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(A) Let the length of each rod be $l$ and the angle between them be $	heta$. When the body is suspended from one end $P$,let the rod $2$ be horizonta

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{1}{3}\right)$

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(A) Let the length of each rod be $l$ and the angle between them be $	heta$. When the body is suspended from one end $P$,let the rod $2$ be horizontal. The rod $1$ makes an angle $	heta$ with the horizontal rod $2$. The weight $mg$ of each rod acts vertically downwards from their respective centers of mass $A$ and $B$. Let $D$ be the point of suspension $P$ projected onto the horizontal rod $2$. The horizontal distance of the center of mass $A$ of rod $1$ from the vertical line passing through $P$ is $x_1 = \frac{l}{2} \cos 	heta$. The horizontal distance of the center of mass $B$ of rod $2$ from the vertical line passing through $P$ is $x_2 = l \cos 	heta - \frac{l}{2}$. For rotational equilibrium about the point of suspension $P$,the torques due to the weights of the two rods must balance each other: $mg \cdot x_1 = mg \cdot x_2$ $\frac{l}{2} \cos 	heta = l \cos 	heta - \frac{l}{2}$ Dividing by $l$ and rearranging: $\frac{1}{2} \cos 	heta = \cos 	heta - \frac{1}{2}$ $\frac{1}{2} = \cos 	heta - \frac{1}{2} \cos 	heta$ $\frac{1}{2} = \frac{1}{2} \cos 	heta$ $\cos 	heta = 1$ (This implies the rods are collinear,which is not a $V$ shape). Re-evaluating the geometry from the provided diagram: The center of mass $A$ is at distance $\frac{l}{2}$ from the vertex along rod $1$. Its horizontal distance from the vertical line through $P$ is $d_1 = \frac{l}{2} \cos 	heta$. The center of mass $B$ is at distance $\frac{l}{2}$ from the vertex along rod $2$. Its horizontal distance from the vertical line through $P$ is $d_2 = l \cos 	heta - \frac{l}{2}$. Equating torques: $mg \cdot d_1 = mg \cdot d_2$ $\frac{l}{2} \cos 	heta = l \cos 	heta - \frac{l}{2}$ $\frac{1}{2} = \frac{1}{2} \cos 	heta \Rightarrow \cos 	heta = 1$. Wait,looking at the diagram,the horizontal distance of $A$ from $P$ is $l \cos 	heta - \frac{l}{2} \cos 	heta = \frac{l}{2} \cos 	heta$. The distance of $B$ from $P$ is $\frac{l}{2} - l \cos 	heta$. Equating: $\frac{l}{2} \cos 	heta = \frac{l}{2} - l \cos 	heta$ $\frac{1}{2} \cos 	heta = \frac{1}{2} - \cos 	heta$ $\frac{3}{2} \cos 	heta = \frac{1}{2} \Rightarrow \cos 	heta = \frac{1}{3}$. Thus,$	heta = \cos^{-1}\left(\frac{1}{3}ight)$.

$A$ rigid body in the shape of a $V$ has two equal arms made of uniform rods. What must the angle between the two rods be so that when the body is suspended from one end,the other arm is horizontal?

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