Question

A block of base 10 cm $\times$ 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^o.$ Then

• A. at $\theta ={30}^o$, the block will start sliding down the plane
• B. the block will remain at rest on the plane up to certain $\theta$ and then it will topple
• C. at $\theta ={60}^o$ the block will start sliding down the plane and continue to do so at higher angles
• D. at $\theta ={60}^o,$ the block will start sliding down the plane and on further increasing $\theta ,$ it will topple at certain $\theta$

Answer 1

The correct option is B the block will remain at rest on the plane up to certain $\theta$ and then it will topple

The maximum angle when the block will not slide should be equal to the angle of repose.
$\varphi =ta{n}^{-1}\mu =ta{n}^{-1}\sqrt{3}={60}^o$
If the block does not topple, the line of action of weight should pass within the base of the block

And for toppling, maximum angle of inclination,
$\theta =ta{n}^{-1}\frac{b/2}{h/2}$
$\theta =ta{n}^{-1}\frac{\left(10/2\right)}{\left(15/2\right)}=ta{n}^{-1}\frac{2}{3}$
$\theta ={34}^o<{60}^o$
For ${60}^o,$ mg is already outside the base.
$\therefore$ As the angle $\theta$ is increased first it will topple before it starts sliding.