Laws Of Motion Question 389
Question: If the coefficient of friction between A and B is $ m _{2} $ , the maximum horizontal acceleration of the wedge A for which B will remain at rest with respect to the wedge is:
Options:
A) $ m _{1}=m _{2} $
B) $ m _{2} $
C) $ m _{1}=m _{2} $
D) $ m _{2} $
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Answer:
Correct Answer: B
Solution:
FBD of block B with respect to wedge A, for maximum ‘a’ perpendicular to wedge:
$ \sum {f _{y’}}=( mg\cos \theta +ma\sin \theta -N )=0 $ and
$ \sum {f _{x’}}=mg\sin \theta +\mu N-ma\cos \theta =0 $
(for maximum a)
$ T’=2,\text{s}$
$ mg\sin \theta +\mu ( mg\cos \theta +ma\sin \theta)-ma\cos \theta =0 $
$ \Rightarrow a=\frac{( g\sin \theta +\mu g\cos \theta)}{\cos \theta -\mu \sin \theta } $
for $ \frac{T _{1}}{T _{2}}=\sqrt{\frac{g+\frac{g}{4}}{g}}=\frac{\sqrt{5}}{2} $
$ \Rightarrow a=g( \frac{\tan 45{}^\circ +\mu }{\cot 45{}^\circ -\mu } );a=g( \frac{1+\mu }{1-\mu } ) $
BETA
