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enter image description here Can someone explain how to get the Fx for the support A and E

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    $\begingroup$ Please edit your question to include your actual work. What did you do to find $F_y$? $\endgroup$
    – Wasabi
    Oct 8, 2018 at 11:16

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As always, let's start with the global equilibrium equations:

$$\begin{align} \sum F_x &= A_x + E_x = 0 \\ \therefore A_x &= -E_x \\ \sum F_y &= A_y + E_y - 3\times40 = 0 \\ \therefore A_y &= 120 - E_y \\ \sum M_A &= 5E_x - 40(5 + 10 + 15) = 0 \\ \therefore E_x &= 240\text{ kN} \\ \therefore A_x &= -240\text{ kN} \end{align}$$

So that's the horizontal reactions, trivially found by using global equilibrium equations.

For completeness, we can also directly find the vertical components since $E$ is only connected to one truss member, therefore $E$'s resultant reaction must be parallel to that member.

$$\begin{align} \dfrac{E_y}{E_x} &= \dfrac{5}{15} \\ \therefore E_y &= \dfrac{E_x}{3} = 80\text{ kN} \\ \therefore A_y &= 40\text{ kN} \end{align}$$

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