04

Physics Factsheet

September 2000

Number 04

Moments & Equilibrium A force may have a variety of effects on an object; these include changing its velocity, changing its shape or causing it to turn. This Factsheet will: • Explain how forces have a turning effect (called a moment) and work through the calculations involved. • Examine what happens when a body is acted upon by many forces, but does not move - this is called being in equilibrium. In this Factsheet, the examples considered will mainly involve the use of moments; cases involving more complicated resolving will be dealt with later.

Turning Forces

Exam Hint : Sometimes there is more than one way to measure the distance. Don't worry - it will work whichever way you do it!

Imagine a perfectly balanced see-saw. If you go and push on one end of it, the see-saw turns about its pivot (one end goes up and the other down) - the force you exert on the see-saw is having a turning effect. The size of this turning effect depends on two things: • how hard you push - the harder you push, the more it turns.

Calculating moments To avoid always having to resolve to find the perpendicular component, you may like to use the following to save time:

•

where you push - if you push near the middle, it has less effect than if you push right at the end. When the see-saw turns, it turns about the support at its middle - so we are interested in the turning effect of forces about this point, which is called the axis. It can turn clockwise or anticlockwise about the axis.

F θ d If we have the situation as shown in the diagram, the perpendicular component of F will be Fsinθ. (You should check this by resolving). So we get: axis

Now we will look at the direction you push in. The diagrams below show three different directions you could push in.

axis In the first diagram, you would not manage to turn the see-saw at all - all you might manage to do is to move the whole thing along! In the second diagram, you would manage to turn it, but your push would not be that effective, since only part of the force you were exerting woud be going to turn the see-saw. In the third diagram, all your effort would be going to turn the see-saw - this is the way you would push if you just wanted to turn it!

Moment = Fdsinθ θ F: d:

θ: d

θ

F

Magnitude of force (newtons) Distance away from the axis that the force is applied (metres) Angle between force and line distance is measured along

Exam Hint : If you are going to use the above formula, you must be very careful to ensure that you use the correct angle for θ. Do not just put in any angle give you in the question - you may have to use angle facts to work out the correct one (see Factsheet 07 - Maths - Geometry).

In each case, the part of the force that is having a turning effect is its component at right angles (or its perpendicular component) to the see-saw. (See Factsheet 02 - Vectors and Forces - for more details on components). In the first diagram, there is no component at right angles to the see-saw - so there is no turning effect. In the third diagram, the entire force is at right angles to the see-saw - so we get maximum turning effect.

• •

We can now find a formula for the turning effect (called a moment or a torque) of any force about any axis. From above, we know it depends on distance from the axis and the perpendicular component of the force:

The unit of moment is Nm. Moment is a vector, as it can tend to rotate the body clockwise or anti-clockwise about the given point. Be sure to always state its direction.

Typical Exam Question

axis

Calculate the moment of the force about the axis shown below [3]

Moment = F⊥ × d F⊥: Perpendicular component of force (in newtons) d: Distance away from the axis that perpendicular the force is applied (in metres).

dis tan ce

Axis 30o

1.5m

component of force

force

10 N y

Note that the angle we are given is not the angle between the force and the line the distance is measured along - so we cannot just put it into the formula! The angle we want is angle y. By using "angles in a triangle = 180o", we get y = 180o−90o − 30o = 60o So moment = Fdsinθ = 10 × 1.5 × sin60o 4 = 13Nm4 anti-clockwise4

If you are required to define the moment of a force, use this: "moment = force (N) × perpendicular distance of the line of action of the force from the axis" This also gives an alternative way to work out moments - use either!

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Physics Factsheet

Moments & Equilibrium Couples

Equilibrium

The diagram below shows an example of a couple acting on an object - two forces, of the same size, acting in different places in opposite directions.

A F = 3N

B

A body that is in equilibrium neither accelerates along a straight line nor changes its rate of rotation. A system of forces in equilibrium satisfies both the two conditions for equilibrium of co-planar forces: (1) total forces = 0 (2) total moments = 0

F = 3N C

0.3m •

The total moment of a couple is always the same, whatever axis you measure it about! To see this, we will look at what happens when we take points A, B and C as the axis in turn: Point A:One force has no moment about A, since d = 0 The other force has moment 0.3 × 3 = 0.9 Nm anticlockwise So total moment = 0.9Nm anticlockwise

The first condition: Total Forces = 0, requires there to be no resultant (i.e total) force acting on the body. This comes from Newton’s First Law, which states that: for a body to remain stationary or to move with a constant velocity (i.e. with no linear acceleration) when several forces act on it, then the total force acting on the body must be zero (i.e. the forces ‘cancel each other’). The total, or resultant, force acting on a body can be shown to be zero in two ways. You may use either method: (1) The Polygon of Forces Method: a tip-to-tail vector diagram of all the forces acting on the object will form a closed polygon; or (2) Resolving Forces into Components: all the forces are resolved into perpendicular components, which are then separately equated to give zero. Both of these methods are covered in more detail in the Factsheets "Vectors" and "Solving Mechanics Problems".

