Answer to Question #100598 in Mechanics | Relativity for Aaliyah Hope

Question

A 27 kg chair initially at rest on a horizontal
floor requires a 176 N horizontal force to set
it in motion. Once the chair is in motion, a
148 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

Let the coefficient of static friction and the coefficient of kinetic friction is $\mu_s$ and $\mu_k$ .

N=mg

The condition to set in the motion, the applied force = friction force

i)

$F=\mu mg=\mu_s\times 27\times 9.81$

$\mu_s=\dfrac{176}{264.87}=0.66$

ii)

$\mu_k=\dfrac{F}{mg}=\dfrac{148}{27\times 9.81}=0.56$