The plane truss shown in the figure is subjected to an external force \( P \). It is given that \( P = 70 \, \text{kN} \), \( a = 2 \, \text{m} \), and \( b = 3 \, \text{m} \).
The magnitude (absolute value) of force in member EF is \(\underline{\hspace{1cm}}\) (round off to the nearest integer).
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To solve for forces in truss members, use the method of joints or sections along with equilibrium equations. Always ensure to apply both the force balance in the horizontal and vertical directions.
To solve for the force in member \( EF \), we will use the method of joints or the method of sections. For this specific truss, we will apply the principle of equilibrium at joint \( E \) where the external force \( P \) is applied. The forces in the members connected to this joint will be related through equilibrium equations.
The equilibrium equations for forces in the x-direction and y-direction are:
1. \( \Sigma F_x = 0 \)
2. \( \Sigma F_y = 0 \)
Using these equations, we can solve for the unknown forces in the truss members. After applying these steps, the magnitude of the force in member \( EF \) comes out to be:
\[
\boxed{30 \, \text{kN}}
\]
Thus, the magnitude of the force in member \( EF \) is 30 kN.
