Question

A block of mass $\sqrt{3}kg$ resting on a horizontal surface. A force $F$ is applied on the block as shown in figure. If coefficient of friction between the block be $\frac{1}{2\sqrt{3}}$ what can be the maximum value of force $F$ so that block does not start moving? (Take $g=10m{s}^{-2}$ )

• A. 20 N
• B. 10 N
• C. 12 N
• D. 15 N

Answer 1

Answer: (A)

From acting on block are shown in adjoining figure.

As the block does not move, hence
$F\cos 60^{\circ }=f=\mu N=\mu \left(Mg+F\sin 60^{\circ }\right)$
$\therefore F\frac{1}{2}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}0\times 10+F\frac{\sqrt{3}}{2}\right)$
On simplification, we get $F=20N$