Question

A uniform beam is pivoted at its centre. Two weights are placed on the beam in the positions shown

and the beam is balanced by an upward force F.

What is the size of F?

A. 6 N B 12 N C 30 N D 60 N​

Answer 1

Topic

Equilibrium

Given

A uniform beam is pivoted at its centre.

Then two weights are placed.

Beam is balanced by a force F.

To Find

Value of F.

Concept Used

Torque

It is the measure of Force which causes rotation.

Torque = r × F

where

r is distance from Centre of Mass and

F is applied Force.

Solving

We can observe, that due to three forces rotation can take place.

It is given that beam is in equilibrium.

So, Torque in anticlockwise direction and in clockwise direction will be equal.

F * 50 cm + 30N * 40 cm = 60N * 30 cm

50F + 1200 = 1800

50F = 600

F = 12 N.

( Note :- Distances will be from Centre of Mass of beam that is centre of beam. )

Answer

So, the magnitude of F is 12 N that is B option.