Question

Two blocks $m_1$ and $m_2$ are resting on a rough inclined plane of inclination ${37}^{\circ }$ as shown in figure. The contact force between the two blocks is
(Given $m_1=4\text{kg},m_2=2\text{kg},$ ${\mu }_1=0.8,{\mu }_2=0.5,g=10{\text{m/s}}^2,\sin {37}^{\circ }=3/5$)

• A. $3.2\text{N}$
• B. $3.6\text{N}$
• C. $7.2\text{N}$
• D. Zero

Answer 1

The correct option is D Zero

For a mass sliding down on inclined plane with friction coefficient $\mu$, the acceleration is given by,

$a=g\sin \theta -\mu g\cos \theta$

As ${\mu }_1>{\mu }_2$, $a_1<a_2$

i.e the acceleration of $m_2$ is greater than that of $m_1$. Hence, there will not be any contact. i.e $N=0$

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

Why this Question? Note: Even if it seems that the two bodies are in contact, when their acceleration is same, the contact force between them is zero.