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~max= σA, where σ (sigma) is a proportionality constant. σ is about the same
for the muscles of all animals and has a numerical value of 3.0 × 105kg m−1s−2.
(a) What are the units of σ in newtons and meters?
(b) In one set of experiments, the average maximum force that the gastrocnemius muscle in the back
of the lower leg could exert was measured to be 755 N for healthy males in their mid-twenties.
What does this result tell us was the average cross-sectional area, in cm2, of that muscle for the
people in the study?
a) N/m^2
b) A = F/sigma
(a) The units of σ in newtons and meters can be determined by rearranging the given equation F~max = σA and solving for σ:
σ = F~max / A
Since force (F~max) is measured in newtons (N) and area (A) is measured in meters squared (m^2), the units of σ will be in newtons per square meter (N/m^2) or pascals (Pa).
(b) To find the average cross-sectional area of the gastrocnemius muscle, we can use the equation σ = F~max / A and rearrange it to solve for A:
A = F~max / σ
Given that F~max = 755 N and σ = 3.0 × 10^5 kg m^−1s^−2, we can substitute these values into the equation:
A = 755 N / (3.0 × 10^5 kg m^−1s^−2)
To convert the units from square meters to square centimeters, we need to multiply the result by 10,000 (since there are 10,000 square centimeters in a square meter):
A (in cm^2) = (755 N / (3.0 × 10^5 kg m^−1s^−2)) * 10,000
Simplifying the expression, we get:
A (in cm^2) = 25.1667 cm^2
So, the average cross-sectional area of the gastrocnemius muscle for the people in the study is approximately 25.1667 cm^2.
(a) To determine the units of σ in newtons and meters, we can consider the formula F~max = σA. The force F~max is measured in newtons (N), and the area A is measured in square meters (m^2). Therefore, the units of σ can be found by rearranging the formula:
σ = F~max / A
Substituting the units, we get:
σ (N m^−2) = (N) / (m^2)
So, the units of σ in newtons and meters are N m^−2.
(b) To find the average cross-sectional area of the gastrocnemius muscle for the people in the study, we can rearrange the formula F~max = σA and solve for A:
A = F~max / σ
Given F~max = 755 N and σ = 3.0 × 10^5 kg m^−1s^−2, we can substitute these values into the equation:
A = 755 N / (3.0 × 10^5 kg m^−1s^−2)
Simplifying the units, we get:
A (m^2) = 755 N / (3.0 × 10^5 kg m^−1s^−2)
Converting from square meters to square centimeters, we can use the conversion factor 1 m^2 = 10000 cm^2:
A (cm^2) = (755 N / (3.0 × 10^5 kg m^−1s^−2)) * (1 m^2 / 10000 cm^2)
Evaluating the expression:
A ≈ 2.52 cm^2
Therefore, the average cross-sectional area of the gastrocnemius muscle for the people in the study is approximately 2.52 cm^2.
(a) To determine the units of σ in newtons and meters, let's analyze the given equation:
F_max = σA
The unit of force (F_max) is the newton (N). The unit of cross-sectional area (A) is square meters (m^2). The proportionality constant (σ) should have units that make the equation dimensionally consistent.
Let's rearrange the equation to isolate σ:
σ = F_max / A
Substituting the unit of force (N) and the unit of area (m^2) into the equation:
σ = (N) / (m^2)
This means that the units of σ in newtons and meters are N/m^2 (newton per square meter).
(b) To find the average cross-sectional area (A) of the gastrocnemius muscle for the people in the study, we can rearrange the equation:
F_max = σA
Solving for A:
A = F_max / σ
Substituting the given values:
F_max = 755 N
σ = 3.0 × 10^5 kg m^−1 s^−2
A = (755 N) / (3.0 × 10^5 N/m^2)
Simplifying:
A = 2.52 × 10^−3 m^2
To convert square meters to square centimeters, we multiply by 10,000 (since there are 10,000 square centimeters in a square meter):
A = (2.52 × 10^−3 m^2) × (10,000 cm^2/m^2)
A = 25.2 cm^2
Therefore, the average cross-sectional area of the gastrocnemius muscle for the people in the study is 25.2 cm^2.