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3D moment couple problem

  1. Oct 5, 2017 #1

    yecko

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    1. The problem statement, all variables and given/known data
    螢幕快照 2017-10-06 上午10.41.35.png

    2. Relevant equations
    M=r x F = d * F (for moment couple)

    3. The attempt at a solution
    2da2c19d-3110-470a-baba-0c2bd604f44d.jpeg

    This is my attempt to the question, yet I have got quite a different answer from my classmates... (especially the moment by the 50N force couple)
    Can anyone point out is there anything wrong with my answer please?

    Moreover, I am not sure how to replace the moment with force couple...
    M=d*F, as no d and F given, are there many answers giving out the same moment?

    Thank you very much for any help!
     
  2. jcsd
  3. Oct 6, 2017 #2

    BvU

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    I'm sorry to not even understand your first line. What is zero ?
    Can you distinguish the 3-4-5 rectangular triangles ? Do you think that's a coincidence ?
     
  4. Oct 6, 2017 #3

    yecko

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    I think Mx is zero because in Fig 1a, two forces are on the same line woth different direction, which cancel each other out in yz plane
     
  5. Oct 6, 2017 #4

    yecko

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    3-4-5 triangle is BCD, But i am not sure the use of it...
     
  6. Oct 6, 2017 #5

    Orodruin

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    This is true about the projection on the yz-plane. Just note that it will not be generally true for all moment directions.

    When you compute the ##M_y## moment, you seem to be assuming that there is a 50 N force in the xz-plane. This is a mistake as it ignores the component of the 50 N couple in the y-direction. The same goes for the force in the xy-plane when you compute ##M_z##.
     
  7. Oct 6, 2017 #6

    yecko

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    I thought i dont need to consider the y direction in xz plane... as y direction is already considered in yz and xy plane...
     
  8. Oct 6, 2017 #7

    Orodruin

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    You do not, you should only consider the components in the x- and z-directions. Your problem is that you are assuming that those components together make a force of total magnitude 50 N, they do not since the 50 N force also involves the y-component. When you just consider the x- and z-components, the total force will be smaller. You need the projection of the force onto the xz-plane.

    Also note that, since you have couples, the total force is zero and you can compute the torque of each couple relative to some conveniently located point.
     
  9. Oct 6, 2017 #8

    yecko

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    image.jpg
    I corrected this problem.. am i right?
    Moreover, How to find single equivent force?

    Thank you for all your help!
     
  10. Oct 6, 2017 #9

    Orodruin

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    To be honest, it is quite unclear what you are doing with the moments now. What is the "horizontal plane"? Why is it important? Which point do you take as the point of reference when computing your moments? If you stop to think and be smart about it, you will be able to make several simplifications and perhaps use different moment arms which simplify the problem significantly.

    Also, it would be good if you posted your work in text instead of as a photo as it is not possible to quote your photo and also impossible to see the photo properly on a small phone screen.
     
  11. Oct 6, 2017 #10

    yecko

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  12. Oct 6, 2017 #11

    Orodruin

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    It does not really help with the quotability issue. Anyway, I am on a laptop now and my comments in #9 still apply.

    Also, are you aware of how to use the vector cross product?
     
  13. Oct 6, 2017 #12

    yecko

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    OK... I have just backed home... Let me type my steps here:

    ΣMx=0N
    let Θ be the angle of 50N to the YZ plane
    tanΘ=(144/2) / sqrt ( (192/2)^2 +160^2 ) ==> Θ=21.1 deg
    d1 in Fig. 1b = 0.008232
    ΣMy = 50 cos 21.1 * 0.008232 - 12.5 * 192e-3 = -2.0160Nm
    d2 in Fig 1c = 0.6566 = 1.1819Nm
    M(resultant) = sqrt( (-2.0160)^2 +1.1819^2) = 2.3369Nm
    tan θ = 2.0160/ 1.1819 ==> Θ=59.6190 deg

    I didnt use any reference point as moment couple is calculated as d*F regardless where the axe is...
    Should I use point C because for no need to consider 2 forces? (please wait a minute for me to redo it...)

    Thanks
     
  14. Oct 6, 2017 #13

    yecko

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    Moreover, How to find single equivalent force?
    I really have no idea in this... though it is indeed what the question asking for...
     
  15. Oct 6, 2017 #14

    yecko

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    image.jpg
    I have done it with C as the axis, yet the answer i have got is different from the one i got above...

    Just let me type the key formula here:
    <0.160,0.144,-0.192> cross <-40,-18,24>+<0.160,0,-0.192> cross <12.5,0,0>
    = <0, 3.83, 2.88> + <0, -2.4, 0>
    = <0, 1.44, 2.88>

    Is there any calculation mistake there??
     
  16. Oct 6, 2017 #15

    Orodruin

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    This agrees with what I got (assuming your units are Nm).
     
  17. Oct 6, 2017 #16

    yecko

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    Thank you very much for helping to solve the problem!!!

    But then, if we use the 'expected' method of the chapter, whats wrong with my steps??

     
  18. Oct 6, 2017 #17

    haruspex

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    You are asked for the single equivalent torque, which you have done.
    That's what I get.
    Edit: that's the trouble with taking 20 minutes to work through it, you end up posting too late to be of help.
     
  19. Oct 6, 2017 #18

    yecko

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    No but the couple of force??

    As the question mentioned two couples meaning two pairs of forces, i think i need to find equivent force having same net force amd moment...
     
  20. Oct 6, 2017 #19

    haruspex

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    The term "couple" can mean just a torque. It does not necessarily imply an actual pair of forces. Certainly you cannot deduce a pair forces from knowing the torque. They could be small forces a long way apart or larger forces closer together.
     
  21. Oct 6, 2017 #20

    yecko

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    As the question mentioned two couples meaning two pairs of forces, i think i need to find equivent force having same net force and moment...
     
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