# In both cases, a free-body diagram for the elevator would look like the adjacent diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton's second law for the elevator. 16. mg F ="la + mg = m(a+g)-m(0.0680g+g)-(4850 kg)(1 0680)(9.80m/s2) 5.08x10'N To find the MINIMUM tension, assume that the acceleration is down. Then Newton's second law for the elevator becomes the following. - (4850 kg) (0.9320) (9.80m/s2)-4.43x10 N

Question

an elevator (mass 4850kg) is to be designed so that the max acceleration is 0.0680g. what are the max and min forces the motor should exert on the suporting cable?

1. why did they add 1. wouldn't it just be Fn=m(a+g), which would be 4850(0.680g+ 9.8)? help_outlineImage TranscriptioncloseIn both cases, a free-body diagram for the elevator would look like the adjacent diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton's second law for the elevator. 16. mg F ="la + mg = m(a+g)-m(0.0680g+g)-(4850 kg)(1 0680)(9.80m/s2) 5.08x10'N To find the MINIMUM tension, assume that the acceleration is down. Then Newton's second law for the elevator becomes the following. - (4850 kg) (0.9320) (9.80m/s2)-4.43x10 N fullscreen

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### Newtons Laws of Motion 