Using the hoist-to-hoist transfer operation with a load of 6969 pounds and dimensions D1 = 10 ft, D2 = 18 ft, S1 = 9.5 ft, S2 = 14 ft, H = 4 ft, calculate the tension in chain B (nearest pound).

• A. 8500 LBS
• B. 8711 LBS
• C. 8600 LBS
• D. 8900 LBS

Answer: (B)

In a hoist-to-hoist transfer, the load is supported by two chains that pull in different directions, so you must use static equilibrium to balance both vertical and horizontal forces. The two chain tensions must satisfy the horizontal balance (the horizontal components cancel each other) and the vertical balance (the sum of the vertical components supports the weight).

First determine the angles each chain makes with the vertical from the given geometry. Each chain angle comes from the horizontal offset of that chain (how far the hoist is from the load) and the vertical distance between the hoist and the load, which are defined by the dimensions D1, D2, S1, S2, and H. Denote the angles to the vertical as theta_A for chain A and theta_B for chain B. Then the force balance equations are:

• Horizontal: T_A sin(theta_A) = T_B sin(theta_B)

• Vertical: T_A cos(theta_A) + T_B cos(theta_B) = W, where W is the load weight (6969 lb)

From the horizontal equation you get T_A = T_B sin(theta_B) / sin(theta_A). Substituting into the vertical equation and solving for T_B gives:

T_B = W / [ (sin(theta_B) cos(theta_A) / sin(theta_A)) + cos(theta_B) ].

Plugging in the geometry-derived sines and cosines for theta_A and theta_B (from D1, D2, S1, S2, and H) yields a tension in chain B of approximately 8711 pounds (nearest pound). The result reflects how the angles skew the load more into one chain when the hoist positions create a more horizontal pull in that chain, even though the total weight is 6969 lb.