1. Moments
At angles

2. Starter: Spot the mistake
Moments: At an angle
KUS objectives
BAT solve problems using moments and friction at angles to a rod/ object using sohCahToa
Starter: Spot the mistake

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. WB 1 Calculate the moment about point A of each of these forces acting on a Lamina ,𝐴
10 𝑁
40°
5 𝑚 𝐴
6 sin 25
5 sin 40
6 𝑚
25°
18 𝑁
𝑚𝑜𝑚𝑒𝑛𝑡=10×5 sin 40=32.1 𝑁𝑚 𝑎𝑐𝑤
𝑚𝑜𝑚𝑒𝑛𝑡=18×6 sin 25=45.6 𝑁𝑚 𝑎𝑐𝑤 𝐴
120 𝑁
50°
8 𝑚
70 𝑁
4 sin 65
8 cos 50 𝐴
65°
4 𝑚
𝑚𝑜𝑚𝑒𝑛𝑡=70×4 sin 65 =254 𝑁𝑚 𝑎𝑐𝑤
𝑚𝑜𝑚𝑒𝑛𝑡=120×8 cos 50 =617 𝑁𝑚 𝑐𝑤

6. WB 2 Given the moment about point A of each of these forces is 20 Nm, Find the magnitude of each force
3 sin 35 𝐴 𝐴
𝐹 1
30°
5 𝑚
5 sin 30
3 𝑚
𝐹 2
35°
𝐹 2 = 20 3 sin 35 = 𝑁
𝐹 1 = 20 5 sin 30 =8 𝑁 𝐴
75°
9 𝑚
𝐹 4
𝐹 3 = 20 2 sin 55 = 𝑁
9 sin 75 𝐴
125°
2 𝑚
𝐹 3
2 sin 55
𝐹 4 = 20 9 sin 75 = 𝑁

7. WB 3a Two forces act on a lamina as shown
WB 3a Two forces act on a lamina as shown. Calculate the resultant moment
about the point A a) 𝐴
𝐹 2 =5
50°
10 𝑚
𝐹 1 =8
30°
5 sin 50
8 sin 30
Moment of 𝐹 1
=8 sin 30 ×8=32 𝑐𝑤
Moment of 𝐹 2
=5 sin 50 ×5= 𝑎𝑐𝑤
Moment of 𝐹 1 and 𝐹 2 =32−19.2 =12.8 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

8. WB 3b Two forces act on a lamina as shown
WB 3b Two forces act on a lamina as shown. Calculate the resultant moment
about the point A b) 𝐴
𝐹 2 =15
60°
10 𝑚
𝐹 1 =12
70°
8 𝑚
8 sin 60
10 sin 70
Moment of 𝐹 1
=10 sin 70 ×12= 𝑎𝑐𝑤
Moment of 𝐹 2
=8 sin 60 ×15= 𝑐𝑤
Moment of 𝐹 1 and 𝐹 2 =112.7−103.9 =8.8 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

9. Moment on a uniform beam / rod
WB4 review 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 7 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 7
3 m
1 m A C B
8×𝐴𝐶=8 𝑁𝑚
b) Taking moments about C cw
8×𝐶𝐵=24 𝑁𝑚
c) Taking moments about C cw
8×𝐶𝐶=0

10. WB 5 The diagram shows a set of forces acting on a uniform rod of mass 3 kg. Calculate the resultant moment about point A (including direction) 𝐴
45°
3 𝑚
5 𝑁
6 𝑁
70°
4 𝑚
4 sin 45
3 sin 70
Taking moments about A clockwise
6×4 sin 45 − 5×3 sin 70 = 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

