- #1
Siune
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Homework Statement
We have a steel rod with density [itex]ρ = 7,90 kg/m^3[/itex]. When it's horizontally on the floor, it's length is [itex]L = 6,00m[/itex]. The rods surface area [itex]A[/itex] is a circle with radius [itex] r=0,04m[/itex]. Steel has Young modulus [itex]E=2,1 \cdot 10^{11}[/itex]
Now the rod is lifted up so it's vertically straight. How much does the rod shorten [itex]\Delta L = ?[/itex] under it's own weight.
Homework Equations
Longitudinal deformation: [itex]\frac{F}{A} = E \frac{\Delta L}{L}[/itex]
The Attempt at a Solution
Problem is easy I assume. I'm not just sure is the force that causes the shortening [itex]F= Mg[/itex] where [itex]M=ρ\pi r^2 L[/itex]
I was thinking that we have infinitely small disc(s) with area [itex]A[/itex] and height [itex]dL[/itex]. Now each disc would feel a force due to it's weight [itex]dm \cdot g[/itex].
And now intergrating through the whole rod (From 0 to [itex]L[/itex]) we would get
[tex]∫dm \cdot g = ∫ (ρ \pi r^2 g) dL = ρ\pi r^2 g L = Mg[/tex]
Now we get [tex]\Delta L = \frac{FL}{AE} = \frac{ρgL^2}{E} = 1,33 \cdot 10^-5 m[/tex].
So the shortening would be independent of the area which I hugely doubt? Especially as we are given the radius. And our teacher has never put "too much info" in the description to trick us. If anyone could expain it with words or thru math I would really appericate it.
Sincerely yours,
Siune