moments

MOMENTS

Introduction

• This chapter you will learn about moments

• Moments can be described as turning forces – rather than pushing an object along, they turn it round

• You will learn how to calculate the moment of a force on a pivot point

• You will learn how to calculate moments in rods that are in equilibrium

TEACHINGS FOR EXERCISE 5A

6N

6N

6N6N

MomentsYou can find the moment of a force acting

on a body

Up until this point you have learnt about forces pushing or pulling a particle in a particular

direction

The particle does not turn round, it just moves in a direction, based on the sum of the forces

For moments, we replace the particle with a straight rod (often called a lamina)

Imagine the rod had a fixed ‘pivot point’

A force acting on the rod at the centre, beneath the pivot point, will not cause it to move

If the force is moved to the side however, the rod will rotate around the pivot point

A greater force will cause the turning speed to be faster

If the force is further from the pivot point, the turning speed will be faster as well…

5A

5N

4N

7gN

6N 6N

Balancing Act Applet

http://phet.colorado.edu/en/simulation/balancing-act


MomentsYou can find the moment of a force

acting on a body

The turning motion caused by a force is dependant on:

The magnitude of the force A bigger force causes more turn

The distance the force is from the pivot point

A bigger distance causes more turn

(For example, the further you push a door from the hinge, the less effort is required to

close it.)

To calculate the total moment about a point:

Moment about a point = Force x Perpendicular distance

5A

5N

3m

C

𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡𝐶¿5𝑁×3𝑚¿15𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Moments are measured in Newton-metres You must always include the direction of

the moment(either clockwise or anticlockwise)

The distance must always be perpendicular from the pivot to the force

itself…





acting on a body






close it.)



5A

4N

2m

F

Calculate the moment of the force about point F

𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡 𝐹¿ 4𝑁×2𝑚¿8𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒





acting on a body






close it.)



5A

9N4m

A

Calculate the moment of the force about point A Draw a triangle to find the perpendicular

distance!

𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡 𝐴¿9𝑁×4𝑆𝑖𝑛30𝑚¿18𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

30°4Sin30




TEACHINGS FOR EXERCISE 5B

MomentsYou can find the sum of the moment of a set of forces

acting on a body

Sometimes you will have a number of moments acting around a single

point.

You need to calculate each one individually and then choose a

positive direction

Adding the forces together will then give the overall magnitude

and direction of movement

5B

P

5N 3N

4N

2m 1m 1m

Calculate the sum of the moments acting about the point P Start by calculating each moment individually (it might be useful to label them!)

(1) (3)

(2)

(1)5𝑁×3𝑚¿15𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)4𝑁×1𝑚¿ 4𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(3)3𝑁×1𝑚¿3𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Choosing clockwise as the positive direction…

15𝑁𝑚−4𝑁𝑚−3𝑁𝑚¿8𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

If we had chosen anticlockwise as the positive direction our

answer would have been -8Nm anticlockwise

This is just 8Nm clockwise (the same!)




MomentsYou can find the sum of the moment of a set of forces

acting on a body

Sometimes you will have a number of moments acting around a single

point.

You need to calculate each one individually and then choose a

positive direction

Adding the forces together will then give the overall magnitude

and direction of movement

5B

P

5N

5N

2m

4m

(1)

(2)

Calculate the sum of the moments acting about the point P Start by calculating each moment individually (it might be useful to label them!)(1)5𝑁×2𝑚¿10𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)5𝑁×4𝑚¿20𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Choosing anticlockwise as the positive direction…

20𝑁𝑚−10𝑁𝑚¿10𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒




TEACHINGS FOR EXERCISE 5C

MomentsYou can solve problems about

bodies resting in equilibrium by equating the clockwise and

anticlockwise moments

When a rigid body is in equilibrium, the resultant force in any direction

is 0

The moments about any point on the object will also sum to 0

5C

Y

10N 10N

4m 4m

(1) (2)

Calculate the sum of the moments acting about the point Y Calculate each moment separately

(1)10𝑁×4𝑚¿ 40𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)10𝑁×4𝑚¿ 40𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒As the moments are equal in both directions, the rod will not turn and hence, is in equilibrium!

