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Static Equilibrium Door question: Calculate Horizontal Forces

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A door, essentially a uniform rectangle of height 2.00 m, width 0.87 m, and weight 130.0 N, is supported at one edge by two hinges, one 35.1 cm above the bottom of the door and one 164.9 cm above the bottom of the door. Calculate the horizontal components of the forces on the two hinges.

    Lower Hinge: Magnitude, direction
    Upper Hinge: Magnitude, direction

    2. Relevant equations

    COM for uniform mass should be right in the middle (coordinates .435, 1)
    Through pythagorean theorem, D distance from COM to Lower hinge: (1-.351)=.649 (.649^2+.435^2)^1/2=.7813 m

    Same for D to upper hinge.

    M=W/g=130/9.81=13.25 N

    I com= 1/2mh^2
    Moment of inertia for door using parallel axis theorem:
    1/2 mh^2+ md^2

    In static equilibrium, all forces must add up to zero.

    3. The attempt at a solution

    First I tried calculating the Moment for each hinge, and that didn't work.

    Conceptually, I know that the the Weight should be counterbalanced in some way by the moment of inertia on the door by the hinges. (I don't think the normal force on the hinges factors in here)

    You can't get a horizontal force from the weight of the door, and I'm having trouble deriving one from the Moment.

    Sorry there isn't more work here---a hint or a push in the right direction would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Jul 29, 2012 #2

    Im thinking if you look at one hinge, its going to have a force that can be broken into two components horizontal and vertical. Im thinking of using the lower hing as the pivot, as if the upper hinge was absent. If the ground was not present the whole door would rotate if the lower hinge worked like a pivot point. Or I guess you could use the upper hinge and pretend like there was no door jamb or wall and the door would rotate.

    So now you use the upper hinge to balance the lower hinge, or vice versa

    try this approach...
     
  4. Jul 29, 2012 #3
    I think I see what you mean, and it makes sense. I'll try that and post again with the results.
     
  5. Jul 29, 2012 #4
    So I tried the method:

    If the pivot pt is at .351, I plugged in .7813mg sin 56.2

    I got r= .7813 from the distance between center of mass and hinge. I got angle 56.2 from arctan y/x (in between hinge and COM)

    Doing a "plug and chug" with these didn't work.

    Then I tried cutting the mg in half, since half will be supported by the other pivot point.

    This didn't work either. Could you give me a hint on finding the x and y force? I think that is where I'm going wrong
     
  6. Jul 29, 2012 #5
    I have not put pencil to paper yet so just thinking.

    So if we have a door with the hinges on the left hand side of the door...

    Seems like the top hinge would get pulled to the right by the door trying to rotate clockwise, so the hinge would have to apply a force to the left. But since the door does not rotate counter clockwise, the bottom hinge would apply a force to the right and it must be equal to the pull left (forces in the x-direction).

    Now for the y direction... Im having trouble seeing the top hinge bear as much weight as the bottom hinge. Oh shucks, they dont want this in the problem as I read it?
     
    Last edited: Jul 29, 2012
  7. Jul 29, 2012 #6
    yes, but I didn't factor the I in on my second attempt. I was using the rXF definition of torque, not using I*α since α would be zero in this case...

    Or is that not correct? Because if we could find angular acceleration we could set

    I*α = r X F to get F

    I'm looking through my book and I'm not seeing any Static equilibrium problems that use angular accel.

    I do agree with what you're saying. The top hinge will have a force in the positive x direction (assuming usual coordinates) and the bottom hinge will have a force in the negative x direction.

    How would I factor in ?
    Btw, my professor used a similar example in class and used I=1/2 mh^2 for the door then mr^2 for the p.a.th
     
  8. Jul 29, 2012 #7
    Well one classic problem is a rod perfectly horizontal attached to a wall by a hinge. They ask what is the force on the hinge at the exact moment the rod (at rest) is perfectly horizontal. The rod will obviously start to rotate and then things change. But, Using the The Iα = Fr and then finding α in terms of g and L (length of rod) you can equate this to tangential acceleration of the center of mass and then use this to find the force on the hinge. It turns out the force on the hinge (pivot) is 1/4 Mg of the rod. In this problem we are only concerned with the y-direction. I am trying to apply this to your situation.

    I am now going to have to put pencil to paper.
     
  9. Jul 29, 2012 #8
    Ok. Organize, I tell myself.

    Net torque = zero ;Forces in the horizontal must also balance no net force in the x- direction.

    So lets work on the net torque. And use one hinge as the point of rotation. If you dont want to use angles the Fxr = Iα but r is just the distance from a hinge to the center of the door horizontally (or half the width of the door), dont need the angle or pythag and all this.

    Now there is another force that works counter clockwise to the one mentioned above so that there is no net torque. That is the force... Lets use the bottom hinge as our point of rotation, that is the force that pulls to the left (key point). These two torques balance out. So you got the torque the top hinge supplies F x (height of the Door).

    Is for you to do the numbers now. But I would be interested to see how much weight each hinge bears... I am lazy and dont like punching in numbers, just looking at ideas.
     
  10. Jul 29, 2012 #9
    Thank you, that makes sense. Again, I'm going to give it a try.

    As for the "thin rod through the axis of rotation" I think that's beyond the scope of the problem.

    Thanks again, I'll check, and repost with thanks/questions
     
  11. Jul 29, 2012 #10
    It is indeed. Just thinking. I need to be careful and organize.
     
  12. Jul 29, 2012 #11
    Btw, you helped me yesterday with a different problem, thanks :)


    Ok, I'm running into a little problem. I'm using "practice mode which has different figures and shows you answers when you are wrong.

    Maybe bc of all the questions I've done already, my brain is having trouble with this:

    Are you saying:

    "F"= the weight, 130 N? Or equals the I moment of inertia?

    If we set FxR=Iσ without the rod part, do we say α=9.81 (height)?

    Thanks
     
  13. Jul 29, 2012 #12
    net torque = 0

    So -F(c) x 1/2width + F(cc) x h = 0 The F for the width arm is mg. c is clockwise cc is counter
     
  14. Jul 29, 2012 #13
    But what is F(c)? and F(cc)? They're the same, but you can't solve for them. Using I in their place didn't work.
     
  15. Jul 29, 2012 #14
    mg x 1/2 width = F x h if net torque = 0

    The clockwise force is mg, we know that. What we dont know is the horizontal force, F, from the pivot above the pivot on the bottom (which I chose as the point of rotation).
     
  16. Jul 29, 2012 #15
    Ok, I see what you're saying. Sorry, this is all tough for me
     
  17. Jul 29, 2012 #16
    I made it tougher with my first suggestions.

    Caffeine was not available this morning...:redface:
     
  18. Jul 29, 2012 #17
    So we get mg x .5 width= Fxh => F = (mg x .5w)/height

    This doesn't give me the right answer, but its in the ball park. I tried multiplying this by the moment of inertia, and that didn't work.

    Maybe the thin rod was meant to be included after all....
     
  19. Jul 29, 2012 #18
    And no problem lol, I'm running on loads and its not helping much
     
  20. Jul 29, 2012 #19
    It seems in static equil problems we almost always use the sin/cos to give us the vectors for force...
     
  21. Jul 29, 2012 #20
    what did you get for F?
     
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