determine the reactions at the supports.

Explanation:

Step1: Calculate total load

The uniformly - distributed load (w = 800\ N/m) and the length of the loaded part (L=3\ m). The total load (P=wL). [P = 800\times3=2400\ N] The load acts at the centroid of the uniformly - distributed load, which is at a distance of (x = 1 + \frac{3}{2}=2.5\ m) from point (A).

Step2: Take moment about point (A)

Let the reaction at (B) be (R_B). The sum of moments about point (A), (\sum M_A=0). [R_B\times4 - 2400\times2.5=0] [4R_B=2400\times2.5] [R_B=\frac{2400\times2.5}{4}=1500\ N]

Step3: Calculate reaction at (A)

The sum of vertical forces (\sum F_y = 0). Let the reaction at (A) be (R_A). [R_A+R_B - 2400 = 0] [R_A=2400 - R_B] Substitute (R_B = 1500\ N) into the equation. [R_A=2400 - 1500=900\ N]

Answer:

The reaction at support (A) is (R_A = 900\ N) and the reaction at support (B) is (R_B = 1500\ N)