A) \[2W\,\sin \alpha \]
B) \[W\,\sin \,\alpha \]
C) \[\frac{\sqrt{3}}{2}W\,\sin \,\alpha \]
D) \[W\sqrt{3}\,\sin \,\alpha \]
Correct Answer: D
Solution :
[d] Net force: \[F=\sqrt{{{(Wsin\alpha )}^{2}}+{{H}^{2}}}\] When the particle is about to slip: \[F=\mu N\] \[\Rightarrow \sqrt{{{(Wsin\alpha )}^{2}}+{{H}^{2}}}=(2tan\alpha )Wcos\alpha \] \[\Rightarrow H=\sqrt{3}W\sin \alpha \]You need to login to perform this action.
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