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Cartesian Vector Form - Door with 2 Chains

  1. Mar 17, 2017 #1
    1. The problem statement, all variables and given/known data

    The door is held open by the means of 2 chains. If the tension in AB and CD is Fa = 300 N and Fc = 250 N, respectively, express each of these in Cartesian Vector Form

    2. Relevant equations

    Sin / cos / tan

    3. The attempt at a solution

    The angle of FA at B is atan(1.5sin30/(1+1.5cos30 )) then +y direction force A is 300cos that angle and -z direction force is 300sin that angle

    There are 2 forces in 2D but I dont understand the 3D. There are 3 forces x, y and z

    See Image Attached
     

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  3. Mar 18, 2017 #2

    haruspex

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    Take it in stages. Drop a vertical from C to a point G below it on the ground. The tension has a vertical component and a horizontal one from G to D.
     
  4. Mar 18, 2017 #3
    Thanks but Im still a bit confused with the (x) force. Im really not sure about the 3D calculations. The horizontal length for C to D is 2 meters with a Fc of 200N

    +y direction force is tan(1.5sin30/(1+1.5cos30) = X then 300 CosX

    -z direction force is tan(1.5sin30/(1+1.5cos30) = X then 300 SinX
     
  5. Mar 18, 2017 #4

    haruspex

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    No, that's just the X component. What is the whole horizontal distance? Hence what is tan(∠CDG)?
     
  6. Mar 18, 2017 #5
    ok I admit I have no idea how to do the 3D x, y, z forces for CD. I have no idea what formula to use or where to start

    The 2D y, z forces for AB I understand that, see the calculations in the image attached below, Im sure thats correct

    17311394_10211780872818684_1088220014_o.jpg
     
  7. Mar 18, 2017 #6
  8. Mar 18, 2017 #7

    haruspex

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    C, D, G lie in a vertical plane. That's just 2D.
    Use Pythagoras to find the distance DG, hence find ∠CDG.
     
  9. Mar 18, 2017 #8
    cool thanks but we dont know the distance DC to find out the length DG using Pythagoras p6.jpg
     
  10. Mar 18, 2017 #9
    ok got it!!!

    p7.jpg
     
  11. Mar 18, 2017 #10
    So does that look correct? If so I can then work out the x, y, z forces. Also how do I express the answer in cartesian vector form?
     
  12. Mar 18, 2017 #11

    haruspex

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    Yes, that's it. You might want to keep one more digit of precision during the calculation... I have 3.14 where you have 3.1.
    Cartesian vector form probably means as three numbers in parentheses: (Fx, Fy, Fz).
     
  13. Mar 19, 2017 #12
    Great, thanks for the help, I think I've got it now. Can you please double check the calculations and answers and let me know if correct, Cheers

    cvf1.jpg
     
  14. Mar 19, 2017 #13
  15. Mar 19, 2017 #14
  16. Mar 19, 2017 #15

    haruspex

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    I'm getting rather different numbers. Where does the 2.23 come from? 1+0.75(√3)/2=2.3.
    But you've basically cracked it, so I'll show you an easier way. You do not need to calculate the angles in degrees.
    In 2D, the components of a unit force in the direction (x,y) are x/h, y/h where h=√(x2+y2), right?
    In 3D, that extends simply to x/h, y/h, z/h where h=√(x2+y2+z2).
     
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