Question

An angular structure $\text{ABC}$ with side $AB=BC=L$ is made up of two thin rods $\text{AB}$ of mass $m$ and $\text{BC}$ of mass $2m$. The structure when hanged from point $A$ stays in equilibrium with side $BC$ horizontal. Find the value of angle $\theta$.

• A. ${30}^{\circ }$
• B. ${\cos }^{-1}\left(\frac{2}{5}\right)$
• C. ${\sin }^{-1}\left(\frac{3}{5}\right)$
• D. $\frac{\pi }{2}-{\sin }^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{2}{5}\right)$


For equilibrium, the net torque about point $A$ should be zero.

Now, the torque due mass $m$ and $2m$ about $A$ are,

${\tau }_m=mg\frac{L}{2}\cos \theta$

${\tau }_{2m}=2mg(\frac{L}{2}-L\cos \theta )$

For equilibrium,

${\tau }_{mg}={\tau }_{2mg}$

$\Rightarrow mg\frac{L}{2}\cos \theta =2mg(\frac{L}{2}-L\cos \theta )$

$\Rightarrow \frac{5}{2}\cos \theta =1$

$\therefore \theta ={\cos }^{-1}\left(\frac{2}{5}\right)$

Hence, option $\left(B\right)$ is the correct answer.