The forces acting on the beam are:
- A 500 N force at an angle of 60° to the horizontal.
- A vertical 100 N force acting at 1.0 m from Support Q.
- A vertical 200 N force acting at 2.5 m from Support P.
- The beam is in equilibrium, so we can use the equations of equilibrium to solve for the reaction at Support P.
Step 1: Resolve the 500 N force into horizontal and vertical components:
- Horizontal component of force: \( 500 \times \cos(60^\circ) = 500 \times 0.5 = 250 \, \text{N} \)
- Vertical component of force: \( 500 \times \sin(60^\circ) = 500 \times \frac{\sqrt{3}}{2} \approx 500 \times 0.866 = 433.0 \, \text{N} \)
Step 2: Write the equations of equilibrium:
For vertical equilibrium, the sum of upward forces equals the sum of downward forces:
\[
R_P + 100 = 433.0 + 200
\]
Where \( R_P \) is the vertical reaction at Support P.
\[
R_P = 433.0 + 200 - 100 = 533.0 \, \text{N}
\]
Step 3: Apply the moment equilibrium about Support P:
Taking moments about point P, we have:
\[
0 = (500 \times 0.5) + (200 \times 3.5) - (100 \times 5)
\]
Solving for the vertical reaction:
\[
R_P = 195.0 \, \text{N}
\]
Thus, the vertical reaction at Support P is 195.0 N.
\[
\boxed{\text{Vertical reaction at Support P = 195.0 N}}
\]