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Homework Help: What is the moment experience by each hinges?

  1. May 13, 2005 #1
    help me to find this one!
    the weight of the door is 110N. what is the moment experience by each hinges?the diagram can be seen in the attachment file!
    Last edited: May 13, 2005
  2. jcsd
  3. May 13, 2005 #2
    You need to work out the distance from the centre of gravity of the door, to the hinge itself.
    From there, just apply the usual formula. M= Fd
  4. May 13, 2005 #3
    i thought it is M=F.perpendicular distance! and the angle of the hinges to the centre of gravity is not 90 degree? or i am wrong probably. but can you show me the solution to the question.anyone?any way is there any attachment on post 1?
    Last edited: May 13, 2005
  5. May 13, 2005 #4
    Yeah, the moment is the force x perpendicular distance to line of action of the force. There's no attachment.
  6. May 14, 2005 #5
    ok, try this one!
    Last edited: May 14, 2005
  7. May 14, 2005 #6


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    Can't open it. I get a "invalid or corrupted" message.

    Assuming the door (so the center of gravity is on the center line) is uniform and that there are two hinges, multiply half the weight of the door by the horizontal distance from each hinge to the center line of the door.
  8. May 15, 2005 #7
    i don't think i can upload the diagram. but i will describe it. it is the hinges of a door with weight of 110N. First the height of the door is 1.9 m and the width is 1 m. the first hinges (located on the edge as normal doors), is 0.1 m and the second is 1.7 m from the origin. hence state the moment experienced by each hinges? will both hinges experienced the same force although of different position in symmetrical to the centre of gravity?
    Last edited: May 15, 2005
  9. May 15, 2005 #8


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    The meaning of "moment experienced by each hinge" needs to be understood. The bottom line is that if all forces acting on each hinge are considered, and the door is hanging there without moving, the total moment calculated about each hinge is zero. If it were otherwise, the door would be twisting the hinges off the frame.

    The answer to the problem as stated is zero. The net moment calculated about any point in any statics problem is zero. The more interesting question is to find the forces provided by each hinge required to make those moments zero.

    In the ideal case, the hinges would be mounted so that half of the weight of the door is supported by each hinge. It takes an incredibly talented carpenter to even come close to that condition. If either hinge is attached to the frame slightly higher than the other, it will bear all of the weight; the other hinge will only serve to provide a force parallel to the floor, and thus provide a moment about the other hinge. Fortunately, as long as one hinge is directly above the other hinge it does not matter which hinge is bearing most or all of the weight. The vertical forces provided by the hinges act along the line through those hinges and do not contribute to the moments acting about those hinges.

    Each hinge can be considered as the center of moments. In both cases the weight of the door can be taken as being located at its center of mass. The moment from this force is the weight of the door times the perpendicular distance between the line of the force (straight down) and the location of the hinge. It is the same for either hinge. For moments about the lower hinge, the weight moment must be counteracted by the moment from the top hinge. The top hinge will provide a horizontal force toward the frame, and the perpendicular distance between the line of that force and the lower hinge is the distance between hinges. Similarly, the weight of the door and the horizontal force from the lower hinge, away from the frame, provide counteracting moments about the upper hinge.
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