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Quick Tension Problem

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A 5 kg mass is attached to two ropes as indicated in the figure below. Find the tension in both ropes.

    2. Relevant equations

    Ʃy= 0 Ʃx = 0

    3. The attempt at a solution

    Mass of object: 5 * 9.81 = 49.05

    T1 sin 30 - T2 cos 20 = 49.05

    T1 cos 30 + T2 sin 20 = 0

    Upon solving I get the T1 to be 17 Newtons and the T2 to be -43 Newtons.

    It doesn't seem right to me.

    Thank you.
     

    Attached Files:

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  2. jcsd
  3. Dec 4, 2013 #2
    Why does the T2 term have the minus sign in the first equilibrium equation, but the plus sign in the second?
     
  4. Dec 4, 2013 #3
    Well in the second equation, the sum equals zero.

    For the first one, I was unsure, and this may sound funny, whether the y component in T2 was pulling down (-) or helping keep the mass up (+).
     
  5. Dec 4, 2013 #4
    I guess it should be positive? What is the reason?
     
  6. Dec 4, 2013 #5
    With every physical problem one starts by assigning positive directions. It can be done arbitrarily, but it needs to be done consistently. You have two directions here: vertical and horizontal. The usual convention is "up" is positive, and "right" is positive, too.

    Then, you consider the forces of tension where the mass is attached. The force of tension exerted by an end of a rope on something is always toward the other end of the rope, because ropes can pull, but never push. Then you look at the vertical and horizontal components of the forces. Are they up or done? Left or right? You assign signs to them accordingly.
     
  7. Dec 4, 2013 #6
    If we say that up is positive and right is positive, then...

    T1 Sin 30 - T2 Cos 20 = 0 ?

    My reasoning for this is that the (T2) rope is pulling in the downward, so it must be negative?
     
  8. Dec 4, 2013 #7
    Correct!
     
  9. Dec 4, 2013 #8
    Thank you. I am assuming my initial answer, then, is also correct!

    Once again, I greatly appreciate your help.

    P.S: I have another question related to a free body diagram for multiple masses. Should I start a new thread or continue this one?
     
  10. Dec 4, 2013 #9
    No, wait. I misread your #6. I thought you wrote T1 cos 30 - T2 sin 20 = 0, for horizontal direction, even though you said "downward".

    For the vertical components, where is the weight of the ball?

    Please show both equations.
     
  11. Dec 4, 2013 #10
    Whoops. Sorry. For the vertical direction, it should equal the weight of the ball.

    Vertical:

    T1 sin 30 - T2 cos 20 = (5 * 9.81)

    Horizontal:

    T1 cos 30 - T2 sin 20 = 0
     
  12. Dec 4, 2013 #11
    The left hand sides of both equations are identical. Not good.
     
  13. Dec 4, 2013 #12
    Really? I don't see it.
     
  14. Dec 4, 2013 #13
    Oops. My vision doing tricks on me. Hope this time I do see it properly :) So, what is the solution?
     
  15. Dec 4, 2013 #14
    Ok.

    T1 = -25.74 N
    T2 = -65.87 N
     
  16. Dec 4, 2013 #15
    Does it look right?
     
  17. Dec 4, 2013 #16
    The negative magnitudes for both tensions indicate something is very wrong with this problem.

    And when looking at the picture again, I understand what the problem is. The angles shown in the picture are greater than 20 and 30 degrees, so the configuration seems plausible. But the correct configuration is show in the attached picture.

    The red lines indicate where the ropes should be according to the description. It is clearly impossible, because the ropes would have to be compressed, which is what negative tensions indicate, and ropes cannot do that. Instead, the ropes would transform to the green configuration.

    So either the problem is not correctly specified, or we misunderstand it. Could it be that the angles are both taken relatively to the horizontal line? Then there is a solution with positive tensions.
     

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