Statics of rigid bodies

Introduction• In this chapter you will revisit Moments from M1

• You will learn to use these alongside the resolving of forces

• You will be able to solve more complicated problems regarding rods and laminas in equilibrium

• As with previous chapters, you will also get more practice at algebraic questions!

Teachings for exercise 5A

Statics of rigid bodiesYou can calculate the moment of

a force acting on a body

The moment of a force acting on a body is the product of the

magnitude of the force and its perpendicular distance from the

point P.

Moments are measured in Newton-metres (Nm), and you should always state the direction, clockwise or anti-

clockwise

Find the sum of the moments about point P in the diagram shown.

5A

𝑀𝑜𝑚𝑒𝑛𝑡=𝐹𝑑

1.2m

0.8m

7N

5N

65°

(1)

(2)

P

Taking moments about P

(1)5×0.8¿ 4𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

(2)7×1.2𝑆𝑖𝑛65¿7.612…𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Moment (2) has a distance labelled which is not the perpendicular Draw a line from point P to the force which meets

it at a right angle Use trigonometry to find an expression for this

distance

1.2Sin65m

7.612…−4=3.612𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒Compare the sizes of the forces

acting clockwise and anti-clockwise

Statics of rigid bodiesYou can calculate the moment of a

force acting on a body

Find the sum of the moments about point P in the diagram shown to the right.

Mark on the perpendicular distances on first

(Big diagrams help with this!)

Calculate each moment individually

5A

6N

6N

4m

2m

(1)

P

40°

50°3N

4Sin40

2Sin50(2)(3)

Taking moments about P

(1)6×4𝑆𝑖𝑛 40¿15.43𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)6×2𝑆𝑖𝑛50¿9.19𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(3) The force of 3N passes through point P, and

hence it will not have any turning motion about P (perpendicular distance is 0) Imagine sitting in the middle of a seesaw!

15.43−9.19=6.23𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒Compare the sizes of the forces

acting clockwise and anti-clockwise

Statics of rigid bodiesYou can calculate the moment of a

force acting on a body

Find the sum of the moments about point P in the diagram shown to the right.

Mark on the perpendicular distances on first

(Big diagrams help with this!)

Calculate each moment individually

5A

6N

6N

4m

2m

(1)

P

40°

50°3N

4Sin40

2Sin50(2)(3)

Taking moments about P

(1)6×4𝑆𝑖𝑛 40¿15.43𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)6×2𝑆𝑖𝑛50¿9.19𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(3) The force of 3N passes through point P, and

hence it will not have any turning motion about P (perpendicular distance is 0) Imagine sitting in the middle of a seesaw!

15.43−9.19=6.23𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

An alternative method which you will sometimes use is to

make keep the distance diagonal and make the force

perpendicular instead!

Teachings for exercise 5b

Statics of rigid bodiesYou need to be able to solve problems about rigid bodies

that are resting in equilibrium

If a body is resting in equilibrium:

There is no resultant force in any direction, so the horizontal

and vertical forces sum to 0

The sum of moments about any point is 0 (the point does not have to be on the body itself)

5B

Statics of rigid bodiesYou need to be able to solve problems about rigid bodies

that are resting in equilibrium

A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A.

A light inextensible string is attached to the rod at a point C

where AC = 3m, and the point D, which is vertically above point A. If

the string is keeping the rod in equilibrium in a horizontal position and the angle between the string

and the rod is 40°, calculate:

a) The tension in the string

b) The magnitude and direction of the reaction at the hinge.

5B

A BC

D

3m 1m

T

40°H

V

6g

40° TSin40TCos40

Start with a diagram and label on all the forces – split the tension into horizontal and vertical components

At the hinge there will be a vertical reaction and a horizontal reaction

Take moments about point A (as we have 3 unknown forces, 2 will be eliminated by doing this!)

These must be equal as the rod is in equilibrium

(1)

(2)

2m

(1)6𝑔×2 ¿12𝑔𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)𝑇𝑆𝑖𝑛40×3¿3𝑇𝑆𝑖𝑛 40𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

3𝑇𝑆𝑖𝑛 40=12𝑔𝑇=

12𝑔3𝑆𝑖𝑛 40

𝑇=61𝑁

Divide by 3Sin40

Calculate

𝑇=61𝑁

Statics of rigid bodiesYou need to be able to solve problems about rigid bodies

that are resting in equilibrium

A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A.

