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 19.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.418, and the coefficient of static friction between the two boxes is 0.788.
What force do you need to exert to accomplish this?
What is the magnitude of the friction force on the upper box?
What is the direction of the friction force on the upper box?

Explanation / Answer

there is no fig. so i just write the formula for calculation

speed = v

ramp of length = d

height of ramp = h

uk = cofficient of kinetic friction

us = static friction

M = mass of lower box

m = mass of upper box

cos(theta) = d / sqrt(d^2 + h^2)

sin(theta) = h / sqrt( d^2 + h^2)

then

N = normal force = m*g

Fx = 0 = - (M+m)* g * sin(theta) + F + Fr

Fy = N - ( M+m) * g * cos(theta) = 0

Fr = uk * N

F = ( M + m) * g * sin(theta) - uk * (M + m) * g * cos(theta)

force do you need to exert to accomplish this is calculated by putting the values in this formula F = ( M + m) * g * sin(theta) - uk * (M + m) * g * cos(theta)

then

Fx = Fb - m * g * sin(theta) = 0

Fy = Nb - m * g *cos(theta) = 0

Fb = m * g * sin(theta)

the magnitude of the friction force on the upper box Fb = m * g * sin(theta)

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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