A 13 cm long animal tendon was found to stretch 3.8 mm by a force of 13 N. The tendon was approximately round with an average diameter of 9.0 mm. Calculate the elastic modulus of this tendon.
4 answers
URGENT
*First, calculate the area:
Area = (pi)r^2 = (π)4.5^2 = 63.6173 mm^2 = 0.0000636173 m^2
*Now, use this equation to find E, the elastic modulus:
E = (F/A) ÷ (ΔL/L0)
*Convert all length measurements to m and then plug and chug:
E = (13/0.0000636173) ÷ (0.0038/0.13) = 6 990 816.05 = 7.0 x 10^6
*Elastic modulus is measured in N per m^2, so the final answer is 7.0 x 10^6 N/m^2
Area = (pi)r^2 = (π)4.5^2 = 63.6173 mm^2 = 0.0000636173 m^2
*Now, use this equation to find E, the elastic modulus:
E = (F/A) ÷ (ΔL/L0)
*Convert all length measurements to m and then plug and chug:
E = (13/0.0000636173) ÷ (0.0038/0.13) = 6 990 816.05 = 7.0 x 10^6
*Elastic modulus is measured in N per m^2, so the final answer is 7.0 x 10^6 N/m^2
Further notes:
*Found r by taking diameter and dividing by two
*F = force = 13 N, listed in the problem
*A = area, which was found above (to convert from mm to m, multiply the mm figure by 10^-3; to convert from mm^2 to m^2, multiply the mm^2 figure by 10^-6, or twice the exponent of the conversion from mm to m)
*ΔL = 3.8 mm, listed in the problem
*L0 (a.k.a. "L initial") = 13 cm, listed in the problem
*Found r by taking diameter and dividing by two
*F = force = 13 N, listed in the problem
*A = area, which was found above (to convert from mm to m, multiply the mm figure by 10^-3; to convert from mm^2 to m^2, multiply the mm^2 figure by 10^-6, or twice the exponent of the conversion from mm to m)
*ΔL = 3.8 mm, listed in the problem
*L0 (a.k.a. "L initial") = 13 cm, listed in the problem
Oh, and the equation for Young's modulus (the elastic modulus), E = (F/A) ÷ (ΔL/L0), is actually stress divided by strain:
Stress = F/A
Strain = ΔL/L0
Stress = F/A
Strain = ΔL/L0
1 answer