572 TemperatureP19.73Let 2θrepresent the angle the curved rail subtends. We haveLLRLTii+==+∆∆21θαafandsin==LiiRLR22Thus,θαα=+=+LRTTi211sinFIG. P19.73and we must solve the transcendental equation=+=0000055∆Tbgsin.sinHoming in on the non-zero solution gives, to four digits,°0 018 161 040 5..radNow,hRRLi=−=−coscossin12This yields h=454.m , a remarkably large value compared to ∆L=550. cm.*P19.74(a)Let xLrepresent the distance of the stationary line below thetop edge of the plate. The normal force on the lower part of theplate is mgx1−afcosand the force of kinetic friction on it isµθkmgx1−afcosup the roof. Again, kmgxcos acts down theroof on the upper part of the plate. The near-equilibrium of theplate requires Fx=∑0−+−−=−=−µθθµθθµµkkkmgxmgxmgmgxmgmgxxcoscossincossincostantan1022122motionfktfkbxLtemperature risingFIG. P19.74(a)and the stationary line is indeed below the top edge by xLLkFHG
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