Question

A block of mass $3\, kg$ is pressed against a vertical wall by applying a force, $F$, at an angle $30^{\circ}$ to the horizontal as shown in the figure. As a result, the block is prevented from falling down. If the coefficient of static friction between the block andwall is $\sqrt{3}$, then the value of $F$ is (use $g=10 \,m / s$ )

• A. $30\, N$
• B. $15 \sqrt{3} \,N$
• C. $60 \sqrt{3} \,N$
• D. $60\, N$

Answer 1

Answer: (A)

The given situation is shown below

Normal reaction is
$N=F \cos \theta=F \cos 30^{\circ}$
Friction force when block is just on the verge of sliding is
$f=\mu N=\mu F \cos 30^{\circ}=\frac{3}{2} F $
$[\therefore \mu=\sqrt{3}]$
As block is in equilibrium,
friction= weight of block $(m g)+$ vertical component of force $(F \sin \theta)$
$\Rightarrow \frac{3}{2} F=3 \times 10+F \times \frac{1}{2} $
$\Rightarrow F=30 \,N$