Physics problem: elevator force

  • Thread starter lmf22
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  • #1
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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.
 

Answers and Replies

  • #2
382
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Upward direction positive; downward direction negative
[tex]\begin{array}{cc}
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\\
\end{array}[/tex]
 

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