EXAMPLE 1.3 The 500-kg engine is suspended from the crane boom in Fig. 1-6a. Determine the resultant internal loadings acting on the cross section of the boom at point E. torsion 1.5 m +ΣΜΑ = 0; FCD SOLUTION Support Reactions. We will consider segment AE of the boom so we must first determine the pin reactions at A. Notice that member CD is a two-force member. The free-body diagram of the boom is shown in Fig. 1-6b. Applying the equations of equilibrium, - FCD()(2 m) [500(9.81) N](3 m) = 0 = 12 262.5 N -Im- E -1 m 1 m +2F, 0: - A, (12 262.5 N)(3) = 0 (a) FCD A, =9810 N +1EF, = 0: -A,+ (12 262.5 N)(3) - 500(9.81) N=0 2 m 1 m Ay =2452.5 N A Free-Body Diagram. The free-body diagram of segment AE is shown in Fig. 1-6c. Equations of Equilibrium. (b) M 9810 N 500(9.81) N