JEE Main & Advanced
Physics
NLM, Friction, Circular Motion
Question Bank
Self Evaluation Test - Laws of Motion
question_answer
The minimum force required to start pushing a body up rough (frictional coefficient u) inclined plane is \[{{F}_{1}}\]while the minimum force needed to prevent it from sliding down is\[{{F}_{2}}\]. If the inclined plane makes an angle \[\theta \] from the horizontal such that\[\tan \theta =2\mu \] then the ratio \[\frac{{{F}_{1}}}{{{F}_{2}}}\] is
A)1
B)2
C)3
D)4
Correct Answer:
C
Solution :
[c] For the upward motion of the body \[\operatorname{mg} sin \theta + {{f}_{1}} = {{F}_{1}}\] or, \[{{F}_{1}} = mg sin \theta +\mu \,\,mg cos \theta \] For the downward motion of the body, \[\operatorname{mg}\,\,sin\,\,\theta -{{f}_{2}}={{F}_{2}}\] \[or\text{ }{{F}_{2}}=\text{ }mg\text{ }sin\text{ }\theta \text{ }-\text{ }\mu \,\,mg\text{ }cos\text{ }\theta \] \[\therefore \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{sin\theta +\mu \,\,cos\,\,\theta }{sin\theta -\mu \,\,cos\,\,\theta }\] \[\Rightarrow \frac{\tan \theta + \mu }{\tan \theta - \mu }= \frac{2\mu +\mu }{2\mu -\mu }=\frac{3\mu }{\mu }=3\]