Free-body diagrams for a particle and for an extended body, with the weight acting at the centre of gravity

Weight, mass and gravitational field strength: g = F/m and W = mg

Newton's first law and equilibrium, including resolving forces

Forces on slopes: components of the weight along and perpendicular to the slope

Newton's second law, ΣF = ma, for a body of constant mass

Connected bodies (towing) and the forces on a person in a lift

Terminal velocity, explained with forces

Newton's third law and the properties of the pair of forces in an interaction

1. One diagram — the free-body diagram — and one equation — ΣF = ma — link force to motion, from a sign hanging still to a skydiver opening a parachute.

2. A free-body diagram shows every force acting on one body, drawn as arrows starting on that body. The weight of an extended body acts at its centre of gravity.

3. Weight W = mg. Mass (kg) is the same everywhere; weight (N) depends on g, which is 9.81 N kg⁻¹ at the Earth's surface.

4. First law: zero resultant force ⇔ at rest or constant velocity. Resolve in two perpendicular directions and set each sum to zero.

5. Second law: ΣF = ma. The acceleration is in the direction of the resultant force.

6. Terminal velocity: drag grows with speed until it equals the weight; the resultant force, and so the acceleration, become zero.

7. Third law: if A pushes or pulls B, B pushes or pulls A with a force of equal size, opposite direction and the same type — acting on the other body.

Almost every mistake in a forces question comes from a force that is missing, doubled or acting on the wrong body. A free-body diagram prevents this by isolating one body and asking a single question: what is pushing or pulling this object?

Drawing one well

Draw only the body you are interested in — a dot for a particle, an outline for an extended body.

Every arrow starts on the body and points the way the force acts; label it with its name or symbol and value.

Make arrow lengths roughly proportional to the sizes of the forces — balanced forces get equal arrows.

For an extended body, draw the weight from the centre of gravity: the single point at which the whole weight can be taken to act. For a uniform body it is at the geometric centre.

Never include forces that the body exerts on other things.

In everyday speech "weight" is often measured in kilograms, but in physics the two are different quantities. Mass tells you how much matter there is and how hard the body is to accelerate. Weight is a force: the pull of a gravitational field on that mass.

At the Earth's surface g = 9.81 N kg⁻¹. Since 1 N = 1 kg m s⁻², this is the same as a free-fall acceleration of 9.81 m s⁻².

WORKED EXAMPLE · QUESTION

A planetary rover has a mass of 1025 kg. The gravitational field strength at the surface of Mars is 3.71 N kg⁻¹. Calculate the weight of the rover on Earth and on Mars, and state its mass on Mars.

SOLUTION

Step 1 Weight on Earth: W = mg = 1025 ×9.81 = 10 055 N = 1.01 × 10⁴ N

Step 2 Weight on Mars: W = 1025 ×3.71 = 3803 N = 3.80 × 10³ N

Step 3 Mass does not depend on the field: mass on Mars = 1025 kg

The law says that motion does not need a cause; changes in motion do. Zero resultant force means zero acceleration: the body is at rest or moving in a straight line at constant speed. It is the special case a = 0 of the second law. A body with zero resultant force is in equilibrium (translational equilibrium; turning effects are on the moments page). Constant speed round a bend is not equilibrium: the direction of the velocity changes, so there is a resultant force.

Solving an equilibrium problem

1. Draw the free-body diagram.

2. Choose two perpendicular directions (horizontal and vertical, or along and perpendicular to a slope).

3. Resolve every force and write: sum of components = 0 in each direction.

4. Or draw the forces tip to tail — for three forces in equilibrium they form a closed triangle.

WORKED EXAMPLE · QUESTION

A café sign of mass 15.0 kg hangs at rest from two cables. Each cable makes an angle of 40° with the horizontal, and the arrangement is symmetrical. Calculate the tension in each cable.

SOLUTION

Step 1 The sign is in equilibrium. Horizontally, the two horizontal components (Tcos 40°) cancel, so the tensions are equal.

Step 2 Vertically, the upward components balance the weight: 2Tsin 40° = 15.0 ×9.81 = 147.2 N

Step 3 Rearrange: T = 147.2/2sin 40° = 114.5 N ≈ 114 N

Each tension is bigger than half the weight (73.6 N), because only its vertical component, 73.6 N, holds the sign up.

