Question

A uniform $L$ shaped rod each of side $A$ is held as shown in the figure. The angle $\theta$ such that the rod remains stable will be

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

Answer 1

The correct option is B

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

Step 1. Given data

Given, an uniform $L$ shaped rod each of side $a$ is held as shown in the below figure, and $\theta$ is the angle that $L$ shaped rod is hanging.

Step 2. Finding torque on a single side of rod of $L$ shaped rod.

Now let us consider the one side of the $L$ shaped rod of length $a$, with an angle $\theta$.

And we know that weight will act at the center of the rod, say point $R$ in the below diagram, and weight is given by

$$\begin{gathered} Weight=mass\times gravity \\ \Rightarrow W=m\times g \end{gathered}$$

Draw a line connecting between $R$ and projection of $P$ and name it as $Q$ as shown in the figure to form a right angle triangle

In $△PQR$, $RQ$ is given by

$\begin{array}{rcl}sin\left(\theta \right)& =& \frac{RQ}{PR}\\ RQ& =& sin\left(\theta \right)\times PR\\ RQ& =& sin\left(\theta \right)\times \frac{a}{2}\end{array}$

Since, $R$ is at the center of length $a$, $PR$ (Hypotenuse) is of length $\frac{a}{2}$.

Therefore, the counterclockwise direction of torque acting is given by,

$\tau =rF\sin \theta$

Where, $\tau =$torque, $r=$ radius, $F=$ force, $\theta =$ angle between the $F$ and lever arm

Torque $\tau =mg\times \frac{a}{2}\sin \left(\theta \right)$

Step 3. Finding torque of another rod of $L$ shaped rod

Now consider remaining single $L$ shaped rod of length $a$ as shown in figure

In $△PQR$, $PQ$ is given by

$\begin{array}{rcl}cos\left(\theta \right)& =& \frac{PQ}{PR}\\ PQ& =& cos\left(\theta \right)\times PR\\ PQ& =& cos\left(\theta \right)\times \frac{a}{2}\end{array}$

In $△PQR$, $PR$ is given by

$\begin{array}{rcl}sin\left(\theta \right)& =& \frac{RQ}{PR}\\ RQ& =& sin\left(\theta \right)\times PR\\ PR& =& a\sin \left(\theta \right)\end{array}$

Here the torque is acting on a clockwise direction and it is given by

$torque=-mg(PQ-PR)$

$torque=-mg\left(\frac{a}{2}\cos \right(\theta )-a\sin (\theta \left)\right)$

Step 4. Finding the net torque of the $L$ shaped rod

We know that for stable bodies the net torque acting on it is equal to zero

Therefore the net torque is given by

$\begin{array}{rcl}mg\left(\frac{a}{2}\sin \right(\theta \left)\right)-mg\left(\frac{a}{2}\cos \right(\theta )-a\sin (\theta \left)\right)& =& 0\\ \frac{\sin \left(\theta \right)}{2}-\frac{\cos \left(\theta \right)}{2}+\sin \left(\theta \right)& =& 0\\ 3\frac{\sin \left(\theta \right)}{2}& =& \frac{\cos \left(\theta \right)}{2}\\ \tan \left(\theta \right)& =& \frac{1}{3}\\ \theta & =& {\tan }^{-1}\left(\frac{1}{3}\right)\end{array}$

Hence, the correct option is (B).