Answer to Question #100596 in Mechanics | Relativity for Jarell banman

Question

A 20 kg chair initially at rest on a horizontal
floor requires a 161 N horizontal force to set
it in motion. Once the chair is in motion, a
133 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:

Let the coefficient of static friction between the chair and the floor be $\mu$.

If the Reaction force (R) = $m$$g$

Where m - mass of the chair and g - acceleration of gravity

Then, $R = 20*9.81 = 196.2N$

From the low of motion,

$F = \mu R$

Since it is required to give 161 N horizontal force to set the chair in motion,

$161 = \mu$$*196.2$

$\mu =$$0.8206$

Therefore, the coefficient of static friction between the chair and the floor is 0.8206