CHAPTER 4 - 3 - 15 Wewrite?F= ycomponent:F mg =ma T...

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m g F T y +  y F T m g a 15. We write ? F  =  m from the force diagram for the bucket: y -component:   F T   –  mg   =  ma ;   63.0 N – (7.50 kg)(9.80 m/s 2 ) = (7.50 kg) a which gives        a  = – 1.40 m/s 2  (down) . 16. The maximum tension will be exerted by the motor when the elevator is  accelerating upward.   We write ? F  =  m from the force diagram for the elevator: y -component:   F Tmax   –  mg   =  ma ,   or   F Tmax  =  m ( a  +  g ) = (4125 kg)(0.0600 + 1)(9.80 m/s 2 ) =        4.29 × 10 4  N . The minimum tension will be exerted by the motor when the elevator is  accelerating downward.  We write ? F  =  m from the force diagram for the car: y -component:   F Tmin   –  mg   =  ma ,   or   F Tmin  =  m ( a  +  g ) = (4125 kg)(– 0.0600 + 1)(9.80 m/s 2 ) =        3.80 × 10 4  N . 17.
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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CHAPTER 4 - 3 - 15 Wewrite?F= ycomponent:F mg =ma T...

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