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The problem is this

You are lowering two boxes, one on top of the other, down the ramp with length 4.75m and height 2.5m (The ramp has its highest point at the right. Even though this isn't relevant to the problem i wanted to include it to give the clearest picture of the situation considering i cannot post the figure) 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 0.410, and the coefficient of static friction between the two boxes is 0.777.

The mass of the upper box is 32kg and the mass of the lower box is 48kg

Attempt at a solution:

I was successful in finding the force needed to accomplish this task is 80.7N

The trouble area is in the second part of the question where they ask "What is the magnitude of the friction force on the upper box?"

So using trig i can see that the angle of elevation is around 27.759degrees

Thus the normal force acting on the upper box would be the force of the weight of the box multiplied by the cosine of the angle of elevation giving me the perpendicular-to-the-plane component which = 32kg(9.8m/s^2)COS27.759

The coefficient of static friction between the two boxes is .777 so the magnitude of friction acting on the upper box should be .777(32kg(9.8m/s^2)COS27.759) = 215.62N

This answer is wrong and I have become incredibly frustrated as everywhere I look it seems i'm doing this correctly, so i must be missing something. Once again, any help is greatly appreciated.