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Tension Force

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data

    A 20 kg mass is suspended by two ropes. Rope 1 goes to the wall horizonatlly from the left side of the mass. Rope 2 goes from the top right corner of the mass to the wall at an angle of 30 degrees from the horizontal top of the mass.

    a- If the mass is at rest, what is the tension force in rope 2?
    b- determine the tension in rope 1.
    c- Find acceleration of the mass immidiately after rope 1 is cut.

    2. Relevant equations

    EF = ma

    3. The attempt at a solution

    1) T2y = w
    T = w / sin 30 = mg / sin 30

    = 392 N

    2) ?

    If I have done part 1 correctly, it seems like the tension in rope 1 would be 0?
  2. jcsd
  3. Sep 23, 2008 #2


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    If I have the picture correctly there is a rope on each side pulling correct?

    While the net horizontal forces may be 0, there is Tension in each.
  4. Sep 23, 2008 #3
    Yes, there is a rope on each side, but rope 1 is halfway down the left side and pulled straight out, horizontally, to the wall. Rope 2 is coming from the right top corner of the mass and pulling to the wall diagonally. The angle of 30 is measured from the horizontal level of the top of the mass to the rope 2. I hope that makes sense. Wish I had a scanner!
    Last edited: Sep 23, 2008
  5. Sep 23, 2008 #4


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    I think I have it.

    If you look at the Tension of the angled rope it has 2 components x and y.

    You know the y component must be m*g because that's all there is holding it up. The Tension then by dividing sin30 is 2*m*g.

    So the horizontal component is 2*m*g*cos30 = (.866)*2*m*g
    and the vertical of course is (1/2)*2*m*g = m*g.

    So the horizontal Tension of one must equal the horizontal Tension of the other.
  6. Sep 23, 2008 #5
    Why is the horizontal 2*m*g*cos30?

    I get as far as T2y = 392 N. Isn't that independent of T2x?
  7. Sep 23, 2008 #6


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    You know that because there is no other vertical force component, ALL of the weight of the block MUST be carried by the vertical component.

    Doesn't that translate at 30 degrees to the Horizontal the Tension in the rope as being m*g/.5 = 2*m*g?

    The Tension vector is the Vector sum of the 2 components.
  8. Sep 23, 2008 #7
    OK. So how do I state the total tension in rope 2? Add the components together?
  9. Sep 23, 2008 #8


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    Wait a minute. You already know the Total Tension. (You gave it already in your first post.) And you know the vertical component of the tension (Ty = m*g). So to find the horizontal component, that's cosθ * T2 = Cosθ *(2mg) = Tx
  10. Sep 23, 2008 #9
    I thought the 392N was the tension in just the y component of rope 2. Is that wrong?
  11. Sep 23, 2008 #10


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    That looks like the total Tension.

    M*g = 20 *9.8 = 196N is the Ty
  12. Sep 23, 2008 #11
    Oh! OK, I see what I was doing. My notes for this class are a mess because the professor rushes through everything. You've explained more to me today than he has all week. Thanks again.
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