Jane is dragging a 1.2 lbf suitcase across the floor by tying a rope to it and a

Question

Jane is dragging a 1.2 lbf suitcase across the floor by tying a rope to it and applying a force at an angle of 7.4 degree above the floor. She found that it takes 1.6 N of force to keep it moving at a constant velocity. What is the coefficient of dynamic friction between the floor and the box? Andrew is moving a W=3.4 lbs box by pushing the box forward and down with a force of F=4.2 N. He has found that the box will be moving with constant velocity only if the angle at which he is pushing the box is exactly alpha = 4.2 degree with respect to the horizontal floor. What is coefficient of dynamic friction between the floor and the box?

Explanation / Answer

Mass of suitcase = 1.2 lbf = 0.5443 kg
Now the force applied on the suitcase is 1.6 N
since angle made by the force with the floor is 7.4 degree
Now we resolve the forces along the vertical and horizontal direction
Taking all the forces in vertical direction
N + 1.6Sin7.4 = 0.5443*9.81
N = 5.1335
Friction force = uN =5.1335u
Since the net force is zero therefore
1.6Cos7.4 = 5.1335u
u = 0.309

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