Physics · 1 Motion, forces and energy

1.5.2 Turning effect of forces

Before you start

A force is a push or a pull, measured in newtons (N), and it has a direction.
Weight is the force of gravity on a mass: in the exam, a mass of $1.0$ kg weighs $9.8$ N.
The resultant force is the single force that has the same effect as all the forces on an object together.
Perpendicular means at right angles (90°).
Converting lengths: $100$ cm $= 1$ m, so $25$ cm $= 0.25$ m.
Rearranging a three-letter equation such as $M = F d$ to make any one letter the subject.

Syllabus objectives covered (6)

Describe the moment of a force as a measure of its turning effect and give everyday examples
Define the moment of a force as moment = force × perpendicular distance from the pivot; recall and use this equation
Apply the principle of moments to situations with one force each side of the pivot, including balancing of a beam
State that, when there is no resultant force and no resultant moment, an object is in equilibrium
Apply the principle of moments to other situations, including those with more than one force each side of the pivot
Describe an experiment to demonstrate that there is no resultant moment on an object in equilibrium

Part 1 of 6

What a moment is: the turning effect of a force

A force can make an object turn about a fixed point. That fixed point is the pivot. When you push a door, it turns about its hinges, so the hinges are the pivot.
The moment of a force is a measure of its turning effect. A force with a large moment turns an object more strongly than a force with a small moment.
The turning effect depends on two things: the size of the force, and how far from the pivot it acts. Push a door at the handle, far from the hinges, and it opens easily. Push the same door close to the hinges and you need a much larger force to open it.
forces diagram — not drawn yet
A door seen from above, hinged at the left end (pivot marked), with one force arrow pushing at right angles at the handle end far from the hinge and a second, longer force arrow pushing at right angles close to the hinge, each labelled with its distance from the hinge.
Close to the hinge you need a bigger push for the same turning effect, because the distance from the pivot is smaller.
A spanner is another example. The pivot is the centre of the nut. A longer spanner lets you undo a tight nut with a smaller force, because your hand is further from the pivot.
A see-saw turns about the pivot in its middle. A child who sits further from the pivot has a bigger turning effect than the same child sitting near the middle.
A wheelbarrow turns about the axle of its wheel when you lift the handles. The handles are long so that your hands are far from the axle, and you can lift a heavy load with a smaller force.
Every moment has a direction of turning: clockwise or anticlockwise. To find it, imagine that force acting on its own and ask which way it would turn the object about the pivot.
When you give an everyday example, name the object, say where the pivot is, and say how the distance from the pivot changes the force needed. 'A door' on its own is a name, not an example of a turning effect.

Part 2 of 6

Calculating a moment

Here is the definition. Learn it word for word: the moment of a force is the force multiplied by the perpendicular distance from the pivot.
$moment = force \times perpendicular distance from the pivot$
Perpendicular distance needs care. Imagine the force arrow extended into a long straight line in both directions. That is the force's line of action. The perpendicular distance is the distance from the pivot to that line, measured at right angles to it.
On a horizontal beam with vertical forces, such as weights hanging down or a support pushing up, the perpendicular distance is just the distance along the beam from the pivot to where the force acts. A see-saw and a balanced metre rule are both like this.
If a force acts at an angle, the perpendicular distance from the pivot to its line of action is shorter than the distance to the point where you push. So the moment is smaller. That is why you get the biggest turning effect from a spanner or a door by pushing at right angles to it.
forces diagram — not drawn yet
Two views of a spanner gripping a nut: on the left a force arrow at right angles to the handle end with a dashed line marking the perpendicular distance from the centre of the nut to the arrow; on the right the same spanner with the force at an angle, its line of action extended as a dotted line, and a shorter dashed perpendicular from the centre of the nut to that line.
The distance that counts is measured at right angles to the force's line of action. Tilting the force shortens it, so the moment drops.
If the line of action of a force passes through the pivot, the perpendicular distance is zero, so the moment is zero. Pushing a door straight towards its hinges does not turn it at all, however hard you push.
Units: with the force in N and the distance in m, the moment is in newton metres, written N m. If the distance is in cm, the moment comes out in N cm. Since $1$ m $= 100$ cm, $1$ N m $= 100$ N cm.
N m means newtons multiplied by metres. Do not write N/m. That is the unit of the spring constant, which is a different quantity.
Paper 4 has no formula sheet, so you must recall this equation. Write it as $M = F \times d$ . To find the force, divide both sides by $d$ , because $d$ is what multiplies $F$ . To find the distance, divide both sides by $F$ .
$F = \frac{M}{d} d = \frac{M}{F}$

