To find the support force at point $A$ in the x-direction, we can apply the principles of static equilibrium, specifically the sum of forces and moments being zero.

Given Information:

• Force $F_1 = 50 \, \text{N}$ acting horizontally to the right at the top-right point of the truss.
• There is also a vertical reaction force at the middle joint pointing downwards (typically representing load forces).

The support at $A$ is a pin support, meaning it can resist forces in both the x and y directions.

Step-by-step Approach:

1. Sum of Forces in the x-direction: The truss is in equilibrium, so the sum of forces in the horizontal direction must equal zero: $\Sigma F_x = 0$ Let $A_x$ be the support reaction at $A$ in the x-direction. The only other force in the horizontal direction is $F_1$, acting to the right. Thus: $A_x + F_1 = 0$ Substituting $F_1 = 50 \, \text{N}$: $A_x + 50 = 0$ Solving for $A_x$: $A_x = -50 \, \text{N}$ The negative sign indicates that the reaction force at $A$ is directed to the left.

Final Answer:

The support force at $A$ in the x-direction is $A_x = -50 \, \text{N}$, meaning it acts to the left.

Would you like details on other forces or further explanations?

Here are 5 relative questions to expand:

1. What are the forces in the y-direction at support $A$?
2. How do the moments about point $A$ affect the system?
3. How does the 4 m vertical distance impact the force distribution?
4. What would happen to the support reactions if $F_1$ increased to 100 N?
5. How can the internal forces in the truss members be calculated?

Tip: When solving trusses, remember that reactions at supports must balance all external forces to ensure equilibrium.