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Force analysis help

  1. Jul 24, 2017 #1
    Hi, I'm trying to figure out the forces Fx in the drawing I have uploaded. If I have a vertical force Fy acting on the bracket I have drawn. I know from experience that this force is going to pull my bracket horizontally but I can't figure out how I work out a horizontal force from a vertical one? Please help??

    20170724_205348.jpg
     
  2. jcsd
  3. Jul 24, 2017 #2

    jambaugh

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    Since you're dealing with a rigid body with one pivot point you will want to resolve the total torque about the pivot. These will be functions of the dimensional parameters and the applied forces. Since the object is static the net torque is zero.
     
  4. Jul 25, 2017 #3
     
  5. Jul 25, 2017 #4
    Can I translate the torque in to force? As the fixings I have at Fx are rated in kN?
     
  6. Jul 26, 2017 #5

    jambaugh

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    The problem with working with forces is that there are more unknown forces than force equations since the pivot point is applying a force. However the pivot point cannot apply a torque about itself by the nature of it being a pivot point. Now your diagram does not show the total length in the y direction but calling that L, the total torque (which adds to 0) is (taking the CCW direction as positive) (and as I read the diagram):
    [tex] 0 = F_x \cdot (L-y) - F_y \cdot x [/tex]
    and the units assuming distances are in meters is kNm as you indicated the forces are give in kilonewtons.

    A Torque times the perpendicular distance off center by which it acts. In a more general setting one can express it in terms of cross products but units will end up being force times distance.
     
  7. Jul 26, 2017 #6

    Merlin3189

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    Yes.
    torque = force x distance, so force = torque/distance
    The distance being the distance from the pivot, perpendicular to the line of action of the force.
    Examples below. (Lxd is the torque caused by the load.)

    bracket.png
     
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