150294
A boy and a man carry a uniform rod of length \(L\), horizontally in such a way that boy gets \(\frac{1}{4}\) th load. If the boy is at one end of the rod, the distance of the man from the other end is
1 ) \(\frac{\mathrm{L}}{3}\)
2 ) \(\frac{L}{4}\)
3 ) \(\frac{2 \mathrm{~L}}{3}\)
4 ) \(\frac{3 \mathrm{~L}}{4}\)
Explanation:
A Exp: (A) : Let, weight of the \(\operatorname{rod}=\mathrm{w}\), reaction of boy \(\left(R_{B}\right)=\frac{w}{4}\), reaction of \(\operatorname{man}\left(R_{m}\right)=w-\frac{w}{4}=\frac{3 w}{4}\) Rotational equilibrium will be equal then, \(\tau_{\mathrm{B}}=\tau_{\mathrm{m}}\) \(\frac{\mathrm{w}}{4} \times \frac{\mathrm{L}}{2}=\frac{3 \mathrm{w}}{4} \times \mathrm{x}\) \(\mathrm{x}=\frac{\mathrm{L}}{6}\) Distance of man from other end \((y)=\frac{L}{2}-x\) \(y=\frac{L}{2}-\frac{L}{6}\) \(y=\frac{L}{3}\)
