Question

What is the maximum value of the force F such that the block shows in the arrangement, does not move?

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

Answer 1

The correct option is A 20 N

Let F be the maximum value of force applied when the block of m=$\sqrt{3}kg$ does not move on the rough surface.

R=normal reaction
Normal to surface.
or $R=Fsin{60}^0+mg$

f=force of friction

$\mu R=Fcos{60}^0$

$\mu (Fsin{60}^0+mg)=Fcos{60}^0$

or $\mu Fsin{60}^0+mg)=Fcos{60}^0$

or $F=\frac{\mu mg}{cos{60}^0-\mu sin{60}^0}$

$\frac{5}{\frac{1}{2}-\frac{1}{4}}$

$=5\times 4=20N.$

Maximum value of force = 20N.