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Basic Question in Mechanical Engineering

  1. Feb 17, 2012 #1
    In this question and solution:

    http://imgur.com/Z8DMo

    Why is it that when we examine the bar BCD, the force exerted by the bar AD, TAD , is not considered in the free body diagram, but when you examine the bar DFG, it is considered in the free body diagram?
     
  2. jcsd
  3. Feb 17, 2012 #2
    Because T is an internal force on the object when you are considering ADG. When you consider BCD or DFG, it is external and you would consider it.
     
    Last edited: Feb 17, 2012
  4. Feb 17, 2012 #3
    Thanks for the reply,

    Am I missing something, or is it not being considered as an external force for the case of BCD. All I see is Dx and Dy, and I'm wondering why T isn't there.
     
  5. Feb 19, 2012 #4

    PhanthomJay

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    It gets confusing when 3 or more members frame into a single joint. At joint D, when looking at a free body of DFG, T_AD is the force from AD and Dx, Dy are the force components from member BCD, acting at Joint D. The force at joint D on member DFG is thus, in the x direction , T_AD sin alpha + Dx, and in the y direction , T_AD cos alpha + Dy

    Now in looking at member DCB, at joint D, T_AD is the force from AD and , and the force from DFG, from above, in the x direction, is (T_AD sin alpha + Dx), and in the y direction, (T_AD cos alpha + Dy). Thus the force acting on DCB in the x direction is (T_AD sin alpha + Dx) - T_AD sin alpha = Dx, and in the y direction, (T_Ad cos alpha + Dy) - T_AD cos alpha = Dy.

    It is important to note that Dx and Dy are the forces in member DCB at joint D. They are NOT the forces in member DFG at joint D.

    Also it should be noted that member AD is a 2-force member.

    Watch directions of forces per Newton 3.
     
  6. Feb 20, 2012 #5
    Wow thank you for replying, I thought this thread was dead.

    So, I'm trying to generalize what you said. When 3 or more members frame into a single joint, the force contributed by each member is considered in the free body diagram of every other member (with flipped directions)?

    Edit:
    And I'd imagine forces often "cancel" out in these cases, like it did with there being no apparent T_AD on the free body diagram of member BCD.
     
    Last edited: Feb 20, 2012
  7. Feb 20, 2012 #6

    PhanthomJay

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    I should emphasize that the solution given in Post 1 is correct.

    When you first draw the FBD of horizontal member DFG, the forces at Joint D acting on it are the member forces from AD, which the solution calls T_AB sin alpha and T_AB cos alpha, and the forces from vertical member BCD , which the solution calls Dx and Dy.
    Now when you look at a FBD of member BCD, you have already defined the forces on it as Dx and Dy, so there is no need to look at T_AD acting on it plus the forces in DFG acting on it (which i'll call Px and Py). If you did look at all individual forces acting on it, you'd have T_AD sin alpha and T_AD cos alpha from member AD, and Px and Py from member DFG. But the net force from those 2 members is just Dx and Dy, so don't go through that extra step.
     
  8. Feb 20, 2012 #7
    Thank you =)
     
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