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Biceps muscle. A relaxed biceps muscle requires a force of \(25.0 \mathrm{~N}\) for an elongation of \(3.0 \mathrm{~cm} ;\) under maximum tension, the same muscle requires a force of \(500 \mathrm{~N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of \(0.200 \mathrm{~m}\) and a cross-sectional area of \(50.0 \mathrm{~cm}^{2}\)

Short Answer

Expert verified
Relaxed modulus: \(3.33\times10^{4} \,\mathrm{Pa}\), Maximum tension modulus: \(6.67\times10^{5}\,\mathrm{Pa}\).

Step by step solution

01

Understanding the Problem

We need to calculate Young's modulus for a biceps muscle modeled as a cylinder under two different forces that produce the same elongation. We have the force and elongation values for both relaxed and maximum tension conditions.
02

Given Data Identification

For both scenarios, the initial length of the muscle is \(0.200\,\mathrm{m}\) and the cross-sectional area is \(50.0\,\mathrm{cm}^2 = 5.0\times10^{-3}\,\mathrm{m}^2\). The elongation is \(3.0\,\mathrm{cm} = 0.03\,\mathrm{m}\).
03

Using Young's Modulus Formula

Young's modulus \(E\) is given by the formula:\[E = \frac{F \cdot L_0}{A \cdot \Delta L}\]where \(F\) is the force, \(L_0\) is the original length, \(A\) is the cross-section area, and \(\Delta L\) is the elongation.
04

Calculating Young's Modulus for Relaxed Condition

For the relaxed condition, \(F = 25.0\,\mathrm{N}\). Substitute the values into the formula:\[E = \frac{25.0 \cdot 0.200}{5.0\times10^{-3} \cdot 0.03}\]Evaluate this expression to find \(E\.\)
05

Simplify for Relaxed Condition

Simplifying the equation:\[E = \frac{5.0}{1.5\times10^{-4}}\]This results in:\[E = 3.33\times10^{4}\,\mathrm{Pa}\]
06

Calculating Young's Modulus for Maximum Tension Condition

For the maximum tension condition, \(F = 500\,\mathrm{N}\). Substitute the values into the formula:\[E = \frac{500 \cdot 0.200}{5.0\times10^{-3} \cdot 0.03}\]Evaluate this expression to find \(E\.\)
07

Simplify for Maximum Tension Condition

Simplifying the equation:\[E = \frac{100}{1.5\times10^{-4}}\]This results in:\[E = 6.67\times10^{5}\,\mathrm{Pa}\]

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Key Concepts

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

Biceps Muscle
The biceps muscle, one of the primary muscles involved in the movement of the arm, helps to flex the elbow and rotate the forearm. It consists of muscle fibers that have unique mechanical properties. These properties can vary depending on whether the muscle is relaxed or under tension.
When studying the mechanical response of a biceps muscle, it's essential to understand how it reacts to forces that cause elongation. The biomechanical properties, such as elasticity and tensile strength, govern these reactions. These can be quantified using measures like Young's modulus. By modeling the muscle as a cylinder, we can simplify these calculations and gain insights into how muscles work in different conditions.
Force and Elongation
Force and elongation are key concepts in understanding muscle behavior under stress. When a muscle is subjected to a force, it tends to lengthen, a process known as elongation.
In our example, the relaxed biceps muscle requires a force of 25 N for an elongation of 3 cm, whereas under maximum tension, the same elongation requires a force of 500 N. This shows how the muscle鈥檚 ability to handle force changes dramatically from a relaxed to a tense state.
Force (\(F\)) is measured in newtons (\(\mathrm{N}\)), and elongation (\(\Delta L\)) is the change in length resulting from the applied force. Understanding this relationship helps in calculating the Young's modulus, which defines the stiffness or elasticity of the muscle under different conditions.
Cylinder Model
A cylinder model is often used in anatomy and physics to simplify the complex structure of a muscle into a more manageable form. By approximating the biceps muscle as a cylinder, calculations of elastic properties become easier.
A cylinder is characterized by its length and cross-sectional area, allowing us to use standard equations to find parameters like Young's modulus. This simplification helps in determining how the muscle behaves under different loads and elongations.
The uniform cylinder model assumes the muscle has a constant cross-section throughout its length, providing an average view of the muscle鈥檚 behavior. While it's an approximation, it can effectively describe the overall response of the muscle to forces.
Cross-Sectional Area
The cross-sectional area of a muscle is crucial for understanding its mechanical properties. It is the area of a slice through the muscle perpendicular to its length.
In calculations regarding Young鈥檚 modulus, the cross-sectional area (\(A\)) plays a significant role. It influences how much force is needed to achieve a given elongation. For the biceps modeled as a cylinder, this area is the basis for calculating how the muscle distributes force internally.
In our exercise, the cross-sectional area is given as 50 cm虏 or 5.0 x 10鈦宦 m虏. A larger cross-sectional area typically indicates a greater capability of the muscle to withstand forces before significant elongation occurs, highlighting the importance of this parameter in biophysics and muscle physiology.

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Most popular questions from this chapter

Crude oil with a bulk modulus of \(2.35 \mathrm{GPa}\) is leaking from a deep- sea well \(2250 \mathrm{~m}\) below the surface of the ocean, where the water pressure is \(2.27 \times 10^{7} \mathrm{~Pa}\). Suppose 35,600 barrels of oil leak from the wellhead; assuming all that oil reaches the surface, how many barrels will it be on the surface?

A \(2.50 \mathrm{~kg}\) rock is attached at the end of a thin, very light rope \(1.45 \mathrm{~m}\) long and is started swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after \(10 \frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

Weight lifting. The legs of a weight lifter must ultimately support the weights he has lifted. A human tibia (shinbone) has a circular cross section of approximately \(3.6 \mathrm{~cm}\) outer diameter and \(2.5 \mathrm{~cm}\) inner diameter. (The hollow portion contains marrow.) If a \(90 \mathrm{~kg}\) lifter stands on both legs, what is the heaviest weight he can lift without breaking his legs, assuming that the breaking stress of the bone is \(200 \mathrm{MPa}\) ?

A mass is oscillating with amplitude \(A\) at the end of a spring. (a) How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy? (b) How far is the mass from the equilibrium position when the kinetic energy is \(\frac{1}{10}\) of the total energy?

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