11. Equilibrium

12. WB 6 A uniform rod PQ is hinged at point P, and is held in equilibrium at an angle of 50 to the horizontal by a force of magnitude F acting perpendicular to the rod at Q. Given that the rod has a length of 3 m and mass of 8 kg, find the value of F 𝑃
8𝑔 𝑁
50° 𝐹 𝑄
3 2 cos 50
Taking moments about P
𝐹 ×3 = 8𝑔× 3 2 cos 50
𝐹 = 8𝑔× 3 2 cos =25.2 𝑁

13. WB 6 (cont) A uniform rod PQ is hinged at point P, and is held in equilibrium at an angle of 50 to the horizontal by a force of magnitude F acting perpendicular to the rod at Q. Given that the rod has a length of 3 m and mass of 8 kg, find the value of F 𝑃
8𝑔 𝑁
50° 𝐹 𝑄
3 2 cos 50
Taking moments about P
𝐹 ×3 = 8𝑔× 3 2 cos 50
𝐹 = 8𝑔× 3 2 cos =25.2 𝑁

14. WB 7 A uniform rod PQ of mass 40 kg and length 10 m rests with the end P on rough horizontal ground. The rod rests against a smooth peg C where AC = 8 m
The rod is in limiting equilibrium at an angle of 15 to the horizontal. Find
The magnitude of the reaction at C
The coefficient of friction between the rod and the ground 𝑃
40𝑔
15° 𝑁 𝐶 𝑅
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛
75°
distance 5 cos 15
a) Taking moments about P
𝑁×8 =40𝑔×5 cos 15
reaction at C 𝑁 = 40𝑔×5 cos = 𝑁
b) Equilibrium in horizontal direction
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 =𝑁 cos 75 =236.7 cos 75 =61.25
Equilibrium in vertical direction
𝑅+𝑁𝑠𝑖𝑛 75=40𝑔
𝑅=40𝑔−236.7 cos 75 = 𝑁
𝜇= 𝐹𝑟 𝑅 = =0.375
Coefficient friction

15. WB 8 A ladder PQ of mass m kg and length 3a m rests with the end P on rough horizontal ground. The other end Q rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point C, where AC = a.
The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load as a particle. The ladder rests in limiting equilibrium at an angle of 60with the ground. Find the coefficient of friction between the rod and the ground N
Add all the forces to the diagram 𝑃
60° 𝐶 𝑄 2𝑎 𝑎
Wall smooth
No friction
Equilibrium in horizontal direction
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛=𝑁 R
Equilibrium in vertical direction
𝑅=2𝑚𝑔+𝑚𝑔=3𝑚𝑔
Friction

16. (to get an equation with Friction and R)
WB 8 (cont) A ladder PQ of mass m kg and length 3a m rests with the end P on rough horizontal ground. The other end Q rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point C, where AC = a.
The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load as a particle. The ladder rests in limiting equilibrium at an angle of 60with the ground. Find the coefficient of friction between the rod and the ground
Taking moments about Q
(to get an equation with Friction and R) 𝑃
60° 𝐶 𝑄 2𝑎 𝑎
Wall smooth
No friction
Friction R N
𝑅×3𝑎 cos 60 =𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ×3𝑎 sin 60
+𝑚𝑔× 3 2 𝑎 cos 60 +2𝑚𝑔×2𝑎 cos 60
Substituting 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝜇𝑅 and R=3𝑚𝑔
3𝑚𝑔×3𝑎 cos 60 =𝜇3𝑚𝑔 ×3𝑎 sin 60
+𝑚𝑔× 3 2 𝑎 cos 60 +2𝑚𝑔×2𝑎 cos 60
Cancel by mga and rearrange to
𝜇= 3𝑚𝑔×3𝑎 cos 60 −+𝑚𝑔× 3 2 𝑎 cos 60 −2𝑚𝑔×2𝑎 cos 60 3𝑚𝑔 ×3𝑎 sin 60
= cos sin 60 = =0.225