A light inextensible string is attached to the rod at a point C

where AC = 3m, and the point D, which is vertically above point A. If

the string is keeping the rod in equilibrium in a horizontal position and the angle between the string

and the rod is 40°, calculate:

a) The tension in the string

b) The magnitude and direction of the reaction at the hinge.

5B

A BC

D

3m 1m

61

40°H

V

6g

40° 61Sin4061Cos40

𝑇=61𝑁

Resolve Horizontally (set left and right forces equal to each other)𝐻=61𝐶𝑜𝑠40

Resolve Vertically (set upwards and downwards forces equal to each other)

𝑉 +61𝑆𝑖𝑛 40=6𝑔

𝐻=46.72𝑁

𝑉=6𝑔−61𝑆𝑖𝑛 40𝑉=19.6𝑁

Calculate

Subtract 61Sin40

Calculate𝐻=46.72𝑁 𝑉=19.6𝑁

Statics of rigid bodiesYou need to be able to solve problems about rigid bodies

that are resting in equilibrium

A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A.

A light inextensible string is attached to the rod at a point C

where AC = 3m, and the point D, which is vertically above point A. If

the string is keeping the rod in equilibrium in a horizontal position and the angle between the string

and the rod is 40°, calculate:

a) The tension in the string

b) The magnitude and direction of the reaction at the hinge.

5B

A BC

D

3m 1m

61

40°H

V

6g

40° 61Sin4061Cos40

𝑇=61𝑁

𝐻=46.72𝑁 𝑉=19.6𝑁

H

V

A H

V

A

R

The resultant force will be somewhere between V and H Use a right-angled triangle to help

46.72

19.6

𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=√46.722+19.62𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁

Calculate𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁

R

Statics of rigid bodiesYou need to be able to solve problems about rigid bodies

that are resting in equilibrium

A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A.

A light inextensible string is attached to the rod at a point C

where AC = 3m, and the point D, which is vertically above point A. If

the string is keeping the rod in equilibrium in a horizontal position and the angle between the string

and the rod is 40°, calculate:

a) The tension in the string

b) The magnitude and direction of the reaction at the hinge.

5B

A BC

D

3m 1m

61

40°H

V

6g

40° 61Sin4061Cos40

𝑇=61𝑁

𝐻=46.72𝑁 𝑉=19.6𝑁

H

V

A H

V

A

R

You also need to calculate the angle above the horizontal

46.72

19.6

𝐴𝑛𝑔𝑙𝑒=𝑇𝑎𝑛−1( 19.646.72 )Calculate

𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁

R

𝐴𝑛𝑔𝑙𝑒=22.8°

θ

,22.8 °𝑎𝑏𝑜𝑣𝑒 h𝑡 𝑒h𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙

Statics of rigid bodiesYou need to be able to solve

problems about rigid bodies that are resting in equilibrium

A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in

equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the

diagram.

a) Find the magnitudes of the tensions in the two strings

b) Find the distance of the centre of mass from A

Let the centre of mass be a distance ‘x’ from A

5B

T1

T2

3g

x 5 - xA B

40° 55°T1Sin40

T1Cos40

T2Sin55

T2Cos55

Split the 2 tensions into horizontal and vertical components The horizontal and vertical components will each cancel themselves outResolve Horizontally𝑇 1𝐶𝑜𝑠 40=𝑇 2𝐶𝑜𝑠55

𝑇 1=𝑇2𝐶𝑜𝑠55  𝐶𝑜𝑠 40

Resolve Vertically𝑇 1𝑆𝑖𝑛40+𝑇2𝑆𝑖𝑛55=3𝑔𝑇 1𝑆𝑖𝑛40+𝑇2𝑆𝑖𝑛55=3𝑔𝑇 1=

𝑇2𝐶𝑜𝑠55  𝐶𝑜𝑠 40

Divide by Cos40

Statics of rigid bodiesYou need to be able to solve

problems about rigid bodies that are resting in equilibrium

A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in

equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the

diagram.