Part 3 of 6

The principle of moments: one force each side

Put a child on each end of a see-saw. The weight of the child on the left turns the see-saw anticlockwise about the pivot. The weight of the child on the right turns it clockwise.
If the two moments are equal, their turning effects cancel and the see-saw balances. If one moment is larger, the see-saw turns that way: the side with the larger moment goes down.
This is the principle of moments. When an object is balanced (in equilibrium), the total clockwise moment about a point equals the total anticlockwise moment about the same point.
$clockwise moment = anticlockwise moment$
With one force on each side of the pivot, write it with the force and distance of each side.
$F_{1} \times d_{1} = F_{2} \times d_{2}$
Balance depends on moments, not on forces. A heavier child balances a lighter one by sitting nearer the pivot. A $300$ N child at $2.0$ m balances a $400$ N child at $1.5$ m, because $300 \times 2.0 = 600$ N m and $400 \times 1.5 = 600$ N m.
forces diagram — not drawn yet
A horizontal see-saw on a central triangular pivot with a 300 N downward weight arrow 2.0 m to the left of the pivot and a 400 N downward weight arrow 1.5 m to the right, both distances marked from the pivot.
Unequal weights, equal moments: 300 N × 2.0 m and 400 N × 1.5 m are both 600 N m, so the see-saw balances.
Measure every distance from the pivot. On a metre rule pivoted at its $50$ cm mark, a load hung at the $20$ cm mark is $50 - 20 = 30$ cm from the pivot. Using $20$ cm is the trap.
Keep the units the same on both sides of the equation. If every distance is in cm, both moments are in N cm and the equation still balances. Convert to N m only when the question asks for a moment in N m.
Now balancing a beam. A uniform beam pivoted at its centre balances on its own, because its weight acts at its centre, which is at the pivot. A force through the pivot has no moment about it, so the beam's own weight does not appear in the moment equation.
To balance such a beam with one load on each side, use $F_{1} d_{1} = F_{2} d_{2}$ . If you know three of the four values, rearrange to find the fourth.

Part 4 of 6

Equilibrium: no resultant force and no resultant moment

An object is in equilibrium when there is no resultant force on it and no resultant moment on it. Both conditions are needed.
No resultant force means the forces in opposite directions cancel. For an object at rest, it does not start to move along.
No resultant moment means the clockwise and anticlockwise moments cancel. For an object at rest, it does not start to turn.
Look at a balanced see-saw. The pivot pushes up on it with a force equal to the total weight of the see-saw and the children, so there is no resultant force. The clockwise and anticlockwise moments are equal, so there is no resultant moment. Both conditions hold, so it is in equilibrium.
One condition on its own is not enough. Turn a steering wheel with two hands: one hand pushes up on the left of the wheel and the other pushes down on the right with an equal force. The two forces cancel, so there is no resultant force. But both forces turn the wheel clockwise, so there is a resultant moment. The wheel turns, and it is not in equilibrium.
Equilibrium does not mean that no forces act. A balanced see-saw has several large forces on it. They cancel, and their moments cancel.
When you are asked for the conditions for equilibrium, state both: 'no resultant force' and 'no resultant moment'. 'The forces are balanced' on its own gives only the first condition.

Part 5 of 6

More than one force each side, and other situations

With several forces on each side, the principle of moments is the same idea. Add up all the clockwise moments. Add up all the anticlockwise moments. For balance, the two totals are equal.
$sum of clockwise moments = sum of anticlockwise moments$
Work out each moment on its own, from its own force and its own distance, and then add the moments. Do not add the forces on one side first and multiply by one distance. The forces act at different distances, so each has a different turning effect.
forces diagram — not drawn yet
A horizontal beam on a central pivot with downward arrows of 20 N at 0.40 m and 10 N at 0.10 m to the left of the pivot, and 15 N at 0.30 m and an unknown W at 0.25 m to the right, each distance marked from the pivot.
Four forces, four distances, four moments. Each moment is worked out separately before the totals are compared.
The forces do not have to be on opposite sides of the pivot. What matters is which way each force turns the object. In a wheelbarrow the pivot is the axle of the wheel. The load pushes down between the wheel and your hands. Your hands lift the handles further out on the same side. The two forces turn the barrow in opposite directions about the axle.
A plank resting on two supports has no single pivot. When an object is in equilibrium, the principle of moments works about any point you choose. So choose one of the supports.
The force from that support acts through the point you chose, so its moment about that point is zero. It drops out of the equation, which leaves only one unknown: the force from the other support.
Then use the other condition for equilibrium. There is no resultant force, so the total upward force equals the total downward force. This gives the force from a pivot, or from the second support, once the moments have given you the first.
Include the beam's own weight whenever it does not act through the point you take moments about. For a uniform beam the weight acts at its centre. If you take moments about one end, the beam's weight is half the beam's length from that point.

