To solve this problem, we need to perform the following steps:

1. Free Body Diagram (FBD):

We'll break down the forces and moments acting on the beam.

• At point A: A pin support is located at point A. This means the reaction forces at point A are vertical ($A_y$) and horizontal ($A_x$).

• Moment at A: A 200 kN·m moment is applied at point A.

• Distributed load: Between points A and B, there is a uniformly distributed load of 25 kN/m over a span of 7 meters.

  • Total load from the distributed load = $25 \, \text{kN/m} \times 7 \, \text{m} = 175 \, \text{kN}$.
  • The load acts at the centroid of the distributed load, which is at the midpoint of 7 meters (3.5 meters from A).

• At point B: There is a roller support with a 150 kN force acting at a 60° angle from the horizontal. This force has two components:

  • Horizontal component: $F_x = 150 \cos(60^\circ) = 75 \, \text{kN}$
  • Vertical component: $F_y = 150 \sin(60^\circ) = 129.9 \, \text{kN}$

• Point Load at B: A point load of 150 kN at B is given, angled at 60° from the horizontal.

2. Equilibrium Equations:

Now, we apply the equilibrium equations to solve for the reactions at points A and B.

• Summation of forces in the x-direction: $\Sigma F_x = 0 \quad \Rightarrow \quad A_x - 75 = 0 \quad \Rightarrow \quad A_x = 75 \, \text{kN}$

• Summation of forces in the y-direction: $\Sigma F_y = 0 \quad \Rightarrow \quad A_y + 129.9 - 175 = 0 \quad \Rightarrow \quad A_y = 45.1 \, \text{kN}$

• Summation of moments about point A: (Taking counterclockwise moments as positive) $\Sigma M_A = 0 \quad \Rightarrow \quad -200 + 175 \times 3.5 + 150 \sin(60^\circ) \times 10 = 0$ Solving this for the moment balance equation will give the moment reaction at A.

Once we solve for the reactions and draw the Free Body Diagram (FBD), it should include all these forces.

I'll compute the necessary values next.### Final Results:

1. Reactions at Point A:

   • Horizontal reaction at A: $A_x = 75 \, \text{kN}$
   • Vertical reaction at A: $A_y = 45.1 \, \text{kN}$

2. Force Components at Point B:

   • Vertical component of the force at B: $F_{yB} = 129.9 \, \text{kN}$

3. Moment at Point A:

   • The resulting moment balance at A gives $M_A = 2111.54 \, \text{kN·m}$, which represents the internal balancing moment to ensure equilibrium.

Next Steps:

You can now use this information to complete the Free Body Diagram (FBD), ensuring the forces and moments are correctly represented. If you'd like assistance with more details or have any questions, feel free to ask!

Here are 5 relative questions you might explore next:

1. How does changing the distributed load affect the reactions at points A and B?
2. What is the significance of the angle at point B, and how does it change the force components?
3. How would adding a moment at point B change the equilibrium equations?
4. Can the system remain in equilibrium if the support at B is removed?
5. How do we calculate the shear force and bending moment diagrams for this beam?

Tip: Always double-check the direction of force components when working with angled forces to ensure accurate equilibrium analysis.