The turning effect of a force about a point: force multiplied by the perpendicular distance from the point to the line of action of the force, in newton metres.

Where does the weight of a uniform rod act?

At its midpoint (its centre of mass). For a non-uniform rod it acts at the centre of mass, which must be found or given.

How do I choose where to take moments?

Take moments about a point where an unknown force acts, because that force then has zero moment and drops out of the equation.

What happens when a beam is about to tilt?

The reaction at the support that is about to lift off is zero, so you can find the limiting position or load by taking moments about the other support.

Year 13 moments revision notes to download as PDF

This Year 13 moments revision notes is written for pupils aged 17-18 and follows the National Curriculum. In short, it shows you how to solve equilibrium problems with moments. Before you start, it helps to revise projectiles and variable acceleration and forces in two dimensions and friction, because this lesson builds on them.

Key points

The moment of a force about a point = force \(\times\) perpendicular distance from the point to the line of action of the force, in newton metres.
Only the component of a force perpendicular to the rod gives a moment: for a force \(F\) at angle \(\theta\) to a rod at distance \(d\), the moment is \(Fd\sin\theta\).
A rigid body is in equilibrium when the resultant force is zero and the total clockwise moment about any point equals the total anticlockwise moment.
A uniform rod has its weight acting at its midpoint; a light rod has no weight; for a non-uniform rod the centre of mass must be found from a balance condition.
For beam problems, take moments about the point where an unknown force acts, then resolve vertically for the other force.
If a beam is on the point of tilting about one support, the reaction at the other support is zero.

Year 13 moments revision notes: what you will learn

A force can make an object turn as well as move. This chapter defines the moment of a force and uses moments to solve problems about beams, rods and tilting.

Starter activity

Balance a ruler on your finger. Place a coin at one end and a 3-coin pile at the other end. Where do you need to move the pivot? Write down a rule that connects the weights and the distances. Try again with the coin pile at 10 cm and the single coin at 25 cm from the pivot. What do you notice about the product of weight and distance on each side? Explain in a sentence why a long spanner makes it easier to loosen a tight bolt.

The moment of a force

Definition: moment of a force

The moment of a force about a point is its turning effect: \[\text{moment}=\text{force}\times\text{perpendicular distance from the point to the line of action of the force},\] measured in newton metres (N m). It can be clockwise or anticlockwise: choose one direction as positive and use it for every force in the question.

Worked example 1

Find the moment of (a) a 15 N force 0.8 m from a hinge at right angles to the rod, (b) a 20 N force at \(30^\circ\) to a rod, acting 0.5 m from the pivot.

(a) \(15\times0.8=12\) N m. (b) Use only the perpendicular component: \(20\sin30^\circ\times0.5=5\) N m.

Equilibrium of a rigid body

Rule: two conditions for equilibrium

A rigid body is in equilibrium when (1) the resultant force is zero, so the forces in two perpendicular directions each balance, and (2) the sum of the moments about any point is zero: the total clockwise moment equals the total anticlockwise moment.

Worked example 2

A child of mass 30 kg sits 2 m from the pivot of a seesaw. Where must a 40 kg child sit to balance?

\(30g\times2=40g\times x\), so \(x=1.5\) m. Mass can replace weight when \(g\) cancels, but say so clearly.

Models for beams and rods

Definition: rod, uniform, light and non-uniform

A beam or plank is modelled as a rod: a rigid straight object with negligible thickness. A uniform rod has its weight acting at its midpoint. A light rod has negligible mass. A non-uniform rod has a centre of mass that is not at the midpoint. A smooth support or peg exerts a force perpendicular to the rod.

Method for a beam problem

Method: beam problems

Draw a large diagram. Mark all forces: weights (at the centre of mass of each object), reactions or tensions at the supports, and any applied forces.
Take moments about a point where an unknown force acts, because its moment is then zero.
Write one moment equation and solve it.
Resolve vertically to find the other force.
Check by taking moments about a different point.

Worked example 3

A uniform beam \(AB\) of length 4 m and mass 20 kg rests on supports at \(A\) and \(B\). A 10 kg load is placed 1 m from \(A\). Find the reactions.

Moments about \(A\): \(R_B\times4=20g\times2+10g\times1=50g\), so \(R_B=12.5g=122.5\) N. Resolving vertically: \(R_A+R_B=30g\), so \(R_A=294-122.5=171.5\) N.

Worked example 4

A uniform rod \(AB\) of length 6 m and mass 12 kg rests on supports at \(C\), 1 m from \(A\), and \(D\), 1.5 m from \(B\). Find the reactions.

Moments about \(C\): \(R_D\times3.5=12g\times2\), so \(R_D=67.2\) N and \(R_C=12g-67.2=50.4\) N.

Worked example 5

A uniform rod of length 1.2 m and mass 3 kg hangs on vertical strings at 0.2 m from each end, with a 4 kg mass 0.3 m from \(A\). Find the tensions.

About the first string (at 0.2 m): \(T_2\times0.8=3g\times0.4+4g\times0.1=1.6g\), so \(T_2=2g=19.6\) N and \(T_1=7g-19.6=49\) N.

Tilting and limiting cases

Rule: about to tilt

When a beam is on the point of tilting about one support, the reaction at the other support is zero. Take moments about the support it is tilting about.

