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, \[\operatorname{mg}\,\,sin\,\,\theta -{{f}_{2}}={{F}_{2}}\] \[or\text{ }{{F}_{2}}=\text{ }mg\text{ }sin\text{ }\theta \text{ }-\text{ }\mu \,\,mg\text{ }cos\text{ }\theta \] \[\therefore \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{sin\theta +\mu \,\,cos\,\,\theta }{sin\theta -\mu \,\,cos\,\,\theta }\] \[\Rightarrow \frac{\tan  \theta  + \mu }{\tan  \theta  - \mu }= \frac{2\mu  +\mu }{2\mu  -\mu }=\frac{3\mu }{\mu }=3\]