If the 1500-lblb boom ABAB, the 220-lblb cage BCDBCD, and the 169-lblb man have centers of gravity located at points G1G1, G2G2 and G3G3, respectively, determine the resultant moment produced by all the weights about point A.

 (M)_a = -8171 lb-ft

Step-by-step explanation:

Step 1:

- We will first mark each weight from left most to right most.

               Point:                      Weight:                             Moment arm r_a

               G_3                          W_g3 = 169                     30*Cos(75) + 4.25

               G_2                          W_g2 = 220                    30*Cos(75) + 2.5

               G_1                           W_g1 = 1500                    10*cos(75)

Step 2:

- Set up a sum of moments about pivot point A, the expression would be as follows:

(M)_a = -W_g3*(30cos(75) + 4.25) - W_g2*(30*Cos(75) + 2.5) - W_g1*10*cos(75)

Step 3:

- Plug in the values and solve for (M)_b, as follows:

    (M)_a = -169*(30cos(75) + 4.25) - 220*(30*Cos(75) + 2.5) - 1500*10*cos(75)

    (M)_a = -2030.462559 -2258.205698 - 3882.285677

    (M)_a = -8171 lb-ft

explanation:

remove parenthesis:

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apply rule

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multiply fractions:

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divide:

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mordancy.