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Falling Elevators

  1. Jun 10, 2004 #1
    edit::
    nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.
    ::edit

    I hit a brick wall over yet another problem concerning energy methods, basically a mismatch between my answer and the book's. I would appreciate whoever has the time to look this over and guess where I might be going wrong.

    Anyway, here's the problem verbatim:

    Begin with work-energy formula:
    [tex] \begin{align*}
    \Delta K =& \, W_\text{gravity} + W_\text{friction} + W_\text{spring} \\
    0 - \frac{1}{2}mv^2 =&\, mgX - fX - \frac{1}{2}kX^2
    \end{align*} [/tex]

    Quadratic formula to find compression distance X (neglect neg ans):
    [tex]
    X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 })
    [/tex]

    If static friction will hold elevator at X, then it follows:
    [tex] F = kX \leq f \text{ or } X \leq \frac{f}{k} [/tex]
    [tex]
    X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 }) \leq \frac{f}{k} [/tex]

    Manipulate to find assertion about spring constant:
    [tex] k \leq \frac{f}{mv^2} (3f - 2mg) [/tex]
    [tex]
    k \leq \frac{[17000 \text{ N}]}{[2000 \text{ kg}][25 \text{ m/s}]^2} (3[17000 \text{ N}] - 2[2000 \text{ kg}][9.80 \text{ m/ss}])
    [/tex]
    [tex] k \leq 160 \text{ N/m} [/tex]

    So, answer is not the same.

    edit::
    nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.
    ::edit
     
    Last edited by a moderator: Jun 10, 2004
  2. jcsd
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