Question

Two touching blocks $1$ and $2$ are placed on an inclined plane forming an angle $60^{\circ }$ with the horizontal. The masses are $m_1$ and $m_2$ and the coefficient of friction between the inclined plane and the two blocks are $1.5\mu$ and $1.0\mu$, respectively. The force of reaction between the blocks during the motion is $(g=$ acceleration due to gravity)

• A. $\left(m_2-m_1\right)\mu g$
• B. $\left(m_2+m_1\right)\mu g$
• C. $\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}\mu g$
• D. $\frac{1}{4}\frac{m_1{m}_2}{m_1+m_2}\mu g$

Answer 1

Answer: (D)

According to the question, there are two blocks of masses $m_1$ and $m_2$ which are kept in contact on inclined plane of inclination $60^{\circ }$ as shown in the figure,

Coefficient of friction between the surface of plane and surface of first block, ${\mu }_1=1.5\mu$ Coefficient of friction between the surface of plane and surface of second block, ${\mu }_2=1\mu$
Now, free body diagram for the first block,

If $a$ be the common acceleration, then
$m_{1}g\sin 60^{\circ }-{\mu }_1\cdot R_1+R=m_{1}a$
$m_{1}g\frac{\sqrt{3}}{2}-1.5\mu m_{1}g\cos 60^{\circ }+R=m_{1}a$
$m_{1}g\frac{\sqrt{3}}{2}-\frac{1.5\mu m_{1}g}{2}+R=m_{1}a...(i)$
Similarly, for second body,

$m_{2}g\sin 60^{\circ }-{\mu }_2{R}_2-R=m_{2}a$
$m_{2}g\frac{\sqrt{3}}{2}-\mu m_{2}g\cos 60^{\circ }-R=m_{2}a$
$m_{2}g\frac{\sqrt{3}}{2}-\frac{\mu m_{2}g}{2}-R=m_{2}a...(ii)$
From Eqs. (i) and (ii), we get
$\frac{m_{1}g\frac{\sqrt{3}}{2}-\frac{1.5\mu m_{1}g}{2}+R}{m_{2}g\frac{\sqrt{3}}{2}-\frac{\mu m_{2}g}{2}-R}=\frac{m_{1}a}{m_{2}a}$
$\Rightarrow \frac{\sqrt{3}}{2}{m}_1{m}_{2}g-0.75\mu m_1{m}_{2}g+m_{2}R$
$=\frac{\sqrt{3}}{2}{m}_1{m}_{2}g-0.5\mu m_1{m}_{2}g-m_{1}R$
$\Rightarrow \left(m_1+m_2\right)R=0.25\mu m_1{m}_{2}g$
$\Rightarrow R=\frac{\mu m_1{m}_{2}g}{4\left(m_1+m_2\right)}$
Hence, the force of reaction between the blocks doing the motion is, $R=\frac{1}{4}\frac{m_1{m}_2}{\left(m_1+m_2\right)}\mu g$.