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Torque of hinges on a door

  1. Jul 14, 2005 #1
    A door 2.30m high and 1.30m wide,has a mass of 13.0kg. A hinge 0.40m from the top and another hinge from the bottom each support half the door`s weight. By assumming that the center of gravity is at the geometrical center of the door,determine the horizontal and vertical force components exerted by each hinge on the door.

    How to solve this question? I have no idea at all to start doing.
     
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  3. Jul 14, 2005 #2

    arildno

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    Remember that since the door is in equilibrium, then the sum of external torques about ANY point must be zero.
    It is simplest to start with studying the equation you get when calculating the net external torque about one of the hinges.
     
  4. Jul 14, 2005 #3
    I dont know the direction of the force
     
  5. Jul 14, 2005 #4

    arildno

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    Quite so;
    But what you DO know, is that only the HORIZONTAL component of the hinge force will give rise to a torque about the other hinge (agreed?)
    In addition, of course, the force of gravity will provide a torque about the hinge you're calculating equilibrium of torques with respect to.
    Thus, you will be able to determine the horizontal component of the hinge force from the equation.

    Okay?
     
  6. Jul 14, 2005 #5
    Sorry..I am kinda stupid. I am not really understand.
    Some more, how about the vertical force?
     
  7. Jul 14, 2005 #6

    Pyrrhus

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  8. Jul 14, 2005 #7

    arildno

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    Okay.
    The lever arm from one hinge to the other is vertical, right?
    Thus, a vertical force acting on a hinge can't produce a net torque around the other hinge, because the lever arm is parallell to that vertical force.
     
  9. Jul 14, 2005 #8
    you first need to draw the FBD of the system and find the force components in the x and y dir. then you can sum the moments around either or both of the hinges. use statics for these equations.
     
  10. Jul 14, 2005 #9

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  11. Jul 14, 2005 #10

    Pyrrhus

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    Because there's also a torque produced by the weight of the door with a lever arm of 1.3/2.
     
  12. Jul 15, 2005 #11
    i have a question about torque: A force F_vec of magnitude F, making an angle theta with the x axis, is applied to a particle located at point A, at Cartesian coordinates (0, 0) in the figure. The vector F_vec and the four reference points (i.e., A, B, C, and D) all lie in the xy plane.

    The torque tau_vec of a force F_vec acting on a particle having a position vector r_vec with respect to a reference point (thus r_vec points from the reference point to the point at which the force acts) is equal to the cross product of r_vec and F_vec, \vec{\tau} = \vec{r} \times \vec{F}. The magnitude of the torque is \tau = r F \sin(\alpha), where alpha is the angle between r_vec and F_vec; the direction of tau_vec is perpendicular to both r_vec and F_vec. For this problem \vec{\tau} = \tau \hat{z}; negative torque about a reference point corresponds to clockwise rotation. You must express alpha in terms of theta, phi, and/or pi when entering your answers.

    heres the question:What is the torque tau_A due to force F_vec about the point A?
    Express the torque about point A at Cartesian coordinates (0, 0).?
     

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  13. Jul 15, 2005 #12

    Pyrrhus

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    Next time make a new thread for your questions.

    if you want the torque about A, where there's only a force F acting on the "particle", you need to draw the position vector from the point you want to calculate the torque to the point of application of the force vector, then do the cross product, if you notice the position vector from A to A will have a magnitude of 0 therefore the torque vector will be 0.
     
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