Q: Two blocks of masses m1 and m2 connected by a string and placed on an rough inclined plane having coefficient of friction μ as shown in figure. The ratio of masses m1/m2 so that the block m1 starts moving downward.
(a) $\displaystyle \frac{m_1}{m_2} > (\mu cos\theta – sin\theta ) $
(b) $ \displaystyle \frac{m_1}{m_2} > ( sin\theta +\mu cos\theta ) $
(c) $ \displaystyle \frac{m_1}{m_2} > ( cos\theta – \mu sin\theta ) $
(b) $ \displaystyle \frac{m_1}{m_2} > ( sin\theta -\mu cos\theta ) $
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Ans: (a)
Sol: As block m1 starts moving downwards , so
$ \displaystyle m_2 g sin\theta > \mu N + m_1 g $
$ \displaystyle m_2 g sin\theta > \mu m_2 g cos\theta + m_1 g $
$ \displaystyle m_2 g (sin\theta – \mu cos\theta ) > m_1 g $
$ \displaystyle m_2 (sin\theta – \mu cos\theta ) > m_1 $
$ \displaystyle (sin\theta – \mu cos\theta ) > \frac{m_1}{m_2} $
$ \displaystyle – ( \mu cos\theta – sin\theta ) > \frac{m_1}{m_2} $
$ \displaystyle ( \mu cos\theta – sin\theta ) < \frac{m_1}{m_2} $