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anonymous
 4 years ago
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:
aY
bY^2
c1/Y
anonymous
 4 years ago
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: aY bY^2 c1/Y

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0If the mass were on the end of a wire of this metal, what would say the answer is, out of a,b,c?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c? I am lost at this question

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0What's the definition of Young's Modulus?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Go 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/801physicsiclassicalmechanicsfall1999/videolectures/lecture26/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{F}{A} = Y \frac{\Delta \ell}{\ell} \] 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
 4 years ago
Best ResponseYou've already chosen the best response.0Intuitively, 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
 4 years ago
Best ResponseYou've already chosen the best response.0and what about this formula, how do we get this e=12MgL^3/3bd^3Y how do we get this?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know what that is. What is this formula and what are all the variables?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0e=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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, you have a formula for it. Why didn't you tell me?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Anyway, you can see immediately that E is proportional to 1/Y. So the answer remains the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, but how do we get this formula?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0anyways, I like you answer better, sweet and simple.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, but it'll come from manipulation of the basic formula for Young's modulus one way or another.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or experiment, I found this: http://amrita.vlab.co.in/?sub=1&brch=74&sim=550&cnt=1
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