3 Blocks Create a Tension Force- Find Acceleration

1. Oct 4, 2008

Phoenixtears

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. 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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2. Oct 5, 2008

Staff: Mentor

The tension is different for each rope. (If not, there would be no net force on the middle block.)

3. Oct 5, 2008

Phoenixtears

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?

4. Oct 5, 2008

Staff: Mentor

Right, I misinterpreted what T you were talking about.
Because there are two tension forces on the middle block.

You left out one of the tension forces.

5. Oct 5, 2008

Phoenixtears

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.