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Tension and force

  1. Nov 7, 2007 #1
    Hi, this is my first post, so please tell me if I'm doing anything incorrectly...thanks.

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

    Three blocks are connected, as shown in Fig. 5-47, on a horizontal frictionless table and pulled to the right with a force T3 = 64.9 N.

    [​IMG]

    Assume that m1 = 12.1 kg, m2 = 24.5 kg, and m3 = 32.7 kg.

    2. Relevant equations

    a = (net force) / (total mass)
    T = (a) x (mass)

    3. The attempt at a solution

    I'm supposed to find the tensions at T1 and T2. I got the answer to T1 by multiplying the acceleration (0.94 m/s^2) by the mass of m1 (12.1 kg) and got a correct answer. However, when I tried doing the same thing for T2 (multiplying 0.94 m/s^2 by the mass of M2, 24.5 kg), the answer was wrong.

    My question: why can't I find T2 the same way I found T1? Is there a difference between T1 and T2?
     
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2
    m1 only has one tension force acting on it, m2 however has two tension forces, this is why you can't apply the exact same method. Does that make sense?
     
  4. Nov 7, 2007 #3
    I think so. Does that mean I have to multiply acceleration by the sum of m2 and m3?
     
  5. Nov 7, 2007 #4
    Well lets be careful about what we're doing. You know T1, the mass of block2, and the acceleration of the system. Think of the forces that are acting on m2. You'll want to use Newtons Second law to find T2. Can you show me how you would set up that equation?
     
  6. Nov 7, 2007 #5
    T - m2(g) = m2 (a)

    (Is this it?)
     
  7. Nov 7, 2007 #6
    You're very close, but you're not specifying what tension you're solving for. The right side of the equation is correct, but what are the forces acting in the horizontal direction? .....Think about it............ it's the two tension forces, correct?

    Edit: I see that you also have "g" in your equation. If you meant that to be the acceleration due to gravity, keep in mind that we're not considering vertical forces, and therefore we don't need to use "g."
     
    Last edited: Nov 7, 2007
  8. Nov 7, 2007 #7
    Since I have 2 tensions, should I write 2 equations?

    Net force - T2 = m2 (a)
    Net force - T2 = m3 (a)

    ...Oh no, I'm afraid that doesn't make much sense, does it?....
     
  9. Nov 7, 2007 #8
    You could write two equations, but we only need one. You're including the "net force" in your equation, however, F = ma IS the net force. For example, in your first equation, you have m2 (a). This IS the net force on m2, Fnet,2 = m2 (a).

    I want you to look at the picture you posted, and look at m2 only. What two horizontal forces are acting on m2?

    Edit: Keep in mind that the sum of the forces acting on m2 in the horizontal direction is going to be equal to the net force in the horizontal direction acting on m2.
     
    Last edited: Nov 7, 2007
  10. Nov 7, 2007 #9
    T2 and T3 are acting on m2?

    Thanks for correcting my mistake about the net force. I think I wanted to write 'applied force' instead but somehow screwed up.
     
  11. Nov 7, 2007 #10
    T3 is only in contact with m3. If I thought of m2 as a single point, as we typically do in free body diagrams (and it may be helpful for you to draw one for m2), then I consider the forces acting on that point. Any force and any body that is not in contact with that point need not be considered. Look at m2 once more, and tell me what two horizontal forces are in contact with m2.
     
  12. Nov 7, 2007 #11
    T1 and T2, then?
     
  13. Nov 7, 2007 #12
    Correct. So if we say that the motion of the block is in the positive direction, can you show me how you would set up Newton's Second Law for m2?
     
  14. Nov 7, 2007 #13
    T1 - T2 = m2 (a)
     
  15. Nov 7, 2007 #14
    That works if you take the motion to the right to be negative, which is fine, as long as you keep track of your coordinate system. You should end up with a negative tension if you use the equation in your post. It's not wrong, but you should make it a positive value to stay consistent with the other tension. Does all of that make sense?
     
  16. Nov 7, 2007 #15
    So it should be the other way around? As in:

    T2 - T1 = m2 (a)
     
  17. Nov 7, 2007 #16
    Not that it should, but since your other tensions have been positive values, it would be appropriate to make T2 positive, so the equation you just posted would be the best one to use.
     
  18. Nov 7, 2007 #17
    That's it? I finally got it right??!! YESSSS!!

    Thank you so much!!!! You have been so incredibly patient and nice and wonderful to a hopeless case like me!! Thank you! Thank you!!!!
     
  19. Nov 7, 2007 #18
    LOL, you're quite welcome. You did a great job yourself, and I admire your persistence :)
     
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