Answer:
In
limiting equilibrium, force of friction \[f=\mu R\]
In equilibrium along the horizontal, \[F\sin
\theta =\mu R\]
and along the vertical, \[F\cos
\theta +mg=R\]
\[=\frac{F\sin \theta }{\mu }\]
\[\therefore \] \[mg=F\left( \frac{\sin \theta }{\mu
}-cos\theta \right)\] or \[F=\frac{\mu \,mg\,}{\sin \theta -\mu \cos \theta
}\]
If \[\tan \theta <\mu \]
\[\left( \sin \theta -\mu \cos
\theta <0 \right)\]
\[\therefore \] F is negative.
So, for angles less than \[{{\tan }^{-1}}\mu \],
one cannot push the block ahead, howsoever large the force may be.
You need to login to perform this action.
You will be redirected in
3 sec