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Tension in a massless wire

  1. Sep 29, 2012 #1
    So I'm just supposed to calculate the tension in the wire (see attachment) and get to the formula shown. What I've done so far is to draw free body diagrams of all the boxes, and then I've just tried to set up some equations and manipulate them.

    However, I can't seem to end up with the expression I'm supposed to end up with. I've attached my attempts at deriving a formula for S. The problem seems to be eliminating the acceleration-components. Any help would be very appreciated!
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2012 #2

    TSny

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    Hi gralla55. I see the attachment showing the figure. But I don't see any attachment where you show your work. The answer stated in the figure does not look correct to me.
     
  4. Sep 29, 2012 #3
    Thank you for your reply! Here it is:

    What I did was just to solve for the Tension Force in each of my free-body diagrams. Still, each of them contained an acceleration component, which does not show in the final equation I'm supposed to get to. It does look weird to me too, but I'm pretty sure it's correct.
     

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  5. Sep 29, 2012 #4

    TSny

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    Thanks for showing your work. You need to consider the direction of the friction forces on m1 and m2 in relation to the direction of the tension forces. Looks like you might have to change some signs. Likewise for the relative directions of the weight and tension forces on m3.

    You'll need to take into account that the total length of string stays constant. You can use that to get an additional equation relating the three accelerations.
     
  6. Sep 29, 2012 #5
    All right, so maybe I can write:

    S - Fr1 = m1a1

    S = m1a1 + um1g

    And since I wrote this direction of increasing x as the positive one, this should mean:

    S = -m2a2 + um2g

    And finally:

    S = (m3a3 + m3g) / 2

    I didn't think about the chord length ting... but that should give:

    a3 = a1+a2

    Does this look right? Thanks again!
     
  7. Sep 29, 2012 #6
    I tried eliminating the accelerations by using the last equation and substituting. I ended up with something that started too look like the final answer, but it still looks like I've made some errors...
     

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  8. Sep 29, 2012 #7

    TSny

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    I suggest that you take the positive directions for x1, x2, and y in the directions shown by the arrows in the diagram. This will require some changes in signs in some of your equations.

    a3 ≠ a1+ a2. A good way to get the correct relation is to write the total length of the string in terms of x1, x2, and y and then note that the total length must remain constant as the masses move.
     
  9. Sep 29, 2012 #8
    As shown in this attachment, I'm getting really close... but some signs are wrong and I've got a factor of 2 which should not be there...

    What is confusing me I think, is that when I'm writing a or g etc, I think of them as positive scalars. So -a cannot be a positive number since a itself is positive.

    When I think of it that way, a3 should equal a1+a2, but maybe I need to change signs or something again? Anyway, I think I will get the right answer if the equations I start with are correct now, I just can't seem to find the correct ones, haha. Thank you so much!
     

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  10. Sep 29, 2012 #9
    Is the correct relation a3 = (a1+a2) / 2?

    I figured if one of the boxes moved, and the other stood still, the length in the y-direction would only be incremented by half because of the pulley.
     
  11. Sep 29, 2012 #10

    TSny

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    Basically that's correct. But there's still the pesky signs to think about. Suppose you choose the positive directions of motion of each mass to be in the direction of the arrows in the original figure. Then if a1 and a2 are both positive, what would be the sign of a3?

    Of course you can choose your positive directions anyway you want, but then you've go to make sure all of your expressions are consistent in sign with your choice. So, if you're not choosing the positive directions according to the arrows in the diagram, it would help if you would specify your choice.
     
  12. Sep 29, 2012 #11
    All right, now I just need to figure out why I've got a factor of 2 in excess somewhere.

    a3 = (2S - m3g) / m3

    Does this equation look correct to you?
     
  13. Sep 29, 2012 #12

    TSny

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    That's correct if you're taking upward as the positive direction for m3.
     
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