Question

A block of mass $50kg$ can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is $0.6$. The least force of pull acting at an angle of ${30}^{\circ }$ from the horizontal which causes the block to just slide is

• A. $600$
• B. $\frac{\sqrt{3}+0.6}{600}$
• C. $\frac{600}{\sqrt{3}+0.6}$
• D. $\frac{600}{\sqrt{3}-0.6}$

Answer 1

The correct option is C $\frac{600}{\sqrt{3}+0.6}$


Net normal force acting will be $(mg-F\sin {30}^{\circ })$
Limiting friction $F=\mu N=0.6(mg-F\sin {30}^{\circ })$
Horizontal component of the applied force is $F\cos {30}^{\circ }$
To just start sliding, applied force= friction force,
$F\cos {30}^{\circ }=0.6(mg-F\sin {30}^{\circ })$
$F\frac{\sqrt{3}}{2}=(0.6\times 50\times 10)-\frac{0.6F}{2}$
$F(\frac{\sqrt{3}}{2}+\frac{0.6}{2})=300$
$F=\frac{600}{\sqrt{3}+0.6}$