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 10.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.491, and the coefficient of static friction between the two boxes is 0.754

What force T do you need to exert to accomplish this?

What is the magnitude f of the friction force on the upper box?

What is the direction of the friction force on the upper box?

Explanation / Answer

theta = tan-1(2.5/4.75) = 27.76 degrees

force of gravity on top block = 32*gsin(theta) = 146 N

the only other force acting on the top block is the friction force between the blocks

it must balance teh gravity for it it to move at constant speed

friction on the top block = 146 N up the incline

so forces on bottom block are

friction force between blocks = 146 N downwards on bottom block

friction force between bottom block and incline f => acts upwards =u*N = 0.491 * (m1+m2)*9.8*cos(theta)

f + T = 48*9.8*sin(27.7) + 146

T = 364.73 - 340.64 =

T = 24.1 N

firction force on upper box = 146 N

direction = upwards

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