We are tasked with finding the magnitude of the reaction force in the horizontal direction at support $A$ required for equilibrium. The system has the following given forces:

• $P = 218 \, \text{N}$ (acting at an angle of 30°),
• $F = 102 \, \text{N}$ (acting vertically),
• Several dimensions of the system (lengths of 1.5 m for each section),
• Two inclined sections of the structure, each angled at 30°.

Steps to solve the problem:

1. Resolve forces: Since we are dealing with equilibrium, we need to ensure that the sum of forces and moments acting on the system is zero. There will be forces in both horizontal and vertical directions, so we'll resolve the forces accordingly.

2. Sum of forces in the horizontal direction:

   • Horizontal component of $P$: $P_x = P \cos(30^\circ) = 218 \times \cos(30^\circ)$.
   • Reaction force at $A$ in the horizontal direction $A_x$.
   • The sum of horizontal forces should satisfy equilibrium: $A_x - P_x = 0.$

3. Sum of forces in the vertical direction:

   • Vertical component of $P$: $P_y = P \sin(30^\circ) = 218 \times \sin(30^\circ)$.
   • The applied force $F = 102 \, \text{N}$ acts vertically downwards.
   • The reaction force at $A$ in the vertical direction $A_y$.
   • The sum of vertical forces: $A_y + P_y = F.$

4. Sum of moments about point $A$:

   • Taking moments about point $A$, we ensure rotational equilibrium. Each force has a moment arm:
     • The force $F = 102 \, \text{N}$ has a moment arm of 1.5 m.
     • The vertical component of $P$ also contributes, with a total length of $3 \times 1.5 = 4.5 \, \text{m}$. $\sum M_A = 0: (F \times 1.5) - (P_y \times 4.5) = 0.$ This will allow us to solve for the vertical reaction $A_y$.

Calculation:

Let me now calculate the individual components and then determine the reaction force at $A$.Here are the results based on the given data and equilibrium conditions:

1. Horizontal component of force $P$: $P_x = 188.79 \, \text{N}$ Therefore, the horizontal reaction force at $A$ (denoted $A_x$) must also be: $A_x = 188.79 \, \text{N}$

2. Vertical component of force $P$: $P_y = 109 \, \text{N}$

3. Vertical reaction force at $A$: From the moment equilibrium, the vertical reaction force $A_y$ is: $A_y = 34.0 \, \text{N}$

Thus, the horizontal reaction force at $A$ is approximately $188.79 \, \text{N}$.

Would you like further details on any part of the process? Here are 5 questions that build on this concept:

1. How would the reaction forces change if the angle of $P$ were different?
2. How can we calculate the reaction forces if the applied force $F$ were placed at a different point?
3. What is the significance of taking moments about different points in a system like this?
4. Can the reaction forces be negative? What would that imply in a physical system?
5. How would additional supports in the system change the analysis?

Tip: When analyzing equilibrium problems, always start by resolving forces into components and applying the sum of forces and moments equations in a systematic way.