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A therapist tells a 74-kg patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg\(-\)cast system (\(\textbf{Fig. P11.51}\)). To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5% of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. The patient also reads that the two lower legs (including the feet) are 14.0% of body weight, with a center of mass 69.0 cm from the hip joint. The cast has a mass of 5.50 kg, and its center of mass is 78.0 cm from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

Short Answer

Expert verified
The strap should be attached 44.96 cm from the hip joint.

Step by step solution

01

Determine the weight of leg parts and cast

The body weight of the patient is 74 kg. First, calculate the weight of the thighs: 21.5% of body weight. Convert it to force by multiplying by gravitational acceleration (9.8 m/sĀ²).Weight of thighs: \(74 \times 0.215 \times 9.8 = 156.11 \text{ N}\)Next, calculate the weight of the lower legs: 14% of the body weight.Weight of lower legs: \(74 \times 0.14 \times 9.8 = 101.08 \text{ N}\)Finally, calculate the weight of the cast: 5.50 kg as a force.Weight of cast: \(5.50 \times 9.8 = 53.9 \text{ N}\)
02

Identify position of center of mass for each part

Identify the position of the center of mass for the thighs, the lower legs, and the cast relative to the hip. - Thighs center of mass: 18.0 cm from hip joint - Lower legs center of mass: 69.0 cm from hip joint - Cast center of mass: 78.0 cm from hip joint
03

Calculate moments about hip joint

Calculate the moment (torque) about the hip joint for each segment of the leg and the cast by multiplying the weight of each by its respective distance from the hip:- Moment of thighs: \(156.11 \times 0.18 = 28.0998 \text{ Nm}\)- Moment of lower legs: \(101.08 \times 0.69 = 69.7452 \text{ Nm}\)- Moment of cast: \(53.9 \times 0.78 = 42.042 \text{ Nm}\)
04

Sum the moments to find total moment

Sum all moments to get the total moment of the leg-cast system:\[ M_{total} = 28.0998 + 69.7452 + 42.042 = 139.887 \text{ Nm} \]
05

Calculate total weight of leg-cast system

Add the forces (weights) of the thighs, lower legs, and the cast to get the total weight:\[ F_{total} = 156.11 + 101.08 + 53.9 = 311.09 \text{ N} \]
06

Determine strap position using center of mass formula

Determine the distance from the hip joint where the strap should be positioned for equilibrium using the center of mass equation. Set total moment equal to the force times distance (the unknown, \(x\)):\[ 139.887 = 311.09 \times x \]Solve for \(x\):\[ x = \frac{139.887}{311.09} = 0.4496 \text{ m} \]Convert to cm: \(0.4496 \times 100 = 44.96 \text{ cm}\)
07

Conclusion

Thus, the supporting strap should be attached 44.96 cm from the hip joint.

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Key Concepts

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

Mass Distribution
Understanding how mass is distributed is essential, especially when dealing with systems like a leg in a cast. Mass distribution refers to how a total mass is spread out over a certain body or structure. In the context of a human leg, different parts have different weights and centers of mass, which affect balance and support.
- The upper legs, or thighs, together make up 21.5% of body weight, reflecting their significance in overall mass distribution.
- The lower legs and feet account for 14% of body weight. Although lighter, their placement affects the overall balance.
- The cast itself, although a separate entity, also contributes to the total mass of the leg system.
Recognizing how these parts contribute to the total weight and their respective center of mass positions is crucial for proper support, as it impacts how the entire system behaves when suspended.
Torque
Torque is a rotational force that can cause an object to turn or rotate. It's crucial when determining the balance of a suspended object like a leg in a cast. Torque is calculated by multiplying a force by a distance from a pivot point, usually measured in newton-meters (Nm).
- Each part of the leg system (thighs, lower legs, and cast) generates its own torque relative to the hip joint.
- The torque for each part depends on its weight and how far its center of mass is from the hip.
- For example, the thighs exert a torque of approximately 28.10 Nm, while the cast contributes a torque of around 42.04 Nm.
By summing these individual torques, we find the total torque, which helps in determining the balance point or equilibrium position for the leg.
Equilibrium
To achieve equilibrium in a system means that all forces and torques are balanced, resulting in a system that is stable and at rest. For the leg-cast system, equilibrium is achieved by positioning the supporting strap so that the total torque around the hip joint is balanced.
- The total moment or torque about the hip must equal the product of total weight and distance to ensure equilibrium.
- This helps in determining the exact position for the strap attachment by using the torque balance equation: \[ M_{total} = F_{total} \times x \] - Solving for the position of the strap ensures the system is at rest without any net rotational force causing discomfort to the patient.
Achieving this balance ensures that the patient's leg is supported with minimal discomfort, proving how vital understanding equilibrium is in practical scenarios.
System of Particles
A system of particles involves multiple distinct masses working together or acting as a single entity. In the context of the leg-cast system, they're treated collectively to find how the system behaves when subjected to forces. The critical aspects of a system of particles include its total mass, center of mass, and combined forces and moments.
- By treating the thighs, lower legs, and cast as a system, we compute the total mass and center of mass that inform how to support the leg.
- This aids in calculating combined forces and torques that act upon the system.
- Using the system of particles approach simplifies complex problems, allowing us to find crucial points like the center of mass for the system.
Understanding a system of particles allows us to see how each part contributes to the whole, ensuring proper support and balance in medical or mechanical applications.

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Most popular questions from this chapter

A holiday decoration consists of two shiny glass spheres with masses 0.0240 kg and 0.0360 kg suspended from a uniform rod with mass 0.120 kg and length 1.00 m (\(\textbf{Fig. P11.62}\)). The rod is suspended from the ceiling by a vertical cord at each end, so that it is horizontal. Calculate the tension in each of the cords \(A\) through \(F\).

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post's length when the load is applied?

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

A metal rod that is 4.00 m long and 0.50 cm\(^2\) in crosssectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young's modulus for this metal?

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