Part 6 of 6

The experiment: showing there is no resultant moment

The aim is to show that, when a beam is balanced, the total clockwise moment equals the total anticlockwise moment. Then there is no resultant moment on the beam.
You need a metre rule, a pivot such as a knife edge, some loads of known weight and some thread to hang them with.
apparatus diagram — not drawn yet
A metre rule balanced horizontally on a knife-edge pivot under its 50 cm mark, with one slotted-mass load hanging from a thread loop on the left of the pivot and another on the right, and the distance from the pivot to each thread loop marked with a double-headed arrow.
Each distance is measured from the pivot to the thread, and the rule must be horizontal before you take readings.
First, balance the rule on the pivot with no loads on it, so that it rests horizontally. For a uniform rule this is at or near the $50$ cm mark. Note the reading above the pivot. The rule's own weight now acts through the pivot, so it has no moment about the pivot.
Hang a load on each side of the pivot from a loop of thread. The thin thread means you know exactly where on the rule each force acts.
Slide one of the loads along the rule until the rule balances horizontally again.
Read the position of each thread on the rule. Subtract the pivot reading, or subtract from it, to get each distance from the pivot.
Find the weight of each load. With $1.0$ kg weighing $9.8$ N, a $100$ g mass weighs $0.98$ N.
Calculate force × distance for every load. Add the moments that turn the rule clockwise, and add those that turn it anticlockwise. The two totals are equal, within the limits of your measurements. That is the evidence that there is no resultant moment.
Small differences between the totals come from measurement. It is hard to judge exactly when the rule is horizontal, and each position can be read only to the nearest millimetre or so.
Repeat with different loads and positions, including two loads on the same side. Equal totals each time show that the result holds in general, not for one arrangement only.
A full Describe answer has these points in order: balance the rule on the pivot with no loads; hang known weights on both sides; adjust until the rule is horizontal; measure each distance from the pivot; calculate the moments and show that the total clockwise moment equals the total anticlockwise moment.

Worked examples

Example 1. A mechanic tightens a nut with a spanner. She pushes at right angles to the spanner with a force of $40$ N, at a perpendicular distance of $0.25$ m from the centre of the nut. (a) Calculate the moment of her force about the centre of the nut. (b) She changes to a longer spanner and pushes at right angles at $0.40$ m from the centre of the nut. Calculate the force she needs to produce the same moment. (c) Explain why the longer spanner makes the job easier.

Show the working

Recall the definition: moment = force × perpendicular distance from the pivot. The pivot is the centre of the nut.
$M = F \times d = 40 \times 0.25 = 10 N m$
For (b) you know the moment and the new distance, and you want the force. $d$ multiplies $F$ , so divide both sides by $d$ .
$F = \frac{M}{d} = \frac{10}{0.40} = 25 N$
Check the answer makes sense: the distance went up, so the force needed went down. $25 \times 0.40 = 10$ N m, the same moment as before.
For (c), link the distance to the force: her hand is further from the pivot, so a smaller force produces the same moment.

Answer: (a) $10$ N m. (b) $25$ N. (c) The force acts at a greater perpendicular distance from the pivot, so a smaller force gives the same moment.

Example 2. A see-saw is pivoted at its centre and balances on its own. Amir, of weight $300$ N, sits $2.0$ m to the left of the pivot. Bea, of weight $400$ N, sits on the right. (a) Calculate how far from the pivot Bea must sit for the see-saw to balance. (b) Bea moves so that she is $2.0$ m from the pivot. Predict what happens to the see-saw.