Point B: Each force has moment 0.15 × 3=0.45 Nm anticlockwise about B So total moment = 0.9 Nm anticlockwise Point C: One force has no moment about C, since d = 0 The other force has moment 0.3 × 3 = 0.9 Nm anticlockwise So total moment = 0.9Nm anticlockwise •

A couple consists of: • two forces • with equal magnitudes (i.e. sizes) • acting in opposite directions • on the same body • along different lines of action The moment of a couple is F × d where F = magnitude of each force d = perpendicular distance between them

The second condition: Total Moments = 0, requires there to be no resultant turning force about any axis. This comes from the Principle of Moments, which states: If a body is in equilibrium, then the total clockwise moment about any axis is equal to the total anticlockwise moment about the same axis. Exam Hint : You can see immediately that Total Moments = 0 if all the forces acing on a body pass through a common point. To see that all the forces pass through a common point, draw on their lines of action, which will cross at a point.

Typical Exam Question

Answers

(a) Define what is meant by the ‘moment of a force’. Is it a vector or a scalar? [4]

(a) The moment of a force is the turning effect of a force about an axis and is given by: Moment = Force(N)4 × Perpendicular 4distance of the line of action of the force from the axis 4(m). The moment of a force is a vector 4, with direction given as clockwise, or anti-clockwise.

(b) The figure below shows a spanner in two positions. 15cm

(b) (i) Moment = Fd = 10 × 0.15 4= 1.5 Nm 4 anticlockwise 4 pivot

d (ii) Moment = Fdsinθ θ = angle between force and spanner o o o o o = 180 - 90 - 60 = 30 (angles on a straight line add to 180 ) o moment = 10 × 0.15 × sin30 4 = 0.75 Nm anticlockwise 4

60o 10N

(iii) Moment = Fd = 1.5 Nm anticlockwise 4

10N Calculate the moment about the pivot for the force acting on the spanner in the figure when (i) the spanner is horizontal [3]

(iv) The moment in (ii) is less than that in (iii) 4 as not all the force acts to turn the spanner in (ii), i.e. not all of it is acting at 90o to the spanner, as it is in (iii). 4

(ii) the spanner is at 60o below the horizontal, with the force acting vertically downwards as shown in the figure [2]

Exam Hint : It may be helpful to draw a simplified sketch diagram showing just forces, angles, distances and the pivot. This can make it easier to see what is going on.

(iii) the spanner is at 60o below the horizontal, with the force acting at right angles to the spanner handle. [1] (iv) Describe and explain any differences between your answers to (ii) and (iii). [2]

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Moments & Equilibrium

Physics Factsheet

Using Equilibrium Exam Workshop This is a typical poor student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answer is given below.

If we know that a body is in equilibrium, we can use that fact to find unknown forces by using the two conditions for equilibrium. The strategy is: 1 Draw a clear diagram, putting in all forces, distances and angles. Use letters for unknown forces.

(a) Define the Principle of Moments [4] Total 4moments 4= 0

2 Choose two perpendicular directions and resolve in these directions. (In the examples in this Factsheet, the directions will always be horizontal and vertical). Use "total force = 0" in each direction - or equivalently "total up = total down" and " total left = total right".

• •

3 Choose an axis and take moments. Use "clockwise moments = anticlockwise moments".

Not enough detail for 4 marks Must state this is only true in equilibrium

(b) A vertical force is applied to keep the 4kg lid of a skip open at an angle of 40°°. The lid is 1m long and is uniform. (take g = 9.8ms-2) Draw a free body diagram for the skip lid, clearly labelling each force. [6] force

Exam Hint • A force has no moment about any point on its line of action. • In more complicated equilibrium calculations it is always easiest if you choose your axis to be at a point that an unknown force that you don't need to find passes through. As the distance of the force to the axis is zero, its moment will also be zero and this will simplify matters by reducing the number of unknowns (see the following example).

1m

weight

40o

O

Typical Exam Question A uniform beam of mass 20kg is in equilibrium, as shown in the diagram. Find the mass, m. (Take g = 10ms-2) [4] 4m

1m

• •

1m

• •

m

60 kg

Draw on the forces acting on the beam (don’t forget the reaction force at the pivot or the weight of the beam). R 4m 1m

(c) By taking moments, or otherwise, calculate the force (F) required to keep the lid in this position. [5]

2m

60 kg

600 N

Take moments about O: 4 F(1.0sin40o)6 = W(0.5cos40o)4 F = 4(10)6(0.5cos40o)/(1sin40o) = 24N (2 sig.figs.) 6

m

200 N •

Take moments about the pivot point so as to eliminate the unwanted force R. Anti-clockwise moment = 600(1) = 600 Nm 4 Clockwise moments = 200(2) + 4m(10) 4

•

Principle of Moments: 600 = 400 + 40m 4 ⇒ 200 = 40m ⇒ m = 5kg 4

What? •

The centre of mass of a body is the point at which all the weight of the body can be considered to act. If the body is ‘uniform’ (i.e. it has the same density throughout) then this is at the object’s geometric centre of symmetry (see examples below). •

Trigonometry is wrong for the component of F perpendcular to the lid. g has been taken as 10ms-2, where the question asked for 9.8ms-2. Read the Question!

