Physics - NLM Question with Solution | TestHub

The situation is shown in figure. In the limiting equilibrium the frictional force f will be equal to $\mu \mathrm{N}$. For horizontal equilibrium

$$\mathrm{F}\sin \theta =\mu \mathrm{N}$$

For vertical equilibrium

$$\mathrm{F}\cos \theta +\mathrm{mg}=\mathrm{N}$$

Eliminating N from these equations,

$$\mathrm{F}\sin \theta =\mu \mathrm{F}\cos \theta +\mu \mathrm{mg}$$

or,

$$F=\frac{\mu mg}{\sin \theta -\mu \cos \theta }$$

If $\tan \theta <\mu$ we have $(\sin \theta -\mu \cos \theta )<0$ and then F is negative. So for angles less than ${\tan }^{-1}\mu$, one cannot push the block ahead, however large the force may be.