Show the working

The see-saw's own weight acts at the pivot, so it has no moment. Only the two children's weights matter.
Amir is on the left, so his weight turns the see-saw anticlockwise.
$anticlockwise moment = 300 \times 2.0 = 600 N m$
For balance, the principle of moments says Bea's clockwise moment must also be $600$ N m. Write that with $d$ for her distance.
$400 \times d = 600$
Divide both sides by $400$ , because $400$ is what multiplies $d$ .
$d = \frac{600}{400} = 1.5 m$
Bea is heavier, so she sits nearer the pivot than Amir. That matches the answer: $1.5$ m is less than $2.0$ m.
For (b), Bea's moment becomes $400 \times 2.0 = 800$ N m clockwise. That is larger than Amir's $600$ N m anticlockwise, so there is a resultant clockwise moment.

Answer: (a) $1.5$ m from the pivot. (b) The see-saw turns clockwise: Bea's side goes down, because her moment ( $800$ N m) is now larger than Amir's ( $600$ N m).

Example 3. A uniform beam $1.0$ m long weighs $12$ N and is pivoted at its centre. On the left of the pivot, a $20$ N load hangs $0.40$ m from the pivot and a $10$ N load hangs $0.10$ m from the pivot. On the right, a $15$ N load hangs $0.30$ m from the pivot and a load of unknown weight $W$ hangs $0.25$ m from the pivot. The beam is balanced. (a) Calculate $W$ . (b) Determine the upward force that the pivot exerts on the beam.

Show the working

The beam is uniform and pivoted at its centre, so its own weight acts at the pivot and has no moment about it. Leave it out of the moment equation.
The loads on the left turn the beam anticlockwise. Work out each moment separately, then add them.
$anticlockwise = (20 \times 0.40) + (10 \times 0.10) = 8.0 + 1.0 = 9.0 N m$
The loads on the right turn it clockwise.
$clockwise = (15 \times 0.30) + (W \times 0.25) = 4.5 + 0.25 W$
The beam is balanced, so the totals are equal.
$4.5 + 0.25 W = 9.0$
Subtract $4.5$ from both sides, then divide both sides by $0.25$ .
$W = \frac{9.0 - 4.5}{0.25} = 18 N$
The trap is to add the left-hand loads first, $20 + 10 = 30$ N, and multiply by one distance. $30 \times 0.40 = 12$ N m, not $9.0$ N m. The $10$ N load is much nearer the pivot, so its turning effect is much smaller.
For (b), the beam is in equilibrium, so there is no resultant force. The upward push of the pivot equals the total downward force, and that includes the beam's own weight.
$F = 20 + 10 + 15 + 18 + 12 = 75 N$

Answer: (a) $W = 18$ N. (b) $75$ N upwards.

Example 4. A uniform plank $4.0$ m long weighs $200$ N. It rests horizontally on two supports: A at its left end and B at its right end. A painter of weight $600$ N stands on the plank $1.0$ m from A. (a) Calculate the upward force on the plank from support B. (b) Determine the upward force on the plank from support A.

Show the working

There are two unknown forces, so take moments about one of the supports. Choose A. The force from A acts through A, so its moment about A is zero and it drops out.
About A, both weights turn the plank clockwise. The plank is uniform, so its weight acts at its centre, $2.0$ m from A.
$clockwise = (600 \times 1.0) + (200 \times 2.0) = 600 + 400 = 1000 N m$
The force from B pushes up at the right end, $4.0$ m from A, so it turns the plank anticlockwise about A.
$anticlockwise = F_{B} \times 4.0$
The plank is in equilibrium, so the moments are equal. Divide both sides by $4.0$ .
$F_{B} = \frac{1000}{4.0} = 250 N$
For (b), use the other condition: there is no resultant force, so the upward forces add up to the downward forces.
$F_{A} = (600 + 200) - 250 = 550 N$
Check by taking moments about B instead. The force from A gives $550 \times 4.0 = 2200$ N m. The weights give $(600 \times 3.0) + (200 \times 2.0) = 1800 + 400 = 2200$ N m the other way. They match. The painter stands nearer A, so A takes the larger share of the load.

Answer: (a) $250$ N upwards. (b) $550$ N upwards.

Where people go wrong

Mistake: Reading a distance off the scale of a metre rule, or measuring from the end of a beam, instead of measuring from the pivot.

Instead: Every distance in a moment is the distance from the pivot. On a rule pivoted at $50$ cm, a load at the $80$ cm mark is $30$ cm from the pivot.

The moment depends on how far the force is from the point the object turns about. The end of the rule is not that point.

Mistake: Adding all the forces on one side and multiplying the total by one distance.

Instead: Work out force × distance for each force separately, then add the moments.

Forces at different distances have different turning effects. A load near the pivot turns the beam much less than the same load near the end.

