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A rod hanging by two strings find theta

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform rod is supported by two strings. The string on the left hand side is horizontal. The string on the right hand side is at 30 degrees with respect to a straight line going from the end of the bar to the ceiling. If you had a straight line going from the end of the rod to the ceiling on the left hand side, what would be the angle with respect to the rod.


    2. Relevant equations
    static equilibrium


    3. The attempt at a solution

    I would assume that the angle we are searching for is 60 degrees. Because the opposing angles forces must equal zero when added and they are in static equilibrium. Thus they are equal.
     
  2. jcsd
  3. Mar 25, 2012 #2

    PeterO

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    I think something is wrong

    I understood "30 degrees with respect to a straight line going from the end of the bar to the ceiling" to mean that if you extended a line from the right hand end, along the line of the rod, you would get to the ceiling.
    In that case "If you had a straight line going from the end of the rod to the ceiling on the left hand side" is difficult. The straight line extension of the left will reach the floor - unless it meets the wall first??
     
  4. Mar 25, 2012 #3
    Here is a link with a picture and description.
     
  5. Mar 25, 2012 #4
  6. Mar 25, 2012 #5
  7. Mar 26, 2012 #6

    PeterO

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    OK - pity you didn't use their word vertical, rather than your word straight.

    From the diagram, that answer is clearly not 60o - meaning, the diagram actually has drawn the (supposed) 30o at approximately that size, and the angle is clearly not 60o.

    Consider the the force acting on the rod - including its weight.

    The left hand string [NOTE: you can only pull with a string] has no vertical component, so the right hand string wholly supports the rod, the left hand string only balances the horizontal component of the tension in the right hand string.

    You need to carefully analyse the three forces - two tensions and one weight.
     
  8. Mar 26, 2012 #7
    So are you saying it is a problem of which one would need to choose arbitrary numbers to calculate the angle?
     
  9. Mar 26, 2012 #8



    You do understand in order to analyze those factors, one would need to know lengths and mass right.
     
  10. Mar 26, 2012 #9

    PeterO

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    Generally if specific information is not given, its actual value is irrelevant. What I like to do, if I don't feel confident about using W for the Weight of the rod and L for the length of the rod, is choose some random values, then just in case, repeat the the calculation with some other values to check that the final answer is unchanged.

    I like to choose values like 7 and 13, as you are unlikely to get a co-incidental answer with numbers like that.
     
  11. Mar 26, 2012 #10
    Yes, I understand. I have summed up the forces and chosen arbitrary numbers but the tension on the ropes are specific to a certain mass of the rod. I have multiple answers and am unclear on the correct answer and process. And my teacher starts a problem and says you should be able to figure it out from here to the class, which means he did not know how to do it. I am on this website to find the correct answer. I know how to work the problem this is not new to me, it is unclear about the arbitrary numbers.

    I see the problem as being in static equilibrium, thus the opposing forces are equal in magnitude with angles in the opposite to each other, which is why I would assume that it is a 60 degree angle.
     
  12. Mar 26, 2012 #11

    NascentOxygen

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    What are your answer/s? I'll see how they compare with my own.

    Can you now try it using algebra? As PeterO suggested, draw the diagram, mark the length of the rod as "L" and its weight as "W". Now do the calculations using these values, together with the angles of 0° and 30° for the supporting strings.

    Tell us your working as you go.
    You are sure to get help here, but don't expect to be told the correct answer. That's not how it works.

    See how you go using algebra now. (With any luck, L and W may cancel in the final working, leaving you with something you can evaluate on a calculator. But no promises. :smile: )
    Assumptions have a nasty habit of being proven wrong. I know only too well! :grumpy:
     
  13. Mar 26, 2012 #12
    Alright well this is how I am seeing it.

    T-2 - T-1 = 0 or -T-1 + T-2 = 0 (these are the x components)

    Since there is no forces acting in the Y direction, we are left with the two x components which are supporting the rod, thus the force due to gravity is irrelevant seeing as it is in equilibrium, it cancels out. Which brings me to...

    Tcos60 - Tcos60 = 0 or Tcos60 - "Tsin30" = 0, and 90 - 30 = 60

    There for the angle theta is equal to 60 degrees.
     
    Last edited: Mar 26, 2012
  14. Mar 26, 2012 #13
    Alright well this is how I am seeing it.

    T-2 - T-1 = 0 or T-1 + T-2 + mg = 0

    Since there is no forces acting in the Y direction, we are left with the two x components which are supporting the rod, thus the force due to gravity is irrelevant seeing as it is in equilibrium, it cancels out. Which brings me to...

    Tcos60 - Tcos60 = 0 or Tcos60 - "Tsin30" = 0, and 90 - 30 = 60

    There for the angle theta is equal to 60 degrees.
     
  15. Mar 26, 2012 #14

    NascentOxygen

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    gravity acts vertically and this must be counteracted by the strings
     
  16. Mar 26, 2012 #15

    Yes, I stated that in my response, but since mg is in the Y direction, vertically as you put it and the system is in a 'static equilibrium' there is also a tension T-y in the opposite direction of the force due to gravity. Because of this fact the forces in the y why direction are irrelevant to the solution of the problem. I think you and the other guy are thinking of a pulley system, unless I am incorrect.

    If anyone could provide any sort of backup to their bland explanations that would be dandy.
     
  17. Mar 26, 2012 #16
    Obviously I am approaching this problem from multiple angles. Based on my math and understanding I feel that my answer is correct. I was hoping someone could either correct me or confirm with me. If you are going to correct me please have a valid explanation with a mathematical backup to prove your hypothesis, I say hypothesis because I do not see any foundation to what you are trying to prove, unless you can back it up. We are on the internet here any dummy could make a response. I am not asking for an answer from anyone, just a legitimate response with some mathematical backup. Otherwise it is if I am reading a non-fictional paper without any sources...


    Is there a fellow physics lover out there?
     
    Last edited: Mar 26, 2012
  18. Mar 26, 2012 #17

    PeterO

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    My approach would be:

    Since the length must be irrelevant [or it would have been given] Let it be zero.

    We would then have just two strings on a point mass - one horizontal, the other at 30 degrees to the vertical.

    The vertical component of the angled string's Tension = Weight [in size - opposite in direction of course.

    You can thus work out what fraction or multiple of the weight force, the Tension in that string must be, and thus calculate the horizontal component.
    The Tension in the horizontal string must balance that.

    Those calculations have been based on the Vertical and horizontal directions.

    Now give the rod some length, and consider rotation about some point - I would choose the right hand end of the rod.

    The weight force x its radius of operation must balance the Tension in the left string x its radius of operation.
    Those radii will involve L and θ so should enable you to solve for θ

    Good luck!!

    ps the vertical and horizontal analysis also works the same if the rod has length!!
     
  19. Mar 26, 2012 #18

    PeterO

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    Quote
    T-2 - T-1 = 0 or -T-1 + T-2 = 0 (these are the x components)

    If these are the x-components, why have you listed the full tensions.
    Certainly the Tension in the left hand string is totally in x direction, so should be listed in full (perhaps as TL so you know we are referring to the Left string), but it should only be TRsin30o (I use TR for the Right hand string.

    Quote:
    Since there is no forces acting in the Y direction,

    There are forces in the y-direction
    the weight of the rod, W, down and TRcos30o up. These also must balance.

    You need to consider rotation about a convenient point to complete the solution.
     
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