A 15 kg long tendon was found to stretch 3.7 mm by 13 N .The tendon was approximately round with average diameter of 8.5 mm.Calculate the youngs modulus of this tendon

Answer 1

A kilogram is a unit of mass.

If your question means:

A 15 cm long tendon was found to stretch 3.7 mm by 13 N

then

E = σ / ε = ( F / A ) / ( dℓ / ℓ )

where

E = Young’s modulus

σ = tensile stress

ε = tensile strain

F = Force

A = Area of tendon

dℓ = amount of deformation

ℓ = length

In this case:

F = 13 N

A = d² π / 4 = 8.5² π / 4 = 56.745 mm²

1 m = 1000 mm

1 mm = 1 m / 1000

1 mm² = ( 1 /1000² ) m²

A = 56.745 mm² = ( 56.745 / 1000² ) = 0.000056745 = 5.6745 ∙ 10 ⁻⁵ m²

1 cm = 10 mm²

1 mm = 1 / 10 cm

dℓ = 3.7 mm = 3.7 / 10 cm = 0.37 cm

ℓ = 15 cm

dℓ / ℓ = 0.37 cm / 15 cm = 0.0246 = 2.467 ∙ 10 ⁻²

E = ( F / A ) / ( dℓ / ℓ ) = ( 13 / 5.6745 ∙ 10 ⁻⁵ ) / 2.467 ∙ 10 ⁻²

E = 2.2909 5∙ 10⁵ / 2.467 ∙ 10 ⁻²

E = 0.928638 ∙ 10 ⁵ ⁺ ² = 0.928638 ∙ 10⁷ N / m²

E = 9.28638 ∙ 10⁶ N / m²

1 N / m² = 1 Pa

E = 9.28638 ∙ 10⁶ Pa

where

Pa = Pascal