smoother, cleaner, or lubricated surfaces slide more easily: a lubricant like oil keeps the peaks apart so they never bond. It also explains why friction generates heat, as the constant making and breaking of contacts dumps energy into the material.

Why the first push is the hardest — and why ABS brakes work

Static friction (between surfaces not yet sliding) can grow larger than kinetic friction (between sliding surfaces), because the microscopic contacts have time to settle and bond more firmly before motion begins. That is why a heavy box resists your first shove, then suddenly 'gives' and slides more easily once moving. The same physics is behind anti-lock braking systems (ABS): a rolling tire grips the road with strong static friction, but a locked, skidding tire slides under weaker kinetic friction. By rapidly pulsing the brakes to keep the tires rolling rather than skidding, ABS shortens stopping distances and lets the driver steer while braking.

The surprising rule: friction ignores contact area

Most people expect that a wider tire or a book lying on its large face should feel more friction — but to a good approximation, friction does not depend on the contact area at all. The friction force is f = μN, which contains the normal force and the coefficient of friction, but not area. The reason is a neat cancellation: spreading the same weight over a larger area lowers the pressure at each point by exactly the factor that the extra area adds back. What actually matters is the pair of materials (through μ) and how hard they are pressed together (through N). Wide tires improve grip mostly through heat, wear, and material effects, not raw contact area.

Banked curves and the inclined plane in the wild

The inclined plane is more than a textbook prop — it is one of the classic simple machines and it appears everywhere. Highway on-ramps, wheelchair ramps, and switchback mountain roads all trade a shorter, steeper climb for a longer, gentler one, reducing the force needed at the cost of more distance. Banked curves on racetracks and highways tilt the road so that part of the normal force points inward, helping supply the centripetal force a turning car needs and reducing its reliance on tire friction. Analyzing these real structures uses exactly the weight-component method — mg sinθ along the slope and mg cosθ into it — from this lesson.

In the real world

Anti-lock brakes (ABS) keep tires rolling instead of skidding, which keeps them in the stronger static-friction regime — letting a car stop faster and steer while braking.

Exit check

An object rests on a horizontal floor with a 2N push and max static friction 5N. Find friction.

Circular Motion & Gravitation

Learning goals

• I can calculate centripetal acceleration (a = v²/r) and centripetal force (F = mv²/r) for uniform circular motion.
• I can apply Newton's law of universal gravitation (F = Gm₁m₂/r²) to find the gravitational force between two masses.
• I can explain how gravity supplies the centripetal force that holds a satellite in orbit.

Readings for this unit

• The force on the box is not the force on the floor

An object can move at a perfectly constant speed and still be accelerating — a fact that feels wrong until you remember that acceleration means any change in velocity, and velocity includes direction. A car rounding a curve at a steady 50 km/h is constantly changing direction, so it is constantly accelerating. This is uniform circular motion: constant speed, continuously turning. The acceleration always points toward the center of the circle and is called centripetal acceleration, given by the clean formula a = v²/r, where v is the speed and r is the radius. Notice that tighter turns (smaller r) and higher speeds (larger v) both demand more acceleration — which is exactly why a sharp turn at high speed feels so violent.

By Newton's second law, an acceleration requires a net force in the same direction, so circular motion demands a net inward force called the centripetal force, F = ma = mv²/r. Here is the single most important idea in this topic: *centripetal is not a new kind of force.* The word simply means 'center-seeking' and describes the *direction* of whatever real force is doing the job. For a ball on a string it is tension; for a car on a curve it is friction between tires and road; for the Moon circling Earth it is gravity. Whenever you see circular motion, ask: which real force is pointing toward the center and supplying the mv²/r that the motion requires?

This clears up the most stubborn misconception in the whole unit — the so-called centrifugal 'force.' When a car turns left, you feel flung to the right and swear something is pushing you outward. Nothing is. Your body, obeying the first law, simply wants to continue straight ahead while the car turns underneath you; the door then pushes you inward to make you turn with it. The outward 'force' is an illusion of riding in an accelerating frame. The only real force in play points *inward*, toward the center.

