If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the resultant moment produced by all the weights about point A.

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance

= cos 75⁰ =

    = 2.59 ft

calculate distance x

= cos 75⁰ =

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft

answer:

i cant give me a pic

explanation: