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CHAPTER 3 - 2

# CHAPTER 3 - 2 - 6(a V =6.0,V =0 1x 2x 2y 1y V =V...

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6. ( a ) V 1 x  = – 6.0,      V 1 y  = 0 ;    V 2 x  =  V 2  cos 45 °  = 4.5 cos 45 °  = 3.18 =       3.2 V 2 y  =  V 2  sin 45 °  = 4.5 sin 45 °  = 3.18 =       3.2 . ( b ) For the components of the sum we have   R x V 1 x  +  V 2 x  = – 6.0 + 3.18 = – 2.82;   R y =   V 1 y  +  V 2 y  = 0 + 3.18 = 3.18. We find the resultant from R   = ( R x 2  +  R y 2 ) 1/2  = [(– 2.82) 2  + (3.18) 2 ] 1/2   =          4.3 ; tan  θ  =  R y / R x  = (3.18)/(2.82) = 1.13, which gives   θ  = 48 °  above –  x -axis Note that we have used the magnitude of  R x  for the angle indicated on the diagram. 7. ( a ) ( b ) For the components of the vector we have V x = –  V  cos  θ  = – 14.3 cos 34.8 °  =         – 11.7 ; V y V  sin  θ  =  14.3 sin 34.8 °  =         8.16 . ( c ) We find the vector from V = ( V x 2  +  V y 2 ) 1/2  = [(– 11.7) 2  + (8.16) 2 ] 1/2   =          14.3 ; tan  θ  =  V y / V x  = (8.16)/(11.7) = 1.42, which gives

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