CHAPTER 4 7 - 33(a Withdownpositive,wewrite?F= 1 m gF =m a Withuppositive,wewrite?F= F m g=m a(b ,weget(m m)g=(m m)a,or a 1 T 2 2 2 1 T 1 =(m m)g(m

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F T F air m c g m c g m h g y 33. ( a ) With down positive, we write ? F  =  m from the force diagram for  m 1 : m 1 g  –  F T  =  m 1 a .   With up positive, we write ? F  =  m from the force diagram for  m 2 : F T  –  m 2 g  =  m 2 a ;   ( b ) When we add the two equations, we get ( m 1  –  m 2 ) g  = ( m 1  +  m 2 ) a ,  or   a   = ( m 1  –  m 2 )g/( m 1  +  m 2 = (1150 kg – 1000 kg)(9.80 m/s 2 )/(1150 kg + 1000 kg) = 0.68 m/s 2 . Using one of the equations, we find the tension: m 1 g  –  F T  =  m 1 a ,  or F T  =  m 1 ( g  –  a ) = (1150 kg )(9.80 m/s 2  – 0.68 m/s 2 ) = 10,500 N. 34. ( a ) We select the helicopter and the car as the system. We write ? F  =  m from the force diagram: F air –  m h g  –  m c g  = ( m h  +  m c ) a , which gives F air = ( m h  +  m c )( a  +  g = (7500 kg + 1200 kg)(0.52 m/s
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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 University of California, Berkeley.

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CHAPTER 4 7 - 33(a Withdownpositive,wewrite?F= 1 m gF =m a Withuppositive,wewrite?F= F m g=m a(b ,weget(m m)g=(m m)a,or a 1 T 2 2 2 1 T 1 =(m m)g(m

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