Question

solve it please,

Review The 1650-lb boom AB, the 200-lb cage BCD, and the 166-lb man have centers of gravity located at points G. G2, and G3, respectively (Figure 1). Part A Determine the resultant moment produced by each weight about point A. Follow the sign convention. Express your answers in kip-feet to three significant figures separated by commas. View Available Hint(s) Figure 1 of 1 > 19 AEP vec O ? (MG): (MG): (MG) A = kip ft Submit Previous Answers 2001 HG X Incorrect; Try Again; 5 attempts remaining The answers you have entered equal the magnitude of the moments. Recall that due to the sign convention positive moments are counterclockwise. Dredd Cadhock

Answer 1

[166 4 G2 1.75fte 2.5 soft 200 lb loft & 165006 6750 moment of 1650 to force about point A is clockwise (Mai) a =-(1650 lb) (106575°ft) (Manda = – 4270.51 16.ft Ans. moment of 200 lo force about A is clock wise (Mona) a = - (200 lb) (30 66575 +2.5) Ft (Mar) A = - 2052.91 lb ft Ans. moment of 166 16 force about A is carucwise (Mad) A = - (166 lb) ( 30 60575 +2.5+1.75) (Max) a = -1994.418 e.Ft Ans.