Two men support a uniform horizontal beam at its two ends. …

Physics System of Particles Rotational Motion General MCQ (Single Correct)

Two men support a uniform horizontal beam at its two ends. If one of them suddenly lets go, the force exerted by the beam on the other man will

Remain unaffected.

Increase.

Decrease.

Become unequal to the force exerted by him on the beam.

Text Solution

Verified by Experts

The correct answer is:

When the beam is supported at A and B, the force exerted by each man = mg/2.

When the support at B is withdrawn, taking torque about A,

τ τ = (mg) l /2 = I α α = (m l 2 /3) α α

or α α = 3g/2 l .

The instantaneous linear acceleration of the centre of mass is

a CM = ( α α )(AC) = (3g/2 l ) l /2 = 3g/4.

Let N = force exerted on the beam at A.

∴ ∴ mg – N = ma CM = m(3g/4) or N = 1/4 mg.

Get practice questions, solutions, and test series in one place.

Share your experience with this question and solution.

Send your comment, doubt, correction, or feedback to admin.

What students say about this solution

No reviews yet. Be the first to write a review.