a) Find the magnitudes of the tensions in the two strings

b) Find the distance of the centre of mass from A

Let the centre of mass be a distance ‘x’ from A

5B

T1

T2

3g

x 5 - xA B

40° 55°T1Sin40

T1Cos40

T2Sin55

T2Cos55

𝑇 1𝑆𝑖𝑛40+𝑇2𝑆𝑖𝑛55=3𝑔𝑇 1=𝑇2𝐶𝑜𝑠55  𝐶𝑜𝑠 40

𝑇 1𝑆𝑖𝑛40+𝑇2𝑆𝑖𝑛55=3𝑔𝑇2𝐶𝑜𝑠55𝐶𝑜𝑠 40 𝑆𝑖𝑛 40+𝑇 2𝑆𝑖𝑛55=3𝑔

𝑇 2𝐶𝑜𝑠55𝑆𝑖𝑛 40+𝑇 2𝑆𝑖𝑛55𝐶𝑜𝑠 40=3𝑔𝐶𝑜𝑠 40

𝑇 2(𝐶𝑜𝑠55𝑆𝑖𝑛40+𝑆𝑖𝑛55𝐶𝑜𝑠 40)=3𝑔𝐶𝑜𝑠 40

𝑇 2=3𝑔𝐶𝑜𝑠 40  

𝐶𝑜𝑠55𝑆𝑖𝑛40+𝑆𝑖𝑛55𝐶𝑜𝑠 40Divide by the whole bracket!

𝑇 2=22.6𝑁Calculate

Sub in an expression for T1

Multiply all terms by Cos40

Factorise the left

𝑇 2=22.6𝑁

Statics of rigid bodiesYou need to be able to solve

problems about rigid bodies that are resting in equilibrium

A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in

equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the

diagram.

a) Find the magnitudes of the tensions in the two strings

b) Find the distance of the centre of mass from A

Let the centre of mass be a distance ‘x’ from A

5B

T1

T2

3g

x 5 - xA B

40° 55°T1Sin40

T1Cos40

T2Sin55

T2Cos55

𝑇 1𝑆𝑖𝑛40+𝑇2𝑆𝑖𝑛55=3𝑔𝑇 1=𝑇2𝐶𝑜𝑠55  𝐶𝑜𝑠 40

𝑇 2=22.6𝑁

𝑇 1=𝑇2𝐶𝑜𝑠55  𝐶𝑜𝑠 40 Sub in T2 (remember to

use the exact value!)𝑇 1=

22.6𝐶𝑜𝑠55  𝐶𝑜𝑠 40

𝑇 1=16.9𝑁Calculate

𝑇 1=16.9𝑁

Statics of rigid bodiesYou need to be able to solve

problems about rigid bodies that are resting in equilibrium

A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in

equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the

diagram.

a) Find the magnitudes of the tensions in the two strings

b) Find the distance of the centre of mass from A

Let the centre of mass be a distance ‘x’ from A

5B

T1

T2

3g

x 5 - xA B

40° 55°T1Sin40

T1Cos40

T2Sin55

T2Cos55

𝑇 2=22.6𝑁𝑇 1=16.9𝑁

Take moments about A to create an equation with x in

(1)

(2)

(1)(2)3𝑔× 𝑥¿3𝑔𝑥𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒𝑇 2𝑆𝑖𝑛55×5¿5𝑇 2𝑆𝑖𝑛55𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Remember that moment (2) is acting from the end of the rod!

There is no rotation so these forces must be equal

T2Sin55

3𝑔𝑥=5𝑇 2𝑆𝑖𝑛55

𝑥=5𝑇2𝑆𝑖𝑛55  

3𝑔𝑥=3.1𝑚

Divide by 3g

Calculate using the exact value for T2

So the centre of mass is 3.1m from

A

(2)

5m

Teachings for exercise 5C

Statics of rigid bodiesI have decided to miss out the part in exercise 5C (making a ‘triangle of forces’)

Pupils tend to find this method quite tricky and they can just use other methods anyway!

