Question

Find the maximum mass $m_2$ that can be hanged from the pulley such that the system remains in equilibrium. Coefficient of friction between block $m_1$ & wedge is $mu$.

• A. $m_2=m_1(sin\theta +\mu cos\theta )$
• B. $m_2=m_1(sin\theta -\mu cos\theta )$
• C. $m_2=\frac{m_1}{(sin\theta +\mu cos\theta )}$
• D. $m_2=\frac{m_1}{(sin\theta -\mu cos\theta )}$

Answer 1

The correct option is A

$m_2=m_1(sin\theta +\mu cos\theta )$

So maximum mass is asked that means if I increase the mass even a little the system should start moving. So if I increase $m_2$ little bit more than maximum, then $m_2$ block will go down and $m_1$ will have a tendency to slide up the block so friction on block $m_1$ will be maximum and in downward direction.

Lets draw free body diagram of equilibrium condition

T=$m_{2}g$

N=$m_{1}gcos\theta$ ..........(i)

$f_r=\mu N=\mu m_{1}gcos\theta$

T=$m_{1}gsin\theta +\mu m_{1}gcos\theta$ ........(ii)

solving (i) & (ii)

$m_{2}g=m_{1}gsin\theta +\mu m_{1}gcos\theta$

$m_2=m_1(sin\theta +\mu cos\theta )$