1. **Stating the problem:** Calculate the net torque (moment) about each point A, B, C, D, and E for the given force system on the beam. 2. **Formula for torque:** The torque $\tau$ about a point is given by: $$\tau = F \times d \times \sin(\theta)$$ where $F$ is the force magnitude, $d$ is the perpendicular distance from the point to the line of action of the force, and $\theta$ is the angle between the force and the lever arm. 3. **Important rules:** - Counterclockwise torque is positive. - Clockwise torque is negative. - Forces perpendicular to the beam have $\theta = 90^\circ$, so $\sin(90^\circ) = 1$. - The 30 N force at E acts downward at $30^\circ$ to the horizontal, so its vertical component is $30 \times \sin(30^\circ) = 15$ N downward. 4. **Calculate torques about each point:** - Distances between points: A-B = 1 m, B-C = 2 m, C-D = 1 m, D-E = 2 m. - Forces and directions: - At A: 5 N down - At B: 20 N up - At C: 10 N down - At D: 5 N up - At E: 30 N down at 30°, vertical component 15 N down --- (i) Torque about A: - Force at A: distance 0 m, torque = 0 - Force at B: 20 N up at 1 m, torque = $20 \times 1 = 20$ Nm (CCW) - Force at C: 10 N down at 3 m, torque = $-10 \times 3 = -30$ Nm (CW) - Force at D: 5 N up at 4 m, torque = $5 \times 4 = 20$ Nm (CCW) - Force at E: 15 N down at 6 m, torque = $-15 \times 6 = -90$ Nm (CW) Net torque about A: $$20 - 30 + 20 - 90 = -80 \text{ Nm}$$ (ii) Torque about B: - Force at A: 5 N down at 1 m left, torque = $-5 \times 1 = -5$ Nm (CW) - Force at B: distance 0 m, torque = 0 - Force at C: 10 N down at 2 m right, torque = $-10 \times 2 = -20$ Nm (CW) - Force at D: 5 N up at 3 m right, torque = $5 \times 3 = 15$ Nm (CCW) - Force at E: 15 N down at 5 m right, torque = $-15 \times 5 = -75$ Nm (CW) Net torque about B: $$-5 - 20 + 15 - 75 = -85 \text{ Nm}$$ (iii) Torque about C: - Force at A: 5 N down at 3 m left, torque = $-5 \times 3 = -15$ Nm (CW) - Force at B: 20 N up at 2 m left, torque = $20 \times 2 = 40$ Nm (CCW) - Force at C: distance 0 m, torque = 0 - Force at D: 5 N up at 1 m right, torque = $5 \times 1 = 5$ Nm (CCW) - Force at E: 15 N down at 3 m right, torque = $-15 \times 3 = -45$ Nm (CW) Net torque about C: $$-15 + 40 + 5 - 45 = -15 \text{ Nm}$$ (iv) Torque about D: - Force at A: 5 N down at 4 m left, torque = $-5 \times 4 = -20$ Nm (CW) - Force at B: 20 N up at 3 m left, torque = $20 \times 3 = 60$ Nm (CCW) - Force at C: 10 N down at 1 m left, torque = $-10 \times 1 = -10$ Nm (CW) - Force at D: distance 0 m, torque = 0 - Force at E: 15 N down at 2 m right, torque = $-15 \times 2 = -30$ Nm (CW) Net torque about D: $$-20 + 60 - 10 - 30 = 0 \text{ Nm}$$ (v) Torque about E: - Force at A: 5 N down at 6 m left, torque = $-5 \times 6 = -30$ Nm (CW) - Force at B: 20 N up at 5 m left, torque = $20 \times 5 = 100$ Nm (CCW) - Force at C: 10 N down at 3 m left, torque = $-10 \times 3 = -30$ Nm (CW) - Force at D: 5 N up at 2 m left, torque = $5 \times 2 = 10$ Nm (CCW) - Force at E: distance 0 m, torque = 0 Net torque about E: $$-30 + 100 - 30 + 10 = 50 \text{ Nm}$$ **Final answers:** - Torque about A: $-80$ Nm (clockwise) - Torque about B: $-85$ Nm (clockwise) - Torque about C: $-15$ Nm (clockwise) - Torque about D: $0$ Nm (balanced) - Torque about E: $50$ Nm (counterclockwise)