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 blocks is, $(m_1=4\text{kg},m_2=2\text{kg},{\mu }_1=0.8,{\mu }_2=0.5,g=10{\text{m/s}}^2,\sin {37}^{\circ }=3/5)$

• 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
writting the Newton's law equation on the block we get
$mg\sin \theta -f=ma$
Where $f$ is the friction force
$\Rightarrow f=\mu mg\cos \theta$
$\Rightarrow a=g\sin \theta -\mu g\cos \theta$
Here the acceleration of blocks depends on the $\mu$ as ${\mu }_1>{\mu }_2$
$a_1<a_2$
$m_2$ slips before $m_1$

So, no contact between $m_1$ and $m_2$
$\Rightarrow N=0$