123_Physics ProblemsTechnical Physics

123_Physics ProblemsTechnical Physics - Chapter 5 P5.17 m =...

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Unformatted text preview: Chapter 5 P5.17 m = 1.00 kg 50.0 m mg = 9.80 N α 0.200 m tan α = 25.0 m α = 0.458° 0.200 m T T Balance forces, mg 2T sin α = mg T= 9.80 N = 613 N 2 sin α FIG. P5.17 T3 = Fg (1) T1 sin θ 1 + T2 sin θ 2 = Fg (2) T1 cos θ 1 = T2 cos θ 2 P5.18 (3) θ2 θ1 Eliminate T2 and solve for T1 T1 = bsinθ Fg cos θ 2 1 cos θ 2 + cos θ 1 sin θ 2 Fg g = Fg cos θ 2 b sin θ 1 + θ 2 T3 = Fg = 325 N FG cos 25.0° IJ = 296 N H sin 85.0° K F cos θ IJ = 296 NFG cos 60.0° IJ = =T G H cos 25.0° K H cos θ K P5.19 1 1 2 See the solution for T1 in Problem 5.18. T1 θ1 T1 = Fg T2 g θ2 T3 163 N FIG. P5.18 T2 125 ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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