Question

A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on the box but the box is observed to be at rest. What is the value of the static friction force

Answer 1

Answer:23 N

Explanation:

Given

Weight of box $W=60\ N$

Coefficient of static friction is $\mu _s=0.5$

Applied force $F=23\ N$

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction $F_s$=applied force

$F_s=23\ N$

Although it maximum value can go up to $(F_s)_{max}=\mu _sN$

$F_s=0.5\times 60$

$F_s=30\ N$