3 Blocks Create a Tension Force- Find Acceleration

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 6K views
Phoenixtears
Messages
82
Reaction score
0

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?
 

Attachments

  • Three Blocks with Tnesion.gif
    Three Blocks with Tnesion.gif
    6.3 KB · Views: 604
Physics news on Phys.org
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.