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

Step-by-step 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* 60`

`F_s=30\ N`