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Rotational mechanics .hinge reactions

  1. Dec 3, 2014 #1
    Rotational Mechanics... Hinges and Normal Reactions -- Please help

    Could any one please explain me the reason for two normal reactions on a hinge attached to a door. Also if there are two hinges the direction of the horizontal component of normal reaction is opposite to the direction of the horizontal component for the lower hinge...why is that? It would be a great help if you could answer the above two questions in detail.....

    Also if there is a single hinge with a rod attached to it how do the hinge reactions vary as the rod moves in a vertical circle....
    no one has been able to give a satisfactory reply to these above question's. .It would be a great relief to get these concepts cleared...Thank you..
     
    Last edited by a moderator: Dec 3, 2014
  2. jcsd
  3. Dec 3, 2014 #2
    I think I understand your first question... typical 2 hinge door. What do you think would happen if you pulled the top hinge pin with the door ajar? Replace the top pin and pull the bottom pin?

    I'm not sure what you mean by the second situation. Perhaps the solution for the first example will provide insight?
     
  4. Dec 11, 2014 #3
    okay...I understand your explanation. ..And I am sorry...let me clarify the second question...what I was asking was why are there two components of hinge reactions.. (as shown in the attached file a rod is fixed to a hinge which in turn is attached to a wall...) So in this situation what role do the horizontal and vertical components play..? This is what I wanted to know..Thank you...
     

    Attached Files:

  5. Dec 11, 2014 #4
    Also..In the second figure (refer attached file) I am not sure whether the horizontal component of hinge reaction will be in a direction along the rod(as I have shown it in the picture) or in the opposite direction?
     
  6. Dec 11, 2014 #5
    If the hinge remains flexible in the vertical axis, as in the door example, the same forces apply in one hinge now. Pull one long pin which way does it go?
     
  7. Dec 11, 2014 #6
    I am sorry. ..Could you please be a little more explicit? And you didn't explain the vertical component...Thank you
     
  8. Dec 11, 2014 #7
    The vertical component is essentially negated by the "door hinge" construction.
     
  9. Dec 11, 2014 #8
    But I am no longer talking about the door...Here is a single hinge pinned to a wall about which a rod rotates...What do the two components of reaction stand for..?
     
  10. Dec 11, 2014 #9
    I'm still not sure of the context entirely, you still have the vertical component on which it pivots, if you slice the hinge off the rod falls down?
     
  11. Dec 11, 2014 #10
    Er..yes.
     
  12. Dec 11, 2014 #11
    Yes..It does
     
  13. Dec 11, 2014 #12
    But if you pull the pin what happens? Is it the same every way you rotate the rod?
     
  14. Dec 11, 2014 #13
    I don't understand your impliaction..If you pull the pin the entire system will come off (along with rod)..Think of it as a hook that has been pinned to the wall along with the rod...and the rod can rotate freely a about it..
     
  15. Dec 11, 2014 #14
    I don't understand the orientations in the pictures is that something different from the hook now?
     
  16. Dec 11, 2014 #15
    How about this? Does this make things clear...?
     

    Attached Files:

  17. Dec 11, 2014 #16
    Which part of my question is ambiguous. ...? I will make an effort to explain that...
     
  18. Dec 11, 2014 #17
    The best I can deduce from this is that you see what is on the right, and you need a function to express the force on the hinge in various positions, for example the image on the left that you drew the rod wouldn't fall straight down, but rather swing down to the picture on the right where Nx=0
     
  19. Dec 13, 2014 #18

    Stephen Tashi

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    Science Advisor

    You should describe the direction of the forces you call "normal reactions". What are they normal to? If possible post a diagram or give a link to one.

    ----
    I see the diagram now. In a 2-D problem there will be a 2-D net force vector exerted upon the hinge. A force in 2-D can be broken into two perpendicular components. The direction of these components has been chosen for convenience. The fact there are two components just tells you that the total force on the door is the sum of these components.
     
    Last edited: Dec 14, 2014
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