Newton's greatest leap was realizing that the same force pulling an apple to the ground also holds the Moon in its orbit — and that gravity is universal, acting between *every* pair of masses in the cosmos. He captured it in Newton's law of universal gravitation: F = Gm₁m₂/r². The force grows with each mass and shrinks with the square of the distance r between their centers, and G is the gravitational constant, a very small fixed number, 6.67 × 10⁻¹¹ N·m²/kg². Because G is so tiny, the gravitational pull between everyday objects is utterly negligible; it takes a body as massive as a planet for gravity to become the dominant force we feel every second of our lives.

The r² in the denominator makes gravity an inverse-square law, and its consequences are dramatic. Double the distance between two objects and the force drops not to half but to one-quarter; triple the distance and it falls to one-ninth. This same equation, evaluated at Earth's surface, reproduces the familiar weight W = mg — plug in Earth's mass and radius and out comes an acceleration of about 9.8 m/s² for every object, regardless of its own mass. Universal gravitation and everyday weight are not two separate ideas; they are the same law seen at two different scales.

Orbits are the grand payoff, and Newton explained them with a famous thought experiment: imagine a cannon on a mountaintop firing a ball ever faster. Fired gently, the ball arcs down and lands; fired fast enough, the ground curves away beneath it just as quickly as it falls, and the ball 'falls around' the planet forever. That is an orbit — continuous free-fall in which gravity supplies exactly the centripetal force the circular path requires, so Gm₁m₂/r² = mv²/r. This is also why astronauts on the space station float: they are not beyond gravity's reach (Earth's gravity is still nearly as strong up there) but are in constant free-fall together with their station, endlessly falling around the Earth.

Model check: Centripetal force is the net inward force, not an extra interaction added to gravity or tension. ac=v²/r=8 m/s². Speed can be constant while velocity direction changes.

Centripetal Acceleration

a = v² ÷ r

v is speed in m/s; r is the radius of the circular path in meters; a points toward the center.

Centripetal Force

F = mv² ÷ r

m is mass in kg; this is the net inward force required for circular motion, supplied by a real force like tension, friction, or gravity.

Universal Gravitation

F = G m₁m₂ ÷ r²

m₁ and m₂ are the two masses in kg; r is the distance between their centers in meters; G = 6.67 × 10⁻¹¹ N·m²/kg².

Worked examples

Avoid these mistakes

• Centripetal force is the net inward force, not an extra interaction added to gravity or tension.

Words to know

centripetal acceleration

The acceleration of an object moving in a circle, always directed toward the center; a = v²/r.

centripetal force

The net inward force required to keep an object moving in a circle; F = mv²/r. It is supplied by a real force such as tension, friction, or gravity.

period (T)

The time it takes an object to complete one full revolution around a circle.

Newton's law of universal gravitation

Every pair of masses attracts with a force F = Gm₁m₂/r², where r is the distance between their centers.

gravitational constant (G)

The fixed number in the gravitation equation, G = 6.67 × 10⁻¹¹ N·m²/kg², measured first by Henry Cavendish.

Newton's apple and the unification of the heavens

The famous story of an apple falling in Newton's garden captures a genuine breakthrough: Newton realized that the force pulling the apple down might be the same force holding the Moon in its orbit. This was a stunning claim, because for centuries the heavens and the Earth were thought to obey entirely different rules. His law of universal gravitation, F = Gm₁m₂/r², proposed that every pair of masses in the universe attracts, from apples to moons to distant galaxies. Using it, Newton could explain Kepler's laws of planetary motion, the tides, and the paths of comets with a single equation — one of the greatest unifications in the history of thought.

Cavendish 'weighs the Earth': measuring G

Newton's equation contained the constant G, but he never knew its value. In 1798, more than a century later, the British scientist Henry Cavendish measured it using an exquisitely sensitive torsion balance: two small lead balls on a suspended rod, twisted ever so slightly by the gravitational pull of larger lead spheres brought nearby. From the tiny twist he extracted G ≈ 6.67 × 10⁻¹¹ N·m²/kg². Once G was known, scientists could rearrange the surface-gravity relationship g = GM/r² to calculate Earth's mass for the first time — which is why Cavendish's delicate experiment earned the nickname 'weighing the Earth.'

There is no centrifugal force

When a car turns sharply, you feel thrown against the door and might swear an outward force is pushing you — the so-called centrifugal force. Physically, no such force exists. Your body, obeying Newton's first law, simply tries to continue in a straight line while the car curves underneath you; the door then pushes you inward to make you turn with the car. The outward sensation is an artifact of observing motion from within an accelerating (rotating) frame of reference. In an inertial frame, the only real horizontal force in a turn is the centripetal one, pointing toward the center — supplied here by friction and the car's door.

