A) \[\left( M+m \right)g\]
B) \[\left( M + m \right) g\,sin\,\theta \]
C) \[\left( M+m \right)g\,cos\,\theta \]
D) \[\left( M+m \right)g\,\tan \,\theta \]
Correct Answer: D
Solution :
If we work in frame of wedge the free body diagram for small block is as follows For block to remain stationary \[\operatorname{ma}\,cos \theta =mg\,sin \theta \] \[\Rightarrow ~~a=\,\,g\,\,tan\,\theta \] ... (i) Now, we assume whole system \[\left( M+m \right)\] is moving with acceleration a \[\operatorname{F}=\left( M+m \right)a\,\,\Rightarrow \,\, a=\,\,\frac{F}{(M+m)}\] ... (ii) By Eqs. (i) and (ii), we get \[\operatorname{F}=\,\,\left( M+m \right)g\,tan\,\theta \]You need to login to perform this action.
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