If the 1700-lb boom AB, the 215-lb cage BCD, and the 164-lb man have centers of gravity located at points \( G_1 \), \( G_2 \), and \( G_3 \), respectively, determine the resultant moment produced by all the weights about point A. (Figure 1)

Express your answer to three significant figures and include the appropriate units. Enter a positive value if the moment is counterclockwise and a negative value if the moment is clockwise.

Let's assume the distances between the points G1, G2, and G3 and point A are d1, d2, and d3, respectively. Then, the moment produced by each weight around point A can be calculated as follows:

Moment of AB = 1700 lb x d1

Moment of BCD = 215 lb x d2

Moment of man = 164 lb x d3

The resultant moment produced by all the weights about point A can be found by adding up these individual moments:

Resultant moment about A = (1700 lb x d1) + (215 lb x d2) + (164 lb x d3)

To express the answer in appropriate units, we need to convert pounds (lb) to a standard unit of force such as newtons (N) and inches (in) to meters (m).

Assuming 1 lb = 4.44822 N and 1 in = 0.0254 m, we get:

1700 lb = 7585.74 N

215 lb = 957.5125 N

164 lb = 730.7096 N

1 in = 0.0254 m

Substituting these values into the above equation, we get:

Resultant moment about A = (7585.74 N x d1) + (957.5125 N x d2) + (730.7096 N x d3)