A) \[mg\cos \theta \]
B) \[mg\sin \theta \]
C) \[mg\]
D) \[mg/\cos \theta \]
Correct Answer: D
Solution :
When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right. For the condition of equilibrium \[mg\sin \theta =ma\cos \theta \] Þ \[a=\frac{g\sin \theta }{\cos \theta }\] \[\therefore \] Force exerted by the wedge on the block \[R=mg\cos \theta +ma\sin \theta \] R\[=mg\cos \theta +m\left( \frac{g\sin \theta }{\cos \theta } \right)\sin \theta \]\[=\frac{mg({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta }\] R\[=\frac{mg}{\cos \theta }\]You need to login to perform this action.
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