What is the acceleration of the upper crate when the top crate begins to slide?

  • Thread starter Anne Armstrong
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In summary, the force of friction acting on the top crate is 868 N and the acceleration is 6.17 m/s2.
  • #1
Anne Armstrong
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Homework Statement


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?

Homework Equations


Ffr=μmg
F=ma

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
 
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  • #2
Anne Armstrong said:
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
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
Hm... I suppose it would be the force of friction? In that case, would I use Ffr=μmg using μk?
 
  • #4
Anne Armstrong said:
Hm... I suppose it would be the force of friction? In that case, would I use Ffr=μmg using μk?
You got it.
 
  • #5
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
Anne Armstrong said:
Alright so I used Ffr=μm1g using μk=0.63 and got Ffr=(0.63)(20 kg)(9.8 m/s2) = 123.48 N
Good.

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

Anne Armstrong said:
I don't see another way to incorporate the new tension
Does the rope pull on the top crate?
 
  • #7
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
Anne Armstrong said:
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?
Correct.

Anne Armstrong said:
So the sum of the forces in the horizontal direction would be... only the force of friction?
Exactly.
 
  • #9
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
Anne Armstrong said:
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
Exactly.
 
  • #11
Great, thank you so much for all your help (and your patience...)!
 
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Related to What is the acceleration of the upper crate when the top crate begins to slide?

1. How do you define acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity and can be positive or negative depending on the direction of the change in velocity.

2. What is the difference between constant and variable acceleration?

Constant acceleration means that the rate of change of velocity remains the same over time, while variable acceleration means that the rate of change of velocity changes over time.

3. How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. This can be represented by the equation a = (v2 - v1) / (t2 - t1), where v2 and v1 are the final and initial velocities and t2 and t1 are the final and initial times.

4. What is the relationship between force and acceleration?

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be represented by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration.

5. How does the acceleration of the first crate affect the acceleration of the second crate?

The acceleration of the first crate can affect the acceleration of the second crate if there is a force acting between them. If the first crate exerts a force on the second crate, the second crate will experience a change in velocity and therefore, acceleration. However, if the crates are not in contact or if the forces acting between them are balanced, the acceleration of the first crate will not affect the acceleration of the second crate.

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