570_Physics ProblemsTechnical Physics

570_Physics ProblemsTechnical Physics - 572 P19.73...

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572 Temperature P19.73 Let 2 θ represent the angle the curved rail subtends. We have LLR L T ii += = + ∆∆ 21 θα af and sin == L i i R L R 2 2 Thus, θαα =+= + L R TT i 2 11 sin FIG. P19.73 and we must solve the transcendental equation =+ = 0 0 0 0 0 5 5 T bg sin . sin Homing in on the non-zero solution gives, to four digits, ° 0 018 16 1 040 5 .. rad Now, hRR L i =− = cos cos sin 1 2 This yields h = 454 . m , a remarkably large value compared to L = 550 . c m . *P19.74 (a) Let xL represent the distance of the stationary line below the top edge of the plate. The normal force on the lower part of the plate is mg x 1 a f cos and the force of kinetic friction on it is µθ k mg x 1 a f cos up the roof. Again, k mgx cos acts down the roof on the upper part of the plate. The near-equilibrium of the plate requires F x = 0 −+ = −= µ θθ µθθ µµ kk k mgx mg x mg mgx mg mg x x cos cos sin cos sin cos tan tan 10 2 2 1 22 motion f kt f kb xL temperature rising FIG. P19.74(a) and the stationary line is indeed below the top edge by xL L k F H G
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