Question

For a three-bar truss loaded as shown in the figure, the magnitude of the force in the horizontal member AB is \_\_\_\_ N (answer in integer).
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Hint:

For trusses with symmetrical loading, simplify the analysis by assuming that forces in symmetrical members are equal, and use equilibrium equations to calculate the unknown forces.

Answer 1

We are asked to determine the force in the horizontal member \( AB \) of the truss, which can be done using the method of joints or the method of sections. Since the truss is symmetrical, we can simplify the analysis by using equilibrium equations. Step 1: Identify the forces and equilibrium.
Let \( F_{AB} \) be the force in the horizontal member \( AB \). The truss is loaded with a 50 N force applied vertically at joint \( C \). Since the truss is symmetrical, the forces in members \( AC \) and \( BC \) will be equal. At joint \( C \), we can set up the equilibrium equations for both the horizontal and vertical directions. The vertical force at joint \( C \) is balanced by the force components of members \( AC \) and \( BC \). The horizontal force at joint \( C \) is balanced by the force in the horizontal member \( AB \). Step 2: Use equilibrium equations.
For vertical equilibrium at joint \( C \): \[ 2 F_{AC} \sin 45^\circ = 50 \quad \Rightarrow \quad F_{AC} = \frac{50}{2 \sin 45^\circ} = \frac{50}{\sqrt{2}} \approx 35.36 \, {N} \] For horizontal equilibrium at joint \( C \): \[ F_{AB} = F_{AC} \cos 45^\circ = 35.36 \times \cos 45^\circ \approx 35.36 \times 0.707 \approx 25 \, {N} \] Thus, the magnitude of the force in the horizontal member \( AB \) is 25 N.