QuestionDetails: Three equal-length straight rods, of aluminum, Invar, andsteel,

Question

QuestionDetails: Three equal-length straight rods, of aluminum, Invar, andsteel, all at 17.0°C, form an equilateral triangle with hingepins at the vertices. At what temperature (in Celsius) will theangle opposite the Invar rod be 59.95°? See Appendix E forneeded trigonometric formulas and Table 18-2 for needed data. the change in angle is = (60o - 59.95o) =0.05o = 0.05 * 0.01745 radian = 8.7 * 10^-4 radian the change in length of the Invar rod is l = lo* ----------------(1) We know from the relation l = lo* (1 + t) or l = lo + lo* t or l - lo=  lo*t or l = lo* t--------------(2) from (1) and (2) lo* = lo*t or = t = * (t2 - t1) or (/) = t2 - t1 or t1 = t2 - (/) t2 = 17.0 oC and = 0.7 * 10^-6/oC the change in length of the Invar rod is l = lo* ----------------(1) We know from the relation l = lo* (1 + t) or l = lo + lo* t or l - lo=  lo*t or l = lo* t--------------(2) from (1) and (2) lo* = lo*t or = t = * (t2 - t1) or (/) = t2 - t1 or t1 = t2 - (/) t2 = 17.0 oC and = 0.7 * 10^-6/oC

Explanation / Answer

   from the theiry we can see that    from the law of cosines we get that     = 59.5o    so that we get    Linvar2= Lalum2 + Lsteel2- 2 Lalum Lsteel cos    where    L = Lo (1 + T)    dividing by Lo and ignoring terms of order(T)2 or higher we get that    1 + 2 invar T = 2 + 2(alum + alum) T - 2 (1 +(alum + alum) T)cos    on rearranging we get that    T = [cos - (1 / 2)] /[(alum + alum) (1 - cos) -invar]          =........oC    T = 17.0oC + T       =........oC

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