On a slope, horizontal and vertical are the wrong axes. Resolve along the slope and perpendicular to it: the normal contact force and friction then each lie along one direction, and only the weight needs resolving. For a slope at angle θ to the horizontal, the weight mg has components

Perpendicular to the slope there is no acceleration, so R = mgcos θ (when no other force has a component in that direction). A quick check on which component gets sin: if the slope were flat (θ= 0) there would be no pull down the slope, and sin 0 = 0.

WORKED EXAMPLE · QUESTION

A crate of mass 8.00 kg rests without sliding on a loading ramp inclined at 20° to the horizontal. Calculate (a) the frictional force on the crate and (b) the normal contact force on the crate.

SOLUTION

Step 1 Weight: W = 8.00 ×9.81 = 78.5 N

(a) Along the slope, friction balances the component of the weight down the slope: F = 78.5sin 20° = 26.8 N up the slope

(b) Perpendicular to the slope, R balances the other component: R = 78.5cos 20° = 73.7 N perpendicular to the slope

Check: √(26.8² + 73.7²) = 78.5 N — friction and the normal contact force together exactly cancel the weight.

When the forces do not balance, the body accelerates — and the second law tells you by how much.

A reliable method

1. Draw the free-body diagram and choose the positive direction — the direction of the acceleration is usually best.

2. Find the resultant force along that direction: forces that way count positive, forces the other way negative.

3. Write ΣF = ma and solve; then use the equations of motion if you need a speed, time or distance.

WORKED EXAMPLE · QUESTION

An electric car of mass 1450 kg has a forward driving force of 3200 N. At one instant, the total resistive force on it is 850 N. (a) Calculate the acceleration of the car. (b) State and explain the resistive force when the car reaches a steady top speed with the same driving force.

SOLUTION

(a) Resultant force, forwards positive: ΣF = 3200 - 850 = 2350 N. Then a = 2350 ÷1450 = 1.62 m s⁻² ≈ 1.6 m s⁻² forwards

(b) Steady speed means constant velocity, so the resultant force is zero (first law): resistive force = 3200 N, equal and opposite to the driving force.

WORKED EXAMPLE · QUESTION

A child on a toboggan, total mass 42 kg, slides from rest down a straight snow slope at 18° to the horizontal. A constant frictional force of 60 N acts up the slope; air resistance is negligible. Calculate (a) the acceleration and (b) the speed after 4.0 s.

SOLUTION

Step 1 Component of the weight down the slope: 42 ×9.81 ×sin 18° = 127.3 N

(a) Resultant force down the slope, then ΣF = ma: a = (127.3 - 60)/42 = 1.60 m s⁻² ≈ 1.6 m s⁻² down the slope

(b) Constant force, so constant acceleration: v = u + at = 0 + 1.603 ×4.0 = 6.4 m s⁻¹

The normal contact force does not appear: it is perpendicular to the slope, so it has no component along the motion.

Connected bodies

Bodies joined by a taut rope or rigid tow bar have the same acceleration. Treat the whole system first: the tension is internal (it pulls the two bodies equally in opposite directions), so it cancels and ΣFₑₓₜₑᵣₙₐₗ = (total mass) ×a. Then take one body on its own to find the tension: its free-body diagram now includes the tension. Pick the body with fewer forces for the second step, and use the other one as a check.

WORKED EXAMPLE · QUESTION

A car of mass 1250 kg tows a trailer of mass 350 kg along a straight, level road. The driving force on the car is 2800 N. The resistive forces are 520 N on the car and 280 N on the trailer. Calculate (a) the acceleration and (b) the tension in the tow bar.

SOLUTION

(a) Whole system — the tension cancels: a = (2800 - 520 - 280)/(1250 + 350) = (2000/1600) = 1.25 m s⁻²

(b) Trailer alone — forward tension, backward resistance: T - 280 = 350 ×1.25, so T = 717.5 N ≈ 718 N

Check with the car alone: 2800 - 520 - 717.5 = 1562.5 N = 1250 ×1.25 ✓

A person in a lift

Two forces act on the person: weight mg down and the normal contact force R from the floor (or scales) up. Scales read R, the apparent weight — what you feel through your feet. Taking upwards as positive, R - mg = ma, so R = m(g + a).

