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redpanda

  • one year ago

Determine the moment of each of the three forces about point B. Please see attached figure.

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  1. redpanda
    • one year ago
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  2. redpanda
    • one year ago
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    I know how to find it about point A, but I am having difficulty seeing how to find it for point B.

  3. Yahoo!
    • one year ago
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    Seperately Solve the hinge forces...:)

  4. redpanda
    • one year ago
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    I don't understand what you mean.

  5. azolotor
    • one year ago
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    He means, solve for the force at each of the points where there is an angle and then sum them to see where the net force is

  6. JamesJ
    • one year ago
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    You guys have missed the question. What is being asked is what the moment due to each force. Can you help the questioner?

  7. JamesJ
    • one year ago
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    I'll do it then. What is the definition of the moment of a force?

  8. redpanda
    • one year ago
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    The moment of a force is its ability to create rotation.

  9. JamesJ
    • one year ago
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    Yes, but mathematically and precisely, with a formula?

  10. redpanda
    • one year ago
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    M = F*d, where d is the distance between the point of application and the pivot

  11. redpanda
    • one year ago
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    or M = r x F

  12. redpanda
    • one year ago
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    (btw thanks for your help, I have been lost for hours)

  13. JamesJ
    • one year ago
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    Yes. Now, let's start with F3. What's the moment due to that force?

  14. JamesJ
    • one year ago
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    ..the moment of F3 about B? Well, what's the distance of the application of the force from B? Zero. Can F3 make something rotate *around* B? No. Therefore the moment due to F3 is zero.

  15. redpanda
    • one year ago
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    ah!! that's what I thought! but I thought it would be wrong bcs it was 0

  16. JamesJ
    • one year ago
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    Now, what about F2?

  17. redpanda
    • one year ago
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    bcs the position vector at B is <0,0,0> and crossing it with the F3 vector would give zero.

  18. JamesJ
    • one year ago
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    yes

  19. redpanda
    • one year ago
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    But one question, would not the perpendicular component of F3 try to produce some moiton that vertical bar?

  20. JamesJ
    • one year ago
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    The question at a qualitative level is this: can a force applied at a point cause other things to rotate about that point? And the answer is no. It might make that point and the solid attached to it move, but it doesn't cause rotation around that point per se. Hence: A force applied to a point causes no rotation about that point.

  21. redpanda
    • one year ago
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    ahh i see

  22. redpanda
    • one year ago
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    its the same as if I go to the center of the bolt and try to make it rotate applying a force right on the center of rotations. no rotation would happen!

  23. JamesJ
    • one year ago
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    yes

  24. redpanda
    • one year ago
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    I have to move at least an infinitesimally small amount away from that center of rotation for it to actually rotate (no matter how large the force). Motion may occur in the form of displacement, but no actual rotation about that point :D

  25. JamesJ
    • one year ago
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    yes

  26. redpanda
    • one year ago
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    awesome.

  27. redpanda
    • one year ago
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    for f2. I think it goes like this:

  28. redpanda
    • one year ago
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    the position vector would be : <-3, 4 , 0>

  29. JamesJ
    • one year ago
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    No, you want the positive vector RELATIVE TO B

  30. JamesJ
    • one year ago
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    Which will be just r = <0,4,0>

  31. redpanda
    • one year ago
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    hmmm would not that be moving to the right of B and up from B?

  32. redpanda
    • one year ago
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    oh yes! sorry I was doing F1 mistakenly !

  33. JamesJ
    • one year ago
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    oh right. yes, F2. Doing them backwards because F1 is the most complicated.

  34. redpanda
    • one year ago
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    r = <0, 4 0 > ; F2 = < -300*sin(60) , -300 *cos(60) , 0 >

  35. JamesJ
    • one year ago
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    check your sin and cos there

  36. redpanda
    • one year ago
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    ah, i see that I did. I was putting it on the wrong side

  37. redpanda
    • one year ago
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    < -300*cos(60) , -300 *sin(60) , 0 >

  38. JamesJ
    • one year ago
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    Yes. Ok, you've got it. I'll leave you with F1 now by yourself.

  39. redpanda
    • one year ago
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    I have dyslexia and its hard to see things sometimes! Thanks for your patience

  40. redpanda
    • one year ago
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    for F1, r = < -3 , 4 , 0 >

  41. JamesJ
    • one year ago
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    yes

  42. redpanda
    • one year ago
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    F1 = < 250*sin(30) , -250*cos(3), 0 >

  43. redpanda
    • one year ago
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    r x F1 = 149.519

  44. JamesJ
    • one year ago
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    Looks right (30 obviously). Now calculate the cross products

  45. redpanda
    • one year ago
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    r2 x f2 = 600

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