Question

The maximun value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \,ms ^{-2}$ )

• A. 20 N
• B. 15 N
• C. 25 N
• D. 10 N

Answer 1

Answer: (A)

Applied force on the block on a horizontal surface, as shown in the figure, we have

As we know that, force of friction, $f=\mu R$
From the diagram,
$ f=\mu\left(W+F \sin 60^{\circ}\right) $
$F \cos\, 60^{\circ}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \sin 60^{\circ}\right) $
$ \Rightarrow F \times \frac{1}{2}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \times \frac{\sqrt{3}}{2}\right) $
$\left\{ \because \cos 60^{\circ}=\frac{1}{2} \\ \sin 60^{\circ}=\frac{\sqrt{3}}{2} \right\} $
$ \Rightarrow \frac{F}{2}=\frac{1}{2 \sqrt{3}} \times \sqrt{3}\left(10+\frac{F}{2}\right) $
$ \Rightarrow \frac{F}{2}=\frac{1}{2} \frac{(20+F)}{2}$
$ \Rightarrow F=\frac{20+F}{20} $
$ \Rightarrow 2 F=20+F $
$\Rightarrow 2 F-F=20 $
$\therefore F=20 N $