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A circular steel wire 1.95 m long must stretch no more than 0.20 cm when a tensile force of 440...

Question:

A circular steel wire {eq}\displaystyle 1.95 m {/eq} long must stretch no more than {eq}\displaystyle 0.20 cm {/eq} when a tensile force of {eq}\displaystyle 440 {/eq} Nis applied to each end of the wire. What minimum diameter is required for the wire?

Hooke's Law:

Hooke's law helps to correlate the relationship between the stress-induced in an object, and the strain developed when a force is applied at the end of that object. As the value of force increases the magnitude of stress increases.

Answer and Explanation: 1


Given Data


  • The length of the wire is; {eq}l = 1.95\;{\rm{m}} {/eq}
  • The change in length of the wire is; {eq}\Delta l = 0.20\;{\rm{cm}} = 0.20 \times {10^{ - 2}}\;{\rm{m}} {/eq}
  • The tensile force is; {eq}F = 440\;{\rm{N}} {/eq}


The standard value of the modulus of elasticity for the steel is {eq}E = 200\;{\rm{GPa}} {/eq}.


The formula of the modulus of elasticity is,

{eq}\begin{align*} E &= \dfrac{\sigma }{\varepsilon }\\ E &= \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}\\ E &= \dfrac{{Fl}}{{\Delta lA}}\\ E &= \dfrac{{Fl}}{{\Delta l\left( {\dfrac{\pi }{4}{d^2}} \right)}}\;\;...\left( 1 \right) \end{align*} {/eq}

Here, {eq}\sigma {/eq} is the stress induced in the wire, {eq}\varepsilon {/eq} is the strain, {eq}A {/eq} is the cross-sectional area of the wire, and {eq}d {/eq} is the diameter of the wire.


Substitute the given value in equation (1) to find the diameter of the wire.

{eq}\begin{align*} 200\;{\rm{GPa}}\left( {\dfrac{{{\rm{1}}{{\rm{0}}^{\rm{9}}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}{{{\rm{1}}\;{\rm{GPa}}}}} \right)& = \dfrac{{440\;{\rm{N}} \times 1.95\;{\rm{m}}}}{{0.20 \times {{10}^{ - 2}}\;{\rm{m}} \times \left( {\dfrac{\pi }{4}{d^2}} \right)}}\\ {d^2} &= \dfrac{{440\;{\rm{N}} \times 1.95\;{\rm{m}}}}{{0.20 \times {{10}^{ - 2}}\;{\rm{m}} \times 200 \times {\rm{1}}{{\rm{0}}^{\rm{9}}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}\left( {\dfrac{\pi }{4}} \right)}\\ d& \approx 1.65 \times {10^{ - 3}}\;{\rm{m}} \end{align*} {/eq}

So, the minimum diameter of the wire should be {eq}1.65 \times {10^{ - 3}}\;{\rm{m}} {/eq}.


Learn more about this topic:

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Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.


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