A block of mass M is placed on a rough horizontal surface. The...

Let the force $F$ be applied at an angle $\theta$ with the horizontal.

The normal reaction $N$ from the surface is given by equating vertical forces:

$N + F \sin\theta = Mg \Rightarrow N = Mg - F \sin\theta$

The maximum static frictional force is:

$f = \mu N = \mu (Mg - F \sin\theta)$

For the block to just slide, the horizontal component of the applied force must be equal to the maximum static friction:

$F \cos\theta = \mu (Mg - F \sin\theta)$

$F \cos\theta + \mu F \sin\theta = \mu Mg$

$F (\cos\theta + \mu \sin\theta) = \mu Mg$

$F = \dfrac{\mu Mg}{\cos\theta + \mu \sin\theta}$

To find the minimum force, the denominator $\cos\theta + \mu \sin\theta$ must be maximum. The maximum value of $a \cos\theta + b \sin\theta$ is $\sqrt{a^2 + b^2}$.

Maximum value of $\cos\theta + \mu \sin\theta = \sqrt{1 + \mu^2}$

This maximum occurs when $\tan\theta = \dfrac{\mu}{1} = \mu \Rightarrow \theta = \tan^{-1}\mu$

Substituting the maximum value of the denominator, the minimum force is:

F_\min = \dfrac{\mu Mg}{\sqrt{1 + \mu^2}}