Asked by Bryce
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.
Answers
Answered by
Henry
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
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
Answered by
ty
no
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