A) \[W\]
B) \[\frac{W}{2}\]
C) \[\frac{3W}{4}\]
D) \[\frac{W}{4}\]
Correct Answer: D
Solution :
[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 \[\tau =I\alpha =Mg\frac{l}{2}\] \[\Rightarrow \frac{M{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] \[[AsI=M{{l}^{2}}/3]\] \[\Rightarrow \]Angular acceleration, \[\alpha =\frac{3}{2}\frac{g}{l}\] And linear acceleration \[a=\frac{1}{2}\alpha =\frac{3g}{4}\] Now if the new normal force at A is F' then \[Mg-F'=Ma\] \[\Rightarrow F'=Mg-Ma=Mg-\frac{3Mg}{4}=\frac{Mg}{4}=\frac{W}{4}.\]You need to login to perform this action.
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