A woman at an airport is towing her 20.5 kg suitcase at constant speed by pullin

Question

A woman at an airport is towing her 20.5 kg suitcase at constant speed by pulling on a strap at an angle of ? above the horizontal. She pulls on the strap with a 25.0 N force, and the friction force on the suitcase is 20.0 N.


What angle does the strap make with the horizontal?
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Explanation / Answer

force the woman pulls with on the strap: F1 = 38.0N horizontal part of this force: F1x = Cos(?) * F1 constant speed => F1x = friction force = 20.0N angle of strap with the horizontal: ? Cos (?) = F1x / F1 = 20.0N / 38.0N ? = arccos (F1x / F) = arccos (20.0N / 38.0N) = 58.2° normal force: N vertical part of F1: F1y = Sin (?) * F1 weight: G = mass of suitcase * g = 22.5 kg * 9.81N/kg N = G - F1y = 22.5kg * 9.81N/kg - Sin (?) * F1 = 188.4N

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