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\(\bullet$$\bullet\) Tension in a muscle. Muscles are attached to bones by means of tendons. The maximum force that a muscle can exert is directly proportional to its cross-sectional area \(A\) at the widest point. We can express this relationship mathematically as \(F_{\text { max }}=\sigma A,\) where \(\sigma(\) sigma) is a proportionality constant. Surprisingly, \(\sigma\) is about the same for the muscles of all animals. and has the numerical value of \(3.0 \times 10^{5}\) in SI units. (a) What are the SI units of \(\sigma\) in terms of newtons and meters and also in terms of the fundamental quantities \((\mathrm{kg}, \mathrm{m}, \mathrm{s})\) (b) In one set of experiments, the average maximum force that the gastrocne- mius muscle in the back of the lower leg could exert was meas- ured to be 755 \(\mathrm{N}\) for healthy males in their midtwenties. What does this result tell us was the average cross-sectional area, in \(\mathrm{cm}^{2},\) of that muscle for the people in the study?

Short Answer

Expert verified
The SI unit of \(\sigma\) is \(\text{Pa} = \text{N/m}^2\); the average area is 25.2 \(\text{cm}^2\).

Step by step solution

01

Determine SI Units of \(\sigma\)

The equation given is \( F_{\text{max}} = \sigma A \). Here, \( F_{\text{max}} \) is measured in newtons (N) and \( A \) is measured in square meters \( \text{m}^2 \). Therefore, the units of \( \sigma \) can be determined by rearranging to \( \sigma = \frac{F_{\text{max}}}{A} \). So, \( \sigma \) has units \( \frac{N}{\text{m}^2} \) (newtons per square meter). This is equivalent to pascals (Pa). Also, in terms of fundamental quantities: \( N = \text{kg}\cdot\text{m/s}^2 \). So, \( \sigma \) is \( \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2} = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} \).
02

Calculate the Cross-Sectional Area A

Given the maximum force \( F_{\text{max}} = 755 \) N and the proportionality constant \( \sigma = 3.0 \times 10^5 \) Pa, we can find \( A \) using the formula \( F_{\text{max}} = \sigma A \). Solving for \( A \) gives: \( A = \frac{F_{\text{max}}}{\sigma} = \frac{755}{3.0 \times 10^5} \text{ m}^2 \).
03

Convert A from Square Meters to Square Centimeters

First, calculate \( A \) in square meters: \( A = \frac{755}{3.0 \times 10^5} = 2.52 \times 10^{-3} \text{ m}^2 \). To convert this to square centimeters, note that \( 1 \text{ m}^2 = 10^4 \text{ cm}^2 \). So, \( A = 2.52 \times 10^{-3} \times 10^4 \text{ cm}^2 = 25.2 \text{ cm}^2 \).

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Key Concepts

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

Tendons and Bones: The Bridge Connecting Forces
Muscles play a crucial role in movement by generating force, which is transmitted to bones through tendons. Tendons, acting as robust connective tissues, are vital in attaching muscles to their respective bones.
The interaction between muscles, tendons, and bones is integral in achieving efficient movement and stability.
  • Tendons: These are the resilient bands of connective tissue that connect muscle to bone. Their elasticity allows them to absorb some of the shock and stress placed on muscles during movement.
  • Bones: Acting as levers, bones facilitate movement when muscles exert tension upon them through tendons.
This harmonious collaboration allows our bodies to perform various activities, from walking and running to lifting and pushing objects. Tendons help in maintaining posture and play a regenerative role since some can repair themselves over time. However, the health and functionality of these components must remain optimal to prevent injuries such as tendonitis, which is the inflammation of a tendon.
Cross-Sectional Area and Its Importance in Muscle Function
The cross-sectional area of a muscle is a fundamental property that correlates with the muscle's ability to generate force. In terms of mechanics, a larger cross-sectional area means a muscle can exert a greater force. This is due to the increased number of muscle fibers that can contract simultaneously.
When considering muscle strength or the study of biomechanics, looking at the muscle's cross-sectional area provides critical insights.
  • Mathematical Relationship: The maximum force a muscle exerts, denoted as \( F_{\text{max}} \), is directly proportional to its cross-sectional area \( A \). This is expressed by the equation \( F_{\text{max}} = \sigma A \), where \( \sigma \) is the proportionality constant.
  • Measuring: Typically, the cross-sectional area is measured in units of square meters (\( \text{m}^2 \)), or converted to square centimeters (\( \text{cm}^2 \)) for more granular analyses.
Understanding this relationship helps in areas such as diagnosing muscle atrophy or monitoring muscle development. In physical training or rehabilitation, increasing a muscle's cross-sectional area through resistance training can enhance its force-generating capacity, thereby improving overall muscle strength and function.
Proportionality Constant: Bridging the Muscle-to-Force Equation
In the equation \( F_{\text{max}} = \sigma A \), \( \sigma \) symbolizes the proportionality constant. This constant is pivotal because it allows for the conversion of cross-sectional area into a quantifiable force.
For muscles, this constant remains surprisingly consistent across different species, including humans, and provides a means of comparison.
  • SI Units: The units of \( \sigma \) are derived from rearranging the equation into \( \sigma = \frac{F_{\text{max}}}{A} \). This results in units of newtons per square meter (\( \text{N/m}^2 \)), also known as pascals (Pa).
  • Fundamental Quantities: In terms of basic units: \( \sigma \) is expressed as \( \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2} \), showcasing its roots in physics fundamentals like mass, distance, and time.
The consistency in \( \sigma \) implies that no matter the size or species, muscles produce a similar force per unit area, making it an essential tool in comparing muscle physiology across different organisms. This characteristic allows biomechanists and physiologists to study muscles under varying conditions and design appropriate training or treatment regimes.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Air-bag safety. According to safety standards for air bags, the maximum acceleration during a car crash should not exceed 60\(g\) and should last for no more than 36 ms. (a) In such a case, what force does the air bag exert on a 75 kg person? Start with a free-body diagram. (b) Express the force in part (a) in terms of the person's weight.

\(\bullet$$\bullet\) A winch is used to drag a 375 \(\mathrm{N}\) crate up a ramp at a con- stant speed of 75 \(\mathrm{cm} / \mathrm{s}\) by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at \(33^{\circ}\) above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25 . (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the accel- eration of the crate immediately after the rope broke?

\(\bullet$$\bullet\) An object is dropped from rest and encounters air resist- ance that is proportional to the square of its speed. Sketch qualitative graphs (no numbers) showing (a) the air resistance on this object as a function of its speed, (b) the net force on the object as a function of its speed, (c) the net force on the object as a function of time, (d) the speed of the object as a function of time, and (e) the acceleration of the object as a function of time.

\(\bullet$$\bullet\) A toboggan approaches a snowy hill moving at 11.0 \(\mathrm{m} / \mathrm{s}\) . The coefficients of static and kinetic friction between the snow and the tobogan are 0.40 and 0.30 , respectively, and the hill slopes upward at \(40.0^{\circ}\) above the horizontal. Find the acceleration of the tobogan (a) as it is going up the hill and (b) after it has reached its highest point and is sliding down the hill.

\(\bullet$$\bullet\) Force during a jump. An average person can reach a maximum height of about 60 \(\mathrm{cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 \(\mathrm{cm} .\) To keep the calculations simple and yet get a reason- able result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 \(\mathrm{cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(W,\) what force does the ground exert on him or her during the jump?

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