Accelerating upwards (speeding up going up, or slowing down going down): R > mg.

Constant velocity, in either direction: R = mg.

Accelerating downwards (speeding up going down, or slowing down going up): R < mg.

It is the direction of the acceleration, not of the velocity, that decides.

WORKED EXAMPLE · QUESTION

A student of mass 60.0 kg stands on bathroom scales in a lift. The lift accelerates upwards at 1.50 m s⁻², then moves at constant velocity, then accelerates downwards at 1.50 m s⁻² as it slows to a stop on a higher floor. (a) Calculate the force the scales exert on the student in each stage. (b) The scales are marked in kilograms (they divide R by 9.81). Calculate their reading in the first stage.

SOLUTION

(a) R = m(g + a), upwards positive: 60.0 ×(9.81 + 1.50) = 679 N; 60.0 ×9.81 = 589 N; 60.0 ×(9.81 - 1.50) = 499 N. 679 N, 589 N, 499 N

(b) Reading = R ÷9.81 = 678.6 ÷9.81 = 69.2 kg. The student's mass is still 60.0 kg — the scales only measure the contact force.

The key fact is that drag (air resistance) increases with speed, while the weight stays the same. Follow a skydiver from the moment of jumping:

At release the speed is zero, so there is no drag: the resultant force is mg and the acceleration is g.

As the speed rises the drag rises, the resultant force mg - D falls, so the acceleration falls — the v–t graph gets less steep.

When D = mg the resultant force is zero, so the acceleration is zero (first law): the body falls at terminal velocity.

Opening a parachute greatly increases the drag at that speed: D > mg, the resultant force is upwards and the skydiver decelerates.

As the speed falls the drag falls, until D = mg again at a new, lower terminal velocity.

The same idea applies horizontally: a vehicle reaches top speed when the resistive forces equal the driving force.

Forces always come in pairs, because a force is an interaction between two bodies. The two forces of a pair have four properties:

They act on different bodies — so they never cancel each other.

They are equal in magnitude and opposite in direction, along the same line.

They are the same type of force (both gravitational, both contact, both electrostatic…).

They act at the same time, for the same duration.

The third law explains how anything starts moving. Walking: your foot pushes backwards on the ground; the ground pushes forwards on you with an equal friction force — that is the force that accelerates you. A rocket engine pushes exhaust gas backwards; the gas pushes the rocket forwards. A swimmer's hands push water backwards; the water pushes the swimmer forwards.

1 Motion, vectors and projectiles

A constant resultant force gives the constant acceleration the equations of motion need, and resolving vectors is the tool for every slope and equilibrium problem here.

3 Momentum

Newton's second law can also be written as force = rate of change of momentum; together with the third law it leads to conservation of momentum in collisions.

4 Moments, Materials and Unit 4

Extended bodies in equilibrium also need balanced turning effects, using the centre of gravity met here. Drag and upthrust on a sphere in a fluid are calculated with Stokes' law in the Materials topic. In Unit 4, a body moving in a circle at constant speed needs a resultant force towards the centre — the second law again.

1. An astronaut has a mass of 72 kg. State her mass on the Moon and calculate her weight there (g = 1.62 N kg⁻¹).

2. A swimmer's hands push backwards on the water. State the Newton's third law partner of this force.

3. A 900 kg lift car accelerates downwards at 0.60 m s⁻². Calculate the tension in its cable.

4. Explain why a falling raindrop reaches a terminal velocity.

5. A 2.0 kg block slides down a smooth (frictionless) slope at 30° to the horizontal. Calculate its acceleration and the normal contact force on it.

1. Mass is unchanged: 72 kg. Weight = 72 ×1.62 = 116.6 N ≈ 117 N.

2. The water pushes forwards on the swimmer's hands, with a force of equal size — a contact force, acting on the swimmer.

3. Upwards positive: T - 900 ×9.81 = 900 ×(-0.60), so T = 900 ×9.21 = 8.29 × 10³ N.

4. As its speed increases, the air resistance on it increases while its weight stays constant, so the resultant force and the acceleration decrease. When the air resistance equals the weight the resultant force is zero, so the acceleration is zero and the velocity stays constant.

5. a = gsin 30° = 4.9 m s⁻² down the slope (the mass cancels). R = 2.0 ×9.81 ×cos 30° = 17 N.