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Torques on a door with hinges

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a do or, hanging on two hinges (s ee figure). The do or is L tall and
    d wide, and the hinges are placed at heights L/4 (point A) and 3 L/4 (point
    B) from the floor. The door is uniform, and so its centre of mass is in the
    middle of the door (point C)

    b) What is the torque from each of the 3 forces around point A?
    2. Relevant equations



    3. The attempt at a solution
    I'm not qute sure what i'm going to do. Are the forces from the two hinges pointing upwards or 45 degrees north east?

    When i'm going to find the torques around point A, should i just use T=fx ? The torque to point B will then be straight forward, but for find the torque to point C i have to decomposed the direction 45 degrees north east?
     
  2. jcsd
  3. Nov 5, 2014 #2

    haruspex

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    They will have an upwards component, but until you've done the algebra you won't know what the horizontal components are or what direction the net force is in. What equations can you write?
    That depends on what T, f and x are ;). Seriously, nobody can answer that until you define them.
     
  4. Nov 5, 2014 #3
    Byj0zHB.png

    Is the forces right on that drawing?


    When i'm going to calculate the torque around A, the torque of A will be 0 (because the arm is 0). Is it right that when I shall find the two others torques around i have to use the arm times the force?
     
  5. Nov 5, 2014 #4
    When i'm going to find the torque around A from B:

    The lenght to B from A = 3L/4 - L/4 = L/2

    T=L/2*Cosx*B

    is this correct?
     
  6. Nov 5, 2014 #5
    What is B here? The force at B? What is x?
     
  7. Nov 5, 2014 #6
    B is the force at B. X is the angle, because it's unknown...
     
  8. Nov 5, 2014 #7
    Cos x may be or not correct, depending on which angle you call x in your drawing. The general formula for torque is a cross product, so the magnitude is written in terms of the sin of the angle between the two vectors. But if you use the complement of the angle, it may be cos.
    You also can write the torque by using the components of B. It will be simpler. Anyway you will need to write the balance of forces and teh balance of torques by using components.
     
  9. Nov 5, 2014 #8

    haruspex

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    Further to nasu's reply, you should not assume the angle of the net force is the same at both hinges. Just put in four unknowns for the horizontal and vertical components of force at each hinge.
     
  10. Nov 6, 2014 #9
    but shouldnt the forces from hing A and B have opposite directions in the x-axis to get static equilibrium?
     
  11. Nov 6, 2014 #10
    Yes, they should.
     
  12. Nov 6, 2014 #11

    haruspex

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    Yes, the horizontal components must be equal and opposite, but you don't yet know whether the vertical components are equal. It is likely from the symmetry of the hinge positions that they will be, but it needs to emerge from the equations.
     
  13. Nov 6, 2014 #12
    is it wrong to write that the Torque from hing A to point C = -d/2*Mg ? where d/2 would be the lever-arm and (-) because of the clockwise rotation ? ?
     
    Last edited: Nov 6, 2014
  14. Nov 6, 2014 #13

    PhanthomJay

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    Watch terminology. The torque about A from the vertical force at C is not the same as the torque about C from the vertical and horizontal forces at A. In fact, it is not possible to determine the vertical force at A from the equilibrium equations. Let's not lose sight of what the original problem is asking: what is the torque about A from each of the three forces C_y, B_x, and B_y? (The sum of which add to 0).
     
  15. Nov 7, 2014 #14
    the Torque around A from the force B_y should then be equal to 0,because of that the vectorproduck = 0, and therefor the Torque around A from the forces could be that
    r1*FBx*sinθ = r2*FGysinΦ
    where r1 should be ½*L and r2 should be √((¼*l)2+(½*d)2) ?
     
    Last edited: Nov 7, 2014
  16. Nov 7, 2014 #15

    PhanthomJay

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    yes
    well OK using cross product rule
    yes, and where θ is 90 degrees and thus sin θ is 1
    well, yes, but where sin Φ = 1/2 d/r2, so r2 cancels out , no need to calculate it.

    When breaking up forces into x and y components, instead of using the cross product rule, note that the moment of a force or force component about a point is the product of that force or force component times the perpendicular distance from its line of action to the point. Makes things a lot simpler. The moment of the gravity force at C about A has already been correctly calculated somewhere in the above posts.
     
  17. Nov 7, 2014 #16
    Thanks for a good answer! :)
     
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