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Find the axial forces of the members 2-3, 9-3 of the truss for the given loads.

**Solution**

From the equilibrium of the forces on the truss we have:

ΣF_{x} = 0 -> H_{1} = 0

ΣM_{1} = 0 -> 200 * 4 - B_{5} * 8 = 0 -> B_{5} = 100KN

ΣF_{y} = 0 -> B_{1} + B_{5} = 200KN -> B_{1} = 100KN

We consider the cross section T so as to calculate the axial forces on member 2-3 and 3-9. We consider the equilibrium of moments to the node 9.

ΣM_{9} = 0 -> 2 * 100 - S_{2-3} * 2 = 0 -> S_{2-3} = 100KN

From the quilibrium of forces at Y direction

ΣF_{y} = 0 -> -S_{9-3} * sin45 + 100 = 0 -> S_{9-3} = 100 / sin45 KN

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