(d) From your answer in (c), calculate the Reaction Force at the pivot of the skip lid. [2]

Centre of Mass In the above example, we used the concept of centre of mass when we put theweight of the beam acting at the middle of it.

•

Force upwards is OK 44 Weight is downwards 4, but not shown to be at the midpoint of the lid 6 The reaction force at the pivot is absent 66. This diagram cannot be in equilibrium, as the forces have different lines of action – always have a quick ‘common sense’ check once you’ve answered a question!

• •

‘Oh no! I’ve got it all wrong!’. If this is how you feel at this point - don’t panic (here we have 7 out of 17 marks already). Re-read the question and try to see where you’ve gone wrong If you can’t see your mistake, move on to the next question and return to this one when you have completed the exam.

• See overleaf for the examiner's answers • = centre of mass

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Physics Factsheet

Moments & Equilibrium

Quantitative Test

Examiner’s Answer

Time for test 25 min (Total marks: 21)

(a) For a body in equilibrium 4, the total 4clockwise moments equal 4 the total anti-clockwise moments 4.

1. Determine the magnitude and direction of the turning forces about O produced in the diagrams. [6] 2m O (i)

(b) There are three forces: force

8N 0.5 m

6m

(ii) 2m O

0.5 m weight

reaction

30 N

40o

O F: force applied to keep the lid open at the end of the lid 4, acting vertically upwards 4 W: weight of the lid. This acts from the centre of the lid 4, as it is uniform, and is vertically downwards 4 R: The reaction force at the pivot of the skip lid 4. This must act vertically upwards 4, if the lid is to be in equilibrium.

2. A rod OA of length 1.3m is hinged at O and a force F, of magnitude 10N is applied successively in the directions 1,2,3 and 4 as shown. State the magnitude of the torque F about O in each case. [7] 1 2 1.3m

30 o

O

this angle is 50o (vertically opposite angles are equal)

3

A

(c) As we do not know anything about the reaction force acting at the pivot O, by working out the moments about O we eliminate this reaction force 4. Then by using the Principle of Moments, we can find F.

4

3. If the beam is in balance; (i) draw a free body diagram for the beam; (ii) determine the mass m; and (iii) determine the reaction force at the pivot. (take g=10ms-2) [8]

F

0.8 m

this angle is 50o (angles in a triangle add to 180o)

8 kg

this angle is 50o (angles in a triangle add to 180o) W

m

1m 1 kg

Hint: use the condition total moments = 0, then total forces = 0)

R 40o

0.4 m

O

Quantitative Test Answers 1. (i) T=Fd=8(2) = 16Nm 4anti-clockwise4; (ii) Torque = 30(2)4 –20(6) 4clockwise = - 60Nm clockwise = 60Nm anti-clockwise44

Moment of weight = mg × sin50°4 = 15.0Nm4, Moment of F = F × 1 × sin50° = 0.77F4 Total clockwise moment = total anticlockwise moment: 15 = 0.77F So F = 15/0.77 = 20N (2 SF)4

2. 1 – moment =10 × 1.3 = 13Nm 4anti-clockwise 4; 2 – Angle between force and OA = 60o So moment = 10 × 1.3 × sin60o 4= 11.3Nm 4anticlockwise 3 – The line of action passes through O, so there is no moment4; 4 – Angle between force and OA = 60o as for 2. But it is in the opposite direction, so moment = 11.3Nm 4clockwise 4

(d) Applying the second condition for equilibrium, that total forces = 0, we resolve vertically: F – W + R = 0 4; R = W – F = 4(9.8) – 20 = 19N (2 SF.)4. Exam Hint • If you are unsure about how to answer a question, look back to see what you have just been asked: usually this leads straight on to the next part of the question. • Before answering a question, have a quick look at all parts of the question, as some later parts can help you answer earlier parts of the question.

3. (i) One mark for each correctly placed, directed and labelled force.444 R 0.8 m 0.4 m 1m m 80 N

Acknowledgements: This Physics Factsheet was researched and written by Thane Gilmour. The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

10m N

10 N

(ii) take moments about unknown reaction at the pivot: 80(0.8) = 10m(0.4) + 10(1.4) 4⇒ m = 12.5 kg 4 (iii) Use the condition that total forces = 0 in equilibrium: R = 80 + 125 + 10 4= 215N 4

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