Laws Of Motion Question 400
String is massless and pulley is smooth in the adjoining figure total mass on the left hand side of the pulley is $ m _{1} $ and on right hand side is $ m _{2} $ . Friction coefficient between block B and the wedge is $ \mu =\frac{1}{2} $ and $ \theta =30{}^\circ $ . Select the wrong answer
Options:
A) Block B will slide down if $ m _{1}=m _{2} $
B) Block B may remain stationary with respect to wedge, for suitable values of $ m _{1} $ and $ m _{2} $ with $ m _{1}>m _{2} $
C) Block B cannot remain stationary with respect to wedge in any case
D) Block B will slide down if $ m _{1}<m _{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
If $ m _{1}=m _{2} $ , the block will slide down when
$ \mu g\cos \theta <g\sin \theta $ or$ \frac{1}{2}( \frac{\sqrt{3}}{2} )<\frac{1}{2} $
If $ m _{1}=m _{2} $ , the block will slide down when
$ \mu ( g+a )\cos \theta <( g-a )\sin \theta $
or $ \frac{1}{2}( \frac{\sqrt{3}}{2} )<\frac{1}{2} $ where $ a=( \frac{m _{2}-m _{1}}{m _{1}+m _{2}} )g $
therefore, in all cases, the block will slide down
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