5C

Teachings for exercise 5D

15°40gSin15

Statics of rigid bodiesYou can solve problems about rigid

bodies resting in limiting equilibrium

If a body is on the point of moving it is said to be in limiting equilibrium. In this

case, the frictional force takes its maximum value, µR, where µ is the

coefficient of friction and R is the normal reaction.

A uniform rod AB of mass 40kg and length 10m rests with the end A on

rough horizontal ground. The rod rests against a smooth peg C where AC = 8m.

The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find:

a) The magnitude of the reaction at C

b) The coefficient of friction between the rod and the ground

5D

15°40g 40gCos15

R

F

N

5m

3m

2m

Draw a diagram and label all the forces: Weight, the normal reactions and friction. Split into components if needed The rod will have a tendency to slide downwards, with the

base moving to the left. Hence, friction will oppose this Taking moments about A will mean we can find the normal

reaction at the peg.Taking moments about A(1)(2)

(1)

(2)

40𝑔𝐶𝑜𝑠15×5¿200 𝑔𝐶𝑜𝑠15𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒𝑁×8¿8𝑁 𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

200𝑔𝐶𝑜𝑠 15=8𝑁25𝑔𝐶𝑜𝑠 15=𝑁237=𝑁

Divide by 8

Calculate

A

B

𝑁=237𝑁40gSin15 is NOT included as a moment about A. This is because it actually acts

down the rod and through point A (as opposed to the place where it has been

drawn), therefore it has a perpendicular distance of 0 and hence can be ignored…

15°237

15°40gSin15

Statics of rigid bodiesYou can solve problems about rigid

bodies resting in limiting equilibrium

If a body is on the point of moving it is said to be in limiting equilibrium. In this

case, the frictional force takes its maximum value, µR, where µ is the

coefficient of friction and R is the normal reaction.

A uniform rod AB of mass 40kg and length 10m rests with the end A on

rough horizontal ground. The rod rests against a smooth peg C where AC = 8m.

The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find:

a) The magnitude of the reaction at C

b) The coefficient of friction between the rod and the ground

5D

15°40g 40gCos15

R

FA

B

𝑁=237𝑁

Now you can resolve horizontally and vertically to find the remaining forces

You will need to split the normal reaction at the peg into horizontal and vertical components

The parallel and perpendicular components of the weight will no longer be needed…

237Cos15237Sin15

Resolving Horizontally

𝐹=237𝑆𝑖𝑛15Resolving Vertically

𝑅+237𝐶𝑜𝑠15=40𝑔𝑅=40𝑔−237𝐶𝑜𝑠15

𝐹=237𝑆𝑖𝑛15 𝑅=40𝑔−237𝐶𝑜𝑠15

Rearrange

15°237

Statics of rigid bodiesYou can solve problems about rigid

bodies resting in limiting equilibrium

If a body is on the point of moving it is said to be in limiting equilibrium. In this

case, the frictional force takes its maximum value, µR, where µ is the

coefficient of friction and R is the normal reaction.

A uniform rod AB of mass 40kg and length 10m rests with the end A on

rough horizontal ground. The rod rests against a smooth peg C where AC = 8m.

The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find:

a) The magnitude of the reaction at C

b) The coefficient of friction between the rod and the ground

5D

15°40g

R

FA

B

𝑁=237𝑁

As the rod is in limiting equilibrium, friction is at its maximum value

Use the formula for FMAX and sub in the values we have calculated

237Cos15237Sin15

𝐹=237𝑆𝑖𝑛15 𝑅=40𝑔−237𝐶𝑜𝑠15

𝐹𝑀𝐴𝑋=𝜇𝑅

237𝑆𝑖𝑛15=𝜇(40𝑔−237𝐶𝑜𝑠15)237𝑆𝑖𝑛15

40𝑔−237𝐶𝑜𝑠15=𝜇

Sub in values

Divide by the bracket

Calculate0.37=𝜇

60°

60°

60°2mg

Statics of rigid bodiesYou can solve problems about rigid bodies resting in limiting

equilibrium

A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall.

A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod

and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with

the ground.

Find the coefficient of friction between the ladder and the ground.

