The correct answer is D because:

Apply the principle of moments, taking the contact point of the shelf with the wall as the pivot point
As the beam is uniform its centre of gravity will be at the midpoint, 0.5 m from the pivot. The mass is 2 kg so its clockwise moment is calculated as:
- $f o r c e = m a s s \times a c c e l e r a t i o n d u e t o g r a v i t y$
- $f o r c e = 2 \times 9.8 = 19.6 N$
- $m o m e n t = f o r c e \times d i s t a n c e$
- $m o m e n t = 19.6 \times 0.5 = 9.8 N$
The 5 kg mass is 1 m away from the pivot point and so its clockwise moment is:
- $m o m e n t = m a s s \times a c c e l e r a t i o n d u e t o g r a v i t y \times d i s t a n c e$
- $5 \times 9.8 \times 1 = 49 N$
The anticlockwise moment due to the wire is the upward force, F, exerted by the wire on the beam multiplied by the distance from the pivot point, 0.7 m:
- $m o m e n t = F \times 0.7$
As the system is in equilibrium the clockwise and anticlockwise moments are equal:
- $0.7 F = 9.8 + 49$
- $F = \frac{58.8}{0.7} = 84 N$
This is the upward force that the wire exerts on the beam, which is the vertical component of the tension in the wire.
The tension in the wire is:

cie-4-1-mcq-hard-q1-2

- $\cos (90 - 30) = \frac{a d j a c e n t}{h y p o t e n u s e}$
- $h = \frac{84}{\cos (60)} = 168 N$
The tension in the wire is 168 N

A is incorrect because this is the answer if mass is not converted into force in the calculation

B is incorrect because this is the answer if the tension in the wire is not calculated from its vertical component

C is incorrect because this is the answer if the mass of the beam is neglected from the calculations

Turning Effects of Forces (CIE A Level Physics)

Exam Questions