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Object drawn across two support points

  1. Jul 8, 2016 #1
    1. The problem statement, all variables and given/known data

    A 60kg object placed 1m from B is drawn across a beam laid down across two support points. Support point A can only withstand 600N of force before giving way. Support point B is assumed to be capable of withstanding any amount of force.

    The beam is 6m in length and weighs 20kg. How far can the object be moved across before support point A collapses?

    2. Relevant equations

    upload_2016-7-9_0-27-55.png

    3. The attempt at a solution

    So I think this is a statics question, at least, this is the context within the course material that poses the question.

    So, first I drew a diagram as follows:

    upload_2016-7-9_0-45-4.png

    Then did this:

    upload_2016-7-9_0-42-18.png


    So I can solve the forces for when the object is at 1m just fine - I think at least. But that is not what the question is asking. How do I get varying distances with this without resorting to doing manual calculations one at a time for each varying distance? Is there an alternative way?

    Using algebra I can break down the equation and rewrite as this:

    upload_2016-7-9_0-51-3.png

    But this goes no where as F_B is still an unsolved value so rewriting is meaningless. F_B's final value is linked directly to the d_P is at and so intuitively, I cannot think of a way that either can be isolated as they are coupled.

    Actually... am I approaching this question the wrong way entirely?


    Thanks for any help.
     
  2. jcsd
  3. Jul 8, 2016 #2

    haruspex

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    There's a given number you have not used.
     
  4. Jul 8, 2016 #3
    You mean FA = 600 right?

    Thought about that, but it doesn't make much sense to me since the sum of Fy doesn't involve distances, the distance part is calculated when FB is calculated.

    edit:

    Wait... you're saying do equations in reverse order?
     
  5. Jul 8, 2016 #4

    haruspex

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    When there's an unknown force that you do not care about, there are typically two possible approaches:
    1. Use one equation involving it to eliminate it from the other equations.
    2. Pick an axis for the moments equation such that the uninteresting force has no moment about it.
     
  6. Jul 8, 2016 #5
    ... Oh, well that's... a rather silly oversight on our part. Think got so fixated on the "procedure of B then A that every example just kept using" that lost track of something as basic as this. Afterwards just kept going round and round in circles. Thanks for the reminder.

    So having done the above, arrived at a final answer of 2.9m for distance at the point of collapse.
     
  7. Jul 9, 2016 #6

    haruspex

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    Is that 2.9m from B or from initial position?
     
  8. Jul 9, 2016 #7
    2.9m from B.
     
  9. Jul 9, 2016 #8

    haruspex

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    Well, it asks
    which might mean from its initial 1m position.
     
  10. Jul 9, 2016 #9
    Hmm okay, so suppose it could mean either 1.9m from initial position or 2.9m from B then.

    The wording is somewhat ambiguous and open to interpretation as to whether "moved across" means from B or from the initial value.
     
  11. Jul 9, 2016 #10

    haruspex

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    I agree, but if it means distance from B I don't know why they mention the 1m initial position.
    Incidentally, I get a different answer. Please post your working.
     
  12. Jul 9, 2016 #11
    As follows:

    upload_2016-7-10_0-24-42.png

    Now that we have a figure for FB, we simply plug that in to solve for dP

    upload_2016-7-10_0-25-0.png

    I had opted to convert the 600N to g instead of convert everything to N to reduce number of conversions required.
     
  13. Jul 9, 2016 #12

    haruspex

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    Looks like you are taking moments about A.
    How did the -cgdM turn into +60g?
     
  14. Jul 9, 2016 #13
    Well as my original equation had stated the conditions for CCW motion was to be negative, under the same conditions -- = +

    At least that is the idea, but I tend to mess up on my sign conventions a lot, could you explain what the issue is here?

    Thanks
     
  15. Jul 9, 2016 #14

    haruspex

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    That does not explain how -cgdM becomes +60g. That's just a question about algebra. To explain it, you need either c or dM to be negstive.
     
  16. Jul 9, 2016 #15
    Suppose we should also define an x/y axis then, dM would become negative if it is stated that B is the initial point on the x axis with positive right and negative left. As such, the point cg is located -3 units to the left from the initial point of B and would thus become negative. This make any sense at all?
     
  17. Jul 9, 2016 #16

    haruspex

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    Not really. You are taking moments about A, no? So all the displacements must be from A. You could make the 20kg at minus 3m from A, but then FB would be at -6m etc.
     
  18. Jul 9, 2016 #17
    Ah alright, I see what's going on. So when I used the equations for torque in the OP, I was calculating B, and as such point cg would have been negative, but as we are now calculating point A, cg becomes positive, so in effect it is FB - cg and not + cg.

    This changes the distance to 0.8776m... from point A, or 5.1225m from B.
     
  19. Jul 9, 2016 #18

    haruspex

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    I estimated it as about 5m from B.
    You can get there more quickly by my second approach, just take moments about B. Then you only need that equation, no need to consider vertical sum of forces.
     
  20. Jul 9, 2016 #19
    Alright.

    You've been a great help, thank you.
     
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