375_Physics ProblemsTechnical Physics

375_Physics ProblemsTechnical Physics - Chapter 12 P12.67...

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Chapter 12 377 P12.67 Let θ represent the angle of the wire with the vertical. The radius of the circle of motion is r = 0850 .s i n m af . For the mass: Fm am v r mr Tm rr == = = 2 2 2 ω θθ sin . sin m Further, T A Y =⋅ strain or TA Y strain Thus, AY m ⋅= strain m a f a f 2 . , giving T r m g FIG. P12.67 π = = ×× × −− AY m strain m m N m kg m 2 ej e j e j bg 3 90 10 7 00 10 1 00 10 120 4 2 10 3 . .. . or = 573 . r a d s . P12.68 For the bridge as a whole: τ AA E nn =− + = 01 3 3 1 0 0 2 0 0 0 af a f a f a f kN m m so n E 13 3 100 200 666 . . kN m m kN a f Fn n yA E += 13 3 0 . kN gives AE = 13 3 6 66 kN kN At Pin A: FF B °+ = sin . . 40 0 6 66 0 kN or F AB = ° = 40 0 10 4 . sin . . kN kN compression xA C ° = 10 4 40 0 0 .c o s . kN so F AC = 104 400 794 o s kN kN tension a f a f At Pin B: yB C ° = 10 4 40 0 40 0 0 . sin . sin . kN Thus, F BC = 10 4 t en s ion a f F F F B B C B D BD = = cos . cos . o s . . 40 0 40 0 0 2104 159 kN kN compression a f By symmetry: DE AB 10 4 comp r e s s DC BC 10 4 t s and EC AC k N t e n s i o n a f We can check by analyzing Pin C:
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