Mistake: Thinking that a beam balances only when the forces on each side are equal.

Instead: A beam balances when the moments are equal. A large force near the pivot can balance a small force far from it.

Balance is about turning, and turning depends on force and distance together.

Mistake: Giving the unit of moment as N/m, or mixing cm and m in one equation.

Instead: The unit is N m (or N cm if the distances are in cm). Use the same distance unit throughout one equation.

A moment is a force multiplied by a distance, not divided by it. N/m is the unit of the spring constant.

Mistake: Saying an object is in equilibrium because 'the forces are balanced', or because 'no forces act on it'.

Instead: State both conditions: no resultant force and no resultant moment.

Forces can cancel and still turn an object, as two hands do on a steering wheel. And a balanced see-saw has several forces on it; they cancel rather than being absent.

Check yourself

Door handles are fixed on the side of the door furthest from the hinges. Explain why.
Answer
The hinges are the pivot. A force at the handle acts at a large distance from the pivot, so it has a large moment, and a small force is enough to open the door.
A force of $25$ N acts at a perpendicular distance of $0.60$ m from a pivot. Calculate its moment about the pivot.
Answer
Moment $= 25 \times 0.60 = 15$ N m.
A force of $50$ N acts along a line that passes through the pivot. What is its moment about the pivot, and why?
Answer
Zero. Its line of action passes through the pivot, so the perpendicular distance is zero, and $50 \times 0 = 0$ .
A uniform metre rule balances on a pivot at its $50$ cm mark. A $2.0$ N load hangs at the $20$ cm mark. At which mark must a $3.0$ N load hang for the rule to balance?
Answer
The $2.0$ N load is $30$ cm from the pivot: anticlockwise moment $= 2.0 \times 30 = 60$ N cm. For balance, $3.0 \times d = 60$ , so $d = 20$ cm on the other side of the pivot: the $70$ cm mark.
A driver turns a steering wheel by pushing up on one side and pulling down on the other side with equal forces. Is the wheel in equilibrium? State the two conditions for equilibrium in your answer.
Answer
No. An object is in equilibrium when there is no resultant force and no resultant moment. The two forces cancel, so there is no resultant force, but both turn the wheel the same way, so there is a resultant moment.
A uniform beam is pivoted at its centre. On the left, a $4.0$ N load hangs $30$ cm from the pivot and a $6.0$ N load hangs $10$ cm from the pivot. Calculate how far to the right of the pivot a $5.0$ N load must hang for the beam to balance.
Answer
Anticlockwise moments $= (4.0 \times 30) + (6.0 \times 10) = 120 + 60 = 180$ N cm. So $5.0 \times d = 180$ , and $d = 36$ cm.
In the metre-rule experiment, why do you balance the rule on the pivot before hanging any loads, and why do you hang the loads from thin thread?
Answer
Balancing first puts the pivot where the rule's weight acts, so the rule's own weight has no moment about the pivot. Thin thread shows exactly where each force acts, so each distance from the pivot can be measured accurately.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A uniform see-saw is pivoted at its centre. Child A, of weight $360$ N, sits $1.5$ m to the left of the pivot.

(a) Define the moment of a force. 1 mark No calculator

Mark scheme

B1 force × perpendicular distance from the pivot (point), owtte; 'force × distance' without 'perpendicular' is insufficient

(b) Calculate the moment of the weight of child A about the pivot. 2 marks

Mark scheme

C1 360 × 1.5 seen
A1 540 N m (accept 54 000 N cm); unit required

(c) Child B, of weight $450$ N, sits on the right of the pivot so that the see-saw balances. Calculate the distance of child B from the pivot. 3 marks

Mark scheme

C1 clockwise moment = anticlockwise moment stated or used, e.g. 450 × d = 540, ecf from (b)
C1 d = 540 / 450
A1 1.2 m; unit required

Question 2. A student wants to show that there is no resultant moment on a beam in equilibrium. She has a metre rule, a knife-edge pivot, loads of known weight and some thread.

