A) \[(M+m)g\,\tan \,\beta \]
B) \[g\,\tan \,\beta \]
C) \[mg\,\cos \,\beta \]
D) \[(M+m)\,g\,\cos ec\,\beta \]
Correct Answer: A
Solution :
The different forces acting on mass m are shown in the adjoining figure. Acceleration of the system \[=\frac{P}{M+m}\] \[\therefore \] Force on mass, \[m=\frac{Pm}{M+m}\] Let the reaction of m on M be f. Then, \[f=\frac{Pm}{M+m}\] Here, note that a pseudo force will act on the smaller mass m as f in backward to applied force. According to figure, m will be stationary when \[f\,\cos \,\beta \,=\,mg\,\sin \,\beta \] \[\frac{Pm}{M+m}\cos \,\beta =mg\,\sin \,\beta \] \[\therefore \] \[P(M+m)\,g\,\tan \,\beta \]You need to login to perform this action.
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