Respuesta :
Answer:
F = 2.1 10³ N
Explanation:
Young's modulus is defined as the tensile strength (pressure) between the unit strain, whose equation is as follows:
Y = P / (ΔL / L)
Body pressure is defined as the relationship between force and area over which it is applied.
P = F / A
Let's replace and clear the force
Y = (F / A) / (DL / L
Y ΔL / L = F / A
F = Y A ΔL / L
Let's calculate the area . The tendon is shaped like a cylinder, so the area is the circle
A = π R²
R = d / 2 = 5.90 10⁻³ / 2
R = 2.95 10⁻³ m
A = π (2.95 10-3)²
A = 2,734 10⁻⁵ m²
Let's calculate deformation
ΔL = 0.149 -0.14
ΔL = 0.009 m = 9 10⁻³ m
Let's replace and calculate
F = 1.20 10⁹ 2,734 10⁻⁵ 9 10⁻³ / 0.14
F = 2.1 10³ N
Answer:
[tex]T=2109N[/tex]
Explanation:
The Young's modulus is defined as the quotient of the stress and the strain in a material. So
[tex]E=\frac{\sigma}{\varepsilon}[/tex]
And, how does Young's modulus is related to the tension (in this particular case in the tendon)? We can use the definition of stress and strain:
[tex]\sigma=\frac{F}{S}\\\\\varepsilon=\frac{\Delta L}{L}[/tex],
Where F is the applied force to the material, S is the cross-sectional area of the material, [tex]\Delta L[/tex] is the deformation and L is the original length.
In this case [tex]F=T[/tex], so we can rewrite the Young's modulus expression as follows:
[tex]E=\frac{\frac{T}{S}}{\frac{\Delta L}{L}}[/tex],
now solving for T we get
[tex]T=E(\frac{\Delta L}{L})(S)[/tex].
Before computing the tension we need to compute the cross-sectional area and the deformation.
The cross-sectional area is the cross section of a cylinder (the tendon's form):
[tex]S=\pi(D/2)^{2}[/tex],
where D is de diameter ([tex]D=5.90*10^{-3}m[/tex]).
[tex]S=\pi(5.90*10^{-3}/2)^{2}[/tex],
[tex]S=2.73*10^{-5}(m^{2})[/tex].
The Deformation is simply the change in length, so
[tex]\Delta L=14.9cm-14cm=0.9cm=0.009m[/tex].
Now we can easily compute the tension:
[tex]T=(1.20*10^{9})(\frac{0.009}{0.14})(2.73*10^{-5})[/tex],
[tex]T=2109N[/tex].
