The downward forces acting on the system are the weight of the man (800 N) and the weight of the beam (200 N). Thus, the total downward force is: \(800 \, \text{N} + 200 \, \text{N} = 1000 \, \text{N}\) The total upward forces are represented by reactions \( P \) and \( Q \): \(P + Q = 1000 \, \text{N}\) Since the beam is uniform, its weight acts at the center, which is located 2.5 m from either end. To satisfy the condition of equilibrium, the total clockwise moments must equal the total anticlockwise moments. Taking moments about point \( Q \): \((200 \, \text{N} \times 2.5 \, \text{m}) + (800 \, \text{N} \times 2 \, \text{m}) = P \times 5 \, \text{m}\) Calculating the moments: \(500 \, \text{N}\cdot\text{m} + 1600 \, \text{N}\cdot\text{m} = 5P\) This simplifies to: \(2100 \, \text{N}\cdot\text{m} = 5P\) Solving for \( P \): \(P = \frac{2100}{5} = 420 \, \text{N}\) Now, substituting \( P \) back into the upward force equation: \(420 \, \text{N} + Q = 1000 \, \text{N}\) Solving for \( Q \): \(Q = 1000 - 420 = 580 \, \text{N}\) Thus, the reactions at \( P \) and \( Q \) are: 420N and 580N respectively.