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Chapter 5: Problem 3

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 \(33 \mathrm{~cm}\) 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 \(17 \mathrm{~cm}\) 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.) (answer check available at lightandmatter.com)

Short Answer

Expert verified
He should use a mass of approximately 4.10 kg.

Step by step solution

01

Understanding Torque and Its Formula

Torque ( \tau ) is the measure of the force that can cause an object to rotate about an axis. The formula to calculate torque is \( \tau = F \times r \), where \( F \) is the force applied perpendicular to the arm, and \( r \) is the distance from the pivot point to the point where the force is applied. In this problem, \( F = m \cdot g \), where \( m \) is mass and \( g = 9.8 \, \text{m/s}^2 \), the acceleration due to gravity.
02

Calculate Original Torque

First, we calculate the torque with the original setup where a \( 2.1 \, \text{kg} \) mass is placed at a \( 33 \, \text{cm} \) distance. Convert the distance to meters: \( 33 \, \text{cm} = 0.33 \, \text{m} \). The force is \( F = 2.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 20.58 \, \text{N} \). Therefore, the original torque is \( \tau_{original} = 20.58 \, \text{N} \times 0.33 \, \text{m} = 6.7904 \, \text{Nm} \).
03

Set Up the Equation for New Torque

He proposes to move the weight to \( 17 \, \text{cm} \) from his elbow. Convert this distance to meters: \( 17 \, \text{cm} = 0.17 \, \text{m} \). Let \( m \) be the mass to find such that the torque will be the same. Thus, the equation is \( \tau_{new} = m \times 9.8 \, \text{m/s}^2 \times 0.17 \, \text{m} = 6.7904 \, \text{Nm} \).
04

Solve for the New Mass

Rearrange to solve for \( m \): \( m \times 9.8 \, \text{m/s}^2 \times 0.17 \, \text{m} = 6.7904 \, \text{Nm} \). Therefore, \( m = \frac{6.7904 \, \text{Nm}}{9.8 \, \text{m/s}^2 \times 0.17 \, \text{m}} \approx 4.10 \, \text{kg} \).
05

Conclusion

The therapist should tell him to use a mass of approximately \( 4.10 \, \text{kg} \) to exert the same torque as the \( 2.1 \, \text{kg} \) mass placed \( 33 \, \text{cm} \) from the elbow.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elbow Rehabilitation
Rehabilitation for an elbow injury is crucial for restoring strength and flexibility. A common method involves using resistance exercises, like lifting weights.
In the context of elbow rehabilitation, torque plays a significant role in determining the effectiveness of these exercises. Torque is the rotational equivalent of linear force.
When a patient lifts a weight, they're applying a force that causes rotation around the elbow joint.
  • This rotational force can either aid or strain the muscles involved.
  • By adjusting the distance of the weight from the elbow, therapists can tweak the torque to match the patient's capabilities.
  • Proper torque is important to avoid discomfort and potential re-injury.
Using a larger mass closer to the elbow can maintain the necessary torque while reducing strain on compromised structures. This strategic approach is vital in tailoring rehabilitation programs to individual needs.
Physics of Rotation
Understanding the physics of rotation is key in both daily activities and physical therapy. With rotation, we consider objects moving around a pivot point, which, in this case, is the elbow joint.
The fundamental principle is that torque is the force that causes rotation.
  • The amount of rotation depends on both the magnitude of the force and the distance from the pivot point.
  • This is why lifting a weight involves calculating torque to ensure it's safe and within the patient's capability.
  • The change in distance requires adjusting mass to maintain torque.
This understanding helps in designing rehabilitation strategies that maximize recovery without exceeding the patient's limitations. By carefully controlling these factors, patients can improve movement and strength without adverse effects.
Force and Distance
When dealing with forces in physics, especially regarding rotational movement, two pivotal components come into play: force and distance.
Force is generated by the muscle when lifting a weight, and distance is the length between the point of force application and the pivot.
  • In the scenario of elbow exercises, the force exerted by the forearm can vary based on the weight and how far it is from the elbow.
  • A larger distance from the pivot point results in greater torque.
  • This is why a weight further from the elbow feels heavier than the same weight positioned closer.
To ensure that the exercise remains beneficial and not harmful, physical therapists calculate the appropriate force and distance settings precisely. This maintains the necessary motivational challenge while safeguarding against injury.

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Most popular questions from this chapter

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Find three vectors with which you can demonstrate that the vector cross product need not be associative, i.e., that \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})\) need not be the same as \((\mathbf{A} \times \mathbf{B}) \times \mathbf{C}\).
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