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Homework Help: 3 Blocks Create a Tension Force- Find Acceleration

  1. Oct 4, 2008 #1
    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)

    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?

    Attached Files:

  2. jcsd
  3. Oct 5, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The tension is different for each rope. (If not, there would be no net force on the middle block.)
  4. Oct 5, 2008 #3
    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!
  5. Oct 5, 2008 #4

    Doc Al

    User Avatar

    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.
  6. Oct 5, 2008 #5
    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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