(a) Describe how she can use this apparatus to show that there is no resultant moment on the rule when it is balanced. 4 marks No calculator

Mark scheme

B1 balances the rule on the pivot with no loads (at or near the 50 cm mark)
B1 hangs loads (known weights) on both sides of the pivot, e.g. from thread loops
B1 moves a load until the rule balances horizontally, then measures the distance of each load from the pivot
B1 calculates force × distance for each load and shows that the total clockwise moment = the total anticlockwise moment; also credit repeating with different loads or positions if the comparison of moments has been stated

(b) With no loads, the rule balances on the pivot at the $50.0$ cm mark. She then hangs a $2.0$ N load at the $10.0$ cm mark and a $1.0$ N load at the $30.0$ cm mark. The rule balances horizontally when a $4.0$ N load hangs at the $75.0$ cm mark. Deduce whether there is a resultant moment on the rule. Use calculations to justify your answer. 3 marks

Mark scheme

B1 anticlockwise moment = (2.0 × 40.0) + (1.0 × 20.0) = 100 N cm (accept 1.0 N m)
B1 clockwise moment = 4.0 × 25.0 = 100 N cm (accept 1.0 N m)
B1 the two moments are equal, so there is no resultant moment; ecf from the student's own two moments

Question 3. A gardener holds a loaded wheelbarrow at rest with its handles raised. The axle of the wheel is the pivot. The total weight of the wheelbarrow and its load is $720$ N. This weight acts at a horizontal distance of $0.40$ m from the axle. The gardener lifts the handles vertically upwards at a horizontal distance of $1.2$ m from the axle.

(a) Calculate the upward force that the gardener exerts on the handles. 3 marks

Mark scheme

C1 moment of the weight = 720 × 0.40 (= 288 N m) seen
C1 F × 1.2 = 720 × 0.40, or F = 288 / 1.2
A1 240 N; unit required

(b) Determine the upward force that the ground exerts on the wheel. Give a reason for your answer. 2 marks

Mark scheme

B1 480 N; ecf: 720 − their (a)
B1 the wheelbarrow is in equilibrium / there is no resultant force, so the total upward force = the total downward force

(c) Explain why long handles make it easier for the gardener to lift the load. 2 marks No calculator

Mark scheme

B1 the gardener's force acts at a greater (perpendicular) distance from the pivot / axle
B1 so a smaller force gives the moment needed to balance the moment of the weight, owtte

How this is examined

Balancing a beam with more than one force on a side: a Calculate for a missing force or distance on a beam, plank or lever with two or more forces on one side, which may be followed by a Determine for the force at the pivot or at a support Each moment worked out separately as force × its own distance from the pivot; 'sum of clockwise moments = sum of anticlockwise moments' written with the numbers in; the unknown with its unit; then the pivot or support force from 'total upward force = total downward force'.
Describing how to show there is no resultant moment on a balanced rule: a four-mark Describe of the metre-rule experiment, which may come with a set of readings to check Rule balanced on a pivot with no loads; known weights hung on both sides; positions adjusted until the rule is horizontal; each distance measured from the pivot; force × distance calculated for each load and the total clockwise moment shown equal to the total anticlockwise moment. Each is a separate point.
Finding the missing distance or force when one force each side balances: a two- or three-mark Calculate on a see-saw, a balanced beam or a metre rule $F_{1} \times d_{1} = F_{2} \times d_{2}$ written with the numbers in; every distance measured from the pivot, not from the end of the rule; the answer with its unit.
The two conditions for equilibrium: a two-mark State, or the reason given in a Deduce about an object that is balanced 'No resultant force' and 'no resultant moment', both stated. 'The forces are balanced' on its own is only the first condition.
Why a longer handle, spanner or lever makes turning easier: a two-mark Explain about an everyday object such as a door, a spanner or a wheelbarrow The force acts at a greater (perpendicular) distance from the pivot, so a smaller force gives the same moment. Name the pivot as well.
What a moment is, in words and as an equation: a one-mark Define, where the unit may also be asked for 'Force × perpendicular distance from the pivot.' Both 'perpendicular' and 'from the pivot' (or 'from the point') are needed. The unit is N m.

Words to know

Pivot: The fixed point that an object turns about, such as the hinges of a door or the centre of a see-saw.
Moment: The turning effect of a force: force × perpendicular distance from the pivot. Measured in N m.
Perpendicular distance: The distance from the pivot to the force's line of action, measured at right angles to that line.
Line of action: The straight line along which a force acts, found by extending the force arrow in both directions.
Newton metre (N m): The unit of moment: a force of $1$ N acting at a perpendicular distance of $1$ m from the pivot. $1$ N m $= 100$ N cm.
Principle of moments: For an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Resultant moment: The overall turning effect left when the clockwise and anticlockwise moments are combined. It is zero when they are equal.
Equilibrium: The state of an object when there is no resultant force and no resultant moment on it.

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