Question

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?

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{2}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{4}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{6}\right)$

Answer 1

Answer: (A)

Let length of each of rod is $l$ and angle between them is $\theta$.

Let the lower rod is horizontal and upper rod makes $\theta$ angle with horizontal.
Weights of rods acts vertically downwards from their centres $A$ and $B$ as shown in the above figure.
Now, perpendicular distance of weight acting through $A$ from point $D$ is
$CD=l\cos \theta -\frac{l}{2}\cos \theta$
$CD=\frac{l}{2}\cos \theta$
and perpendicular distance of weight acting through $B$ from point $D$ is
$BD=\frac{l}{2}-l\cos \theta =\frac{l}{2}(1-2\cos \theta )$
At equilibrium torque of these two weights about $D$ must balance each other.
i.e. $mg\times \frac{l}{2}\cos \theta =mg\times \frac{l}{2}(1-2\cos \theta )$
$\Rightarrow \frac{3}{2}\cos \theta =\frac{1}{2}$
$\Rightarrow \cos \theta =\frac{1}{3}$
or $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$