Questions: CH IS Problem 2 A physical therapist wants her patient to rehabilitate his injured elbow by laying his arm flat on a table and then lifting a 2.1 kg mass by bending his elbow. In this situation, the weight is 33 cm from his elbow. He calls her back, complaining that it hurts him to grasp the weight. He asks if he can strap a bigger weight onto his arm, only 17 cm from his elbow. How much mass should she tell him to use so that he will be exerting the same torque? (He is raising his forearm itself, as well as the weight.) Instruction Draw free-body diagrams and write down net force and net torque equations. (Make sure to show your coordinate system.) From the equations of motion, answer the question.

Solution Steps

Let's establish a coordinate system where the elbow joint is at the origin. The forearm extends along the positive x-axis, and the y-axis points upward. We'll define:

• Original mass: \(m_1 = 2.1 \text{ kg}\)
• Original distance from elbow: \(r_1 = 33 \text{ cm} = 0.33 \text{ m}\)
• New distance from elbow: \(r_2 = 17 \text{ cm} = 0.17 \text{ m}\)
• New mass to be determined: \(m_2\)
• Gravitational acceleration: \(g = 9.8 \text{ m/s}^2\)

For the original setup:

• The forearm pivots around the elbow joint
• The 2.1 kg mass exerts a downward force \(F_1 = m_1g\) at distance \(r_1\) from the pivot
• This creates a torque \(\tau_1 = r_1 \times F_1\) around the elbow

For the new setup:

• The new mass \(m_2\) exerts a downward force \(F_2 = m_2g\) at distance \(r_2\) from the pivot
• This creates a torque \(\tau_2 = r_2 \times F_2\) around the elbow

For the original setup, the torque is: \(\tau_1 = r_1 \times m_1g = 0.33 \text{ m} \times 2.1 \text{ kg} \times 9.8 \text{ m/s}^2 = 6.79 \text{ N·m}\)

For the new setup, the torque should be the same: \(\tau_2 = r_2 \times m_2g = 0.17 \text{ m} \times m_2 \times 9.8 \text{ m/s}^2 = 1.666 \times m_2 \text{ N·m}\)

Since we want the same torque in both cases: \(\tau_1 = \tau_2\) \(6.79 \text{ N·m} = 1.666 \times m_2 \text{ N·m}\) \(m_2 = \frac{6.79 \text{ N·m}}{1.666 \text{ N·m/kg}} = 4.08 \text{ kg}\)

Final Answer