1. (a) The center of mass is given by xcom = [0 + 0 + 0 + (m)(2.00) + (m)(2.00) + (m)(2.00)]/6.00m = 1.00 m. (b) Similarly, ycom = [0 + (m)(2.00) + (m)(4.00) + (m)(4.00) + (m)(2.00) + 0]/6m = 2.00 m. (c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have xcog =
x1m1g1 + x2m2g2 + x3m3g3 + x4m4g4 + x5m5g5 + x6m6g6 = 0.987 m. m1g1 + m2g2 + m3g3 + m4g4 + m5g5 + m6g6
(d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m.