Question
A 55 kg box is pulled with constant speed across the floor with a rope that is at an angle of 40 degrees above the horizontal. If the tension in the rope is 125n what is the coefficient of friction between the box and the floor?
Answers
pull force = 125 cos 40
lift force = 125 sin 40
Fdown on floor= 55*9.81 - 125 sin 40
friction force back = Fdown*mu
F = m a = 0
so
125 cos 40 = Fdown * mu
lift force = 125 sin 40
Fdown on floor= 55*9.81 - 125 sin 40
friction force back = Fdown*mu
F = m a = 0
so
125 cos 40 = Fdown * mu
95.75 N correct??
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