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Roller (Particle) Equilibrium

  1. Jul 7, 2004 #1

    (see attached picture for problem)

    "Three identical smooth rollers are positioned in a groove as shown. Knowing that each roller (with radius 2.5 ft) weighs 500 lb, determine the minimum angle [tex]\alpha[/tex] required to make the pile not collapse (all surfaces are assumed frictionless)."

    I know this is a particle equilibrium problem. My idea is that since the system is symmetric, we should be able to balance half of the system inorder to get a solution (in fact, I believe we have too, otherwise the balances will yield 0=0). For example, taking the right half of the system, I believe the that the only forces we have to worry about are the N2 pointing left, the N1 pointing up, the two weights, and the N1 normal between the two balls should be ignored. These are the equations I've come up with (though I think the geometic relations are wrong and variable with the value of [tex]\alpha[/tex]), with O being the low point where the two slopes intersect:

    [tex]\sum {F_Y} & = & \frac{1}{2}w + w + N_1 \cdot sin(\alpha)=0[/tex]
    [tex]\sum {M_O} & = & 2.5 \cdot w \cdot cos(\alpha) + 2.5 \cdot N_1 + N_2 \cdot 2.5 (1 + sin(\alpha))=0[/tex]
    [tex]\sum {F_X} & = & N_2 - N_1 \cdot cos(\alpha)=0[/tex]

    (sorry, my latex is bad)

    and of course, w=-500

    Solving those equations, I fiind alpha = -112.8 (deg) (along with several imaginary answers)
    Any help would be greatly appreciated!


    Attached Files:

  2. jcsd
  3. Jul 8, 2004 #2
    I don't know if you agree with me, but I work this ploblem like this:
    First, I work on the roller on top, so--> (N4+N5)sin(60)=mg and N4cos(60)=N5cos(60)---> N4=N5--->2(N5)sin(60)=mg--->N5=mg/(2sin(60))........(1).
    Then, I start working with the roller on the right.
    N5cos(60)+N2-N1sin(alpha)=0....(2) and N1cos(alpha)-mg-N5sin(60)=0.....(3).
    Substituting (1) on (3)----> N1cos(alpha)=mg+(1/2)mg=(3/2)mg......(1)*
    In equation (2) in order to get a minimum (alpha), N2 must be zero----->
    --->N1sin(alpha)=N5cos(60) and from (1)--->N1sin(alpha)=mg/(2tan(60).....(2)*
    Finally, dividing (2)* by (1)*---> tan(alpha)=(1/3tan(60))-----> alpha~10.8933....
    Due to certain geometrical properties, I used 60 degrees.

    I hope this help and forgive me because I don't know how to use Latex wisky40
  4. Jul 8, 2004 #3
    That makes perfect sense. The critical condition that I couldn't come up with was that N2=0. And your solution, being independent of mass or dimension, also makes sense. The work looks solid. Thanks a lot!!
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