Question

A 21 kg chair initially at rest on a horizontal floor requires a 175 N horizontal force to set it in motion. Once the chair is in motion, a 137 N horizontal force keeps it moving at a constant velocity. The acceleration of gravity is 9.81 m/s 2 . a) What is the coefficient of static friction between the chair and the floor?

Answer 1

Answer:$\mu =0.85$

Explanation:

Given

mass of chair m=21 kg

Force required to set chair in motion is 175 N

Once the chair is in motion 137 N is require to move it with constant velocity

i.e. 175 N is the amount of force needed to just overcome static friction and 137 is the kinetic friction force

thus

$f_s(static\ friction)=\mu \cdot N$

where $\mu$is the coefficient of static friction and N is Normal reaction

$N=mg$

$f_s=\mu mg$

$\mu mg=175$

$\mu =\frac{175}{21\times 9.8}$

$\mu =0.85$