5D

A

B

C

RW

RG

F mg

mgCos60

Start with a diagram and label all forces – both masses should be split into parallel

and perpendicular components

We will now take moments about point A to give us the

value of RW

60°

RW Sin60

a

0.5a

1.5a

(1)

(2)

(3)

(1)2𝑚𝑔𝐶𝑜𝑠60×𝑎¿2𝑎𝑚𝑔𝐶𝑜𝑠60𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(2)𝑚𝑔𝐶𝑜𝑠 60×1.5𝑎¿1.5𝑎𝑚𝑔𝐶𝑜𝑠 60𝑁𝑚𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒(3)𝑅𝑊 𝑆𝑖𝑛60×3𝑎¿3 𝑎𝑅𝑊𝑆𝑖𝑛60𝑁𝑚𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

3𝑎 𝑅𝑊𝑆𝑖𝑛60=2𝑎𝑚𝑔𝐶𝑜𝑠 60+1.5𝑎𝑚𝑔𝐶𝑜𝑠603𝑅𝑊𝑆𝑖𝑛60=2𝑚𝑔𝐶𝑜𝑠60+1.5𝑚𝑔𝐶𝑜𝑠603𝑅𝑊𝑆𝑖𝑛60=3.5𝑚𝑔𝐶𝑜𝑠60

𝑅𝑊=3.5𝑚𝑔𝐶𝑜𝑠60  3𝑆𝑖𝑛60

𝑅𝑊=3.5𝑚𝑔𝐶𝑜𝑠60  3𝑆𝑖𝑛60 Cancel a’s

Group terms

Divide by 3Sin60

𝑅𝑊= 7𝑚𝑔√3  18

Calculate in terms of mg

2mgCos60

60°

60°

60°2mg

Statics of rigid bodiesYou can solve problems about rigid bodies resting in limiting

equilibrium

A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall.

A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod

and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with

the ground.

Find the coefficient of friction between the ladder and the ground.

5D

A

B

C

RW

RG

F mg

mgCos60

Now we can resolve horizontally and vertically

This will allow us to find expressions for RG and F, and

hence, the coefficient of friction

60°

RW Sin60

a

0.5a

1.5a

(1)

(2)

(3)

𝑅𝑊=7𝑚𝑔√3  18

2mgCos60

Resolving Horizontally𝐹=𝑅𝑊

𝐹=7𝑚𝑔√3  18

We already know RW so therefore also

know F!

𝐹=7𝑚𝑔√3  18

60°

60°

60°2mg

Statics of rigid bodiesYou can solve problems about rigid bodies resting in limiting

equilibrium

A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall.

A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod

and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with

the ground.

Find the coefficient of friction between the ladder and the ground.

5D

A

B

C

RW

RG

F mg

mgCos60

Now we can resolve horizontally and vertically

This will allow us to find expressions for RG and F, and

hence, the coefficient of friction

60°

RW Sin60

a

0.5a

1.5a

(1)

(2)

(3)

𝑅𝑊=7𝑚𝑔√3  18

2mgCos60

Resolving Vertically𝑅𝐺=2𝑚𝑔+𝑚𝑔

Simplify

𝐹=7𝑚𝑔√3  18

𝑅𝐺=3𝑚𝑔

𝑅𝐺=3𝑚𝑔

60°

60°

60°2mg

Statics of rigid bodiesYou can solve problems about rigid bodies resting in limiting

equilibrium

A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall.

A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod

and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with

the ground.

Find the coefficient of friction between the ladder and the ground.

5D

A

B

C

RW

RG

F mg

mgCos60

As the ladder is in limiting equilibrium, we can use the

formula for friction

60°

RW Sin60

a

0.5a

1.5a

(1)

(2)

(3)

𝑅𝑊=7𝑚𝑔√3  18

2mgCos60

𝐹=7𝑚𝑔√3  18

𝑅𝐺=3𝑚𝑔

𝐹𝑀𝐴𝑋=𝜇𝑅7𝑚𝑔√3  18

=3𝑚𝑔𝜇

7√3  18

=3𝜇

7√3  54

=𝜇

0.22=𝜇

Sub in values

Cancel mg’s

Divide by 3

Calculate

Summary• You have had a reminder of moments

• You have seen how these can be used alongside your other skills in resolving forces

• You have solved problems involving laminas and rods, and seen some very tricky algebraic questions!