**Problem 6.39****Consider the truss shown in (Figure 1). Suppose that \( F_1 = 4 \, \text{kN} \) and \( F_2 = 5.5 \, \text{kN} \). After the truss is sectioned use a single equation of equilibrium for the calculation of each of the required forces.**---**Diagram Explanation:**The diagram shows a truss with various members labeled from A to H. It has five vertical forces acting on it:- A 2 kN force at point A.- Two 4 kN forces labeled \( F_1 \) between points B and C.- A 5.5 kN force labeled \( F_2 \) at point D.- A 3 kN force at point E.The truss is supported horizontally from points A to E, with horizontal distances of 5 m between each subsequent support (A-B, B-C, etc.).Additionally, there are vertical measurements marked on the right side for the height of the truss, indicating 3 m and 2 m in different sections.---**Part B****Determine the force in member \( HC \). State if this member is in tension or compression.**Express your answer to three significant figures and include the appropriate units. Use a negative value in the case of compression and positive value in the case of tension.\[ F_{HC} = 1.70 \, \text{kN} \]Status: Incorrect, try again; 3 attempts remaining.---This problem involves analyzing the forces on a truss, determining equilibrium, and calculating the forces within individual members.

The given figure is shown below:The given data is shown below: F1=4 kNF2=5.5 kNThe free body diagram of truss is shown below:Consider forces in x-direction and applying equilibrium condition ∑Fx=0Ax=0Taking moment about E and applying equilibrium condition ∑ME=0-Ay×20+2×20+F1×15+F1×10+F2×5=0-Ay×20+2×20+4×15+4×10+5.5×5=0-20 Ay+40+60+40+27.5=0Ay=8.375 kNThe free body diagram of the section passing through the members BC, HC and HG is shown below':Taking moment about H and applying equilibrium condition ∑MH=0FBC×3+Ay×5-2×5=03 FBC+8.375×5-10=03 FBC+31.875=0FBC=-10.625 kNFBC=10.625 kN CompressionConsider forces in y-direction and applying equilibrium condition Ay-2-F1+FHC×332+52-FHG×222+52=08.375-2-4+334 FHC-229 FHG=0229 FHG-334 FHC=2.375 ............1Consider forces in x-direction and applying equilibrium condition ∑Fx=0FHG×522+52+FHC×532+52+FBC=0529 FHG+534 FHC-10.625=0 ................2 Solving equations (1) and (2) FHC=2.186 kN (Tension)FHG=9.942 kN (Tension)The force in member BC and HC are 10.625 kN (C) and 2.186 kN (T) respectively.