Question

Two blocks of masses $m_1=m$ and $m_2=2m$ are connected by a light string passing over a smooth pulley. The mass $m_1$ is placed on a rough inclined plane of inclination $\theta ={30}^{{}^0}$. It was found that when the system was released, block $m_1$ remained at rest. The frictional force between block $m_1$ and the inclined plane is:

• A. $\frac{3mg}{2}$
• B. $3mg$
• C. $\frac{4mg}{3}$
• D. $\frac{mg}{2}$

Answer 1

The correct option is A $\frac{3mg}{2}$

By applying force balance equation:

On mass, $m_2=2m$

$2mg=Tequation\left(1\right)$

On mass, $m_1=m$

$N=mgcos\theta \left(normalreactionforce\right)f=\mu Nf=\mu mgcos\theta$

$mgsin\theta +\mu mgcos\theta =Tequation\left(2\right)$

By equating equation $1$ and $2$:

$mgsin\theta +\mu mgcos\theta =2mgsin\theta +\mu cos\theta =2put\theta =30weget\mu =\sqrt{3}$

By putting value of $\mu =\sqrt{3}$ in $f=\mu mgcos\theta$, we get:

$f=\sqrt{3}\times \frac{\sqrt{3}}{2}mg=\frac{3}{2}mg$