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Homework Help: Physics problem: elevator force

  1. Sep 16, 2004 #1
    An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0700g.
    What is the maximum force the motor should exert on the supporting cable?
    What is the minimum force the motor should exert on the supporting cable?

    I converted the acceleration to .6867 m/s^2 and plugged that and the mass into F=ma, but the answers doesn't seem to be right.
  2. jcsd
  3. Sep 16, 2004 #2
    Upward direction positive; downward direction negative
    Newton's\ 2nd\ Law\\
    \sum \vec{F}=m\vec{a}\\
    F_{ec} + W =ma \ with \ a= \ The \ elevator's\ acceleration\\
    F_{ec} + (-mg) = ma\\
    F_{ec} = m(a+g)\\
    F_{ec_max}= m(0.0700g+g)\ When\ a=+0.0700g\ ie\ The\ elevator \ accelerates\ upward\\
    F_{ec_min}= m(-0.0700g+g)\ When\ a=-0.0700g\ ie\ The\ elevator \ accelerates\ downward\\

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