A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 76-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 59 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

Answer 1

Wc = m*g = 22kg * 9.8N/kg = 215.6 = Wt.
of crate.

Fc = 215.6N.[0o] = Force of crate.
Fp = 215.6*sin(0) = 0 = Force parallel to floor.
Fv = 215.6*cos(0) = 215.6 N. = Force
perpendicular to floor.

Fs = u*Fv = Force of static friction.

Fap-Fp-u*Fv = m*a
76-0-u*215.6 = m*0 = 0.
u*215.6 = 76
u = 0.353 = Coefficient of static friction.

59-0-u*215.6 = m*a =m*0 = 0
u*215.6 = 59
u = 0.274 = Coefficient of kinetic friction.