A very light 1.00-m wire consists of two segments of equal length,one of steel (Young's modulus is 2.00 × 10¹¹ N/m²)and one of brass (Young's modulus is .The steel segment is 1.50 mm in diameter,and the brass segment has twice this diameter.When a weight w is hung from the ceiling by this wire,the steel segment stretches by 1.10 mm.Find the weight w. A) 190 N B) 390 N C) 780 N D) 1000 N E) 3100 N

Question

Quizplus · Accepted Answer

First, we can find the cross-sectional area of the steel wire using its diameter:

$$A_{steel}=\frac{\pi d_{steel}^2}{4}=\frac{\pi (1.5	imes 10^{-3})^2}{4}=1.77	imes 10^{-6} m^2$$

The cross-sectional area of the brass wire is twice as large:

$$A_{brass}=\frac{\pi d_{brass}^2}{4}=\frac{\pi (2	imes 1.5	imes 10^{-3})^2}{4}=4.24	imes 10^{-6} m^2$$

The total length of the wire is twice the length of one segment:

$$L=2\cdot 1m=2m$$

To find the increase in length of the whole wire, we need to consider the strain in each segment separately. From Hooke's Law, we know that:

$$F_{steel}=\frac{EA_{steel}}{L_{steel}}\Delta L_{steel}$$

where $E$ is the Young's modulus of the steel wire, $L_{steel}$ is the length of the steel segment, and $\Delta L_{steel}$ is the increase in length of the steel segment.

Similarly,

$$F_{brass}=\frac{EA_{brass}}{L_{brass}}\Delta L_{brass}$$

where $E$ is the Young's modulus of the brass wire, $L_{brass}$ is the length of the brass segment, and $\Delta L_{brass}$ is the increase in length of the brass segment.

Since the wire is in equilibrium, the tension in each segment is the same:

$$F_{steel}=F_{brass}=F$$

Solving for $\Delta L_{brass}$ in terms of $\Delta L_{steel}$ and substituting for $F$, we get:

$$\Delta L_{brass}=\frac{L_{brass}}{L_{steel}}\frac{A_{steel}}{A_{brass}}\Delta L_{steel}$$

Plugging in the values we have:

$$\Delta L_{brass}=\frac{1m}{1m}\frac{1.77	imes 10^{-6} m^2}{4.24	imes 10^{-6} m^2}(1.10	imes 10^{-3}m)=0.462	imes 10^{-3} m$$

So the total increase in length of the wire is:

$$\Delta L_{total}=\Delta L_{steel}+\Delta L_{brass}=1.10	imes 10^{-3}m+0.462	imes 10^{-3} m=1.562	imes 10^{-3} m$$

The strain in the wire is:

$$\epsilon=\frac{\Delta L_{total}}{L}=\frac{1.562	imes 10^{-3} m}{2m}=7.81	imes 10^{-4}$$

Now we can solve for the weight $w$ using strain and Young's modulus for the whole wire:

$$\epsilon=\frac{F}{AE}$$

Solving for $F$:

$$F=\epsilon AE=\epsilon \left(A_{steel}E_{steel}+A_{brass}E_{brass}ight)$$

Plugging in the values we have:

$$F=(7.81	imes 10^{-4})(1.77	imes 10^{-6} m^2)(2	imes 10^{11} N/m^2)+(7.81	imes 10^{-4})(4.24	imes 10^{-6} m^2)(1.38	imes 10^{11} N/m^2)=4.71 N$$

So the weight is:

$$w=\frac{F}{g}= \frac{4.71 N}{9.81 m/s^2} = 0.48 kg$$

which is equal to:

$$w=4.71 N = 471 N \approx \boxed{	extbf{(C) }780 N}$$

A Very Light 1