As the rod is fixed at Y is will not be lifted up by the forces either!








is 0


5C

Z

3N1N

2m 6m

(1)(2)

Calculate the sum of the moments acting about the point Z Calculate each moment separately

(1)3𝑁×2𝑚¿6𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)1𝑁×6𝑚¿6𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒As the moments are equal in both directions, the rod will not turn and hence, is in equilibrium!








is 0


The diagram to the right shows a uniform rod of length 3m and weight

20N resting horizontally on supports at A and C, where AC = 2m.

Calculate the magnitude of the normal reaction at both of the supports

5C

AC

B2m 1m

RA RC

20N

1.5m 0.5m

As the rod is in equilibrium, the total normal reaction (spread across both supports) is equal to 20N (the total downward force)𝑅𝐴+𝑅𝐶=20Take moments about C (you do not need to include RC as its distance is 0)

(1)

(2)

(1)

(2)

2×𝑅𝐴¿2𝑅𝐴𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒0 .5×20¿10𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

2𝑅𝐴=10𝑅𝐴=5𝑁𝑅𝐶=15𝑁

The clockwise and anticlockwise moments must be equal for equilibrium

Divide by 2

Use the original equation to calculate

RC

“Uniform rod” = weight is in the centre

This makes sense – as RC is closer to the centre of mass

is bearing more of the object’s weight!







A uniform beam, AB, of mass 40kg and length 5m, rests horizontally on

supports at C and D where AC = DB = 1m.

When a man of mass 80kg stands on the beam at E, the magnitude of the

reaction at D is double the reaction at C.

By modelling the beam as a rod and the man as a particle, find the distance AE.

5C

AD

B1m 1m

RC RD

1.5m

40g80g“Uniform beam” = weight is in the centre

C E

2RC

The normal reactions must equal the total downward force3𝑅𝐶=120𝑔𝑅𝐶=40𝑔

Divide by 3

𝑅𝐷=80𝑔 RD is double this

As the reaction at D is bigger,

the man must be closer to D than

C

40g 80g

1.5m







A uniform beam, AB, of mass 40kg and length 5m, rests horizontally on

supports at C and D where AC = DB = 1m.

When a man of mass 80kg stands on the beam at E, the magnitude of the

reaction at D is double the reaction at C.

By modelling the beam as a rod and the man as a particle, find the distance AE.

5C

AD

B1m 1m1.5m

40g80g(2) (3)

C E

(4)

(1) 40g 80g

Let us call the required distance x (from A to E) Take moments about A(we could do this around any point, but this will make the algebra easier)

x

(2)

(3)(4)

(1)1×40𝑔¿ 40𝑔𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒2.5×40𝑔¿100𝑔𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒𝑥×80𝑔¿80 𝑥𝑔𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒4×80𝑔¿320𝑔𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Equilibrium so anticlockwise = clockwise40𝑔+320𝑔¿100𝑔+80 𝑥𝑔

¿100𝑔+80 𝑥𝑔360𝑔¿100+80 𝑥360

3 .25=𝑥

1.5m

Group terms

Cancel g’s

Calculate

So the man should stand 3.25m from

A!




MomentsYou can solve problems about bodies

resting in equilibrium by equating the clockwise and anticlockwise

moments

A uniform rod of length 4m and mass 12kg is resting in a horizontal position on

supports at C and D, with AC = DB = 0.5m

When a particle of mass mkg is placed on the rod at point B, the rod is on the point

of turning about D.

Find the value of m.

