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(C) Let the total length of the beam be $l$ and its weight $W$ act at the center $C$ (at $l/2$ from either end).Let the positions of the two persons b

Vedclass · Accepted Answer

$5 : 9$

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(C) Let the total length of the beam be $l$ and its weight $W$ act at the center $C$ (at $l/2$ from either end).Let the positions of the two persons be $P_1$ and $P_2$.$P_1$ is at $l/5$ from one end,and $P_2$ is at $l/3$ from the other end.The distance of $P_1$ from the center $C$ is $d_1 = \frac{l}{2} - \frac{l}{5} = \frac{3l}{10}$.The distance of $P_2$ from the center $C$ is $d_2 = \frac{l}{2} - \frac{l}{3} = \frac{l}{6}$.For the beam to be in rotational equilibrium,the net torque about the center of mass $C$ must be zero:$N_A 	imes d_1 = N_B 	imes d_2$$N_A 	imes \frac{3l}{10} = N_B 	imes \frac{l}{6}$$\frac{N_A}{N_B} = \frac{l}{6} 	imes \frac{10}{3l} = \frac{10}{18} = \frac{5}{9}$.Thus,the ratio of the normal reactions is $5 : 9$.

Two persons of equal height are carrying a long uniform wooden beam of length $l$. They are at distances $\frac{l}{5}$ and $\frac{l}{3}$ from the nearest ends of the rod. The ratio of the normal reaction at their heads is:

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