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Homework Help: 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


    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:



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


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


    Then I found FA using the Pythagorean theorem


    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


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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


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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


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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


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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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