You are lowering two boxes, one on top of the other, down the ramp shown in the

Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.454, and the coefficient of static friction between the two boxes is 0.817.

Part A: What force do you need to accomplish this?

Part B: What is the magnitude of the friction force on the upper box

Explanation / Answer

Equations

w = mg

F = uk(m1+m2)g+us*m2g

Solution:

Well i added the two boxes masses to 80 kg and i got the angle to be 27.7 degrees

(1)

w = m(tot)*g = 784,8 N

(2)

F = uk(m1+m2)g +us*m2*g <==> F = 599.4 N

The answer should be 57.1 N

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