365_Physics ProblemsTechnical Physics

365_Physics ProblemsTechnical Physics - 367 Chapter 12...

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Chapter 12 367 P12.44 (a) Sum the torques about top hinge: τ = 0: CD A B 00 2 0 0 3 0 0 0 200 300 300 392 1 50 1 80 a f a f a f a f af ++ ° −+ += N N m m m cos . sin . . .. Giving A = 160 N right bg . 1.50 m 1.50 m 392 N 1.80 m C D T cos 30.0 ° T sin 30.0 ° A B FIG. P12.44 (b) F x = −− ° =−= CA C 200 30 0 0 160 173 13 2 N N N cos . . In our diagram, this means 13 2 . N to the right . (c) F y = 0: ++ − + ° = BD 392 200 30 0 0 N s i n . = 392 100 292 N N u p (d) Given C = 0: Take torques about bottom hinge to obtain AB D T T 0 0 0180 0 392 150 300 180 0 a f a f a f a f + + ° + ° = s i n . . c o s . . m N m m m so T = + = 588 156 192 Nm 1.50 m m N . . P12.45 Using FF xy ∑∑∑ == = 0, choosing the origin at the left end
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