1.1 A beam is 10 meters long. Support A is located 2 meters from the left end, and Support B is located 7.5 meters from the left end. Loads are applied as follows: 450 kg at the left end, 6 kN at 3.5 meters from the left end, 700 kg at 6 meters from the left end, and 5 kN at the right end. Calculate the reactions at Supports \( A \) and \( B \).

To find the reactions at Supports A and B, you will need to use the principles of static equilibrium, meaning that the sum of forces and the sum of moments around any point must be zero. 1. Calculate the total load: - The total load due to the weights can be calculated by converting the masses to forces (using \( g \approx 9.81 \, \text{m/s}^2 \)): - \( 450 \, \text{kg} \) = \( 450 \times 9.81 \, \text{N} \approx 4414.5 \, \text{N} \) - \( 700 \, \text{kg} \) = \( 700 \times 9.81 \, \text{N} \approx 6867 \, \text{N} \) - The external loads also include \( 6 \, \text{kN} = 6000 \, \text{N} \) and \( 5 \, \text{kN} = 5000 \, \text{N} \). - Total load \( W = 4414.5 + 6000 + 6867 + 5000 \approx 22281.5 \, \text{N} \). 2. Solve for reactions \( R_A \) and \( R_B \) using equations of equilibrium: - Sum of vertical forces: \( R_A + R_B - 22281.5 = 0 \) (1) - Sum of moments around point A (taking counterclockwise as positive): \[ R_B \times 5.5 - 4414.5 \times 0 - 6000 \times 1.5 - 6867 \times 4 - 5000 \times 8 = 0 \] \[ R_B \times 5.5 = 0 + 9000 + 27468 + 40000 \] \[ R_B \times 5.5 = 60468, \quad R_B \approx 10905.45 \, \text{N} \] 3. Substitute \( R_B \) back into equation (1) to find \( R_A \): - \( R_A + 10905.45 - 22281.5 = 0 \) - \( R_A \approx 11376.05 \, \text{N} \). Thus, the reactions at Supports A and B are approximately \( R_A \approx 11376.05 \, \text{N} \) and \( R_B \approx 10905.45 \, \text{N} \).