Moments, levers and gears
What you'll learn
- Why a force can make an object rotate about a pivot.
- How to calculate the moment of a force using a recall equation.
- How to use the principle of moments for balanced objects.
- How levers and gears transmit turning effects, and how lubrication reduces wasted energy.
In Edexcel GCSE Physics (1PH0), the moments, levers and gears points here are labelled Separate Physics content, so they are part of your course.
Forces can cause rotation
A force is a push or pull. You already know forces can change an object’s speed, direction or shape. Forces can also make objects rotate, which means turn around a point or axis.
A door is a good example: the hinge stays in place, and the door turns around it. Pushing near the handle turns the door much more easily than pushing close to the hinge.

Pivot and line of action
- A pivot or fulcrum is the fixed point or axis about which an object turns.
- The line of action of a force is an imaginary straight line showing the direction in which the force acts.
A force is most likely to cause rotation when its line of action does not pass through the pivot. The further the line of action is from the pivot, the larger the turning effect.
Deciding whether a force turns a door
- Identify the pivot: for a door, the pivot is the hinge.
- Compare the distances from the hinge: a push at the handle has a larger perpendicular distance from the pivot than a push near the hinge.
- For the same size force, the handle push produces the larger turning effect because it acts further from the pivot.
- If a force’s line of action passed exactly through the hinge, its perpendicular distance would be zero, so it would produce no turning effect about the hinge.
The moment of a force
The moment of a force is the turning effect of that force about a pivot. You are expected to recall and use this equation:
M=F×dM = F \times dM=F×dwhere:
- MMM is the moment of the force, measured in newton metres (N m)
- FFF is the force, measured in newtons (N)
- ddd is the distance normal to the direction of the force, measured in metres (m)
“Normal to the force” means “at right angles to the force”. So ddd is the shortest distance from the pivot to the force’s line of action.
Bigger force or bigger distance
A moment gets larger if the force gets larger, or if the perpendicular distance from the pivot gets larger.
Calculating a moment
A force of 80 N is applied at right angles to a spanner, 0.30 m from the centre of a nut. Calculate the moment about the nut.
- Choose the correct distance: the force is at right angles, so the normal distance is d=0.30 md = 0.30\ \text{m}d=0.30 m.
- Substitute into the equation: M=F×d=80 N×0.30 mM = F \times d = 80\ \text{N} \times 0.30\ \text{m}M=F×d=80 N×0.30 m.
- Calculate the result: M=24 N mM = 24\ \text{N m}M=24 N m, so the turning effect is 24 newton metres about the nut.
Wrong distance, wrong unit
Do not automatically use the full length of the object. Use the perpendicular distance from the pivot to the force’s line of action, and convert centimetres into metres. The unit for moment is N m, not N/m.
Clockwise and anticlockwise moments
Moments have a sense of rotation: they can be clockwise or anticlockwise about the pivot.
For example, on a seesaw:
- a downward force on the left side tends to rotate the seesaw anticlockwise
- a downward force on the right side tends to rotate it clockwise

