CHAPTER 4 - 3

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online