Question

A homogeneous block having its cross section as a parallelogram of sides $a$ and $b$ as shown, is lying at rest and is in equilibrium on a smooth horizontal surface. Then, find the value of acute angle $\theta$ for which it will be in equlibrium

• A. $\theta <{\cos }^{-1}\left(\frac{b}{a}\right)$
• B. $\theta \le {\cos }^{-1}\left(\frac{b}{a}\right)$
• C. $\theta \ge {\cos }^{-1}\left(\frac{b}{a}\right)$
• D. $\theta >{\cos }^{-1}\left(\frac{b}{a}\right)$

Answer 1

The correct option is B $\theta \le {\cos }^{-1}\left(\frac{b}{a}\right)$

From $\theta ={90}^{\circ }$, if we reduce the angle as shown, COM of the parallelogram will shift towards right. As we know, weight ($mg$) force will act at COM.

To balance torque due to $mg$ force, normal force will start shifting towards right.

When block is just about to topple, normal reaction gets shifted to the extreme right point.


In this condition, COM is directly above the extreme point $\text{O}$.
So, $\cos \theta =\frac{b}{a}$

As shown in figure, $mg$ and normal force are acting in the same line. Hence there is no torque and the block is in equilibrium.

From this position, if we reduce $\theta$ more, then COM of the parallelogram will shift even more towards right and $mg$ force will act at the COM of the parallelogram. To balance this, normal force can not shift more towards right. So, due to torque produced by $mg$ force, the parallelogram block will topple.

Hence for equilibrium (not topple),
From figure,
$\cos \theta \le \frac{b}{a}$
$\theta \le$ ${\cos }^{-1}\left(\frac{b}{a}\right)$