A 400 lb crate is on a 20 degree incline. If a 120 lb horizontal force (P), push

Question

A 400 lb crate is on a 20 degree incline. If a 120 lb horizontal force (P), pushes on the crate (upwards), determine the normal and frictional forces acting on the crate. Take friction of surface = 0.3 and friction of k = 0.2? What is "friction of k" and how does it work into the reaction equations?

Explanation / Answer

Force due to P along the incline = PCos20 = 120*Cos20 = 112.8 lb Force due to P perpendicular to incline = PSin20 = 20*Sin20 = 41 lb Component of weight along the incline = 400Sin20 = 136.8 lb Component of weight normal to incline = 400*Cos20 = 375.9 lb Net normal force = 41 + 375.9 = 416.9 lb Net force along the incline = 136.8 - 112.8 = 24 lb Net static frictional force = 0.3*normal force = 0.3*416.9 = 125 lb > 24 lb Since static frictional force > Net force along the incline, the object will not move. "Friction of k" is the kinetic coefficient of friction. Once the net force along the incline becomes greater than 125 lb, then friction force will reduce to 0.2*normal force = 0.2*416.9 = 83.4 lb.

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