Question

A block of mass 'm' is supported on a rough wall by applying a force P as shown in figure. Coefficient of static friction between block and wall is ${\mu }_s$

For what range of values of P, the block remains in static equilibrium?

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

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 P sin$\theta$ just prevents downward fall of the block. Therefore there are two possibilities.

Case (i) If P sin $\theta$> mg, the block has a 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 =P cos$\theta$ .............(i)

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

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

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

Case (ii) If P sin $\theta$ < mg, the block has a tendency to move down.

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

N =P cos$\theta$

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

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

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

Therefore the block will be in static equillibrium for $\frac{mg}{sin\theta +\mu cos\theta }\le P\le \frac{mg}{sin\theta -\mu cos\theta }$