3 Blocks Create a Tension Force- Find Acceleration

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Homework Help Overview

The problem involves determining the acceleration of a 2.6 kg block on a table, considering the effects of kinetic friction and tension in a system of three blocks. The original poster attempts to analyze the forces acting on each block using free body diagrams and Newton's second law.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the role of tension in the system. There is a focus on the differences in tension across the blocks and the implications of these differences on the calculations. Questions arise regarding the validity of substituting tension values and the potential oversight of forces acting on the middle block.

Discussion Status

Participants are actively engaging with each other's reasoning, questioning assumptions about tension and its effects on the blocks. Some guidance has been provided regarding the need to account for multiple tension forces acting on the middle block, indicating a productive direction in the discussion.

Contextual Notes

The original poster expresses concern about the accelerations potentially not being equal and is considering whether to approach the problem with a single equation or separate equations for each block. There is an acknowledgment of a misinterpretation regarding the tension forces involved.

Phoenixtears
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Homework Statement


The coefficient of kinetic friction between the m = 2.6 kg block in Figure P8.35 and the table is 0.28. What is the acceleration of the 2.6 kg block? (The image is attached)
______m/s2


Homework Equations



Fk= (Mu)(N)
2nd Law statements

The Attempt at a Solution


I began by drawing three force diagrams, one for each block. Using the 2.6 and 3 blocks I wrote out 2nd law statements (because the block is shifting right), and substituted in variables for the friction forece equation:

Ca= T- Cg
Ba= T- Fk
Fk=(Mu)(Bg)... (According to my force diagrams , normal force equals weight force)

Then I sovled for T on the first equation. Substituting that in for T in the second, and replacing the Fk with the other equation:

Ba= Ca+ Cg- (Mu)(Bg)

I then solved for a. After plugging in all the numbers, I got a very high number as my a. Near 55, if I recall. What exactly did I do wrong?
 

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Phoenixtears said:
Then I sovled for T on the first equation. Substituting that in for T in the second,
The tension is different for each rope. (If not, there would be no net force on the middle block.)
 
Doc Al said:
The tension is different for each rope. (If not, there would be no net force on the middle block.)

But, according to the 3rd law, the tension between the 3 block and the 2.6 block would be equal. So why would the substitution not work?

Thanks in advance!
 
Phoenixtears said:
But, according to the 3rd law, the tension between the 3 block and the 2.6 block would be equal.
Right, I misinterpreted what T you were talking about.
So why would the substitution not work?
Because there are two tension forces on the middle block.

Phoenixtears said:
Ba= T- Fk
You left out one of the tension forces.
 
Doc Al said:
Because there are two tension forces on the middle block.


You left out one of the tension forces.

Right. That makes sense. I suppose I'm thinking horizontally (a different problem I had to do). Thank you! Now, is it possible to include the third equation? Or should I just do the two separately. My fear is that the accelerations won't equal. I'll try it out right now.
 

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