A woman is pushing a packing case of mass 20 kg up a 37 degree inclined slope. T

Question

A woman is pushing a packing case of mass 20 kg up a 37 degree inclined slope. The coefficient of kinetic friction between the case and the slope is mk=0.100 and the coefficient of static friction is ms=0.200 (note: cos(37)=0.8 and sin(37)=0.6)

1.) If the woman pushes in a direction parallel to the slope, calculate the force she must apply in this direction to keep the case moving at a constant speed

2.) Calculate the MINIMUM force the woman must apply parallel to the slope to keep the case from sliding down the slope

Explanation / Answer

F - Fg - fk = 0

F = Fg + fk = m*g*sin37 + uk*m*g*c

F = (20*9.8*0.6) + (0.1*20*9.8*0.7) = 131.32 N

---------------

Fg -F - fk = 0

F = Fg - fk = m*g*sin37 - uk*m*g*c

F = (20*9.8*0.6) - (0.1*20*9.8*0.7) = 103.88 N

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