CHAPTER 3 - 4 - 12.(a) For the components we have R R x = A...

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Unformatted text preview: 12.(a) For the components we have R R x = A + B + C x x x 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 = 11.6. (b) We find the resultant from R = (R + R ) = [(24.0) + (11.6) ] = 26.7; x y 2 2 1/2 2 2 1/2 tan θ = R /R = (11.6)/(24.0) = 0.483, which gives y x θ = 25.8° above + x­axis. y B x – A R θ 13. (a) For the components we have R = B – A x x x R y = – 26.5 cos 56.0° – 44.0 cos 28.0° = – 53.7; = B – A y y = 26.5 sin 56.0° – 44.0 sin 28.0° = 1.3. We find the resultant from R = (R + R ) = [(– 53.7) + (1.3) ] = 53.7; x y 2 2 1/2 2 2 1/2 tan θ = R /R = (1.3)/(53.7) = 0.0245, which gives y x θ = 1.40° above – x­axis. y A – B x θ x R Note that we have used the magnitude of R for the angle indicated on the diagram. (b) For the components we have R = A – B x x x R y = 44.0 cos 28.0° – (– 26.5 cos 56.0°) = 53.7; = A – B y y = 44.0 sin 28.0° – 26.5 sin 56.0° = – 1.3. We find the resultant from R = (R + R ) = [(53.7) + (– 1.3) ] = 53.7; x y 2 2 1/2 2 2 1/2 tan θ = R /R = (1.3)/(53.7) = 0.0245, which gives y x θ = 1.40° below + x­axis, which is opposite to the result from (a). ...
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CHAPTER 3 - 4 - 12.(a) For the components we have R R x = A...

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