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Engineering statics equilibrium problem

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Here is the prompt/picture
    FBD.png

    The hint given is a FBD of the bead is recommended to being this problem. Find the coordinates of B so that both the magnitude and orientation of the elastic cord force can be properly represented. Also, two mutually orthogonal normal force directions (to bar AC) need to be included to permit a general representation of normal forces acting on the bead. A shortcut to solving for the elastic cord force can be obtained by writing the vector equilibrium equation, then taking the dot product of that equation with a unit vector pointing in the direction of bar AC. Since the normal forces are by definition perpendicular to the bar, their contribution is zero, and a single scalar equations remains for the for P.

    2. Relevant equations
    ΣFx=0
    ΣFy=0
    ΣFz=0
    Basic trigonometry

    3. The attempt at a solution
    Alright so I started off by finding the coordinates of all the points
    A (124, 0, 0)mm
    B (?, 31, 21)mm
    C (0, 62, 42)mm
    D (62, 0, 62)mm

    To find B I made a right triangle with A and C to find the magnitude. Since B is in the middle of AC as stated, I divided the magnitude I found and got 65.5. So B (65.5, 31, 21)mm.

    I found the unit vector of AC like the hint suggested and found that to be (-0.856, 0.428, 0.290).

    I am stuck here, I don't know how to go about finding the vector equilibrium.

    Thank you for any help!
     
  2. jcsd
  3. Sep 23, 2015 #2

    SteamKing

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    Why did you do this? If the y coordinate and z coordinate of point B are both halfway between the y and z coordinates of points A and C, it stands to reason that the x coordinate of B will also be halfway between the x coordinates for these two points.
    Find the correct location for point B.

    Once you know this, you can then find the tension force in the cord BD, given the spring constant and its unstretched length.
     
  4. Sep 23, 2015 #3
    A (124, 0, 0)mm
    B (62, 31, 21)mm
    C (0, 62, 42)mm
    D (62, 0, 62)mm

    TBD=k(l-l0) = 4(20-51.4) = -125.6
     
  5. Sep 23, 2015 #4

    SteamKing

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    Tensions are usually taken to be positive. The stretch in the spring would be calculated as stretched length - unstretched length = 51.4 - 20 = 31.4 mm
    The tension would therefore be 4 N/mm * 31.4 mm = 125.6 N.
     
  6. Sep 23, 2015 #5
    So I add this value with the unit vector in the direction of AC?
     
  7. Sep 23, 2015 #6

    SteamKing

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    The tension in the cord is pulling the bead at B against the bar AC. You want to convert the magnitude of this tension, 125.6 N, into a force vector FBD parallel to BD.

    Then, according to the directions in the problem statement, you want to take the dot product of FBD and the unit vector uAC to find out what magnitude P must be.
     
  8. Sep 23, 2015 #7
    So the unit vector of BD is (0, -0.6031, 0.7977), then multiplied by the tension is (0, -75.75, 100.2)

    The dot product of that and uAC is -3.367 which doesn't seem right...
     
  9. Sep 23, 2015 #8

    SteamKing

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    This looks OK.

    I think it's correct. I think the fact that FBDuAC is negative means that P is in the direction of point C.
     
  10. Sep 23, 2015 #9
    The answer was positive, hm, anyway I got it right. So for finding the reaction between bead B and rod AC, it would be just the force that is perpendicular to P right? I know the hint says nothing about it but wouldn't the cross product between them?
     
  11. Sep 23, 2015 #10
    I know there are infinitely number of vectors perpendicular to bar AC, but I don't know how to find it specifically where bead B is. Thank you so much for your help, I need to figure out these problems within the next hour. :cry:
     
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