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Please help to find normal force of the system.

  1. Dec 17, 2011 #1

    2. Relevant equations: N=mg; M=Fl

    3. The attempt at a solution: i find that the normal force of the rope AD is 8.5N.

    Sorry that in my first post i'm already asking for help, but i'm getting really desperate with this problem. I need to find the Normal force at the point O and direction of it. I tried multiplying the N of the rope times cos of 45° but it's not right.
    The answer should be: 7.2N, 57° vertical.
    Thanks in advance.
    P.S. I apologise for my bad english.
  2. jcsd
  3. Dec 17, 2011 #2
    Maybe I'm missing something here but a normal force is perpendicular to the surface and the only two surfaces involved at O are at 0 and [itex]\frac{\pi}{2}[/itex] so where 57[itex]^o[/itex] comes from I have no idea.
  4. Dec 17, 2011 #3
    That is exactly what i was thinking, but the book says otherwise..
    Well actually it says "find reaction force having in mind that point O car rotate when it's not balanced". And i assume that by "reaction force" it means normal force.
  5. Dec 17, 2011 #4


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    Homework Helper

    If the rod can rotate around O then it is connected to a hinge, and the reaction force comes from the pin through the hinge, fixed to the wall. The reaction force of the pin can be of any direction. I drew the hinge into your picture.


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  6. Dec 17, 2011 #5


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    Staff: Mentor

    I think you are translating the textbook wrongly. On force diagrams, the word "normal" means perpendicular to a line or surface. The force at 0 will have both a vertical and horizontal component, so the sum of these components means the reaction at O can not be exactly perpendicular to the vertical support.

    To solve the problem, you will need to take moments about a couple of convenient points. One of these is point D. So as part of your working, you'll need the vertical height DO.
  7. Dec 17, 2011 #6
    You should draw the forces on the diagram.
    As well as the reaction force at O which you are trying to find there is a Tension in the string, The weight due to the 0.6kg object at B and the weight of the beam itself (presumably this is uniform)
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