Engineering maths help Watch

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Engclasshelp
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Hey, i am studying for my engineering materials exam, first year undergrad.

What is the minimal diameter of a copper wire that can support the mass of 100 kg without breaking?
The Young’ modulus of copper is 120 GPa, yield strength 190 MPa
and tensile strength 250 MPa.


I know to find the maximum diameter i have to use the tensile strength although using stress=force/area and i need to find diameter i'm not actually given force and need to find area which cant be found from the yield stress because i imagine the diameter for yield stress will be different than for tensile stress. Any help would be great, thanks !!
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TeeEm
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(Original post by Engclasshelp)
Hey, i am studying for my engineering materials exam, first year undergrad.

What is the minimal diameter of a copper wire that can support the mass of 100 kg without breaking?
The Young’ modulus of copper is 120 GPa, yield strength 190 MPa
and tensile strength 250 MPa.


I know to find the maximum diameter i have to use the tensile strength although using stress=force/area and i need to find diameter i'm not actually given force and need to find area which cant be found from the yield stress because i imagine the diameter for yield stress will be different than for tensile stress. Any help would be great, thanks !!
engineering forum if there is such place must be better
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samsama
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first find the weight on the copper wire using f=ma.
f = 100kg \cdot 9.8ms^{-2} = 980N

you know that stress = force / area, i.e.
\tau = \tfrac{f}{a}

rearrange to find the area:
 a = \tfrac{\tau}{f}

substitute in what you know:
a = \frac{980}{190*10^6} = 5 \cdot 10^{-6}m^2

area of a circle is:
a = \frac{\pi \cdot d^2}{4}

rearrange to find the diameter:
d = \sqrt{4 \cdot a \cdot \pi}

and so
d = \sqrt{4 \cdot 5 \cdot 10^{-6} \cdot \pi} = 0.0081m = 8.1mm

i'm just guessing so i might be wrong.
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Engclasshelp
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(Original post by samsama)
first find the weight on the copper wire using f=ma.
f = 100kg \cdot 9.8ms^{-2} = 980N

you know that stress = force / area, i.e.
\tau = \tfrac{f}{a}

rearrange to find the area:
 a = \tfrac{\tau}{f}

substitute in what you know:
a = \frac{980}{190*10^6} = 5 \cdot 10^{-6}m^2

area of a circle is:
a = \frac{\pi \cdot d^2}{4}

rearrange to find the diameter:
d = \sqrt{4 \cdot a \cdot \pi}

and so
d = \sqrt{4 \cdot 5 \cdot 10^{-6} \cdot \pi} = 0.0081m = 8.1mm

i'm just guessing so i might be wrong.
Thanks so much!!
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MathsNerd1
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(Original post by Engclasshelp)
Hey, i am studying for my engineering materials exam, first year undergrad.

What is the minimal diameter of a copper wire that can support the mass of 100 kg without breaking?
The Young’ modulus of copper is 120 GPa, yield strength 190 MPa
and tensile strength 250 MPa.


I know to find the maximum diameter i have to use the tensile strength although using stress=force/area and i need to find diameter i'm not actually given force and need to find area which cant be found from the yield stress because i imagine the diameter for yield stress will be different than for tensile stress. Any help would be great, thanks !!
Moved to Engineering forum
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freshestjoz
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thx but your second to the last step, your rearranging is wrong
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