## anonymous 5 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: a-Y b-Y^2 c-1/Y

1. 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?

2. anonymous

c? I am lost at this question

3. JamesJ

What's the definition of Young's Modulus?

4. 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/

5. 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

6. 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.

7. 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 $Y_{iron} > Y_{copper}$

8. anonymous

9. JamesJ

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

10. 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

11. JamesJ

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

12. JamesJ

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

13. anonymous

yeah, but how do we get this formula?

14. anonymous

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

15. JamesJ

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

16. anonymous

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

17. JamesJ

ok