Question

Two blocks $A$ & $B$ are connected by a light inextensible string passing over a fixed smooth pulley as shown the figure. The coefficient of friction between the block $A$ and the horizontal table is $\mu =0.2$. If the block $A$ is just to slip, find the ratio of the masses of the blocks.

Answer 1

Let the masses be $m_A,m_B$ respectively.

For just slipping case, driving force just equals the frictional force.

$T=m_{B}g$,

Driving force$=T\cos \left({60}^0\right)=\frac{m_{B}g}{2}$

Normal acting on $A=m_{A}g-T\sin \left({60}^0\right)=m_{A}g-m_{B}g\sin \left({60}^0\right)$

$f=\mu N=0.2\times g\times m_A-m_B\sin \left({60}^0\right)$

$f=T\cos \left({60}^0\right)$

$2(m_A-m_B\frac{\sqrt{3}}{2})=m_B\times \frac{10}{2}$

$\frac{m_A}{m_B}-\frac{\sqrt{3}}{2}=\frac{5}{2}$

$m_A:m_B=\frac{5}{2}+\frac{\sqrt{3}}{2}$