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Find the magnitude of the horizontal force exerted by the lower hinge

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A symmetrical door of mass 27.5 kg is 5.44 m high and 2.46 wide. The hinges are placed at the very top and very bottom of the door. Find the magnitude of the horizontal force exerted by the lower hinge.


    2. Relevant equations
    Fd=Fd
    sum tau clockwise = sum tau counter-clockwise

    3. The attempt at a solution
    The center of mass is dead center of the door, since it's symmetrical. However, we have only done rotation for "1-dimensional" things like a table. I know that when you apply something like this to a table, you put the counterclockwise on one side, then the clockwise on the other. Since the door is vertical and has 2 points of rotation (the 2 hinges) I'm not sure how to work this out...
     
  2. jcsd
  3. Nov 9, 2008 #2
    Re: Equilibrium


    Gravity acts upon the center of mass of an object. So, I'm assuming the following is legit.:

    [tex]\sum\tau=0.[/tex] Taking the bottom hinge as the axes of rotation, we find:
    [tex]\tau_g=\vec{F}_g\times\vec{r}=m\vec{g}\vec{r}\sin(\theta)[/tex], where [tex]\theta=0.735 rad[/tex]. Now, what next?
     
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