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).

 

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 a truss using equilibrium principles. Let’s break it down step by step:

Step 1: Identify forces and symmetry.
The truss is symmetrical, and there is a vertical load of 50 N applied at the central joint \( C \). Because of symmetry, the force in members \( AC \) and \( BC \) will be equal in magnitude. The member \( AB \) is horizontal, and its force arises due to the horizontal components of \( AC \) and \( BC \).

Step 2: Vertical equilibrium at joint \( C \).
Let \( F_{AC} \) be the force in member \( AC \). The vertical components of forces in \( AC \) and \( BC \) must balance the applied 50 N load:
\[ 2F_{AC} \sin(45^\circ) = 50 \Rightarrow F_{AC} = \frac{50}{2 \cdot \sin(45^\circ)} = \frac{50}{\sqrt{2}} \approx 35.36 \, \text{N} \]

Step 3: Horizontal equilibrium at joint \( C \).
The horizontal components of \( F_{AC} \) and \( F_{BC} \) act in opposite directions and balance the force in \( AB \):
\[ F_{AB} = F_{AC} \cos(45^\circ) = 35.36 \cdot \cos(45^\circ) = 35.36 \cdot \frac{1}{\sqrt{2}} \approx 25 \, \text{N} \]

Final Answer:
The magnitude of the force in the horizontal member \( AB \) is \( \boxed{25 \, \text{N}} \).