Worked example 6

A uniform plank of length 6 m and mass 20 kg rests on supports 1 m from each end. A man of mass 60 kg walks towards one end. How far can he go before it tilts?

Tilting about the nearer support starts when \(60g\times x=20g\times2\), where \(x\) is his distance beyond the support, so \(x=0.667\) m. He can go 0.667 m beyond the support, which is 0.333 m from the end.

Worked example 7

A uniform beam of weight 60 N and length 4 m is supported at one end and at a point 1 m from the other end. What load \(W\) at that end makes it tilt?

It tilts when \(W\times1=60\times1\), so \(W=60\) N.

Non-uniform rods and angled forces

Worked example 8

A non-uniform rod of mass 5 kg and length 2 m balances on a pivot 0.9 m from \(A\) when a 2 kg mass hangs at \(A\). Find the centre of mass.

Let the centre of mass be \(d\) from \(A\). Moments about the pivot: \(2g\times0.9=5g\times(d-0.9)\), so \(d=1.26\) m.

Worked example 9

A uniform rod of mass 5 kg and length 2 m is hinged at \(A\) and held horizontal by a string at \(B\) making \(40^\circ\) with the rod. Find the tension. A wheelbarrow carries 40 kg at 0.6 m from the axle; find the lift \(F\) at 1.4 m.

The perpendicular component of the tension is \(T\sin40^\circ\), so \(T\sin40^\circ\times2=5g\times1\) and \(T=38.1\) N. For the wheelbarrow \(1.4F=40g\times0.6\), so \(F=168\) N.

Watch out

Use the perpendicular distance, not the distance along a slanting line. A uniform rod's weight acts at the midpoint, not at the end. Include the mass of the rod itself. Do not mix clockwise and anticlockwise moments. In the limiting case the reaction at the support about to lift off is zero.

Note: the lever

Worked example 10: Taking moments about the unknown force

A uniform plank of length 3 m and mass 12 kg lies on two supports at its ends. A 20 kg load is 1 m from one end \(A\). Find the reaction at \(B\).

Take moments about \(A\) so that \(R_A\) has no moment: \(R_B\times3=12g\times1.5+20g\times1=38g\), giving \(R_B=12.67g=124.1\) N. Check by moments about \(B\): \(R_A\times3=12g\times1.5+20g\times2=58g\), \(R_A=19.33g=189.5\) N. Then \(R_A+R_B=313.6=32g\) as required.

Worked example 11: A moment with a force at an angle

A 50 N force acts on a door handle at an angle of \(40^\circ\) to the door, 0.9 m from the hinge. Find the moment about the hinge.

Only the component perpendicular to the door contributes: \(50\sin40^\circ=32.1\) N. Moment \(=32.1\times0.9=28.9\) N m. Equivalently, the perpendicular distance from the hinge to the line of the force is \(0.9\sin40^\circ=0.579\) m, and \(50\times0.579=28.9\) N m. Pushing at right angles (\(50\times0.9=45\) N m) is more effective.

Worked example 12: A beam supported at one point

A uniform beam of mass 10 kg and length 4 m is supported at a point 1 m from end \(A\). A mass \(m\) hangs at \(A\). Find \(m\) for balance.

The centre of the beam is 1 m on the other side of the support. Moments: \(m\times1=10\times1\), so \(m=10\) kg. The support force is \(20g=196\) N. If \(m=15\) kg, the beam would rotate with \(A\) going down.

Worked example 13: Two different pivots

A seesaw of length 4 m with masses 30 kg at one end and 50 kg at the other balances at a distance \(x\) from the 50 kg end. Find \(x\) for a light seesaw, and for a seesaw of mass 20 kg.

Light seesaw: \(50x=30(4-x)\), so \(80x=120\) and \(x=1.5\) m. With a uniform seesaw of mass 20 kg, the centre is 2 m from each end. Moments about the pivot: \(50x=30(4-x)+20(2-x)\), giving \(50x=160-50x\) and \(x=1.6\) m. The mass of the seesaw moves the pivot 0.1 m further from the 50 kg child.

Rule: how to take moments

Choose a direction (say anticlockwise) as positive. For each force write force \(\times\) perpendicular distance and give it the right sign. Choose the point about which you take moments so that as many unknown forces as possible pass through it. For a beam in equilibrium, the sum of all moments is zero about every point, so you can always check using a second point.

Watch out: common errors

Using the length along the slanting force instead of the perpendicular distance.
Forgetting the weight of the rod itself, or placing it at the end instead of the centre.
Assuming both supports carry equal loads when there is a load on the beam.
In a tilting problem, taking moments about the wrong support: use the support about which the beam is tilting.

Did you know?

Archimedes of Syracuse (about 287-212 BC) is reported to have said "Give me a place to stand and I will move the Earth", describing the power of the lever. His law of the lever, that two weights balance when their distances from the pivot are in inverse proportion to the weights, is the first example of the principle of moments that engineers still use to design cranes, bridges and playground equipment.

Otto’s tip

Take moments about the unknown force you do not want, then resolve vertically for the other one.

How to use this Year 13 moments revision notes

First, read the key points so that you know where the Year 13 moments revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the moments worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the moments test and mark it with the test answers.

Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 13 moments revision notes gives you everything you need for the chapter.