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Tension in two ropes attached to a beam

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data

    I have attached an electronic image of the problem to this post.

    The problem states that there is a beam (AB), which has length 5 metres and mass 25 kilograms. It also states the beam is suspended in a horizontal position by two vertical ropes. One rope is attached to the beam at A and the second rope is attached at point C on the beam, where the length CB = 0.5 m and the length AC is 4.5 m.

    The problem then states that a particle of mass 60 kilograms is attached to the beam at B and the beam remains at equilibrium in it's horizontal position. The question also says the beam is modelled as a uniform rod and the ropes are modelled as light strings.

    Give this information, the questions asks me to find:

    i) the tension in the rope attached to the beam at A.

    and

    ii) the tension in the rope attached to the beam at C.

    2. Relevant equations

    Tension = mass x acceleration due to gravity

    Moment = force x perpendicular distance

    3. The attempt at a solution

    I believe that the,

    Tension in rope A + Tension in rope C = 85g

    From here, I'm not sure how I'm meant to relate the moments to find the tensions in the strings.

    Any help is greatly appreciated. Thank you for reading.
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2017 #2

    haruspex

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    Do you understand how to choose an axis, and add up moments about that axis?
    That equation is not general enough to be useful.
     
  4. Jun 2, 2017 #3

    scottdave

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    When you say 85g, I assume that you mean (85 kg)*g, where g is the acceleration due to gravity. If that is the case, then yes you have made a good first step.
    Next, you want to pick a point, and sum moments about that point. Since it is in equilibrium, then you know the sum of the moments will equal to zero.
    Just as you found that the sum of vertical forces equalled zero: Ta + Tc + (85kg)*g = 0, where g = -9.8 m/s2. Notice that I added the (85 kg)g but took g as a negative direction. Try picking point A as the point to sum moments about. Do you know how to account for the beam weight?
     
  5. Jun 2, 2017 #4
    Yes, when I say 85g I do mean (85 kg)*g, apologies, I should have made that clearer.

    Also I think I managed to get the answer,

    I chose to take moments about the point A, which, using the principle for moments in equilibrium that anticlockwise moments = clockwise moments, led me to,

    Tension in rope C x 4.5 = (60g) x (25g x 2.5), this led me solve for the Tension in rope C, which gave me Tension(C) = 789.4 N.

    Using that figure, I used the equation,

    Tension(A) + Tension(C) = 85g, rearranging to make Tension(A) the subject and plugging in my value for Tension(C) to give,

    Tension(A) = (85 x 9.8) - (789.4),

    This gave me a value for Tension(A) equal to 43.6 N.

    EDIT:

    I just checked the mark scheme and these are the correct values, but I just realised I did not use g= -9.8 ms-2, so I'm not sure whether the mark scheme is wrong, or for some reason, I do not need to take g to be negative?
     
  6. Jun 2, 2017 #5

    scottdave

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    I put it in like that, when I had everything on the same side and added the (85 kg)*g. If I subtract (85 kg)*g from both sides, then you have Ta + Tc = -(85 kg)*g, so my negative value (for g) cancels the negative sign in front of 85. Did that make sense?
     
  7. Jun 2, 2017 #6
    Yes, that makes sense. Thank you for the help.
     
  8. Jun 2, 2017 #7

    scottdave

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    I am glad you got the right answer, but I was confused by your workings:
    Since you don't have all of the units, it is difficult to follow. Many people seem to think this is being picky, but I've seen way too many workings, which result
    in the wrong answer, and the person cannot see their mistake. But putting units in, it becomes obvious that the dimensions do not balance on each side of the equation, and the source of the error is pretty easy to find.

    Looking at how you typed it in, you have on the left side: Tension {force} x 4.5 {assume meters} = Newton meters (a torque unit).
    On the right side (60g) { force} x 25g {force} x 2.5 {meters assumed}. That's Newton2 meters.
    It should be (60g) x ( a distance) + (25g) x (2.5 meters). I hope this helps.
    If you get in the habit of doing this, it will help you spot errors, and also help your professor figure out what you are doing (if it is being graded, for example).
    If it is a real-world project, then other people may be reading your writeups about forces, etc in a structure.
     
  9. Jun 2, 2017 #8
    Apologies for that, I missed it out when I was copying it from my pen and paper work, you're correct it should be,

    Tension in rope C x 4.5 metres = (60g x 5 metres) + (25g x 2.5 metres)
     
  10. Jun 2, 2017 #9

    CWatters

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    Regarding the sign of "g".... At the start of your working you should make some declarations if they haven't already done so in the problem. For example...

    "I define up, right and clockwise as positive"

    You need to give g a sign that is consistent with your declaration.

    I also think it's best to sum your torques to zero explicitly. Eg write equations in the form...
    A + B - C = 0
    Rather than..
    A + B = C
     
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