Orbits, GPS, and Newton's cannonball

Newton explained orbits with a thought experiment: a cannon fired from a tall mountain shoots a ball that arcs to the ground, but fired fast enough, the ground curves away beneath it as quickly as it falls, so the ball circles the planet — an orbit is simply perpetual free-fall. Real satellites embody this: the International Space Station circles about 400 km up at roughly 7.7 km/s, completing an orbit every 90 minutes, while GPS satellites sit much higher and time their signals so precisely that engineers must even correct for effects of relativity. In every case, gravity supplies exactly the centripetal force the circular path demands, setting Gm₁m₂/r² equal to mv²/r.

In the real world

GPS satellites stay in orbit because Earth's gravity supplies precisely the centripetal force their circular paths require — the same physics that keeps the Moon circling Earth.

Exit check

v=4 m/s,r=2 m. Find inward acceleration.

Springs, Torque & Static Equilibrium

Learning goals

• I can calculate spring force and energy within an elastic model.
• I can apply both force and torque balance about a stated pivot.

Readings for this unit

• The force on the box is not the force on the floor

Hooke’s law models a spring within its linear elastic range: restoring force Fs=−kx, with x measured from the natural equilibrium length and k in N/m. The force magnitude is k times the absolute extension. The minus sign expresses opposition to displacement, not a negative stiffness. Outside this tested range the relation can fail or permanent deformation can occur.

Elastic potential energy is Us=½kx². Work needed to stretch a spring gradually from zero to x is the area under the force-displacement graph. Because force grows from zero to kx, the triangular area is ½kx², not simply final force times total stretch.

A force can also cause rotation. Torque magnitude about a chosen pivot is τ=rF sinθ, with θ between the radius-to-application-point and the force. The perpendicular lever arm version τ=F dperp is equivalent. A force through the pivot produces no torque about it.

A rigid object in static equilibrium needs both zero net force and zero net torque. Balancing upward and downward forces alone does not guarantee it will not rotate. Choose a pivot that removes an unknown support force from the torque equation, then solve remaining force balances.

Ideal levers trade force for motion without creating energy. A longer input lever arm can allow a smaller force to balance a larger load; the input moves farther. Actual mechanisms have friction, flexing, and mass, so evaluate efficiency and stresses before applying an ideal ratio.

A household ruler balance or a gentle small spring is suitable with adult setup and soft light loads. Do not use heavy suspended objects or overstretch an elastic band toward a face. This unit’s equations describe a bounded model, not a structural safety rating.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Use the elastic range for Hooke’s law

Extension x is the change from a spring’s natural length. In its linear elastic range, the restoring force is −kx: its sign opposes displacement from equilibrium. A quasistatic applied force balancing it has magnitude kx, producing the plotted positive slope. Past the elastic range a spring may not return to its original length or obey the same linear law. Distinguish the force exerted by the spring from the external force required to hold it stretched.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What balancing force is needed at x=0.03 m for this model?

Visual reasoning — after your attempt

F=kx=200×0.03=6 N; the spring exerts 6 N in the opposite direction.

An ideal linear spring exerts restoring force F=−kx along a chosen displacement axis. The minus sign indicates opposition to displacement from equilibrium. Real springs follow this approximation only over a suitable range; k is a stiffness, not a force by itself.

A 200 N/m spring stretched 0.050 m has a restoring-force magnitude 10 N in the linear model.

Pause and explain

Why is k not measured in newtons alone?

Check your reasoning after answering

Since force is proportional to extension, k has units N/m so multiplying by displacement gives force.

Part 2 · Use the perpendicular moment arm

Torque measures a force’s turning effect about a chosen pivot. Its magnitude is rFsinθ=F r⊥, where r⊥ is the shortest distance from pivot to the force’s line of action. A large force aimed directly through the pivot can have zero torque even with a long position vector. Assign a turning sign convention, such as counterclockwise positive, and retain N·m units; torque and energy share dimensions but describe different quantities.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What torque results if this 20 N force acts along the bar through the pivot?

Visual reasoning — after your attempt

Zero, because its perpendicular moment arm is zero.

