# 2 boxes ontop of each other on a ramp with friction and Force of pull

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1. Oct 18, 2014

### tanh^-1(x)

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 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 0.435, and the coefficient of static friction between the two boxes is 0.756.

There is a rope attached to the lower box and you are standing higher up the ramp holding the rope tugging on the box.
The length of the ramp is 4.75m and height is 2.5m

The top box is 32kg the bottom one is 48 kg.

This is Mastering Physics Homework.

What force do you need to exert to accomplish this? <------please answer this one 1st
What is the magnitude of the friction force on the upper box?
What is the direction of the friction force on the upper box?

2. Relevant equations
F=ma
Ffriction=mu*Fnormal

3. The attempt at a solution
SO I calculated the angle of the ramp from the horizontal to be 27.76 degrees.
I do:
Net Force X = -80*9.8*sin(angle) + Fpull + Fn*.756 = 0 (1)
Net Force Y = Fn - 80*9.8cos(angle) = 0 (2)
I solved through the 2nd equation to find Fn to be 693.77 N
I solve for Fpull to be -365.16 N and I plugged in 365 but it is wrong, and the negative sign usually is not the problem for Mastering Physics, but I know if I get a negative sign I'm doing something wrong.
What am I doing wrong?

#### Attached Files:

• ###### ramp.jpg
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2. Oct 18, 2014

### BvU

Hello Tan, and welcome to PF :-)

I think you're doing fine, but it looks as if Fn*.756 got lost !
So: what happened to the friction force, which you and I calculate to be 301 N ?

3. Oct 18, 2014

### tanh^-1(x)

HI,
Yes, thank you for the fn*.756! However, I still get -159.3 as my answer, which is also wrong. How do you calculate friction force to be 301 N? I calculated it to be 693 N, which is not what you said. Maybe you mean:
Net Force X = -80*9.8*sin(angle) + Fpull + Fn*.435 = 0 (3)
This way I get:
Net Force X = -365.16 + Fpull + 301.78 = 0 (4)
Fpull = 63.38 N

4. Oct 18, 2014

### tanh^-1(x)

It worked! Thanks for the 301 N! I was calculating Fn * .756 not by .435, so I got the 1st answer right!

5. Oct 18, 2014

### tanh^-1(x)

Ummm... Now how do I do problems #2 and #3?
I made two more equations but they only account for the smaller box of weight 32kg,
Net Force X = 0 = -32*9.8*sin(27.76) + Fn*.756 (5)
Net Force Y = 0 = -32*9.8*cos(27.76) + Fn (6)
So through (6) I get Fn= 277.5 N, plugging that into (5) doesn't equal 0 so its definitely wrong...

6. Oct 19, 2014

### BvU

Sorry about the .756 (I did a copy/paste). As you can reconstruct, I used the .435 for the calculation.

Now, the static coefficient of friction tells you something about the friction force, namely the maximum value. If that isn't exceeded by the net result of the other forces, no acceleration occurs. So this is in fact a two step process: first check, then apply Fnet, total = ma.
(Where, in this case a = 0).