Question

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

• A. remain unaffected.
• B. increase.
• C. decrease.
• D. become unequal to the force exerted by him on the beam.

Answer 1

The correct option is C decrease.

When the beam is supported at A and B, the force exerted by both men are equal which is equal to $\frac{mg}{2}$
When one person suddenly lets go the support, we get the torque about the person holding the beam as
$\tau =mg\frac{l}{2}=I\alpha =(\frac{m{l}^3}{3})\alpha$
or
$\alpha =\frac{3g}{2l}$
The acceleration of the center of mass is given as
$a_{C}M=\alpha (\frac{l}{2})=\frac{3g}{2l}(\frac{l}{2})=\frac{3g}{4}$
Thus force N exerted by man holding the beam
$mg-N=m{a}_{cm}=m(\frac{3g}{4}$
or
$N=\frac{1}{4}mg$
Thus the force decreases.