Question

A triangular prism of mass $m_1$ with a block of mass $m_2$ placed on it is released from rest on a smooth inclined plane of inclination $\theta$. The block does not slip on the prism. Then,

• A. the acceleration of the prism is $g\cos \theta$
• B. the acceleration of the prism is $g\tan \theta$
• C. the minimum coefficient of friction between the block and prism is ${\mu }_{min}=\cot \theta$
• D. the minimum coefficient of friction between block and prism is ${\mu }_{min}=\tan \theta$

Answer 1

The correct option is D the minimum coefficient of friction between block and prism is ${\mu }_{min}=\tan \theta$

As the blocks do not slip over each other, we can say that both the blocks move together with same acceleration downwards along the incline.

$a=\frac{Netdrivingforce}{Totalmass}=\frac{(m_1+m_2)g\sin \theta }{m_1+m_2}=g\sin \theta$

$N=m_{2}g-m_{2}g{\sin }^2\left(\theta \right)=m_{2}g{\cos }^2\left(\theta \right)f=\mu N\Rightarrow f=\mu m_{2}g{\cos }^2\left(\theta \right)\dots \dots \dots \dots \dots \dots \left(1\right)f=m_{2}g\sin \left(\theta \right)\cos \left(\theta \right)\dots \dots \dots \dots \dots \dots \left(2\right)comparingbothequations\mu m_{2}g{\cos }^2\left(\theta \right)=m_{2}g\sin \left(\theta \right)\cos \left(\theta \right)\mu \cos \left(\theta \right)=\sin \left(\theta \right)\tan \left(\theta \right)=\mu$