Question

You drink so much boba tea that you have your own straw. It has an outer diameter of 14 mm, inner diameter of 11 mm, and length of 21 cm. It is made of glass with a Young's modulus of 68 GPa and tensile strength of 7 MPa. What is the effective spring constant of the straw with respect to elongation in N/m?

When you hold either end you can stretch the straw with up to 90 N. How much does the straw elongate in mm?

Answer 1

K = 3.07 x 10⁷ N/m = 30.7 MN/m

Δx = 2.93 x 10⁻⁶ m = 0.00293 mm

Step-by-step explanation:

In order to find out the effective spring constant of the straw, we can use the following equation:

K = AE/L

where,

K = Effective Spring Constant = ?

A = Cross-Sectional Area = π(Do - Di)²/4 = π(0.014² - 0.011²)/4 = 9.5 x 10⁻⁵ m²

L = Length of Straw = 21 cm = 0.21 m

E = Young's Modulus = 68 GPa = 68 x 10⁹ Pa

Therefore,

K = (9.5 x 10⁻⁵ m²)(68 x 10⁹ Pa)/(0.21 m)

K = 3.07 x 10⁷ N/m

Now, we find stress to check the elastic limit:

Stress = Force/Area = 90 N/9.5 x10⁻⁵ m²

Stress = 0.95 MPa < Tensile Strength (7 MPa)

Thus, the straw is still in elastic limit and hook's law can be applied:

Δx = F/K = (90 N)/(3.07 x 10⁷ N/m)

Δx = 2.93 x 10⁻⁶ m = 0.00293 mm