A) w
B) w/2
C) 3w/4
D) w/4
Correct Answer: D
Solution :
Let us first consider linear acceleration of CG. When the man at B withdraws his support, the bar turns about A with an angular acceleration\[\alpha \]given by\[I\alpha =\frac{wl}{2}\] As \[I=\frac{m{{l}^{2}}}{3}\]and\[w=mg\] hence \[\alpha =\frac{3g}{2l}\] Hence, linear acceleration of C.G, \[a=\frac{1}{2}\alpha =\frac{1}{2}\frac{3g}{2l}=\frac{3g}{4}\] Now, if normal reaction at A is N, then \[w-N=ma\Rightarrow mg-N=m\frac{3g}{4}\] \[\Rightarrow \]\[N=\frac{mg}{4}=w/4\]You need to login to perform this action.
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