A man drags a 180 lb crate across a floor by pulling on a rope inclined at 20.0 degrees with the horizontal. If the coefficient of static friction is 0.560 , and the coefficient of kinetic friction is 0.350, find the acceleration (in feet/second^2) of the crate at the instant it breaks loose and starts sliding.

Question

ty · Accepted Answer

no

Henry · Answer

Mass = 180Lb * 0.454kg/Lb = 81.72 kg.

m*g = 81.72 * 9.8 = 801 N = Wt. of crate
= Normal(Fn).

Fs = u*Fn = 0.56 * 801 = 449 N. = Force of static friction. = Hor. component of
applied force(Fx).

Fk = u*Fn = 0.350 * 801 = 280 N. = Force of kinetic friction.

a = (Fx-Fk)/m = (449-280)/81.72 = 2.07
m/s^2 = 6.82 Ft/s^2