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- anonymous

A metal beam with Y as the young's modulus is supported at the two ends and loaded at the center. The depression is directly proportional to:
a-Y
b-Y^2
c-1/Y

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- anonymous

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- JamesJ

If the mass were on the end of a wire of this metal, what would say the answer is, out of a,b,c?

- anonymous

c? I am lost at this question

- JamesJ

What's the definition of Young's Modulus?

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- JamesJ

Go back and understand what it means.
Then answer my question about the wire.
Then think again about the beam.
====
This lecture might be helpful, if a bit advanced:
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-26/

- anonymous

I studied the topic again, I never get this for some reason. But I got the answer for your question
If the mass were on the end of a wire of this metal, what would say the answer is, out of a,b,c?
it would then be Y

- JamesJ

\[ \frac{F}{A} = Y \frac{\Delta \ell}{\ell} \]
So for a given force and length, \( Y \Delta \ell = constant \). Hence \( \Delta \ell \) is inversely proportional to \( Y \). Therefore with the wire alone, the answer will be
c) 1/Y.
The same principle applies with the beam. Here the cross sectional area is a bit more complicated and related to the length of the beam, but the basic physical principles are these same.

- JamesJ

Intuitively, this also makes sense. Y is a measure of 'stiffness'. Y of iron is higher than that of copper say. You'd expect the copper beam to bend more, consistent with the fact that 1/Y for copper > 1/Y for iron as
\[ Y_{iron} > Y_{copper} \]

- anonymous

and what about this formula, how do we get this
e=12MgL^3/3bd^3Y
how do we get this?

- JamesJ

I don't know what that is. What is this formula and what are all the variables?

- anonymous

e=depression
M=mass of load
g=gravitational acceleration
l=length of beam
b is the breadth of the beam
d is the thickness of the beam
Y=modulus

- JamesJ

Oh, you have a formula for it. Why didn't you tell me?

- JamesJ

Anyway, you can see immediately that E is proportional to 1/Y. So the answer remains the same.

- anonymous

yeah, but how do we get this formula?

- anonymous

anyways, I like you answer better, sweet and simple.

- JamesJ

I'm not sure, but it'll come from manipulation of the basic formula for Young's modulus one way or another.

- anonymous

or experiment, I found this:
http://amrita.vlab.co.in/?sub=1&brch=74&sim=550&cnt=1

- JamesJ

ok

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