Question

Chapter 12, Problem 045 In the figure, a lead brick rests horizontally on cylinders A and B. The areas of the top faces of the cylinders are related by AA= 1.8 Ag; the Young's moduli of the cylinders are related by EA= 2.7 Ep. The cylinders had identical lengths before the brick was placed on them. What fraction of the brick's mass is supported (a) by cylinder A and (b) by cylinder B? The horizontal distances between the center of mass of the brick and the centerlines of the cylinders are defor cylinder A and dg for cylinder B. (c) What is the ratio dAldg? com of brick (a) Number 4.86 Units kg (b) Number Units (c) Number Units Click if you would like to Show Work for this question: Open Show Work

Answer 1

a since the brick is now horizontal and the cylinders were initially the same length l, then both have been compressed and equal amount Al. Thes, = Et = n - 1 - 2 - 3 FA = AATA B ABEB. = FA - FB - AREA ABEB = (1.8 AB) (2.7EB) ABGB 4 = 4.86 TAIFB & WL- FA FA - 4.86. FA= 4.86 FB. FA FB. 4.86 FA+ FB =W. FAT FA, I W. W= 5.86 FA 486. FA = 4.86 w 5.86 4.86 FA = 4.86 =) 0.829 This is FA = 4.86 FB Fatto =W. 4.86 FB t FB = W. 5,86 FB =W FB = 0.170w B = 01170

TA=TB FA.de = FB (-) de fe dB dA = FB JB 4.86 FB dA = 0.205 do