Question

$AV-$ shaped rigid body has two identical uniform arms. What must be the angle between the two arms, so that when the body is hung from one end the other arm is horizontal?

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

Answer 1

Answer: (A)

Let length of each of rod is I 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 above figure.
Now, perpendicular distance of weight acting through point $A$ from point $D$ is

$\theta CD=l\cos \theta -\frac{l}{2}\cos \theta$

$CD=\frac{1}{2}\cos \theta$

and perpendicular distance of weight acting through $B$ from point $D$ is

$BD=\frac{l}{2}-l\cos \theta =\frac{l}{2}\left(1-2\cos \theta \right)$

At equilibrium torque of these two weights about $D$ must balance each other

$i.emg\times \frac{l}{2}\cos \theta =mg\times \frac{l}{2}\left(1-2\cos \theta \right)$

$\Rightarrow \frac{3}{2}\cos \theta =\frac{1}{2}$

$\Rightarrow \cos \theta \frac{1}{3}$

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