Question

A block of mass m is supported on a rough wall by applying a force P as shown in Fig. Coefficient of static friction between block and wall is $\mu$.
For what range of value of P, the block remains in static equilibrium?

Answer 1

We can make components of P in vertical (up) and horizontal (right).

The block under the influence of P sin$\theta$ may have a tendency to move upward or it may be assumed that $Psin\theta$ just prevents downward fall of the block. Therefore, there are two possibilities.

Case I: If $Psin\theta >mg$, the block has tendency to move up. In this case, force of friction is downward.

From conditions of equilibrium of block,

$\sum F_x=N-Pcos\theta =0$

$N=Pcos\theta$

$\sum F_y=Psin\theta \mu N-mg=0$

$Psin\theta -\mu Pcos\theta -mg=0$

$P_{max}=\frac{mg}{sin\theta =\mu cos\theta }$

Case II

If $Psin\theta <mg$. the block has tendency to move down. In this case, the friction force acts upward.

$\sum F_x=N-Pco\theta =0$

$N=Pcos\theta$

$\sum F_y=Psin\theta +\mu N=mg=0$

$Psin\theta +\mu Pcos\theta -mg=0$

$P_{min}=\frac{mg}{sin\theta +\mu cos\theta }$

Therefore, the block will be in static equilibrium for

$\frac{mg}{sin\theta +\mu cos\theta }\le P\le \frac{mg}{sin\theta -\mu cos\theta }$