1. **Problem statement:** Calculate the muscle force and joint forces in a forearm holding a weight of 40.0 N, given lever arms and forces. 2. **Given data:** - Weight force: $F_g = 40.0$ N - Distance from elbow to muscle insertion (forearm lever arm): $r_f = 50$ cm = 0.50 m - Distance from elbow to weight (load lever arm): $r = 35.0$ cm = 0.35 m - Additional lever arm (possibly hand or finger): $V = 15.0$ cm = 0.15 m - Force to calculate: muscle force $F_i$ 3. **Step 1: Calculate muscle force $F_i$ using torque equilibrium** The forearm is in static equilibrium, so the sum of torques about the elbow is zero: $$\sum \tau = 0 = F_i \times r_f - F_g \times r$$ Rearranged to solve for $F_i$: $$F_i = \frac{F_g \times r}{r_f}$$ Substitute values: $$F_i = \frac{40.0 \times 0.35}{0.50} = \frac{14.0}{0.50} = 28.0\ \text{N}$$ 4. **Step 2: Calculate force at the elbow joint $F_r$** The elbow joint force balances the vertical forces: $$F_r = F_i + F_g = 28.0 + 40.0 = 68.0\ \text{N}$$ 5. **Step 3: Calculate torque at the elbow joint (if needed)** Torque due to muscle force: $$\tau_i = F_i \times r_f = 28.0 \times 0.50 = 14.0\ \text{Nm}$$ Torque due to weight: $$\tau_g = F_g \times r = 40.0 \times 0.35 = 14.0\ \text{Nm}$$ They balance, confirming equilibrium. **Final answers:** - Muscle force $F_i = 28.0$ N - Elbow joint force $F_r = 68.0$ N - Torques balance at 14.0 Nm