Question

A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface of the block is 0.25. The friction force acting on the block is :

• A. 15 N downwards
• B. 25 N upwards
• C. 20 N downwards
• D. 20 N upwards

Answer 1

Answer: (C)

20 N downwards

Net force in upward direction $f_a=Fsin30-mg$
$=100\times \frac{1}{2}-3\times 10=20N$
Limiting friction $F_l=\mu Fcos{30}^o$
$=\frac{1}{4}\times 100\frac{\sqrt{3}}{2}=21.15N$
Hence block will not slip. Hence
$f_r=f_a=20N$