A uniform rod of density rho is placed in a w | vedclass

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Question

(A) Let $L$ be the total length of the rod. The rod is uniform,so its weight $W = A L \rho g$ acts at its center of mass,which is at a distance $L/2$

Vedclass · Accepted Answer

$\sin 	heta = \frac{1}{2}\sqrt{\frac{ho_0}{ho}}$

Vedclass · Answer

(A) Let $L$ be the total length of the rod. The rod is uniform,so its weight $W = A L ho g$ acts at its center of mass,which is at a distance $L/2$ from the bottom end $P$.Let $l$ be the length of the rod submerged in the liquid. The depth of the liquid is $h = L/2$. From the geometry,$\sin 	heta = h/l = (L/2)/l = L/(2l)$,which implies $l = L/(2 \sin 	heta)$.The buoyant force $F_B = A l ho_0 g$ acts at the center of the submerged portion,which is at a distance $l/2$ from the bottom end $P$.For rotational equilibrium about the bottom end $P$,the torque due to weight must balance the torque due to the buoyant force:$W \cdot (L/2) \cos 	heta = F_B \cdot (l/2) \cos 	heta$$(A L ho g) \cdot (L/2) \cos 	heta = (A l ho_0 g) \cdot (l/2) \cos 	heta$$L^2 ho = l^2 ho_0$$(l/L)^2 = ho / ho_0$Substituting $l = L/(2 \sin 	heta)$:$(1 / (2 \sin 	heta))^2 = ho / ho_0$$1 / (4 \sin^2 	heta) = ho / ho_0$$\sin^2 	heta = \frac{1}{4} \cdot \frac{ho_0}{ho}$$\sin 	heta = \frac{1}{2} \sqrt{\frac{ho_0}{ho}}$

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