A uniform beam of mass 1.4 kg is pivoted at P as shown. The beam has a length of 0.60 m and P is 0.20 m from one end.

 Question 47

A uniform beam of mass 1.4 kg is pivoted at P as shown. The beam has a length of 0.60 m and P is 0.20 m from one end. Loads of 3.0 kg and 6.0 kg are suspended 0.35 m and 0.15 m from the pivot as shown.

 

What torque must be applied to the beam in order to maintain it in equilibrium?

A 0.010 N m                B 0.10 N m                  C 0.29 N m                 D 2.8 N m

 

 

 

Reference: Past Exam Paper – June 2013 Paper 12 Q14

 

 

 

Solution:

Answer: D.

Each of the masses will exert a force downwards on the beam (due to gravity) which will be equal to their weight.

Weight = mg

We also need to consider the weight of the beam of mass 1.4 kg which acts at the centre.

 

For equilibrium,

sum of clockwise moments = sum of anti-clockwise moments

 

The 6.0 kg load exert a clockwise moment.

Moment = Force × perpendicular distance

Clockwise moment = 0.15 × (6.0×9.81) = 8.829 Nm

 

Since the uniform beam has a total length of 0.60 m, its centre of mass will act at the 0.30 m mark (centre of the beam), a distance of (0.30 – 0.20 =) 0.10 m from the pivot. This causes an anticlockwise moment.

The 3.0 kg load also exerts an anticlockwise moment.

Anticlockwise moments = [0.10 × (1.4×9.81)] + [0.35 × (3.0×9.81)] = 11.6739 Nm

 

The torque that must be applied should act in such a way that the 2 moments (clockwise and anticlockwise) calculated would become equal.

Therefore, a (clockwise) torque that must be applied = 11.6739 – 8.829 = 2.8449 ≈ 2.8 Nm.