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Tension between two accelerating blocks

  1. Sep 18, 2012 #1
    Problem text:

    Forces act on block A and block B, which are connected by a massless string. Force [itex]\vec{F}[/itex]A=(12 N)[itex]\hat{i}[/itex] acts on block A, with mass 4.0 kg. Force [itex]\vec{F}[/itex]B=(24 N)[itex]\hat{i}[/itex] acts on block B, with mass 6.0 kg. What is the tension in the string?

    The system looks something like this:

    [A]-----

    Relevant equations:

    F=ma

    Attempt at solution:

    The professor never went over tension in class, so I've been Googling the pants off this problem and combing through the book for a solution, each to no avail. I tried working through it for about 30 minutes and gave up in frustration, after having come up with about 6 different answers.
     
  2. jcsd
  3. Sep 18, 2012 #2

    Doc Al

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    Staff: Mentor

    Show what you've tried.

    Hint: Identify the forces acting on each mass and apply Newton's 2nd law.
     
  4. Sep 18, 2012 #3
    Posting anything I've done wouldn't be of value, since everything I tried was essentially the mathematical equivalent of alchemy.

    If by apply Newton's 2nd law, you mean add up all the forces, then I come up with Fnet=36N, M=10kg, and A=3.6m/s^2.
     
  5. Sep 18, 2012 #4

    Doc Al

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    That's fine. You took as your 'system' both masses together. But to solve for the tension, now analyze the forces on one of the blocks.
     
  6. Sep 18, 2012 #5
    So...

    T=3.6m/s^2*14.4kg=14.4N

    Is it really that simple? I'm gonna have to kick myself for that!

    Anyways, thanks a lot Doc Al!
     
    Last edited: Sep 18, 2012
  7. Sep 18, 2012 #6

    Doc Al

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    Not sure what you are doing here. (Does that equation even make sense? :wink:)

    Do this:
    Pick one of the masses.
    Identify the forces acting.
    Apply Newton's 2nd law: ƩF = ma. (You already found the a!)
     
  8. Sep 18, 2012 #7

    Woops. Meant to write T=3.6m/s^2*4kg=14.4N

    That should be right, no?
     
  9. Sep 18, 2012 #8

    Doc Al

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    What you have is m*a for the 4kg mass. That's the right hand side of the equation ƩF = ma. But what about the left hand side?

    You need to identify the individual forces acting on that mass (one force will be the tension) and then add them to get an expression for ƩF.
     
  10. Sep 18, 2012 #9
    So you're saying I need to take into account the 12 N and subtract it from the 14.4 N I just calculated to get 2.2 N of tension force?

    Edit:

    Nevermind, that was more alchemy. My give up
     
  11. Sep 18, 2012 #10

    Doc Al

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    Do this: Name the two forces acting on the 4kg mass and tell me the direction they act.
     
  12. Sep 18, 2012 #11
    What I know for sure is that there is a 12N force in the positive direction, and I'm fairly sure that a 14.4N force is being applied, but in which direction I don't know.

    Edit:

    Here's another try.

    [itex]\vec{F}[/itex]T + [itex]\vec{F}[/itex]A=ma

    ma-[itex]\vec{F}[/itex]A=[itex]\vec{F}[/itex]T

    4kg*3.6m/s2-12N=[itex]\vec{F}[/itex]T

    14.4N-12N=[itex]\vec{F}[/itex]T=2.4N
     
    Last edited by a moderator: Sep 18, 2012
  13. Sep 18, 2012 #12

    Doc Al

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    Good. That force is given.

    Don't be guessing the answer--you will solve for it. Just label the force and its direction. (What's exerting the force?)
     
  14. Sep 18, 2012 #13
    [itex]\vec{F}[/itex]i=(2.4N)[itex]\hat{i}[/itex]

    Final answer.
     
  15. Sep 18, 2012 #14
    Wait a sec... Isn't that what I did in post #11? Maybe you didn't see; I edited it.
     
  16. Sep 18, 2012 #15

    Doc Al

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    Now you're cooking! :approve:
     
  17. Sep 18, 2012 #16
    Thanks to you. srsly
     
    Last edited: Sep 18, 2012
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