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Magnitude of The Force Supported at Point A

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the magnitude of the force supported by the point at A under the action of the 2.9-kN load applied to the bracket. Neglect friction in the slot.

    2. Relevant equations

    ƩMA=0
    ƩFy=0
    ƩFx=0

    3. The attempt at a solution

    I started by finding the moment around A so that I could solve for the vertical force at B:

    ΔyAC=(135)sin(34)

    MA=0=(2.9)(75.459)-FBy(155)
    FBy=-1.41

    Then I found FBX using trig functions and the angle between the vertical axis and FB:

    (-1.41)tan(56)=-2.09

    Then I summed all the forces to find the components of the force at point A:

    ƩFy=0=-1.41+FAy
    FAy=1.41Kn
    ƩFx=0=-2.09+2.99+FAx
    FAx=0.9Kn

    Then I found FA using the Pythagorean theorem

    FA=√(0.92+1.412)=1.672Kn

    This was wrong, am I missing a force/forces somewhere, or are my calculations incorrect?

    Thanks for any help!!!
     

    Attached Files:

    Last edited: Sep 16, 2013
  2. jcsd
  3. Sep 16, 2013 #2

    PhanthomJay

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    check math!
    0 = -2.09 + 2.9 + FAx
    Check typo and math!
     
  4. Sep 16, 2013 #3
    Good catch Phantom, I fixed it. Doesn't change the final answer though.
     
  5. Sep 16, 2013 #4

    collinsmark

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    [itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

    [itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

    Given what you're tying to do, are you sure you want to use sin(34) for that?
     
  6. Sep 16, 2013 #5
    I changed my calculation to 135*sin(34) and re-did all of my calculations to get a final answer of FA=2.09Kn, but thats still wrong.
     
  7. Sep 16, 2013 #6

    collinsmark

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    What I meant to express is that you are using the sine function. Are you sure that the sine function is the appropriate function here? Maybe a cosine would be better? :wink:
     
  8. Sep 16, 2013 #7
    Gah! Typos are an enemy today. I did use 135*cos(34) and ended up with FA=2.09Kn.
     
  9. Sep 16, 2013 #8

    collinsmark

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    Okay, 2.09 kN is the y-component of the force at point A. But there's still the x-component to figure out.
     
  10. Sep 16, 2013 #9
    The x-component of the force at A = 3.10-2.9 = 0.2

    After Pythagorean theorem the force at A comes to 2.099Kn which the system says is incorrect.
     
  11. Sep 16, 2013 #10
    I figured it out, I just had to round to 2.10.

    Thanks collinsmark!
     
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