A tendon is stretched slightly under the force T of 1000 N. How much will the tendon stretch under this load, in mm to one decimal place, if the diameter is 4 mm, the length is 30 mm, and the Young's Modulus is 1.5 GPa?
To find how much the tendon will stretch under a load of 1000 N, we can use Hooke's Law for axial deformation, which states that the deformation is directly proportional to the applied force.
The formula for axial deformation is given by:
ΔL = F * L / (A * E)
Where:
ΔL is the change in length of the tendon,
F is the force applied to the tendon (1000 N),
L is the initial length of the tendon (30 mm),
A is the cross-sectional area of the tendon, and
E is the Young's modulus (1.5 GPa).
To calculate the cross-sectional area, we can use the formula for the area of a circle:
A = π * r^2
Where:
A is the cross-sectional area of the tendon, and
r is the radius of the tendon (diameter/2).
Let's plug in the values to calculate the cross-sectional area:
r = 4 mm / 2 = 2 mm = 0.002 m
A = π * (0.002 m)^2
A ≈ 0.00001257 m^2
Now, we can substitute all the values into the formula for axial deformation:
ΔL = (1000 N) * (30 mm) / (0.00001257 m^2 * 1.5 GPa)
ΔL ≈ 15000000 mm³ / (1.885 × 10^9 N/m²)
ΔL ≈ 0.00796 mm
Therefore, the tendon will stretch approximately 0.008 mm under the load of 1000 N.
To calculate the stretch of the tendon, we can use Hooke's Law, which states that the amount of deformation in a material is directly proportional to the force applied to it. The formula for calculating the stretch is:
Stretch = (Force * Length) / (π * Diameter^2 * Young's Modulus)
Let's plug in the given values into the formula:
Force = 1000 N
Length = 30 mm
Diameter = 4 mm (radius = 2 mm)
Young's Modulus = 1.5 GPa = 1.5 * 10^9 Pa
Stretch = (1000 * 30) / (π * 4^2 * (1.5 * 10^9))
Simplifying the equation:
Stretch ≈ 0.00857 mm (rounded to one decimal place)
Therefore, the tendon will stretch approximately 0.0086 mm under a force of 1000 N.