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 Post subject: Elevators and Force
PostPosted: Sat, 18 Sep 2004 15:40:50 UTC 
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An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is .0600 g. What are the maximum and minimum forces the motor should exert on the supporting cable?

For the maximum (assuming elevator goes up):

Is the tension force the same as the motor's exerting force (which has opposite direction)?

F_t - mg = T - 4850 kg*9.8 m/s^2 = T- 47530 N
T - 47530 N = 4850 kg*0.0588 m/s^2
T - 47530 N = 2851.8 N
T = 50,381.8 N = F_motor ???

How do I get the minimum force?

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PostPosted: Sun, 19 Sep 2004 05:31:42 UTC 
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F = MA
M = 4850

Max A = 1.06g (gravity + acceleration up) = 10.388
Max F = 50381.8

Min A = .94g (gravity - acceleration down) = 9.212
Min F = 44678.2


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PostPosted: Sun, 19 Sep 2004 15:11:17 UTC 
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Could you explain where the .94g and 1.06g come from?

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"Which is a kind of integrity, if you look on every exit being an entrance somewhere else."
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PostPosted: Mon, 20 Sep 2004 04:47:56 UTC 
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The statement of the problem said that the maximum acceleration of the elevator is .06 g. The tension (=force) in the cable holding the elevator must support the weight of the elevator (mg) as modified my any additional acceleration: +.06mg going up, -.06mg going down.

There is more than one way to attack this problem. (Usual for most problems.)

Instead of writing mg+.06mg {or mg-.06mg} and factoring out mg to get (1+.06)mg {or (1-.06)mg}, I considered acceleration by itself: the cable must provide enough force to counter the acceleration due to gravity plus extra going up, or minus extra going down. Having established the accelerations that must be provided by the cable, I then calculated the required forces.


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