Question

A block of mass $3kg$ 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 $g=10m/s$ )

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

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 $(mg)+$ vertical component of force $(F\sin \theta )$
$\Rightarrow \frac{3}{2}F=3\times 10+F\times \frac{1}{2}$
$\Rightarrow F=30N$