You are lowering two boxes, one on top of the other, down the ramp by pulling on

Question

You are lowering two boxes, one on top of the other, down the ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 20.0 . The coefficient of kinetic friction between the ramp and the lower box is 0.492, and the coefficient of static friction between the two boxes is 0.848. A) What force do you need to exert to accomplish this? B) What is the magnitude of the friction force on the upper box? C) What is the direction of the friction force on the upper box?

Explanation / Answer

The force down the ramp = (m1 + m2) . g . sin ø Friction force up the ramp = µ ( m1 + m2 ) . g . cos ø At constant speed, any constant speed , the force up the ramp = the force down the ramp Tension ( the answer you want ) = (m1 + m2 ) . g ( sin ø – 0.251 cos ø ) The friction force on the upper box will be acting to prevent the box sliding forward. The friction force will be acting up the slope.

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