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Uniform plank resting on two supports.

  1. Jan 2, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform plank, of weight 30N and length 3m, resting on two supports. The supports are 0.5m and 2.0m from the left-hand end of the plank. A weight of 18N is suspended from the left-hand end of the plank.
    206d3dv.jpg

    By how much should the weight on the left-hand en be increased so that the reaction Y becomes zero?


    2. Relevant equations
    The reactions that have been calculated from the first part of the assignment are 107N for point X and 196N for point Y .

    3. The attempt at a solution
    Calculations from part a) of the question, which was about finding the reactions at X and Y:
    X x 1 = Y x 0.5
    mg = 30N x 9.8m/s
    X = mg - Y
    (mg - Y) x 1 = Y x 0.5
    1 x mg - 1 x Y = 0.5 x Y
    1 x mg = 1.5 x Y
    1 x mg / 1.5 = 1 x 30 x 9.8 / 1.5 = 196N

    X + Y = X + 196N = mg
    X = mg - 196 = 30 x 9.8 - 196 = 98N
    98 + (18 x 0.5) = 107N

    X = 107N
    Y = 196N

    How do I rearrange the calculations from calculating the reactions into finding how much weight that needs to be added? I am pretty sure I need to consider the length (d) of the plank - I can't simply just add 89N to make the 107N into 196N, right?

    If there is a general formula/calculation for this, that would be very helpful.


     
    Last edited: Jan 2, 2017
  2. jcsd
  3. Jan 2, 2017 #2

    BvU

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    Hi an,

    Indeed, no. That would change the load on the other guy as well.
    There is. Should be in your list of relevant equations: force balance and moment balance.
    Please show your working for the first part you solved. Hopefully in terms of the variables (i.e the relationships, not the numerical values). Then consider what change is needed for part 2.
     
  4. Jan 2, 2017 #3
    I added my calculations from the first part of the question, but need help in understanding the second part. I have been searching my books and the internet and can't find a formula to calculate the second part of this question.
     
  5. Jan 2, 2017 #4

    haruspex

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    The plank has a weight of 30N, not a mass of 30kg.
     
  6. Jan 3, 2017 #5
    So I don't need to multiply with 9.8? I can just keep it 30N ?
     
  7. Jan 3, 2017 #6
    So mg = 30N ?
     
  8. Jan 3, 2017 #7

    haruspex

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    Yes.
     
  9. Jan 3, 2017 #8
    Thank you. Do you have any insight on how to calculate how much weight should be added to X to make Y zero?
     
  10. Jan 3, 2017 #9
    The answer for X is a negative number when I leave out x9.8, could that be correct?

    The new answers are:
    X = -157N
    Y = 20N
     
  11. Jan 3, 2017 #10

    haruspex

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    Please explain how you get this equation:
     
  12. Jan 3, 2017 #11
    The length from the center of the plank to support X is 1m and to Y is 0.5m, so the force at point X multiplied by the length should be equal to the force multiplied by the length at point Y
     
  13. Jan 3, 2017 #12

    haruspex

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    Are there no other forces on the plank which have a moment about the plank's centre?
     
  14. Jan 3, 2017 #13
    There is a weight of 18N on the left hand end of the plank, which I added on after calculating the force of the plank only on the supports.

    Here is the full first question: A uniform plank, of weight 30N and length 3m, resting on two supports. The supports are 0.5m and 2.0m from the left hand end of the plank. A weight of 18N is suspended from the left hand end of the plank.

    a) Find the reactions, X and Y, at the two supports.
     
  15. Jan 3, 2017 #14

    haruspex

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    That's no excuse for leaving it out of the moments equation.
     
  16. Jan 3, 2017 #15
    So how would that work on support Y? Is it 2m away? Then I have two separate weights with different distances to the two supports, so it will be two separate equations?
     
  17. Jan 3, 2017 #16
    The new equation will be:
    X - 30N x 1m = 30Nm
    18N x 0.5m = 9Nm

    Y - 30N x 0.5m = 15Nm
    18N x 36Nm

    X = 39Nm
    Y = 51Nm

    Somehow it does not add up that there is more force applied on support Y, that is the furthest away from the extra weight...
     
  18. Jan 3, 2017 #17

    haruspex

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    How do you get that? I thought you were taking moments about the mid point of the plank.
     
  19. Jan 3, 2017 #18
    It should say X = 30N x 1m = 30N
    Support X is 1m away from the midpoint
     
  20. Jan 3, 2017 #19

    haruspex

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    There are four vertical forces acting on the beam. If you take moments about an arbitrary point, all four forces will have a moment about that point, so all four should feature in the equation. You can make things a bit simpler by picking the point of action of one of the forces as the axis, so that force has no moment, but the other three forces will still be in the equation. I do not understand why you keep writing moments equations that only include two forces.
     
  21. Jan 3, 2017 #20
    It is because our teacher didn't go through any of this with us and I have no clue what I am doing. I don't even know what and arbitrary point is or how to pick a point :(
     
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