M1: Chapter 5Moments

Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Last modified: 2nd April 2014

Learning Objectives:• Understand what is meant by the moment of a force.• Use moments to solve problems involving unknown

lengths, forces and masses.

Intro

This is a door

Why do you think the handle is put on the other side of the door from the hinge?

Increasing the distance of the force applied from the point of rotation increases the

‘turning effect’ of the force.?

If I double the distance of my finger from the hinge, what happens to the force

required to keep the door open?

As the distance doubles, the force required halves (we’ll see why).?

Moments

The ‘moment’ of a force…… measures the turning effect of the force on the body on which it is acting.

about a point perpendicular distance

10m

m = 70kg

Moment ?

?!

Moments - Examples

𝐹=20𝑁

10𝑚

𝑃

1

Moment of force F about point P clockwise

𝐹=20𝑁

𝑃

2

Moment of force F about point P

There is no ‘turning’ effect.

10𝑁

5𝑚𝑃

3

Moment of force F about point P

clockwise

60 °5 𝑠𝑖𝑛60

? ? ?

We see what force is acting perpendicularly.

10𝑁

2𝑚𝑃

4

Moment:

30 °

?

5𝑁3𝑚

𝑃

5

Moment:

65 °

?

Summing momentsWe can also find the overall moment by summing them – just treat one of the directions (clockwise or anticlockwise) as negative.

20𝑁

10𝑚

𝑃

1

Sum of moments (about )

clockwise?

10𝑁15𝑚

2

2𝑚 1𝑚 1𝑚𝑃

5𝑁

4𝑁

3𝑁

Sum of moments (about ):

?3 1𝑚 3𝑚

𝑚=50𝑘𝑔 𝑚=30𝑘𝑔

Sum of moments (about ):

?4 𝑥4𝑚 𝑚=30𝑘𝑔

Sum of moments (about ):

?

(Rod is light)

𝑃

𝑃

5𝑁

This whole chapter in a nutshell…

10m

m = 70kg

! If a rigid body is in equilibrium then:

The resultant force in any direction is 0.The sum of the moments about any point is 0.

ab

2m

i.e. Forces up = forces down

i.e. Clockwise moments = anticlockwise moments

Example

10m

m = 70kg m = ?

2m

Lewis and Tom are on a uniform seesaw of mass 20kg, where the pivot is not central. Lewis weighs 70kg and is 10m from the pivot. Tom is 2m from the pivot. The seesaw is horizontal.

Determine the reaction force at the pivot of the seesaw.Determine how heavy Tom is.

Bro Tip: First draw on forces. Click to Brosketch

LewisTom

70𝑔 𝑚𝑔20𝑔

𝑅

Method 1:Equating moments about pivot:

Equating forces in vertical direction,

Method 2:Equating moments about Tom:

Equating forces in vertical direction:? ?

Hint: Choose a suitable point to take moments about.

Moments about which point?

10m

m = 70kg m = ?

2mLewis

Tom

70𝑔 𝑚𝑔20𝑔

𝑅

If we take moments about Tom:We don’t have to worry about Tom’s mass because Tom’s weight doesn’t contribute to the moment about Tom.

Key Point: You can choose different points to take your moments about, depending on which variable we wish to ‘ignore’.

If we take moments about the pivot:We don’t have to worry about since it doesn’t have any moment about the pivot.?

?

Test Your UnderstandingA uniform rod of mass 20kg is supported at each end so that it is horizontal.The bar is 4m long and a mass of 10kg is placed 1m from one end.Find the magnitude of the reactions at each support.

𝑅1 𝑅2

10𝑔20𝑔Moments about second support:

Similarly taking moments about first:? Working

4𝑚2𝑚

1𝑚

? Diagram

Q

Finding centre of massA non-uniform rod of length 4m and mass 5kg is in equilibrium in horizontal position resting on two supports at C and D where and . The magnitude of the reaction at is twice the magnitude of the reaction at . Find the distance of the centre of mass of the rod from .

𝐶

5𝑔Let centre of mass be metres from .Forces up = forces down:

Moments about : ? Working

2𝑅

𝐷

𝑅

𝐴 𝐵

1𝑚2𝑚 2𝑚𝑥

𝐶? Diagram

Q

Exam Questions

On provided hand-out.(Solutions on following slides)

Answers

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Answers

? ?

Hint: When the beam is on the verge of tilting about P, there is no reaction force at Q. ?

Answers

? ?

Answers

?

Answers

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Answers

??

SummaryMoment of a force

What are the two crucial things to consider for systems in equilibrium?1. Forces up = Forces down2. Moments clockwise = Moment anticlockwise

Where tends to be best to take moments about?• When there is an unknown distance:

Sometimes about the point where the distance is measured from (but if in doubt, take from the end).

• When there are two unknown forces:About the point where one of the forces is acting, so that we don’t consider this force in our moments equation and thus have less unknowns.

How do we deal with rods with unknown centres of mass?Create a distance say for the centre of mass. Usually best to find moments about the end of the rod.

If a rod is on two supports P and Q, and is on the verge of tilting about P?There is no reaction force at Q.

?

?

?

?

?

? C LO: Understand what is meant by the moment of a force.

C LO: Solve problems involving moments.