Torque about a specified pivot has magnitude rF sinθ, or force times the perpendicular distance from pivot to the force’s line of action. A force through the pivot gives zero torque even if large. Specify clockwise and counterclockwise signs before summing.

For a perpendicular 10 N force at 0.40 m, torque magnitude is 4.0 N·m. An angled force at that radius has a smaller perpendicular contribution.

Pause and explain

Why can a long lever arm still yield zero torque?

Check your reasoning after answering

If the force’s line of action passes through the pivot, perpendicular distance is zero; the physical handle length alone does not determine torque.

Part 3 · Require both force and torque balance

Static equilibrium means the body remains at rest without translational or rotational acceleration. It needs ΣF=0 and Στ=0 about a selected pivot in the rigid-body model. Equal opposing torques alone can coexist with nonzero net force, while balanced forces can still form a turning couple. For this weightless-bar example equal downward loads at equal opposite arms balance torque, and the upward support must balance their combined 60 N force.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Could a zero-torque calculation alone establish static equilibrium?

Visual reasoning — after your attempt

No. The vector force sum must also vanish, and the starting state must be at rest.

Static equilibrium needs zero net external force and zero net torque. Torque balance alone does not guarantee no translation. Choose a convenient pivot to eliminate an unknown support torque, then use the force balance to determine that support.

A beam may have balanced moments but still an unbalanced total downward force unless the support force is included.

Pause and explain

Can force balance alone establish static equilibrium?

Check your reasoning after answering

No. Equal opposite forces at different lines of action can produce a nonzero turning effect; torque must also balance.

Part 4 · Account for distributed weight and assumptions

A center of mass summarizes a body’s mass distribution. In a uniform gravitational field, its total weight can be represented at that point for external force and torque calculations. A uniform bar’s center lies at its midpoint, but an attached load or nonuniform density changes it. The small arrows show distributed weight and the larger arrow is an equivalent alternative representation, not an additional force to add on top of them.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Should the distributed arrows and equivalent mg both enter the same torque sum?

Visual reasoning — after your attempt

No. They are two representations of the same weight; counting both would double it.

For a uniform beam in a uniform gravitational field, weight acts effectively at its midpoint. A nonuniform beam requires its center of mass. Draw each load and support; indicate whether beam weight, friction, or connector mass is neglected.

A light beam assumption can simplify torque, but a real heavy beam can substantially change the support and spring requirement.

Pause and explain

What is needed if beam mass is not negligible?

Check your reasoning after answering

Its weight and center of mass must enter the force and torque account, along with the existing loads and supports.

Apply the whole unit · Balance a load with an ideal spring

Illustrative massless horizontal beam pivots at its left end. A 10.0 N downward load acts 0.40 m from the pivot; an upward linear spring acts 0.20 m from it, k=200 N/m.

Your task

Find spring force and extension, then determine the signed vertical pivot force. Explain why torque balance does not answer the pivot-force question alone.

What your response should show

• Define the system, quantities, units, and relevant conditions before selecting a model.
• Show intermediate calculations or a labeled evidence account, including signs or denominators.
• Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
• Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.

Response guidance — after your attempt

1. Torque balance: Fspring×0.20=10.0×0.40; Fspring=20.0 N upward.

2. Extension magnitude=20.0/200=0.100 m in the linear range.

3. Vertical balance: Rpivot+20.0−10.0=0; pivot force is 10.0 N downward.

The downward pivot force is consistent with the assumed fixed pivot, which can exert force in either vertical direction. A simple unfastened support that can only push upward would not fit this arrangement. Beam weight and an out-of-range spring would require a revised model.

Scientific references and source-model foundations

• OpenStax — Newton laws and friction
• OpenSciEd 8.1 — contact-force foundations
• OpenSciEd public middle-school figure license — CC BY 4.0

Parent / teacher teaching notes

Use separate force and torque sums. Ask whether the support can physically provide the calculated direction rather than treating every algebraic support as feasible. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Worked examples

Avoid these mistakes

• Torque’s N·m unit does not make it the same physical quantity as energy.

Words to know

spring constant

Stiffness parameter in a linear elastic model.

torque

Turning effect of a force about a specified pivot.

lever arm

Perpendicular distance from pivot to force line.

In the real world

Use the chapter investigation to connect this unit to measured evidence and a practical decision.

Exit check

Why does a force through a pivot have zero torque?