A) \[{{\tan }^{-1}}\mu \frac{\mu mg}{\sqrt{(1+{{\mu }^{2}})}}\]
B) \[{{\tan }^{-1}}\mu ,\frac{mg}{\sqrt{(1+{{\mu }^{2}})}}\]
C) \[{{\tan }^{-1}}\mu ,\frac{\mu mg}{\sqrt{(1-{{\mu }^{2}})}}\]
D) \[{{\tan }^{-1}}\mu ,\frac{mg}{\sqrt{(1-{{\mu }^{2}})}}\]
Correct Answer: A
Solution :
\[\mu =\tan \theta \] \[\therefore \] \[\theta ={{\tan }^{-1}}\mu \] \[\therefore \] \[\sin \theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\] and \[\cos \theta =\frac{1}{\sqrt{1+{{\mu }^{2}}}}\] \[{{F}_{\min }}=\frac{\mu mg}{\cos \theta +\sin \theta }\] \[=\frac{\mu mg}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{{{\mu }^{2}}}{\sqrt{1+{{\mu }^{2}}}}}\] \[=\frac{\mu mg}{\sqrt{1+{{\mu }^{2}}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
