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Static problem with three cylinders

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Static problem

    Three identical cylinders of mass m are arranged in a triangle as shown:

    [​IMG]

    The friction coefficient between cylinders is "u1" and between cylinders and ground "u2". What condition must satisfy "u1" and "u2" to remain the system in equilibrium ?

    N1 is normal force between left cylinder and top cylinder (Fr1 friction force in the same point (u1*N1))
    N2 is normal force between right cylinder and top cylinder (Fr2 friction force in the same point (u1*N2))
    N3 is normal force between left cylinder and right cylinder (Fr3 friction force in the same point (u1*N3))
    N4 is normal force between ground and left cylinder (Fr4 friction force in the same point (u2*N4))
    N5 is normal force between ground and right cylinder (Fr5 friction force in the same point (u2*N5))

    2. Relevant equations

    solve({(1/2)*N1*sqrt(3)+(1/2)*N2*sqrt(3)+(1/2)*Fr2-(1/2)*Fr1-m*g = 0, (1/2)*Fr1*sqrt(3)+(1/2)*N1+(1/2)*Fr2*sqrt(3)-(1/2)*N2 = 0, N4+Fr3+(1/2)*Fr1-(1/2)*N1*sqrt(3)-m*g = 0, Fr4-N3-(1/2)*Fr1*sqrt(3)-(1/2)*N1 = 0, N5-m*g-Fr3-(1/2)*N2*sqrt(3)-(1/2)*Fr2 = 0, N3-Fr5+(1/2)*N2-(1/2)*Fr2*sqrt(3) = 0,Fr1+Fr2 = 0, Fr4+Fr3+Fr1 = 0, Fr2+Fr3-Fr5 = 0, Fr1 = u1*N1, Fr2 = u1*N2, Fr3 = u1*N3, Fr4 = u2*N4, Fr5 = u2*N5}, {u1, u2, Fr1, Fr2, Fr3, Fr4, Fr5, N1, N2, N3, N4, N5})

    3. The attempt at a solution

    {Fr2 = 0, Fr3 = 0, Fr1 = 0, Fr4 = 0, u1 = 0, N5 = (3/2)*m*g, N4 = (3/2)*m*g, N3 = -(1/6)*m*g*sqrt(3), N1 = (1/3)*m*g*sqrt(3), N2 = (1/3)*m*g*sqrt(3), u2 = 0, Fr5 = 0}

    :uhh:

    Possible interpretation of the result:

    The solution is correct (i suppose) to frictionless case, so with friction:

    Fr4 + N3 = 0 -> u2 * N4 = - N3 -> u2 * (3/2)*m*g = (1/6)*m*g*sqrt(3) -> u2 = sqrt(3)/9 ~ 0.19 ?

    u2 = 0.19 and u1 = 0 is sufficient to equilibrium !?
     
    Last edited: Jun 3, 2007
  2. jcsd
  3. Jun 3, 2007 #2
    Ok i solved it !!

    If you put u1, u2, u3, u4, u5 as friction coefficients (instead of u1 and u2) in equations:

    2. Relevant equations:

    solve({(1/2)*N1*sqrt(3)+(1/2)*N2*sqrt(3)+(1/2)*Fr2-(1/2)*Fr1-m*g = 0, (1/2)*Fr1*sqrt(3)+(1/2)*N1+(1/2)*Fr2*sqrt(3)-(1/2)*N2 = 0, Fr1+Fr2 = 0, N4+Fr3+(1/2)*Fr1-(1/2)*N1*sqrt(3)-m*g = 0, Fr4-N3-(1/2)*Fr1*sqrt(3)-(1/2)*N1 = 0, Fr4+Fr3+Fr1 = 0, N5-m*g-Fr3-(1/2)*N2*sqrt(3)-(1/2)*Fr2 = 0, N3-Fr5+(1/2)*N2-(1/2)*Fr2*sqrt(3) = 0, Fr2+Fr3-Fr5 = 0, Fr1 = u1*N1, Fr2 = u2*N2, Fr3 = u3*N3, Fr4 = u4*N4, Fr5 = u5*N5},{u1, u2, u3, u4, u5, Fr1, Fr2, Fr3, Fr4,Fr5, N1, N2, N3, N4, N5})

    3. Solutions:

    N1 = (1/2)*m*g
    N2 = (1/2)*m*g
    N3 = 0
    N4 = (3/2)*m*g
    N5 = (3/2)*m*g

    Fr1 = 1/2*(sqrt(3)-2)*m*g
    Fr2 = -1/2*(sqrt(3)-2)*m*g
    Fr3 = 0
    Fr4 = -1/2*(sqrt(3)-2)*m*g
    Fr5 = -1/2*(sqrt(3)-2)*m*g

    u1 = -2+sqrt(3) ~ -0.268 (negative because Fr1 have opposite direction)
    u2 = 2-sqrt(3) ~ 0.268
    u3 anyone
    u4 = -(1/3)*sqrt(3)+2/3 ~ 0.089
    u5 = -(1/3)*sqrt(3)+2/3 ~ 0.089

    So, the solutions are:

    Friction coefficient between cylinders: 0.268
    Friction coefficient between cylinders and ground: 0.089

    I though that friction was more important between ground and cylinders that between cylinders for stability but the result show opposite to my logic. :rofl:

    If someone finds any error in operations, please reply the thread.

    Greetings!
     
    Last edited: Jun 3, 2007
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