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Need help finding the restraining reaction force for a beam

  1. Mar 14, 2019 #1
    1. The illustrated structure is affected by a known couple, and try to figure out the restraining reaction force of the hinge A and hinge E. LW9hyYP.jpg

    We should analyse ECD instead. Since arm CD is a two force members, so N(C) in in the direction where CD connects by these two points. The distance from E to diagonal CD is a/√2. So we have N(C)=√2 m/a. Because N(C)=N(E) ( N(C) and N(E) together form a couple), N(E)=√2 m/a.
    Where did I go wrong?
     
  2. jcsd
  3. Mar 14, 2019 #2

    haruspex

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    I agree with your reasoning and answer. Who says it is wrong? What other answer is given ?
     
  4. Mar 14, 2019 #3
    The official answer just doesn't match mine.
     
  5. Mar 14, 2019 #4

    haruspex

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    Ok, but what is the official answer?
     
  6. Mar 15, 2019 #5
    R_A= \frac m 2a, R_E=\frac \sqrt {{2} m} a,
     
  7. Mar 15, 2019 #6

    haruspex

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    Your latex had some errors. In fixing it up I arrived at
    But that makes the reaction at E the same as you got, so perhaps you meant something else.
     
  8. Mar 16, 2019 #7
    Yes. You did it in the right format. Do u know how to get this official answer?
     
  9. Mar 16, 2019 #8

    haruspex

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    As I wrote, it's the same as your answer, just written differently. They both say ##(\sqrt 2)(\frac ma)##.
     
  10. Mar 17, 2019 at 2:40 AM #9
    How did u get ##N_A## ?
    Can u draw a diagram to illustrate?
     
  11. Mar 17, 2019 at 3:45 AM #10

    haruspex

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    I thought we were discussing NE.
    For NA, take moments about the other hinge.
     
  12. Mar 17, 2019 at 4:03 AM #11
    How can u get it? can u help me out?
     
  13. Mar 17, 2019 at 4:31 AM #12

    haruspex

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    Consider the whole frame as one body. If you take the B hinge as axis, what moments are there?
     
  14. Mar 17, 2019 at 6:00 AM #13
    zcrgSYO.png Because point B has a pulley on the ground, ##N_B## is vertical.
    Choose A be centroid
    $$\sum m_A (F)=0, ~2a N_B + m=0$$
    $$\Rightarrow N_B=\frac {-m} {2a},$$ So it's downward.
    X-axis, $$X_A=0$$
    Y-axis, $$Y_A=0 $$ $$N_B + Y_A=0$$
    $$\Rightarrow Y_A= -N_B = \frac {m} {2a}$$ So ##Y_A## it's upward.
     
    Last edited: Mar 17, 2019 at 6:13 AM
  15. Mar 17, 2019 at 6:12 AM #14

    haruspex

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    Right... except, strictly speaking there could be equal and opposite horizontal forces at A and B, making the problem indeterminate.
     
  16. Mar 17, 2019 at 6:20 AM #15
    You mean $$X_A = X_ B$$ It may not equals to 0.
    We just can't calculate it from the information already given.
    Is that what you are trying to say here?
     
  17. Mar 17, 2019 at 2:41 PM #16

    haruspex

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    Right, except XA=-XB. And if not zero then it changes YA and YB so as to balance the torque.
     
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