Asked by Anonymous
A force of 400 N at an angle of 15 degrees above the horizontal is needed to pull a 180 kg trunk across the floor at constant speed. What is the coefficient of kinetic friction between the trunk and the floor?
Answers
Answered by
Henry
M*g = 180 * 9.8 = 1764 N. = Wt. of trunk.
Fn = 1764 - 400*sin15 = 1661 N. = Normal force.
Fap*Cos A-Fk = M*a.
Fk = Fap*Cos A-M*a = 400*Cos15-M*0 = 386.4 N. = Force of kinetic friction.
Fk = u*Fn = 386.4.
u = 386.4/Fn = 386.4/1661 = 0.233 = Coefficient of kinetic friction.
Fn = 1764 - 400*sin15 = 1661 N. = Normal force.
Fap*Cos A-Fk = M*a.
Fk = Fap*Cos A-M*a = 400*Cos15-M*0 = 386.4 N. = Force of kinetic friction.
Fk = u*Fn = 386.4.
u = 386.4/Fn = 386.4/1661 = 0.233 = Coefficient of kinetic friction.
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