Chapter 8.2, Problem 9P

To determine

To calculate: Amount of stretch in the second wire.

Answer to Problem 9P

The second wire stretches $1.0125\text{ mm}$ .

Explanation of Solution

Given information:

The following variation:

The stretch in a wire under a given tension varies directly as the length of the wire and inversely as the square of the diameter.

For the first wire:

When length is $2\text{ m}$ and diameter is $1.5\text{ mm}$ , stretch is $\text{1}\text{.2 mm}$ .

For the second wire (under the same tension):

Length is $\text{3 m}$ and diameter is $\text{2}\text{.0 mm}$

Calculation:

Let S be the stretch in the wire, l be the length and d be the diameter. Then, S varies directly as l and inversely as $d^2$ ,

  $S=k\frac{l}{d^2}$

where k is a constant.

Substituting $S=1.2\text{ mm}$ , $l=2\text{ m}=2000\text{ mm}$ and $d=\text{1}\text{.5 mm}$ ,

  $\begin{array}{l}S=k\frac{l}{d^2}[\text{Equation]}\\ 1.2=k\frac{2000}{{\left(1.5\right)}^2}[\text{Substitute]}\\ 1.2=k\frac{2000}{2.25}[\text{Find square]}\\ k=\frac{1.2\times 2.25}{2000}[\text{Multiply both sides by 2}\text{.25/2000]}\\ k=\frac{27}{20000}[\text{Simplify]}\end{array}$

Therefore, an equation of the given variation is $S=\frac{27}{20000}\left(\frac{l}{d^2}\right)$

To find S when $l=3\text{ m}=3000\text{ mm}$ and $d=\text{2}\text{.0 mm}$ , substitute in $S=\frac{27}{20000}\left(\frac{l}{d^2}\right)$ .

  $\begin{array}{l}S=\frac{27}{20000}\left(\frac{l}{d^2}\right)[\text{Equation]}\\ S=\frac{27}{20000}\left(\frac{3000}{2^2}\right)[\text{Substitute]}\\ S=\frac{27}{20000}\left(\frac{3000}{4}\right)[\text{Find square]}\\ S=1.0125[\text{Simplify]}\end{array}$