# Strain in different direction

1. Oct 20, 2016

1. The problem statement, all variables and given/known data
In this question , I am not convinced that δAD = δAC cos theta

2. Relevant equations

3. The attempt at a solution
, i think it should be
δAC = δAD cos theta , am i right ?
I think so because we normally get the non-vertical and non -horizontal line and then cos the angle or sin the angle to get the length in x and y direction , right ?

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2. Oct 20, 2016

### Staff: Mentor

No. You're not right. Call L the distance between C and D. Does this distance change?

3. Oct 20, 2016

no
no , the distance doesnt chnage , so ?

4. Oct 20, 2016

### Staff: Mentor

Call x the distance from A to C. Using the Pythagorean theorem, in terms of L and x, what is the distance between A and D?

5. Oct 20, 2016

AD = sqrt ( L^2 + x^2 )

6. Oct 20, 2016

### Staff: Mentor

Good. Now, if the distance between A and C changes by dx, in terms of L, x, and dx, by how much does the distance between A and D change?

7. Oct 20, 2016

can you
can you give some hint , i have no idea

8. Oct 20, 2016

### Staff: Mentor

9. Oct 20, 2016

yes , and ?

10. Oct 20, 2016

### Staff: Mentor

Apparently, that wasn't a big enough hint. What is the derivative with respect to x of $\sqrt{L^2+x^2}$?

11. Oct 20, 2016

0.5(2x) (1/ $\ sqrt{L^2+x^2}$ )

12. Oct 20, 2016

### Staff: Mentor

What is 0.5(2x) equal to?

13. Oct 20, 2016

X

14. Oct 20, 2016

### Staff: Mentor

You really need to learn to use LaTex to display your equations. The current way you do it is totally unacceptable. Here is the link: https://www.physicsforums.com/help/latexhelp/
Use LaTex from now on.

As far as the derivative of $\sqrt{L^2+x^2}$ is concerned, you obtained:$$\frac{x}{\sqrt{L^2+x^2}}$$Is that what you meant?

15. Oct 20, 2016

Yes, so , what are you trying to say?

16. Oct 20, 2016

### Staff: Mentor

You still don't see it, huh?

Well, if dx is the change in the length of AC, then the change in length of AD is $\frac{x}{\sqrt{L^2+x^2}}dx$.
Or equivalently, if $\delta_{AC}$ is the change in length of AC, then the change in length of AD is $\frac{AC}{AD}\delta_{AC}$:
$$\delta_{AD}=\frac{AC}{AD}\delta_{AC}$$
But,$$\frac{AC}{AD}=\cos{\theta}$$
Therefore,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$

17. Oct 20, 2016

Do you mean AC (del AC ) = xdx ?

18. Oct 20, 2016

ok , but , i still dun know why Ac = Ad cos theta , where theta is between AC and AD ..... because normally , we will make the 'slanted line ' to get the horizontal x -direction and vertical (y-direction ) , am i right ?

Last edited: Oct 20, 2016
19. Oct 20, 2016

one more thing, how could AD = AC cos theta , where cos theta = 4/5 ,
it's given that AD = 2 , AC = 1.6

AC cos theta = 1.6(4/5) = 1.28 which is not = AD (2m)

20. Oct 20, 2016

i dont understand this part , we know that δ = PL / AE , we also know that δAD = (4/5)δAC , so , shouldnt (4/5)δAC = (4/5)(1.6) / AE ?

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