Questions: An athlete holds a 7.5-kg shot put in his hand with his lower arm horizontal, as shown in the figure. His lower arm has a mass of 2.8 kg and its center of gravity (or center of mass) is 12 cm from the elbow-joint pivot. How much force must the extensor muscle (which is FM in the figure) in the upper arm exert on the lower arm? 1500 N 100 N 500 N 1000 N

Solution Steps

• The shot put has a mass of 7.5 kg, so its weight \( W_{\text{shot}} = 7.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 73.5 \, \text{N} \).
• The lower arm has a mass of 2.8 kg, so its weight \( W_{\text{arm}} = 2.8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 27.44 \, \text{N} \).
• The distance from the elbow joint to the shot put is 30 cm (0.3 m).
• The distance from the elbow joint to the center of gravity of the lower arm is 12 cm (0.12 m).

• The torque due to the shot put: \( \tau_{\text{shot}} = W_{\text{shot}} \times 0.3 \, \text{m} = 73.5 \, \text{N} \times 0.3 \, \text{m} = 22.05 \, \text{Nm} \).
• The torque due to the lower arm: \( \tau_{\text{arm}} = W_{\text{arm}} \times 0.12 \, \text{m} = 27.44 \, \text{N} \times 0.12 \, \text{m} = 3.2928 \, \text{Nm} \).

• For rotational equilibrium, the sum of the torques around the elbow joint must be zero.
• Let \( F_M \) be the force exerted by the extensor muscle at a distance of 2.5 cm (0.025 m) from the elbow joint.
• The torque due to the muscle force: \( \tau_M = F_M \times 0.025 \, \text{m} \).

• The total torque due to the weights must be balanced by the torque due to the muscle force: \[ \tau_M = \tau_{\text{shot}} + \tau_{\text{arm}} \] \[ F_M \times 0.025 \, \text{m} = 22.05 \, \text{Nm} + 3.2928 \, \text{Nm} \] \[ F_M \times 0.025 \, \text{m} = 25.3428 \, \text{Nm} \] \[ F_M = \frac{25.3428 \, \text{Nm}}{0.025 \, \text{m}} = 1013.712 \, \text{N} \]

Final Answer