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Contact forces

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data
    http://answers.yahoo.com/question/index?qid=20090812013110AAcrnFs


    2. Relevant equations
    Equilibrium - Fx = Fcos(theta), Fy = Fsin(theta)



    3. The attempt at a solution
    I've got the answer however it's different to the one posted the link above - isolated the top pipe and used equilibrium for the forces which were - normal force at C the weight W of the pipe and the force F exerted by the 2nd pipe. By doing this I found that F and C is 1387.344N. Then isolated the bottom pipe - forces present were - normal force at B and A, W and F exerted by 1st pipe. Did equilibrium and ended up with normal force at B = 4162.031N and normal force at A to be 3924N.
    Did I do something wrong or is there an issue with the answer on the link?
     
  2. jcsd
  3. Aug 8, 2011 #2

    PeterO

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    Homework Helper

    Not a very successful link ????
     
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