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
$\tau =I\alpha =Mg\frac{l}{2}$
$\Rightarrow \frac{M{l}^2}{3}\alpha =Mg\frac{l}{2}$ [As $I=Ml3]$
$\Rightarrow$ 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 $Mg-F^{\prime }=Ma$
$\Rightarrow F^{\prime }=Mg-Ma=Mg-\frac{3Mg}{4}=\frac{Mg}{4}=\frac{W}{4}$