Question

A straw 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. a. What is the effective spring constant of the straw with respect to elongation in N/m? b. When you hold either end you can stretch the straw with up to 90 N. How much does the straw elongate in mm? c. By how much can you extend the length of the straw before it breaks?

Answer 1

Given that,

Outer diameter = 14 mm

Inner diameter = 11 mm

Length = 21 cm

Young's modulus = 68 GPa

Tensile strength = 7 Mpa

(a). We need to calculate the effective spring constant of the straw with respect to elongation

Using formula of effective spring constant

`(Y)/(\Delta l)=(YA)/(l)`

`k=(YA)/(l)`

Where, k = effective spring constant

Y= Young's modulus

A = area

l = length

Put the value into the formula

`k=(68*10^(9)*(\pi)/(4)(0.014^2-0.011^2))/(21*10^(-2))`

`k=1.90*10^(7)\ N/m`

(b). Force = 90 N

We need to calculate the stretch the straw

Using formula of stretch

`\Delta l=(F)/(k)`

Where, F = force

k = effective spring constant

Put the value into the formula

`\Delta l=(90)/(1.90*10^(7))`

`\Delta l=0.0000047\ m`

`\Delta l=4.7*10^(-6)\ m`

(c). We need to calculate the extend the length of the straw before it breaks

Using formula of extend length

`E_(max)=(Y\Delta l)/(l)`

`\Delta l=(E_(max)l)/(Y)`

Put the value into the formula

`\Delta l=(7*10^6*21*10^(-2))/(68*10^(9))`

`\Delta l=0.0000216\ m`

`\Delta l=0.0216*10^(-3)\ m`

`\Delta l=0.0216\ mm`

Hence, (a). The effective spring constant of the straw with respect to elongation is
`1.90*10^(7)\ N/m`

(b). The stretch the straw is
`4.7*10^(-6)\ m`

(c). The extend length of the straw before it breaks is 0.0216 mm.