Question

A 50.0 kg person is walking horizontally with a constant acceleration of 0.25 m/s2 inside an elevator. The elevator is also accelerating downward at a rate of 1.0 m/s2. Find the force of friction between his shoes and the floor of the elevator. What minimum force of friction between his shoes and the floor would make this scenario possible?

Answer 1

The normal force is:

$$N=mg-ma_1=m(g-a_1)=50(9.8-1)=440\ N$$

The force of friction between his shoes and the floor of the elevator:

$$F_f=ma_2=(50)(0.25)=12.5\ N$$

The minimum coefficient of friction between his shoes and the floor:

$$\mu =\frac{F_f}{N}=\frac{12.5}{440}=0.028$$