Answer to Question #100590 in Mechanics | Relativity for Parker Freeman

Question

A 21 kg chair initially at rest on a horizontal
floor requires a 172 N horizontal force to set
it in motion. Once the chair is in motion, a
140 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

The mass of chair is 21 kg

so, force due to gravity acting on the block = mg = 21*9.81 = 206 N downwards

for vertical equilibrium equal and opposite force will be applied by surface which will be equal to 206 N in upward direction which is also called as normal force (N)

a.) F = $\mu$_sN , where N is the normal force and $\mu$_s is the coefficient of static friction

between the chair and the floor

F is given 172 N

substituting this in the equation, we get

172 = μ_s*206

μ_s = 172/206

μ_s = 0.835