... the publisher. Engineering Mechanics - Statics Chapter 9Units Used:Mg 10 3 kg=kN 10 3 N=Given:L 8m=ρw1.0Mgm 3 =a 3m=b 2m=g 9.81ms2=Solution:F 3 ρwgabL= F 3 47 0.88 kN=F2ρwgabL= ... integrals, we haveL0ax14b2x2a 4 +⌠⎮⎮⎮⌡d=L 23. 6 63 m=xc1L0axx 14b2x2a 4 +⌠⎮⎮⎮⌡d=xc9.178 m=A 2πxcL= A 1 .36 5 10 3 × m2=Problem 9-1 04 The suspension bunker ... 6in=Solution:A 4a2a22−πa2 4 −= A 97.7 in2=xc1Aa2−22− 3 a⎛⎜⎝⎞⎟⎠πa2 4 a4a 3 π−⎛⎜⎝⎞⎟⎠−⎡⎢⎣⎤⎥⎦= xc0.262− in=yc1Aa2−22a 3 πa2 4 4a 3 πa−⎛⎜⎝⎞⎟⎠−⎡⎢⎣⎤⎥⎦=...