CHAPTER 3 - 5 - y A – B α θ R β x C y – C R B β θ...

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Unformatted text preview: y A – B α θ R β x C y – C R B β θ α A x y x C θ – A α – B β 14. (a) For the components we have R = A – B + C x x x x R R y = 44.0 cos 28.0° – (– 26.5 cos 56.0°) + 0 = 53.7; = A – B + C y y y = 44.0 sin 28.0° – 26.5 sin 56.0° – 31.0 = – 32.3. We find the resultant from R = (R + R ) = [(53.7) + (– 32.3) ] x y 2 2 1/2 2 2 1/2 = 62.7; tan θ = R /R = (32.3)/(53.7) = 0.602, which gives y x θ = 31.0° below + x­axis. Note that we have used the magnitude of R for y the angle indicated on the diagram. (b) For the components we have R = A + B – C x x x x R y = 44.0 cos 28.0° + (– 26.5 cos 56.0°) – 0 = 24.0; = A + B – C y y y = 44.0 sin 28.0° + 26.5 sin 56.0° – (– 31.0) = 73.6. We find the resultant from R = (R + R ) = [(24.0) + (73.6) ] x y 2 2 1/2 2 2 1/2 = 77.4; tan θ = R /R = (73.6)/(24.0) = 3.07, which gives y x θ = 71.9° above + x­axis. (c) For the components we have R = C – A – B x x x x R y = 0 – 44.0 cos 28.0° – (– 26.5 cos 56.0°) = – 24.0; = C – A – B y y y = – 31.0 – 44.0 sin 28.0° – 26.5 sin 56.0° = – 73.6. We find the resultant from R = (R + R ) = [(– 24.0) + (– 73.6) ] x y 2 2 1/2 2 2 1/2 = 77.4; tan θ = R /R = (73.6)/(24.0) = 3.07, y x which gives θ = 71.9° below – x­axis. ...
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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 3 - 5 - y A – B α θ R β x C y – C R B β θ...

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