"You are lowing two boxes, one on top of the other, down a ramp. Both boxes move together at a constant speed. The coefficient of kinetic friction between the ramp and the lower box is 0.430, and the coefficient of static friction between the two boxes is 0.779. The lower box has mass 48.0 kg, the upper box has mass 32.0 kg. What force do you need to exert to accomplish this?"
Here is the picture provided: http://i.imgur.com/rL9pxAo.jpg
F = ma, F(friction) = F(normal) * coefficient of friction
The Attempt at a Solution
I[/B] found the angle from the horizontal up to be 27.25 degrees. Next I made free body diagrams for both blocks with the x axis being aligned with the ramp.
X: (48)(9.8)sin(27.25) = F(tension from me) + F(friction, bottom)
Y: (48)(9.8)cos(27.25) + F(from top block) = F(normal from ramp)
X: (32)(9.8)sin(27.25) = F(friction, top)
Y: (32)(98)cos(27.25) = F(normal from lower box)
Now I am not sure how to accomplish this. I tried to find the normal force that the top box experiences from the lower box, and equate this with the downward force that the lower box experiences. Using that downward force and the downward component of weight, I found what I thought to be a normal force. I then multiplied that by 0.430, the coefficient of kinetic friction, and said that that frictional force summed with the tension would be equal to the x component of the lower boxes weight. However, that didn't seem to work. I've arrived at several wrong answers. 121 N, 299 N, 59.97 N, and 514 N all by trying slightly different things. None worked. I only have one attempt left at the problem online and am in serious need of some help.