1. Problem statement: We analyze static equilibrium of a beam with forces applied at angles and distances from a fulcrum. 2. Key formula: For static equilibrium, the sum of torques (moments) about the fulcrum must be zero: $$\sum \tau = 0$$ Torque $\tau = F \times d \times \sin(\theta)$ where $F$ is force magnitude, $d$ is distance from fulcrum, and $\theta$ is angle between force direction and beam. 3. Part a) Given: - Force 1: $F_1=100$ N, $d_1=0.5$ m, $\theta_1=40^\circ$ - Force 2: $F_2=50$ N, $d_2=?$, $\theta_2=60^\circ$ Set clockwise torque = counterclockwise torque: $$F_1 d_1 \sin(\theta_1) = F_2 d_2 \sin(\theta_2)$$ Solve for $d_2$: $$d_2 = \frac{F_1 d_1 \sin(40^\circ)}{F_2 \sin(60^\circ)} = \frac{100 \times 0.5 \times \sin(40^\circ)}{50 \times \sin(60^\circ)}$$ Calculate sines: $\sin(40^\circ) \approx 0.6428$, $\sin(60^\circ) \approx 0.8660$ $$d_2 = \frac{100 \times 0.5 \times 0.6428}{50 \times 0.8660} = \frac{32.14}{43.3} \approx 0.742 \text{ m}$$ 4. Part b) Given: - Force 1: $F_1=100$ N, $d_1=0.3$ m, $\theta_1=50^\circ$ - Force 2: $F_2=?, d_2=0.4$ m, $\theta_2=70^\circ$ Set torques equal: $$F_1 d_1 \sin(50^\circ) = F_2 d_2 \sin(70^\circ)$$ Solve for $F_2$: $$F_2 = \frac{F_1 d_1 \sin(50^\circ)}{d_2 \sin(70^\circ)} = \frac{100 \times 0.3 \times \sin(50^\circ)}{0.4 \times \sin(70^\circ)}$$ Calculate sines: $\sin(50^\circ) \approx 0.7660$, $\sin(70^\circ) \approx 0.9397$ $$F_2 = \frac{30 \times 0.7660}{0.4 \times 0.9397} = \frac{22.98}{0.3759} \approx 61.1 \text{ N}$$ 5. Part c) Elbow flexion static equilibrium: - Elbow flexors force $F_f$ at $d_f=0.025$ m, angle $\theta_f=35^\circ$ - Forearm + hand mass $m=5$ kg at $d_m=0.20$ m - Gravity force $F_g = mg = 5 \times 9.81 = 49.05$ N acting downward Torque by gravity: $$\tau_g = F_g d_m = 49.05 \times 0.20 = 9.81 \text{ Nm}$$ Torque by flexors: $$\tau_f = F_f d_f \sin(35^\circ)$$ Set equilibrium: $$\tau_f = \tau_g \Rightarrow F_f = \frac{\tau_g}{d_f \sin(35^\circ)} = \frac{9.81}{0.025 \times 0.574} = \frac{9.81}{0.01435} \approx 683.5 \text{ N}$$ 6. Part d) Holding 10 kg dumbbell at $d_d=0.45$ m: - Dumbbell force $F_d = 10 \times 9.81 = 98.1$ N Total torque by weights: $$\tau_{total} = F_g d_m + F_d d_d = 49.05 \times 0.20 + 98.1 \times 0.45 = 9.81 + 44.15 = 53.96 \text{ Nm}$$ Flexor force: $$F_f = \frac{53.96}{0.025 \times 0.574} = \frac{53.96}{0.01435} \approx 3759.4 \text{ N}$$ Final answers: - a) $d_2 \approx 0.742$ m - b) $F_2 \approx 61.1$ N - c) $F_f \approx 683.5$ N - d) $F_f \approx 3759.4$ N