A) \[\frac{W\,\cos \phi }{\cos \,(\theta -\phi )}\]
B) \[\frac{W\,\sin \phi }{\cos \,(\theta -\phi )}\]
C) \[\frac{W\,\tan \phi }{\sin (\theta -\phi )}\]
D) \[\frac{W\,\sin \phi }{g\sin (\theta -\phi )}\]
E) None of these
Correct Answer: E
Solution :
\[R=(W+P\sin \theta )\] Now \[{{F}_{f}}=\mu R=\mu (W+P\sin \theta )\] But \[\mu =\tan \phi \] \[\therefore \] \[{{F}_{f}}=\tan \phi (W+P\sin \theta )\] Also, \[{{F}_{f}}=P\cos \theta \] Eliminating\[P\], we have \[{{F}_{f}}=\frac{W\sin \phi }{\cos (\theta +\phi )}\] So, none of the options is correct.
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