Question

A block of mass $\sqrt{3}$ kg is kept on a frictional surface with $\mu =1/2\sqrt{3}$ . The minimum force to be applied (as shown) to move the block is:-

• A. $5N$
• B. $20N$
• C. $10N$
• D. $20/3N$

Answer 1

The correct option is D $20/3N$

$f=Fcos60$

Normal force is $N=-F\sin {60}^o+mg$
So, $F\cos {60}^o=\mu (mg-F\sin {60}^0)$
$F=\frac{\mu mg}{\cos {60}^0+\mu \sin {60}^0}=\frac{\frac{1}{2\sqrt{3}}(\sqrt{3}\times 10)}{\frac{1}{2}+\frac{1}{2\sqrt{3}}\frac{\sqrt{3}}{2}}=\frac{20}{3}N$