# Find tension force on 2 boxes on a ramp

1. Feb 29, 2016

### Johnson1704

1. The problem statement, all variables and given/known data
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 19.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.457, and the coefficient of static friction between the two boxes is 0.818.

2. Questions

A What force do you need to exert to accomplish this?
B What is the magnitude of the friction force on the upper box?

3. Relevant equations

X forces: Fpull - Ffr - Fg*sin(theta) = 0
Y forces: Fn - Fg*cos(theta) = 0
Ffr = mu_k * Fn

3. The attempt at a solution (A)
I found theta to be 27.76
Ffr = mu_k * [Fg*cos(theta)]
Ffr = 317.0513 N
Fpull = Ffr + Fg*sin(theta) = 6782.214 N

The solutions i've seen on this forum say to use -Fgsin(theta) but i don't see why it would be negative. I've checked my work so many times

Last edited: Feb 29, 2016
2. Feb 29, 2016

### Johnson1704

bump

3. Feb 29, 2016

### Johnson1704

please brahs, it's been a solid 2 hours i've been on this problem

4. Feb 29, 2016

### PhanthomJay

in what direction is:
the force of the pull on the boxes?
the force of the gravity component acting parallel to the plane?
the direction of motion of the boxes?
the direction of the friction force?

Show how you are arriving at these numbers step by step.

5. Feb 29, 2016

### haruspex

You did not attach the figure, so you need to describe the set up in more detail. "Lowering" suggests you are pulling up the ramp, parallel to it perhaps, to prevent the boxes accelerating down. Or maybe the rope is vertical?
You say you found the slope to be 27.76 degrees - based on what?

6. Mar 8, 2016

### Johnson1704

I figured it out about 30 mins after I posted and deleted the attachment and was gonna delete the post but couldn't figure it out. Thanks though!!