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 w.r.t. 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\sec $

$ 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 } ) $



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