# Two accelerating crates

1. Sep 23, 2015

### Anne Armstrong

1. The problem statement, all variables and given/known data
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 20 kg and the larger bottom crate has a mass of m2 = 80 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 and the coefficient of kinetic friction between the two crates is μk = 0.63. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). Tension is initially 232 N. If the tension is increased in the rope to 1085 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

2. Relevant equations
Ffr=μmg
F=ma

3. The attempt at a solution
Based on previous questions, I found that the maximum acceleration at which the crates can move without the top crate sliding is 7.74 m/s2 and the maximum tension at which the crates can be pulled without the top crate sliding is 46.4 N.
I attempted to use the same approach to solve this problem:
F=m1,2*a --> 1085 N = (80 kg + 20 kg)*a --> a=10.85 m/s2
F=m1*a --> F=(20 kg)*(10.85 m/s2) --> F=217 N =the net force on the top crate
Fnet=FT-Ffr or FT-Fnet=Ffr
so 1085 N - 217 N = Ffr --> Ffr=868 N
Ffr=m1*a --> 868 N = (20 kg)*a --> a=43.4 m/s2

I have a feeling there is an error in the first step (finding the acceleration to be 10.85 m/s2, but I'm not sure. I also realized after I finished that I didn't include any coefficients of friction, which I don't think is correct... I hope my process was clear, I'd appreciate any help! Thanks

2. Sep 23, 2015

### Staff: Mentor

That would be the acceleration if the crates didn't slide with respect to each other. But they do.

Start over. You are told that the crates are sliding. So what's the horizontal force acting on the upper crate?

3. Sep 23, 2015

### Anne Armstrong

Hm... I suppose it would be the force of friction? In that case, would I use Ffr=μmg using μk?

4. Sep 23, 2015

### Staff: Mentor

You got it.

5. Sep 23, 2015

### Anne Armstrong

Alright so I used Ffr=μm1g using μk=0.63 and got Ffr=(0.63)(20 kg)(9.8 m/s2) = 123.48 N
This would be the force of friction acting on the top crate, so would I then use Ffr=Ft-Fnet to find the Fnet, the sum of the forces acting on the top crate? I don't see another way to incorporate the new tension

6. Sep 23, 2015

### Staff: Mentor

Good.

First things first: What forces act on the top crate?

Does the rope pull on the top crate?

7. Sep 23, 2015

### Anne Armstrong

The forces acting on the top crate are the force of friction from the bottom crate, the force of gravity downward, and the normal force upward, correct? So the sum of the forces in the horizontal direction would be... only the force of friction?

8. Sep 23, 2015

### Staff: Mentor

Correct.

Exactly.

9. Sep 23, 2015

### Anne Armstrong

So would I just use F=m1a to find the acceleration?
Aka: Ffr=m1a --> 123.48 N = (20 kg)a --> a=6.17 m/s2

10. Sep 24, 2015

### Staff: Mentor

Exactly.

11. Sep 24, 2015

### Anne Armstrong

Great, thank you so much for all your help (and your patience...)!!