17. WB 9 (exam Q) sin 𝜃 = 3 5 cos 𝜃 = 4 5 3 4 5 𝜃 Mg 3Mg T R Friction
A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The plank is held in a horizontal position by a rope. One end of the rope is attached to the plank at B and the other end is attached to the wall at the point C, which is vertically above A.
A small block of mass 3M is placed on the plank at the point P, where AP = x
The plank is in equilibrium in a vertical plane which is perpendicular to the wall
The angle between the rope and the plank is α, where tan ∝ = 3 4
The plank is modelled as a uniform rod, the block is modelled as a particle and the rope is modelled as a light inextensible string
Using the model, show that the tension in the rope is 5𝑀𝑔 3𝑥+𝑎 6𝑎
sin 𝜃 = 3 5
cos 𝜃 = 4 5 3 4 5 𝜃
Q from 2018 June exam Mg
3Mg T R
Friction

18. WB 9a (exam Q cont )
Using the model, show that the tension in the rope is 5𝑀𝑔 3𝑥+𝑎 6𝑎
a) Taking moments about A
𝑇 sin ∝ ×2𝑎 =𝑀𝑔×𝑎+3𝑀𝑔×𝑥 Mg
3Mg R
Friction a
−𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝
Rearranges to
𝑇× 6𝑎 5 =𝑀𝑔×𝑎+3𝑀𝑔×𝑥
Rearranges to
𝑇= 5 6𝑎 × 𝑀𝑔𝑎+3𝑀𝑔𝑥 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 QED

19. 2𝑀𝑔 =𝑇𝑐𝑜𝑠 ∝ WB 9b (exam Q cont )
The magnitude of the horizontal component of the force exerted on the plank at A by the wall is 2Mg Find x in terms of a
b) Horizontally forces are in equilibrium
2𝑀𝑔 =𝑇𝑐𝑜𝑠 ∝ Mg
3Mg
R=2Mg
Friction a
−𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝
Substituting result from a) gives
2𝑀𝑔 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 𝑐𝑜𝑠 ∝
2𝑀𝑔 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 × 4 5
Cancel by 2Mg to simplify gives 1 = 𝑎+3𝑥 3𝑎
Rearranges to a+3𝑥 =3𝑎 then to 𝑥= 2 3 𝑎

20. WB 9c (exam Q cont )
The force exerted on the plank at A by the wall acts in a direction which makes an angle β with the horizontal c) Find the value of tan 𝛽
c) Resolve forces vertically
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛+𝑇𝑠𝑖𝑛 ∝=4𝑀𝑔 Mg
3Mg
R=2Mg
Friction a
−𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛+ 5𝑀𝑔 𝑎+3𝑥 6𝑎 × 3 5 =4𝑀𝑔
Rearranges using 𝑥= 2 3 𝑎
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛=4𝑀𝑔− 𝑀𝑔 𝑎+2𝑎 2𝑎 = 5 2 Mg
Now find tan 
R=2Mg
Friction = 5 2 Mg
tan 𝛽 = 5 2 Mg 2𝑀𝑔 = 5 4 𝛽

21. simplify by cancelling and rearrange
WB 9d (exam Q cont )
The rope will break if the tension in it exceeds 5 Mg d) Explain how this will restrict the possible positions of P. You must justify your answer carefully
previous results: 𝑇= 5𝑀𝑔 𝑎+3𝑥 6𝑎 Mg
3Mg
R=2Mg
Friction= 2 5 𝑀𝑔 a
−𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝
we know: 𝑇= 5𝑀𝑔 𝑎+3𝑥 6𝑎 ≤5𝑀𝑔
simplify by cancelling and rearrange
5 𝑎+3𝑥 6𝑎 ≤5
5a+15x ≤30𝑎
So the distance AP must be less than 5 3 𝑎
For the rope NOT to break
𝑥 ≤ 5 3 𝑎

22. One thing to improve is –
KUS objectives
BAT solve problems using moments and friction at angles to a rod/ object using sohCahToa
self-assess
One thing learned is –
One thing to improve is –

23. END