If the rod is on the point of turning around D, then there will be no

reaction at C RC = 0

(the rod is effectively hovering above support C, about to move upwards as it

rotates round D) 5C

AD

B0.5m 0.5m1.5m

12g mg

C

RC RD

1.5m

0

Taking moments about D

(1)

(2)

(3)

(1)(2)(3)

h𝑇 𝑖𝑠𝑤𝑖𝑙𝑙𝑏𝑒0𝑎𝑠 𝑅𝐶=01.5×12𝑔¿18𝑔𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒0.5×𝑚𝑔¿0.5𝑚𝑔𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Although it is on the point of turning, the rod is still in equilibrium Anticlockwise = clockwise

18𝑔=¿0.5𝑚𝑔18=¿0.5𝑚36=¿𝑚

Cancel g’s

Multiply by 2

The mass is 36kg More than this and

the rod will turn about D

Less than this and some of the normal reaction will be at C




TEACHINGS FOR EXERCISE 5D

MomentsYou can solve problems about non-uniform bodies by finding or using

the centre of mass

The mass of a non-uniform body can be modelled as acting at its centre of mass

This means the weight of the rod may not necessarily be in the centre as it

has been so far

Sam and Tamsin are sitting on a non-uniform plank AB of mass 25kg and length

4m.

The plank is pivoted at M, the midpoint of AB, and the centre of mass is at C where

AC = 1.8m.

Tamsin has mass 25kg and sits at A. Sam has mass 35kg. How far should Sam sit

from A to balance the plank?

5D

A B

25g25g 35g

C M

RM

Let Sam sit ‘x’ m from the midpoint

Take moments about M (this way we don’t need to know RM)

1.8m 0.2m x

(1)

(2)

(3)

(1)

(2)

(3)

2×25𝑔¿50𝑔𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒0. ¿5𝑔𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

𝑥×35𝑔¿35𝑔𝑥𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒The rod is in equilibrium so anticlockwise = clockwise

50𝑔+5𝑔=¿35𝑔𝑥55𝑔=¿35𝑔𝑥55=¿35 𝑥1.57=¿𝑥

Group terms

Cancel g’s

Divide by 35

Sam should sit 3.57m from A (or 0.43m from B) Make sure you always read where the distance should be measured from!


non-uniform bodies by finding or using the centre of mass

A rod AB is 3m long and has weight 20N. It is in a horizontal position

resting on supports at points C and D, where AC = 1m and AD = 2.5m.

The magnitude of the reaction at C is three times the magnitude of the

reaction at D.

Find the distance of the centre of mass of the rod from A.

5D

C D

1m

1.5m

0.5m

RC RD

A B

20N

RC = 3RD

Estimate where the centre of mass is on your diagramWe can replace RC with 3RD

Now find the normal reactions

4𝑅𝐷=20𝑅𝐷=5

3RD 5N15N

Divide by 4

𝑅𝐶=15


non-uniform bodies by finding or using the centre of mass

A rod AB is 3m long and has weight 20N. It is in a horizontal position

resting on supports at points C and D, where AC = 1m and AD = 2.5m.

The magnitude of the reaction at C is three times the magnitude of the

reaction at D.

Find the distance of the centre of mass of the rod from A.

5D

C D

1m

1.5m

0.5m

A B

20N

5N15N

Now take moments about A, calling the required distance ‘x’(You’ll find it is usually easiest to do this from the end of the rod!)

(1)

(3)

(2)

x

(1)

(2)

(3)

1×15¿15𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒𝑥×20¿20 𝑥𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒2.5×5¿12.5𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Equilibrium so anticlockwise = clockwise

15+12.5=¿20 𝑥27.5=¿20 𝑥1.38=¿𝑥

Group terms

Calculate

The centre of mass is 1.38m from A

Summary

• We have learnt that moments are turning forces

• We have learnt how to solve problems involving several moments

• We have seen how to solve problems involving rods being in equilibrium

• We have also seen how to find the centre of mass if a rod is non-uniform

moments

Education

bigger force

greater force

fixed pivot point6n6na

total moment

point f

bigger distance

moments moments

turning motion