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, $g=10m{s}^{-2}$ )

• 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,
$f=\mu \left(W+F\sin 60^{\circ }\right)$
$F\cos 60^{\circ }=\frac{1}{2\sqrt{3}}\left(10\sqrt{3}+F\sin 60^{\circ }\right)$
$\Rightarrow F\times \frac{1}{2}=\frac{1}{2\sqrt{3}}\left(10\sqrt{3}+F\times \frac{\sqrt{3}}{2}\right)$
$$\begin{gathered} \left\{\because \cos 60^{\circ }=\frac{1}{2} \\ \sin 60^{\circ }=\frac{\sqrt{3}}{2}\right\} \end{gathered}$$
$\Rightarrow \frac{F}{2}=\frac{1}{2\sqrt{3}}\times \sqrt{3}\left(10+\frac{F}{2}\right)$
$\Rightarrow \frac{F}{2}=\frac{1}{2}\frac{(20+F)}{2}$
$\Rightarrow F=\frac{20+F}{20}$
$\Rightarrow 2F=20+F$
$\Rightarrow 2F-F=20$
$\therefore F=20N$