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Uniform plank - Supporting force

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform plank is supported by two equal 120 N forces at X and Y. The support at X is then moved to Z (halfway to the plank center).

    The supporting force at Y then becomes?

    _____________________________
    X...........Z...........c......................Y

    i tried to draw it above. The line is the plank, X and Y are the supports, Z is where X is moved to, and (c) is the center of the plank

    3. The attempt at a solution

    I came across this as i am studying for my test and i am a little lost.

    The solution says to do this:

    (240 N)(1/4L) = Fy(3/4L)

    and the answer is 80 N

    I am confused on where this solution comes from.

    Any explaining would be very helpful :)

    Thank you
     
  2. jcsd
  3. Apr 11, 2010 #2

    rock.freak667

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    Well start off with X and Y being at the ends. If the system is in equilibrium there, what should be the force acting at the center c?
     
  4. Apr 11, 2010 #3

    PhanthomJay

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    you are given the initial reactions; what must be the weight of the plank, and at what point does the resultant force of the plank's weight act?
     
  5. Apr 11, 2010 #4
    The force at the center would be (120N + 120N) = 240N correct?

    Is the equation used for this dealing with torque?

    thanks for the help
     
  6. Apr 11, 2010 #5

    rock.freak667

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    No, now you have the forces. So now you have force at the center. Now move X to Z (I don't know if you have that distance) and just take moments about any point. For example, take moments about the new position of X i.e. Z.
     
  7. Apr 11, 2010 #6
    I dont have the distance so i believe the plank length is considered L.

    I am not sure what you mean by moment.

    When you say moment do you mean Moment of force = Torque:[itex]\mathbf{\tau}= \mathbf{r} \times \mathbf{F}[/itex]
     
  8. Apr 11, 2010 #7

    PhanthomJay

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    the terms torque, moment, or moment of a force, are often used interchangeably.
     
  9. Apr 11, 2010 #8
    Oh ok thanks for the insight on that :)

    does this seem right.

    if i take the torque about point Z then

    tcenter + ty = tz

    (240N)(1/4L) + (Fy)(3/4L) = 0
     
  10. Apr 11, 2010 #9

    rock.freak667

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    Watch your sign convention, if clockwise moments are +ve, then anticlockwise moments are -ve.
     
  11. Apr 11, 2010 #10
    Ok thank you :)
     
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