question_answer

A block placed on a horizontal surface is being pushed by a force F making an angle \[\theta \]with the vertical. If the friction coefficient is u, how much force is needed to get the block just started. Discuss the situation when \[\tan \theta <\mu \], Fig. 3(HT).13.

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.