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(D) Let $T_1$ be the tension in the first rope and $T_2$ be the tension in the second rope.Since the rod is in equilibrium,the sum of vertical forces

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$\frac{2W}{3}$

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(D) Let $T_1$ be the tension in the first rope and $T_2$ be the tension in the second rope.Since the rod is in equilibrium,the sum of vertical forces is zero:$T_1 + T_2 = W$  --- $(1)$Taking the moment of forces about the left end (point $A$):The weight $W$ acts at the center of the rod,which is at a distance $L/2$ from the left end.The second rope is attached at a distance $L - L/4 = 3L/4$ from the left end.For rotational equilibrium,the sum of torques about $A$ must be zero:$W 	imes \frac{L}{2} - T_2 	imes \frac{3L}{4} = 0$$W 	imes \frac{L}{2} = T_2 	imes \frac{3L}{4}$$T_2 = W 	imes \frac{L}{2} 	imes \frac{4}{3L}$$T_2 = \frac{2W}{3}$

$A$ uniform rod of length $L$ and weight $W$ is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached at a distance $L/4$ from the right end. The tension in the second rope is

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