Question

A horizontal heavy uniform bar of weight $W$ is supported at its ends by two men. At the instant, one o f the men lets go off his end of the rod, the other feels the force on his hand changed to

• A. $W$
• B. $\frac{W}{2}$
• C. $\frac{3W}{4}$
• D. $\frac{W}{4}$

Answer 1

Answer: (D)

Let the mass of the rod is $M \Rightarrow $ Weight (W) = Mg
Initially for the equilibrium $F+F=Mg$
$\Rightarrow F=Mg/2$
When one man withdraws, the torque on the rod
$\tau=I \alpha=Mg \frac{l}{2} $
$\Rightarrow \frac{Ml^{2}}{3} \alpha=Mg \frac{l}{2}$ [As $I=Ml 3]$
$\Rightarrow $ Angular acceleration,

$\alpha=\frac{3}{2} \frac{g}{l} $
and linear acceleration,
$a=\frac{l}{2} \alpha=\frac{3g}{4} $
Now if the new normal force at A is F' then $Mg-F' =Ma$
$\Rightarrow F'=Mg -Ma=Mg -\frac{3 Mg}{4} = \frac{Mg}{4}=\frac{W}{4}$