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

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

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

Problem 37P: An elevator filled with passengers has a mass of 1.70103kg . (a) The elevator accelerates upward...

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