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4CP

Expert-verified
Found in: Page 45

Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

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?

(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.

See the step by step solution

Step 1: Given data

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

Step 2: Change in length of the wire

The change in length of a wire of initial length L and cross-sectional area A when force F is applied to it can be expressed as,

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

Here L is Young's modulus of the material of the wire.

Step 3: 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,

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

Thus, the new change in length is 4 mm.

Step 4: 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,

Thus, the new change in length is 16 mm.

Step 5: 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,

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

Thus, the new change in length is 16 mm.