What force must be applied to push a carton weighing 350 N up a 19° incline, if the coefficient of kinetic friction is 0.45? Assume the force is applied parallel to the incline and the velocity is constant. (Enter the magnitude of the force.)

Answer 1

Wc = 350 N. = Wt. of carton.

Fc = 350N @ 19o. = Force of carton.
Fp = 350*sin19 = 113.95 N. = Force parallel to incline.
Fv = 350*cos19 = 330.9 N. = Force perpendicular to incline.

Fk = u*Fv = 0.45*330.9 = 148.9 N. = Force of kinetic friction.

Fap-Fp-Fk = m*a.
Fap-113.95-148.9 = m*0 = 0
Fap-262.85 = 0
Fap = 262.85 N. = Force applied.