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Calculating Moment of a Force

  1. Aug 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Crates are placed on a framework for temporary storage. Calculate the moment of the force exerted by the crate around support A.

    2. Relevant equations

    Moment = force * distance

    3. The attempt at a solution

    The framework is 14 m in length and the crate is 2m away from Support A. What is the force acting down on the crate? Is it just the weight of the crate because of gravity or is there something else i am overlooking? The weight is 200KG
     
  2. jcsd
  3. Aug 19, 2011 #2
    I have attached a photo to help understanding.

    Is the force acting down on the crate:

    F=mg

    F=200*9.81
    F=1962

    Now times this by 2 to get the moment of the force since the crate is 2 metres away from A.

    3924Nm?

    Is this correct as the crate seems to cover 2 metres?

    Thanks in advance.
     

    Attached Files:

  4. Aug 19, 2011 #3

    ehild

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    The question is the moment of force (torque).

    ehild
     
  5. Aug 19, 2011 #4
    Thanks for your reply, isnt that what i have answered? Since moment=force * distance
     
  6. Aug 20, 2011 #5
    Anyone help?

    Cheers, Joe
     
  7. Aug 20, 2011 #6

    ehild

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    Sorry, I did not notice that you multiplied the force by 2 m. Your answer is correct if the centre of mass of the crate is at 2 m horizontal distance from A. So the figure is not quite correct.

    ehild
     
  8. Aug 20, 2011 #7
    What distance would u say this was classed as?

    I could only guess, that the centre of gravity acts through the centre of the crate which would work out as 1 metre, would you say that was correct?

    The course material is a poor form of instruction

    Thanks, Joe
     
  9. Aug 20, 2011 #8

    ehild

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    The problem says nothing about the size of the crate. I think you can take that the middle 2 m from A, so 3924 Nm has to be correct.

    ehild
     
    Last edited: Aug 20, 2011
  10. Aug 20, 2011 #9
    Thankyou very much for your help

    Joe
     
  11. Aug 20, 2011 #10
    Thankyou very much for your help

    Joe
     
  12. Aug 20, 2011 #11
    Is there any more information about how the table is supported by the ground? It seems to me that the question is nearly impossible to answer without more information about the supports A and B. I would assume that A and B are both exerting a verticle force. If A and B are both capable of exerting a moment, then you have 4 unknowns and the system is statically indeterminate. If either is exerting a horizontal force, then I don't think you would even need a moment at A at all. If A (and ONLY A) is able to exert a moment on the system and neither support is giving a horizontal force, then I'm pretty sure your answer is correct, but that seems like a pretty big assumption for them to ask you to make.

    Edit: Actually, now that I think about it, I'm pretty sure you can solve it so that A and B are only exerting upward forces, and there's no moment at all. I think either the answer is zero, or the system is statically indeterminate, or you have some very weird table legs that don't offer any upward force.
     
  13. Aug 20, 2011 #12
    All the information im given is:

    On reaching the next level the crates are placed on a framework for temporary
    storage.
    Calculate the moment of the force exerted by the crate around support A.

    Joe
     
  14. Aug 20, 2011 #13
    Are you sure there's no implied information, or something they said before about table legs and what support they offer? For example, in my statics class, the prof always drew the supports in a way that we would know exactly what support reactions existed at the support so that she wouldn't have to write it out in the question every time.

    I've attached a free body diagram of my interpretation. Even my free body diagram makes some assumptions, and it's still indeterminate. You only have 2 useful equations, but you have three unknowns. The thing is, there's absolutely no reason, based on the information you've given, to assume that the moment at B is zero, so adding that diagram kills it even more. Adding horizontal forces to both legs doesn't really change anything, since they just cancel each other out. That system is indeterminate based on my FBD unless one of the forces/moments shown is zero. Since it seems more reasonable that a table leg would be providing an upward force as opposed to a moment and no upward force, it would make sense to me that the moment at A is zero. That's a pretty poor argument, though, even if it ends up being right.
     

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  15. Aug 20, 2011 #14
    I think the idea of the question is just to test the knowledge to calculate the moment of a force and pretending Support B isnt present in order to do this, i agree in a sense that you have to make presumtions which is disgusting
     
  16. Aug 20, 2011 #15
    If that's the case, I would argue against it. The assumption that the leg is providing no support is a pretty unreasonable one. If they want you to assume that, then they should have made some indication of that. Can you ask your prof if that assumption is supposed to be made?
     
  17. Aug 20, 2011 #16
    Its a distance learning course which doesn't help, so getting a reply from a tutor takes ages, i asked a few quesitons last start of last week, still don't have a response as yet, maybes on holiday.
     
  18. Aug 20, 2011 #17
    Oh wait... hold on. I just re-read the question. They want the moment that the crate exerts on point A, not the total moment reaction at A. Wow... I think I just epically failed there. Terribly sorry about that.

    In that case, it wouldn't matter whether the moment created by the crate is canceled out by a force at B or a moment at A, it just matters what moment is created by the crate, in which case I don't know what's wrong with your answer. Sorry again for not reading the question properly...
     
  19. Aug 20, 2011 #18

    ehild

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    Only the moment of the force exerted by the crate around the support A was the question, not the forces exerted by the supports. The crate exerts a normal force, which is equal to its weight and the line of this force passes through the CM of the crate. The mass of the crate is 200 kg, and it is also given that the crate is at 2 m distance from support A. You can ask, which part of the crate is there - the left side, the right side or the middle. If the middle, then it is all right.

    ehild
     
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