A) 0.5
B) 0.6
C) \[\sqrt{3}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: D
Solution :
For equilibrium, normal to plane \[N=mg\,\cos \,\] ? (1) Net force along the plane downward \[F=mg\,\sin \,\,\text{+}{{f}_{k}}\] ...(2) where\[{{f}_{k}}\]is kinetic friction but\[{{f}_{k}}\]\[\mu N=\mu \,mg\,\cos \,\,\] ...(3) from eq. (1), (2), and (3) we get \[\therefore \]\[F=mg\sin \text{ + }\mu \,\text{me cos }\]You need to login to perform this action.
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