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A block of mass $3 \mathrm{~kg}$ 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 and wall is $\sqrt{3}$, then the value of $F$ is (use, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$30 \mathrm{~N}$
The given situation is shown below
Normal reaction is
$\mathbf{N}=F \cos \theta=F \cos 30^{\circ}$
Friction force when block is just on the verge of sliding is
$f=\mu N=\mu F \cos 30^{\circ}=\frac{3}{2} F \quad[\therefore \mu=\sqrt{3}]$
As block is in equilibrium,
friction= weight of block $(m g)+$ vertical
component of force $(F \sin \theta)$
$\Rightarrow \quad \frac{3}{2} F=3 \times 10+F \times \frac{1}{2} \Rightarrow F=30 \mathrm{~N}$
Normal reaction is
$\mathbf{N}=F \cos \theta=F \cos 30^{\circ}$
Friction force when block is just on the verge of sliding is
$f=\mu N=\mu F \cos 30^{\circ}=\frac{3}{2} F \quad[\therefore \mu=\sqrt{3}]$
As block is in equilibrium,
friction= weight of block $(m g)+$ vertical
component of force $(F \sin \theta)$
$\Rightarrow \quad \frac{3}{2} F=3 \times 10+F \times \frac{1}{2} \Rightarrow F=30 \mathrm{~N}$
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