Let's break down the problem step by step. ## **Step 1: Identify the 3 External Forces** 1. **Weight of Upper Arm (\( W_u \))** - Acts vertically downward at the center of the upper arm. 2. **Weight of Forearm & Hand (\( W_f \))** - Acts vertically downward at the center of the forearm/hand. 3. **External Force (\( P \))** - Acts at an angle of \( 50^\circ \) to the forearm/hand (given as 20 N). --- ## **Step 2: Calculate the Forces** ### **Convert Body Weight to Newtons:** - \( 1 \text{ lb} = 4.448 \text{ N} \) - Body Weight = \( 160 \text{ lb} \times 4.448 \text{ N/lb} = 711.68 \text{ N} \) #### **Assume:** - Upper arm: ~2.71% of body mass - Forearm & hand: ~1.62% of body mass #### **Upper Arm Weight (\( W_u \)):** - \( W_u = 0.0271 \times 711.68 = 19.29 \text{ N} \) #### **Forearm & Hand Weight (\( W_f \)):** - \( W_f = 0.0162 \times 711.68 = 11.53 \text{ N} \) --- ## **Step 3: Position Vectors to Shoulder Joint** - **Shoulder to Upper Arm Center:** \( r_{u} = 0.15 \) m (half of 0.3 m) - **Shoulder to Forearm/Hand Center:** \( r_{f} = 0.3 + 0.2 = 0.5 \) m (center of forearm/hand) - **Shoulder to External Force:** \( r_{P} = 0.7 \) m (full arm + forearm length) --- ## **Step 4: Calculate the Torque/Moment (Magnitude and Direction)** \[ \tau = r \times F \] Magnitude: \(\tau = r F \sin(\theta)\) Where \(\theta\) = angle between r and F ### **A. Torque by \( W_u \):** - \(\theta = 25^\circ\) (vertical force, arm at 25°) - \(\tau_{u} = 0.15 \times 19.29 \times \sin(25^\circ) = 0.15 \times 19.29 \times 0.4226 = 1.22 \, \text{Nm}\) (clockwise) ### **B. Torque by \( W_f \):** - \(\tau_{f} = 0.5 \times 11.53 \times \sin(25^\circ) = 0.5 \times 11.53 \times 0.4226 = 2.44 \, \text{Nm}\) (clockwise) ### **C. Torque by \( P \):** - The force \( P \) acts at 50° to the forearm, but the arm is at 25° from horizontal, so the angle with the horizontal is \( 25^\circ + 50^\circ = 75^\circ \). - \(\tau_{P} = 0.7 \times 20 \times \sin(75^\circ)\) - \(\sin(75^\circ) = 0.9659\) - \(\tau_{P} = 0.7 \times 20 \times 0.9659 = 13.52 \, \text{Nm}\) (counterclockwise) --- ## **Step 5: Resultant Torque** \[ \tau_{net} = (\tau_{P}) - (\tau_{u} + \tau_{f}) \] \[ \tau_{net} = 13.52 - (1.22 + 2.44) = 9.86 \, \text{Nm} \ \text{(counterclockwise)} \] --- ## **Step 6: Resultant Force** Sum the vectors (all forces act downward except \( P \), which acts at an angle). ### **Resolve P into x and y components:** - \( P_x = 20 \cos(50^\circ) = 12.86 \) N (horizontal, away from the shoulder) - \( P_y = -20 \sin(50^\circ) = -15.32 \) N (vertical, downward) ### **Sum forces in x and y:** - \( F_{x, total} = P_x = 12.86 \) N - \( F_{y, total} = -W_u - W_f + P_y = -19.29 - 11.53 - 15.32 = -46.14 \) N ### **Resultant Magnitude:** \[ F_{R} = \sqrt{(12.86)^2 + (46.14)^2} = \sqrt{165.42 + 2130.87} = \sqrt{2296.29} = 47.92 \text{ N} \] ### **Resultant Direction:** \[ \theta = \tan^{-1} \left( \frac{46.14}{12.86} \right ) = \tan^{-1} (3.59) = 74.4^\circ \text{ below the horizontal} \] --- ## **Final Answers** ### **Resultant Force at the Shoulder Joint:** - **Magnitude:** \( 47.9 \) N - **Direction:** \( 74.4^\circ \) below the horizontal ### **Resultant Torque at the Shoulder Joint:** - **Magnitude:** \( 9.86 \) Nm - **Direction:** Counterclockwise --- ### **Summary Table** | Force | Magnitude (N) | Torque (Nm) | Direction | |--------------------|:-------------:|:-----------:|:-----------------:| | Upper Arm (\(W_u\))| 19.29 | 1.22 | Clockwise | | Forearm (\(W_f\)) | 11.53 | 2.44 | Clockwise | | External (\(P\)) | 20 | 13.52 | Counterclockwise | | **Resultant** | **47.9** | **9.86** | Counterclockwise |