Question

Yoda pushes a 23.5 kg chair across the floor with a for F = 17.4N [W]. If the chair moves at constant velocity, what is the coefficient of friction,μ between the chair and the floor ?

Answer 1

First, write Newton's second law: the net force is zero (because there is no acceleration), but two forces are acting on the chair: the force of Yoda's attraction $F$ and the force of friction $f$:

$$0=F-f,\\ f=\mu N,\\ N=mg.\\\space\\ f=\mu mg,\\ 0=F-\mu mg,\\\space\\ \mu=\frac{F}{mg}=\frac{17.4}{23.5·9.8}=0.076.$$