Is strain in direction AD equal to strain in direction AC times cos theta?

[chetzread]

Homework Statement

In this question , I am not convinced that δAD = δAC cos theta

Homework Equations

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 ?
[/B]

 

[Chestermiller]

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

 

[chetzread]

no

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

no , the distance doesn't chnage , so ?

 

[Chestermiller]

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?

 

[chetzread]

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?

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

 

[Chestermiller]

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

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?

 

[chetzread]

can you

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?

can you give some hint , i have no idea

 

[Chestermiller]

Have you had calculus?

 

[chetzread]

Have you had calculus?

yes , and ?

 

[Chestermiller]

yes , and ?

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

 

[chetzread]

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

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

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

 

[Chestermiller]

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

What is 0.5(2x) equal to?

 

[chetzread]

What is 0.5(2x) equal to?

X

 

[Chestermiller]

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?

 

[chetzread]

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?

Yes, so , what are you trying to say?

 

[Chestermiller]

Yes, so , what are you trying to say?

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,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$

 

[chetzread]

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,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$

Do you mean AC (del AC ) = xdx ?

 

[chetzread]

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,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$

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 ?

 

[chetzread]

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)

 

[chetzread]

i don't understand this part , we know that δ = PL / AE , we also know that δAD = (4/5)δAC , so , shouldn't (4/5)δAC = (4/5)(1.6) / AE ?

 

[Tom.G]

@Chestermiller

$\frac{AC}{AD}=\cos{\theta}$

Therefore,

$\delta_{AD}=\delta_{AC}\cos{\theta}$

Isn't there a problem with this step?

 

[Chestermiller]

$$AC=x$$
$$\delta_{AC}=dx$$
$$AD=\sqrt{L^2+x^2}$$
$$\delta_{AD}=d\sqrt{L^2+x^2}=\frac{x}{\sqrt{L^2+x^2}}dx=\frac{AC}{AD}\delta_{AC}=\frac{1.6}{2}\delta_{AC}=\frac{4}{5}\delta_{AC}$$

 

[Chestermiller]

@Chestermiller

Isn't there a problem with this step?

NO.

 

[Chestermiller]

one more thing, how could AD = AC cos theta , where cos theta = 4/5 ,
it's given that AD = 2 , AC = 1.6AC cos theta = 1.6(4/5) = 1.28 which is not = AD (2m)

Who said AD=AC cos theta?
I said $\delta_{AD}=\delta_{AC}\cos{\theta}$