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

Question

jamesj · Answer

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 · Answer

c? I am lost at this question

jamesj · Answer

What's the definition of Young's Modulus?

jamesj · Answer

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 · Answer

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 · Answer

$$ \frac{F}{A} = Y \frac{\Delta \ell}{\ell} $$

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 · Answer

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 · Answer

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

jamesj · Answer

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

anonymous · Answer

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 · Answer

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

jamesj · Answer

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

anonymous · Answer

yeah, but how do we get this formula?

anonymous · Answer

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

jamesj · Answer

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

anonymous · Answer

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

jamesj · Answer

ok