Question

You hang a 10 kg mass from a copper wire, and the wire stretches by .

(a)If you suspend the same mass from two copper wires, identical to the original wire, what happens?

(b)If you suspend the same mass from a copper wire with half the cross-sectional area but the same length as the original wire, what happens?

(c)If you suspend the same mass from a copper wire with the same cross-sectional area but twice the length of the original wire, what happens?

Short answer

(a) The stretch in the wire when two wires are used instead of one is 4 mm.

(b) The stretch in the wire when the cross-sectional area is halved is 16 mm.

(b) The stretch in the wire when the initial length is doubled is 16 mm.

Step-by-step solution

Given data

A 10 kg mass is hung from a copper wire, and the wire stretches by 8 mm.

Change in length of the wire

The change in length of a wire of initial length Land cross-sectional area Awhen forceFis applied to it can be expressed as,

$\mathbf{∆}\mathbf{L}\mathbf{=}\frac{\mathbf{FL}}{\mathbf{AY}}$ (I)

Here Lis Young's modulus of the material of the wire.

Determining the change in length when two wires are used instead of one

a)

When two wires are used, the cross-sectional area doubles. From equation (I), the new change in length can be calculated as,

$\begin{array}{rcl}∆\mathrm{L}& =& \frac{\mathrm{FL}}{2\mathrm{AY}}\\ & =& \frac{1}{2}\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& \frac{1}{2}\times 8\mathrm{mm}\\ & =& 4\mathrm{mm}\end{array}$

Thus, the new change in length is 4 mm.

Determining the change in length when the cross-sectional area is halved

b)

From equation (I), the new change in length when the cross-sectional area is halved can be calculated as,

$\begin{array}{rcl}∆\mathrm{L}& =& \frac{\mathrm{FL}}{{\displaystyle \frac{\mathrm{A}}{2}}\mathrm{Y}}\\ & =& 2\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& 2\times 8\mathrm{mm}\\ & =& 16\mathrm{mm}\\ & & \end{array}$

Thus, the new change in length is 16 mm.

Determining the change in length when the initial length is doubled

c)

From equation (I), the new change in length when the initial length is doubled can be calculated as,

$\begin{array}{rcl}∆\mathrm{L}& =& \frac{\mathrm{F}2\mathrm{L}}{\mathrm{AY}}\\ & =& 2\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& 2\times 8\mathrm{mm}\\ & =& 16\mathrm{mm}\end{array}$

Thus, the new change in length is 16 mm.