$(a)$ A steel wire of mass $\mu $ per unit length with a circular cross section has a radius of $0.1\,cm$. The wire is of length $10\,m$ when measured lying horizontal and hangs from a hook on the wall. A mass of $25\, kg$ is hung from the free end of the wire. Assuming the wire to be uniform an lateral strains $< \,<$ longitudinal strains find the extension in the length of the wire. The density of steel is $7860\, kgm^{-3}$ and Young’s modulus $=2 \times 10^{11}\,Nm^{-2}$.
$(b)$ If the yield strength of steel is $2.5 \times 10^8\,Nm^{-2}$, what is the maximum weight that can be hung at the lower end of the wire ?
$(a)$ A steel wire of mass $\mu $ per unit length with a circular cross section has a radius of $0.1\,cm$. The wire is of length $10\,m$ when measured lying horizontal and hangs from a hook on the wall. A mass of $25\, kg$ is hung from the free end of the wire. Assuming the wire to be uniform an lateral strains $< \,<$ longitudinal strains find the extension in the length of the wire. The density of steel is $7860\, kgm^{-3}$ and Young’s modulus $=2 \times 10^{11}\,Nm^{-2}$.
$(b)$ If the yield strength of steel is $2.5 \times 10^8\,Nm^{-2}$, what is the maximum weight that can be hung at the lower end of the wire ?
$(a)$ Consider an element $d x$ at a distance $x$ from the load $(x=0)$. If $\mathrm{T}(x)$ and $\mathrm{T}(x+d x)$ are tensions on the two cross sections a distance $d x$ apart then,
$\therefore \mathrm{T}(x+d x)-\mathrm{T}(x)=g d m$
$\quad=\mu g d x$
(where $d m=$ mass of wire $d x$ and $\mu=$ is the mass per unit length $=\frac{d m}{d x}$ )
$\therefore d \mathrm{~T}=\mu g d x \quad[\because \mathrm{T}(x+d x)-\mathrm{T}(x)=d \mathrm{~T}]$
Integrating on both side,
$\therefore \mathrm{T}(x)=\mu g x+\mathrm{C}$
but at $x=0$, tension $\mathrm{T}(0)=0+\mathrm{C}$
$\therefore \mathrm{Mg}=\mathrm{C}$ where $\mathrm{M}$ is suspended mass
$\therefore \mathrm{T}(x)=\mu g x+\mathrm{Mg}$
If element $d x$, increases by length $d r$ then strain $=\frac{d r}{d x}$
Young modulus $\mathrm{Y}=\frac{\frac{\mathrm{T}(x)}{\mathrm{A}}}{\frac{d r}{d x}}$
$\frac{d r}{d x}=\frac{\mathrm{T}(x)}{\mathrm{YA}}$
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