## College Physics (4th Edition)

(a) The diameter of the wire is $1.3~mm$ The tensile stress is $8.4\times 10^7~N/m^2$ (b) The maximum weight that can be hung from the wire is $571~N$
(a) $Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length The Young's modulus of copper is $120\times 10^9~Pa$ We can find the radius of the wire: $A = \frac{F~L}{Y~\Delta L}$ $\pi~r^2 = \frac{F~L}{Y~\Delta L}$ $r = \sqrt{\frac{F~L}{\pi~Y~\Delta L}}$ $r = \sqrt{\frac{(120~N)(3.0~m)}{(\pi)(120\times 10^9~Pa)(2.1\times 10^{-3}~m)}}$ $r = 0.674\times 10^{-3}~m$ $r = 0.674~mm$ Since the diameter of the wire is twice the radius, the diameter is $1.3~mm$. We can find the tensile stress: $\frac{F}{A} = \frac{F}{\pi~r^2}$ $\frac{F}{A} = \frac{120~N}{(\pi)~(6.74\times 10^{-4}~m)^2}$ $\frac{F}{A} = 8.4\times 10^7~N/m^2$ The tensile stress is $8.4\times 10^7~N/m^2$ (b) We can find the maximum weight: $\frac{F}{A} = 4.0\times 10^8~N/m^2$ $F = (4.0\times 10^8~N/m^2)~(\pi~r^2)$ $F = (4.0\times 10^8~N/m^2)~(\pi)~(6.74\times 10^{-4}~m)^2$ $F = 571~N$ The maximum weight that can be hung from the wire is $571~N$