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A 46.2-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.566 and 0.440,...Question
A 50.9-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.526 and 0.311, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?
Answers
Henry
M*g = 50.9 * 9.8 = 499 N. = Wt. of crate.
= Normal(Fn).
Fs = u*Fn = 0.526 * 499 = 264 N. = Force of static friction.
Fk = 0.311 * 499 = 155.2 N. = Force of kinetic friction.
a. (Fap-Fs) = M*a.
Fap-264 = M*0 = 0.
Fap = 264 N. = Fore applied.
b. (Fap-Fk) = M*a.
Fap-155.2 = M*0 = 0.
Fap = 155.2 N.
= Normal(Fn).
Fs = u*Fn = 0.526 * 499 = 264 N. = Force of static friction.
Fk = 0.311 * 499 = 155.2 N. = Force of kinetic friction.
a. (Fap-Fs) = M*a.
Fap-264 = M*0 = 0.
Fap = 264 N. = Fore applied.
b. (Fap-Fk) = M*a.
Fap-155.2 = M*0 = 0.
Fap = 155.2 N.