A file cabinet weighing 1500 N is at rest on the floor. The coefficient of static friction between the floor and the cabinet is 0.5. What is the minimum force required to make the file cabinet move horizontally

1 answer

Fc = 1500 N @ 0o. = Force of cabinet.
Fp = 150*sin(0) = 0 = Force parallel to floor.
Fv = 1500*cos(0) = 1500 N. = Force perpendicular to floor = Normal.

Fs = u*Fv = 0.5*1500 = 750 N. = Force of
static friction.

Fap-Fp-Fs = m*a.
Fap-0-750 = m*0 = 0
Fap = 750 N. = Force applied.