Question

In the figure shown if coefficient of friction is $\mu$, then $m_2$ will start moving upwards if

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

Answer 1

The correct option is B $\frac{m_1}{m_2}>\sin \theta +\mu \cos \theta$

$\begin{array}{c}{m}_{1}gactingdownward,andTactingupwardfor{m}_2.\\ m_{2}g\sin \theta actingparalleltotheplanedownward,andTactingupwardparalleltothesurfaceplane.\\ Nowifthereisfriction,addblockisposedtomoveup,thenfrictionforcewillactdownward.\\ Maximumforceworkingdownthheincline\\ m_{2}g\left(\sin \theta +\mu \cos \theta \right)\\ Nowfortheblocktojustmovewemusthave\\ m_{1}g-T>m_{2}g\left(\sin \theta +\mu \cos \theta \right)-T\\ m_{1}g>m_{2}g\left(\sin \theta +\mu \cos \theta \right)\\ or,\frac{m_1}{m_2}>\left(\sin \theta +\mu \cos \theta \right)\end{array}$

Hence, Option $B$ is correct Answer.