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Two Situations of Tension Same Thing?

  1. Feb 27, 2008 #1
    [SOLVED] Two Situations of Tension... Same Thing?

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

    A box is hanging with two wires. (Refer to diagrams). Find the two tensions.
    My first question is, if the two diagrams are different situations that will lead to different answers? Personally I don't think they would... since the x coordinates still have to equal zero, with the same angles... but I'm not exactly sure, can someone please clarify the concept behind this?

    Diagram 1
    [​IMG]

    Diagram 2
    [​IMG]

    2. Relevant equations

    T1*sin56 + T2*sin45 - 35N = 0

    -(T1*cos56) + T2*cos45 = 0

    3. The attempt at a solution

    Prepared for substitution
    T2 = T1(cos56)/cos45

    T1(sin56) + T1(sin45)(cos56)/(cos45) - 35N = 0
    T1(sin56 + sin45*cos56/cos45) = 35N

    Solved for T1 = 25N.

    Plugged in and got T2 = 20N.

    So based on those equations I get T1 = 25N, and T2 = 20N... but the answer key shows a way different answer so I am confused. Could it be that the answer key is wrong? Or maybe it is because of the two different situations I've shown above? I'm use to doing problems like diagram 2, but for this question it is shown like diagram 1?

    Please let me know if I made a mistake, this is driving me crazy.
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2
    The situations are the same. This is because the assumption that the mass is a point is made. Further more, any vector (and so, any force) can be moved parallel to itself and it will remain effectively the same vector.

    Looking at your working out, it looks fine to me, I can't see were you went wrong, unless you made some silly error.

    EDIT: T2>T1, you have made an error in finding T2, perhaps T1 is correct.
    I can see that T2>T1 because it is opposite the larger angle in the triangle. Also from this:
    T2 = T1(cos56)/cos45
    cos56>cos45>0, so cos56/cos45>1, hence T2>T1
     
    Last edited: Feb 27, 2008
  4. Feb 27, 2008 #3
    Ok wait. So if I label the left tension as T1, and the right tension as T2, did I do the problem correctly?
     
  5. Feb 27, 2008 #4
    Can anyone please help me? I don't understand the error I made.
    I think I made an error here

    -(T1*cos56) + T2*cos45 = 0

    Please help me!
     
  6. Feb 27, 2008 #5
    Hell I dunno either, it looked ok to me

    That equation should be ok, it doesn't matter where the negative goes, you could divide both sides by -1, which moves it to the other term, and 0/-1 is still 0, so whichever

    You DID use more accuracy than 25 and 20, right? I got like 25.stuff and 20.stuff
     
  7. Feb 27, 2008 #6
    Are you getting 25 for the left and 20 for the right?
     
  8. Feb 28, 2008 #7
    Look, the string opposite the larger angle MUST be bigger (from trigonometry), if T1 (the string opposite the smaller angle) is 25N, T2 MUST be larger. As far as I can tell, you got T1 correct, you must be substituting incorrectly- sorry I don't have a calculator on me.
     
  9. Feb 28, 2008 #8
    Crud, I made a mistake: cos56<cos45

    You made rounding off errors:

    T1 = 25.21
    T2 = 19.94

    Totally disregard what I said earlier- I was making a fool of myself.
     
  10. Feb 28, 2008 #9
    Yeah thanks for confirming... no idea why this simple problem is causing so much problems.
     
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