The principle of moments
An object is in rotational equilibrium when it is not starting to rotate and is not changing its rate of rotation. For a balanced beam or seesaw, the clockwise and anticlockwise turning effects cancel out.
Principle of moments
For rotational forces in equilibrium:
total clockwise moment=total anticlockwise moment\text{total clockwise moment} = \text{total anticlockwise moment}total clockwise moment=total anticlockwise momentIf there is more than one force causing clockwise moments, add all the clockwise moments together. Do the same for the anticlockwise moments.
Finding an unknown force on a balanced beam
A beam balances on a central pivot. A 12 N force acts 0.50 m to the left of the pivot. An unknown force acts 0.30 m to the right of the pivot. Find the unknown force.
- Identify the moments: the left force produces an anticlockwise moment, and the right force produces a clockwise moment.
- Calculate the known moment: M=12 N×0.50 m=6.0 N mM = 12\ \text{N} \times 0.50\ \text{m} = 6.0\ \text{N m}M=12 N×0.50 m=6.0 N m.
- Apply equilibrium: clockwise moment equals anticlockwise moment, so F×0.30 m=6.0 N mF \times 0.30\ \text{m} = 6.0\ \text{N m}F×0.30 m=6.0 N m.
- Solve for the force: F=6.0 N m÷0.30 m=20 NF = 6.0\ \text{N m} \div 0.30\ \text{m} = 20\ \text{N}F=6.0 N m÷0.30 m=20 N.
Balancing forces instead of moments
A beam can have unequal forces and still balance if the larger force is closer to the pivot. For rotational equilibrium, compare moments, not just forces.
Levers
A lever is a rigid object that turns about a pivot. Levers are used to transmit the rotational effect of a force.
Common examples include:
- a crowbar
- scissors
- pliers
- a bottle opener
- a seesaw
The force you apply is called the effort. The force you are trying to overcome is called the load. The distances from the pivot are sometimes called the effort arm and load arm.
A lever can make a small effort force produce a large load force if the effort acts much further from the pivot than the load does.
Levers trade force for distance
A lever can increase force, but not for free: the effort end usually moves a larger distance than the load end. In real machines, some energy is also transferred wastefully because of friction.
Using a lever to lift a load
A crowbar is used as a lever. A 100 N effort acts 0.60 m from the pivot. The load is 0.15 m from the pivot. Estimate the load force that can be balanced.
- Use the principle of moments for a balanced lever: effort moment equals load moment.
- Calculate the effort moment: 100 N×0.60 m=60 N m100\ \text{N} \times 0.60\ \text{m} = 60\ \text{N m}100 N×0.60 m=60 N m.
- Set the load moment equal to this: Fload×0.15 m=60 N mF_{\text{load}} \times 0.15\ \text{m} = 60\ \text{N m}Fload×0.15 m=60 N m.
- Solve: Fload=60 N m÷0.15 m=400 NF_{\text{load}} = 60\ \text{N m} \div 0.15\ \text{m} = 400\ \text{N}Fload=60 N m÷0.15 m=400 N.
Gears
A gear is a wheel with teeth around its edge. Gears transmit rotational effects from one axle to another because the teeth push on each other.
When two gears mesh:
- they rotate in opposite directions
- the teeth prevent slipping
- a larger gear turns more slowly than a smaller gear, but can give a larger turning effect

The first gear is often called the input gear because it receives the driving force. The gear that is made to turn is the output gear. A gear between them is called an idler gear; it can change the direction of rotation without being the final output.
Predicting gear direction and speed
A 10-tooth input gear rotates clockwise at 6 rotations per second. It drives a 30-tooth output gear. Predict the output direction and speed.
- Meshed gears turn in opposite directions, so the output gear rotates anticlockwise.
- Compare the number of teeth: the output gear has 3 times as many teeth as the input gear, so it makes one rotation for every 3 rotations of the input gear.
- Calculate the output speed: 6 rotations/s÷3=2 rotations/s6\ \text{rotations/s} \div 3 = 2\ \text{rotations/s}6 rotations/s÷3=2 rotations/s.
- Since the output gear is larger and turns more slowly, it gives a larger turning effect in an ideal gear system.
Gear direction shortcut
Each contact between two meshed gears reverses the direction of rotation. Two contacts reverse it twice, so the first and third gears rotate in the same direction.
Lubrication and unwanted energy transfer
Friction is a contact force that opposes motion between surfaces. In levers and gears, friction occurs at pivots, axles and between moving surfaces.
Lubrication means adding a slippery substance, such as oil or grease, between surfaces. The lubricant forms a thin layer that separates the surfaces and reduces friction.
Less friction means less work is done against friction, so less energy is transferred wastefully to the thermal energy store of the surfaces and surroundings. This also reduces wear and heating.
Explaining lubrication in a gear system
A machine with gears becomes hot after running for a long time. Explain how adding oil helps.
- Identify the unwanted force: friction acts between moving gear teeth and at the axles.
- Explain the action of the lubricant: oil forms a thin layer between surfaces, so the surfaces rub less directly against each other.
- Link to energy transfer: with less friction, less energy is transferred to thermal stores, so more of the input energy is used for useful rotation.
Lubrication does not create energy
Lubrication does not make extra energy. It reduces unwanted energy transfers, making the machine more efficient and reducing heating and wear.
In the exam
- Always mark the pivot first, then measure distances from the pivot to each force’s line of action.
- For moment calculations, write M=F×dM = F \times dM=F×d and make sure distance is in metres before substituting.
- For balancing problems, separate clockwise and anticlockwise moments, add each side, then set them equal.
- For gears, state the direction change first, then compare gear sizes to explain speed and turning effect.
- For lubrication questions, mention reduced friction and reduced unwanted energy transfer to thermal stores.
Check yourself
- Why is it easier to open a door by pushing near the handle than near the hinge?
- A 15 N force acts 0.40 m from a pivot. What is the moment of the force?
- If a small gear drives a larger gear, what happens to the output speed and turning effect?