A) \[\theta ={{\tan }^{-1}}\left( \mu \right),F=\frac{\mu W}{\sqrt{1+{{\mu }^{2}}}}\]
B) \[\theta ={{\tan }^{-1}}\left( \frac{1}{\mu } \right),F=\frac{\mu W}{\sqrt{1+{{\mu }^{2}}}}\]
C) \[\theta =0,F=\mu W\]
D) \[\theta ={{\tan }^{-1}}\left( \frac{\mu }{1+\mu } \right),F=\frac{\mu W}{1+\mu }\]
Correct Answer: A
Solution :
[a] Let the force F is applied at an angle \[\theta \] with the horizontal. |
For horizontal equilibrium,\[F\cos \theta =\mu R\] (i) |
For vertical equilibrium, |
\[R+F\sin \theta =mg\]or\[R=mg-F\sin \theta \] (ii) |
Substituting this value of R in eq. (i), we get |
\[F\cos \theta =\mu (mg-F\sin \theta )\] |
\[=\mu \,mg-\mu \,F\sin \theta \] |
or,\[=F(\cos \theta +\mu \sin \theta )=\mu mg\] |
or,\[F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }\] (iii) |
For F to be minimum, the denominator \[(\cos \theta +\mu \sin \theta )\] should be maximum. |
\[\therefore \]\[\frac{d}{d\theta }(\cos \theta +\mu \sin \theta )=0\] |
or\[-\sin \theta +\mu \cos \theta =0\]or\[\tan \theta =\mu \] |
or\[\theta ={{\tan }^{-1}}(\mu )\] |
Then, \[\sin \theta =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\] and\[\cos \theta =\frac{1}{\sqrt{1+{{\mu }^{2}}}}\] |
Hence,\[{{F}_{\min }}\] |
\[=\frac{\mu }{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{1}{\sqrt{1+{{\mu }^{2}}}}}=\frac{\mu w}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec