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Equilibrium Problem - Balancing Torques

  1. Sep 2, 2016 #1
    1. The problem statement, all variables and given/known data

    A door 1.00 m wide and 2.00 m high weighs 330 N
    and is supported by two hinges, one 0.50 m from the top and the
    other 0.50 m from the bottom. Each hinge supports half the total
    weight of the door. Assuming that the door’s center of gravity is at
    its center, find the horizontal components of force exerted on the
    door by each hinge.

    2. Relevant equations

    T = F*l

    3. The attempt at a solution

    This problem is really giving me a hard time, because I know that both the horizontal and vertical components of the force of the hinge are exerting torques

    So, I start off with

    H1cos(a) + H2cos(b) = 0

    H2sin(a) + H2sin(b) - mg = 0

    For balancing the torque I choose the top left hinge as a pivot point

    Then the torque due to gravity would be mg*l which is just 0.5mg, now im not quite sure how to find the lever arm for the hinge force. Any suggestions?
     
  2. jcsd
  3. Sep 2, 2016 #2

    billy_joule

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    Science Advisor

    Have you drawn a diagram?
    What is the distance between the hinges?
    (You don't need any trig functions to answer the question).

    That depends on where the axis is you take moments about. You've chosen to use one of the hinges as the axis, this means the vertical forces at the hinges are irrelevant.
     
  4. Sep 2, 2016 #3
    Why is it the case that the vertical forces at the hinges are irrelevant. I can see why this is the case for the hinge that you choose as your pivot point, but wouldn't the vertical force at the other hinge still exert a torque since it may have a component that is perpendicular to the distance between the two hinges.
     
  5. Sep 2, 2016 #4

    billy_joule

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    Science Advisor

    Look at the door in the room you're in; both hinges lie on the same vertical line, correct? And the vertical forces are, of course, collinear with that line, correct?
    So if we take moments about any point on that vertical line (including at either hinge) the lever arm length for the vertical forces is zero, as their line of action passes through that point.
    If that's not clear, draw a diagram and we can see where you're going wrong.
     
  6. Sep 2, 2016 #5
    Right, ofcourse, that makes sense. For some weird reason I thought the hinges were on either side of the door

    So, now I just have the horizontal force of the bottom hinge, H2cos(b) * 1 = 0.5mg, and plugging that into the first equation gives us H1cos(a) = -0.5mg

    Thank you for your help
     
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