1. Moments

2. Moments Starter: KUS objectives
BAT understand moment and solve problems using moments on a Uniform rod
Starter:

3. Moments
The moment of a force measures the turning effect of the force on the body on which it is acting
The moment of a force F about a point P is the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the point P
Moment of F about P = Fd clockwise
The magnitude of the force is measured in newtons (N) and the distance is measured in metres (m), so the moment of the force is measured in newton-metres (Nm)
Moment of F about P = 15 Nm anticlockwise

4. Using Trigonometry
You may need to use trigonometry to find the perpendicular distance
Moment of F about P = clockwise
e.g. Moment of the force about P =
= clockwise

5. Moment on a uniform beam / rod
WB1 A light rod AB is 4 m long and can rotate in a vertical plane about fixed point C where AC = 1 m. A vertical force F of 8 N acts on the rod downwards.
Find the moment of F about C when F acts a) at A b) at B c) at C
a) Taking moments about C acw
1 m
3 m A C B
b) Taking moments about C cw
c) Taking moments about C cw

6. Sum of Moments You can find the sum of the moments about a point
When doing this, choose whether to take clockwise or anticlockwise moments
WB2 Find the sum of the clockwise moments about P
5 N
3 N
2 m
1 m
1 m
4 N
What will happen here?

7. Moments
Equilibrium

8. Sum of Moments: Equilibrium
If an object is in Equilibrium then it will not turn about any point.
This means the sum of moments about any point must be zero
Also, the resultant forces on the object must be zero
WB3. find the sum of the moments about P
3 N
3 N
P is called a ‘Pivot’
1 m
1 m
6 N
The sum of moments is zero
The resultant force is zero

9. Uniform rods
WB4 A uniform rod AB of length 4 m and mass 5 kg is held up by a pivot at C .
where AC = 1.5 m. Calculate the mass of the particle which must be attached at A to maintain equilibrium with the rod horizontal. RC
RC is the reaction force of the Pivot
1.5 m
3 m
The centre of mass of the rod is in its middle
0.5 m
Wrod
Wparticle
Taking moments about C:
Wparticle × RC × 0 = Wrod × 0.5
Wparticle = 5 × 1.5 × 2 = 15 kg

10. WB5 A uniform beam AB of length 5 m and mass 30 kg rests horizontally on supports at C and D. Where AC = BD = 1 m. A man of mass 75 kg stands on the beam at E where AE = 2 m.
Calculate the magnitude of the reaction at each of the supports C and D. RC RD
1 m
1 m
0.5 m
1.5 m
1 m
Wman
Wrod
We have two unknowns so we want to build two equations. 
First resolve the forces
Secondly, take moments about C, there are 3 forces other than RC
If more needed then take moments about D or another point

11. WB5
1 m
1.5 m
75g
30g RC RD
0.5 m
resolve the forces
RC + RD = g
Take moments about C
75g × g × 1.5 = RD × 3
So RD = 392 N
RC + RD = g
RC = N

12. WB6 A uniform plank AB has mass 40 kg and length 3 m.
A load of mass 20 kg is attached to the plank at B.
The loaded plank is held in equilibrium, with AB horizontal, by two vertical ropes attached at A and C, as shown in the diagram.
The plank is modelled as a uniform rod and the load as a particle.
Given that the tension in the rope at C is three times the tension in the rope at A, calculate
(a) the tension in the rope at C,
(b) the distance CB.
Taking moments about C x

13. x m WB7 A uniform plank AB has mass 40 kg and length 4 m.
It is supported in a horizontal position by two smooth pivots, one at the end A, the other at the point C of the plank where AC = 3 m, as shown in the diagram above.
A man of mass 80 kg stands on the plank which remains in equilibrium.
The magnitudes of the reactions at the two pivots are each equal to R newtons.
By modelling the plank as a rod and the man as a particle, find
(a) the value of R,
2 m
x m
(b) the distance of the man from A.
Taking moments about A

14. The child and the plank are in equilibrium. Calculate:
WB8 A plank AE, of length 6 m and mass 10 kg, rests in a horizontal position on supports at B and D, where AB = 1 m and DE = 2 m.
A child of mass 20 kg stands at C, the mid-point of BD, as shown in the diagram. The child is modelled as a particle and the plank as a uniform rod.
The child and the plank are in equilibrium. Calculate:
(a) the magnitude of the force exerted by the support on the plank at B,
(b) the magnitude of the force exerted by the support on the plank at D.
Taking moments about D
Taking moments about B 1m
1.5m
1.5m 2m 3m 3m

15. WB8 Alternative method for part (b)
The child now stands at a point F on the plank.
The plank is in equilibrium and on the point of tilting about D.
(c) Calculate the distance DF. 1m x
Taking moments about D

16. WB9 A steel girder AB has weight 210 N
WB9 A steel girder AB has weight 210 N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end A. The other cable is attached to the point C on the girder, where AC = 90 cm, as shown in the figure above. The girder is modelled as a uniform rod, and the cables as light inextensible strings.
Given that the tension in the cable at C is twice the tension in the cable at A, find
(a) the tension in the cable at A,
(b) show that AB = 120 cm.
If AB = x, taking moments about A

17. WB9 A small load of weight W newtons is attached to the girder at B
WB9 A small load of weight W newtons is attached to the girder at B. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position.
The tension in the cable at C is now three times the tension in the cable at A.
(c) Find the value of W.
Taking moments about C

18. WB10 Two interlocking gears are in equilibrium
WB10 Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop.
Find the value of M.

19. Weight of right mass is 10g (N)
WB10 Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop.
Find the value of M.
Weight of right mass is 10g (N)
Moment on right gear is force × distance from centre
Moment = 10g × 0.08 = 0.8g (N m)
Force on left gear by right gear is
= 0.8𝑔 0.1
𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑠𝑡𝑎𝑛𝑐𝑒
Moment on left gear is force × distance from centre
= 8g × 0.05 = 0.4g (N m)
𝑤𝑒𝑖𝑔ℎ𝑡= 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑠𝑡𝑎𝑛𝑐𝑒
= 0.4𝑔 0.02 =20𝑔
M = 20g ÷g = 20 (kg)

20. Exam Q: Find rod length and Build up to simultaneous equations
90 cm C B
WB11 A steel girder AB has weight 210 N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end A. The other cable is attached to the point C on the girder, where AC = 90 cm, as shown in the figure above. The girder is modelled as a uniform rod, and the cables as light inextensible strings.
Given that the tension in the cable at C is twice the tension in the cable at A, find
the tension in the cable at A,
show that AB = 120 cm
A small load of weight W newtons is attached to the girder at B. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position.
The tension in the cable at C is now three times the tension in the cable at A.
(c) Find the value of W

21. WB11
(b) show that AB = 120 cm.
If AB = x, taking moments about A

22. WB11 A small load of weight W newtons is attached to the girder at B
WB11 A small load of weight W newtons is attached to the girder at B. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position.
The tension in the cable at C is now three times the tension in the cable at A.
(c) Find the value of W.
Taking moments about C

23. One thing to improve is –
KUS objectives
BAT understand moment and solve problems using moments on a Uniform rod
self-assess
One thing learned is –
One thing to improve is –

24. END