Question

Two blocks of masses $m_1=m_2=m$ are connected by a string of negligible mass which passes over a frictionless pulley fixed on the top of an inclined plane as shown in the figure. When the angle of inclination $\theta ={30}^{\circ }$, the mass $m_1$ just begins to move up the inclined plane. What is the coefficient of friction between block$m_1$ and the inclined plane?

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2\sqrt{3}}$

Answer 1

The correct option is C

$\frac{1}{\sqrt{3}}$

The block $m_1$ will just begin to move up the plane if $m_{1}gsin\theta +f$ equals tension T.
$m_{1}gsin\theta +f=T\Rightarrow m_{1}gsin\theta +\mu m_{1}gcos\theta =T........\left(1\right)$
For block $m_2$, $T=m_{2}g$
Putting $T=m_{2}g$ in equation (1), $m_{2}g=m_{1}g(sin\theta +\mu cos\theta )$
Now, it is given that $m_1=m_2=m$ and $\theta ={30}^{\circ }$.
Therefore, we have $1=sin{30}^{\circ }+\mu cos{30}^{\circ }$
$\Rightarrow \mu =\frac{1}{\sqrt{3}}$
Hence, the correct choice is (c).