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Force Required to be a minimum

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    If the resultant force acting on the bracket is required to be a minimum, determine the magnitude of F1. Set Φ=31°
    XllYL1D.png

    2. Relevant equations
    Fx=F(cos(θ))
    Fy=F(sin(θ))

    3. The attempt at a solution
    First I tried to find ∑Fx and ∑Fy
    ΣFx=0=200+cos(67.38)260+cos(59)F1x
    ΣFy=0=sin(67.38)260+sin(59)F1y

    Solving for F1, I got
    x=-384.06
    y=279.99

    And then to find the magnitude, I found the root of the sum of their squares

    √(-384.06² + 279.99²)=479.335N

    which was wrong.

    A few things: I know the x-component can't be negative, it makes no sense, but I don't know what to change to make it positive. I also don't know what is implied by "required to be a minimum."
     
    Last edited: Feb 2, 2016
  2. jcsd
  3. Feb 2, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Something important missing here?
     
  4. Feb 2, 2016 #3
    I'm not sure what you mean, am I missing a step or did I leave something out? All it says in the prompt is "Set ϕ = 31° ."
     
  5. Feb 2, 2016 #4

    NascentOxygen

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    There's a degree symbol? Typesetting sometimes misses symbols. Do you think it says set F1's angle to be 31°?
     
  6. Feb 2, 2016 #5
    Oh sorry, I see a degree symbol in my original post, not sure why it doesn't show up.

    nAUgaKZ.png

    I did set ϕ = to 31°, but I used 59° in my equations just because it was a little easier to visualize for me. I didn't think it would make a difference because sin(31)=cos(59).
     
  7. Feb 2, 2016 #6

    NascentOxygen

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    The phi isn't showing up on my screen, it's just a blank space.
     
  8. Feb 2, 2016 #7
    Oh that's odd, I even see the phi in your quoted text. I rewrote it using the editor instead of copy-pasting symbols, so hopefully it looks a little more accurate.
     
  9. Feb 2, 2016 #8

    NascentOxygen

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    Your mistake is in equating the sum of the components to zero. That would certainly be ideal, but with the angle fixed all you can aim for is a minimum, not zero.
     
  10. Feb 2, 2016 #9
    What does it mean to find a minimum? A minimum force that would keep it in equilibrium? The smallest possible resultant force?
     
  11. Feb 2, 2016 #10

    NascentOxygen

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    The resultant sum of those 3 external forces is to be its smallest possible value.
     
  12. Feb 3, 2016 #11
    So instead of ∑Fx,y being equal to 0, they should equal the smallest possible value?

    Wouldn't that be Fx=0 and Fy=0?
     
  13. Feb 3, 2016 #12

    NascentOxygen

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    The character set you are using seems problematical. I saw Fy= -∞ in your post, which seemed a puzzling value; yet when I quote that to respond, that -∞ has transformed into a 0. You should be using the characters that appear when you tap ##\Sigma## in the forum editor toolbar, or else use Latex.

    You can't make the resultant horizontal force equal to zero, as all forces have a component to the right---there is not going to be any cancelling in the x-direction.
     
  14. Feb 3, 2016 #13
    I had typed -∞ at first, but after considering it, I realized it didn't make sense so I changed it to 0.

    So I can't set Fx=0, but I want to get this value as small as possible. Do I set up a limit as Fx approaches 0?
     
  15. Feb 3, 2016 #14

    NascentOxygen

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    Just my caching problem, then.
    Probably not.

    You had the right idea back in your opening post:
    So go back to your first post, and improve on that. If force up is +, then downwards will be -
    That won't be F1x, it's just F1 because the cos converts it to the horizontal component.
     
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