Question

A 2000 N force stretches a wire by 1 mm. A second wire of the same material is twice as long and has twice the diameter. How much force is needed to stretch it by 1 mm? Explain

Short answer

The force required to stretched the second wire by $1mm$is$4000N$.

Step-by-step solution

Given information

Force on wire is $F=2000N$

The length of the second wire is role="math" localid="1650289361551" $L_2=2L_1$

The diameter of the second wire is role="math" localid="1650289386744" $d_2=2d_1$

$∆L=1mm$

Area of the second wire

As the diameter of the second wire is $d_2=2d_1$, the radius of the second wire is $r_2=2r_1$.

The area of the second wire is

role="math" localid="1650289304101" $$\begin{gathered} A_2={\mathrm{\pi r}}_2^2 \\ =\mathrm{\pi }{\left(2{\mathrm{r}}_1\right)}^2 \\ =4{\mathrm{\pi r}}_1^2 \\ =4A \end{gathered}$$

Force required to stretched the second wire by 1 mm

Linear stretch and compression is given by $\frac{F}{A}=Y\left(\frac{∆L}{L}\right)$.

The force required to stretch the second wire with the length of role="math" localid="1650288979993" $∆L=1mm$is

role="math" localid="1650289601869" $$\begin{gathered} F_2=A_{2}Y\left(\frac{∆L}{L_2}\right) \\ =4A_{1}Y\left(\frac{∆L}{2L_1}\right) \\ =2\left(A_{1}Y\frac{∆L}{L_1}\right) \\ =2F_1 \\ =2\left(2000\right) \\ =4000N \end{gathered}$$

Therefore, the force to stretch the second wire with the length of$1mm$is$4000N$.