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Q.
What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move?
NTA AbhyasNTA Abhyas 2022
Solution:
The forces acting on the block are shown. Since the block is not moving forward for the maximum force $F$ applied, therefore
$F \cos 60^{\circ}=f=\mu N \ldots$ (i)
(Horizontal Direction) and
$F \sin 60^{\circ}+m g=N$.....(ii)
From (i) and (ii)
$F \cos 60^{\circ}=\mu F \sin 60^{\circ}+m g$
$\Rightarrow F=\frac{\mu m g}{\cos 60^{\circ}-\mu \sin 60^{\circ}}$
$=\frac{\frac{1}{2 \sqrt{3}} \times \sqrt{3} \times 10}{\frac{1}{2}-\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2}}=\frac{5}{\frac{1}{4}}=20 N$
