Finding the depression and the tension of the string

[kaspis245]

Homework Statement

A wire of unstretched length $l_o$ is extended by a distance $10^{-3}l_o$ when a certain mass is hung from its bottom end. If this same wire is connected between two points, A and B, that are a distance $l_o$ apart on the same horizontal level, and the same mass is hung from the midpoint of the wire as shown, what is the depression y of the midpoint, and what is the tension of the wire?

https://imgur.com/niVACGg

Relevant Equations

Hook's Law:
$F=-kx$

I would assume that all the tension in the wire originates from the weight of the mass, and is equal to $T=mg$, but that is incorrect.
Forming a right triangle with a hypothenuse denoted by $x$ and applying Pythagoras theorem:
\begin{cases}x^2=y^2+l_o^2 \\ x+y=l_o+\Delta l\end{cases}
Solving for $y$ I get:
$$y=\frac{0.01l_o⋅ 2.01}{2.02}≈0.01l_o$$
which is too incorrect.

The answers given by my textbook:
$y=l_o/20$
$T=5mg$

How can I arrive at those?

 

[BvU]

Forming a right triangle with a hypothenuse denoted by $x$ and applying Pythagoras theorem:

That is extremely confusing if your relevant equation is $F=-kx$.

$$\begin{cases}x^2=y^2+l_o^2 \\ x+y=l_o+\Delta l\end{cases} $$

Do that again...

Oh, and don't use Imgur -- cut and paste directly

 

[kaspis245]

$x=l_o+\Delta l-y$
$x^2=(l_o+\Delta l-y)(l_o+\Delta l-y)=l_o^2+y^2+(\Delta l)^2+2l_o\Delta l-2yl_o-2y\Delta l$

Putting this into the first equation:

$x^2=y^2+l_o^2=l_o^2+y^2+(\Delta l)^2+2l_o\Delta l-2yl_o-2y\Delta l$
$(\Delta l)^2+2l_o\Delta l-2yl_o-2y\Delta l=0$
$y=\frac{2l_o\Delta l+(\Delta l)^2}{2l_o+2\Delta l}=\frac{(2+10^{-3})⋅10^{-3}}{2(1+10^{-3})}l_o=\frac{2.001⋅0.001}{2.002}l_o$

I noticed there were some zeros missing in my solution above. Still, I don't seem to get the right answer.

 

[BvU]

What on Earth is the $x$ in your 'other' equations ? (I know what it is in you relevant equation)

I still don't see where you relevant equation is put into action ...

 

[sojsail]

Please explain your equation:

x=lo+Δl−y

You said in your original post that x is the hypotenuse. Perhaps you should draw the triangle of which x is the hypotenuse. Then explain what is lo, Δl, and y. Perhaps just draw it for yourself (if posting a photo would be a problem) and see if you do not see an error. Hint: if I am guessing right, you have 2 triangles. Edit: 2 right triangles.Remember, only right triangles have a hypotenuse.

What is lo?

 

[haruspex]

$x^2=y^2+l_o^2$, $x+y=l_o+\Delta l$

The mass is suspended from the midpoint of the wire. Both of those equations are as though it is directly underneath one end of the wire.

 

[kaspis245]

I assumed that the depression $y$ does not change no matter where the mass is located, so I formed a right triangle like this one:

everything else followed from that. Was that a wrong assumption to make?

I don't really know how Hook's equation could be used here, was hoping someone would give me a hint. I should point out that the $x$ in the diagram does not have anything to do with the Hook's law, just a poor choice of variables. Sorry about that.

 

[haruspex]

Was that a wrong assumption to make?

Any assumption you cannot justify is wrong, and you would certainly struggle to justify that one.

Your first diagram is a bit deceptive in that it shows the mass somewhat off centre. Can you see how to construct a right triangle with the mass positioned correctly?

 

[BvU]

Do I need to remind you of your relevant equation $F= - kx$ ? How's that in relation to your $x$ in the pictures ?

 

[kaspis245]

I still don't quite understand why the triangle I formed would be any different, but never mind that.

Assuming that the mass is at the midpoint of the wire and that the length of one side of the formed triangle is $a$, these are the equations I got:
\begin{cases} a^2=y^2+l_o^2/4 \\ 2a=l_o+\Delta l \end{cases}
Solving for $y$ I get:
$$y=0.022l_o$$
Not quite the correct answer.

As mentioned before, $x$ in the pictures is not related to Hook's law - just a poor choice of variables.

 

[BvU]

Because $l_0^2$ is not the same as twice $\left ({1\over 2} l_0\right )^2$, but more importantly: you can never make a horizontal force balance that way ! And you really need one ...

Now finally get rid of one of the $x$ to end your own confusion...

 

[haruspex]

I still don't quite understand why the triangle I formed would be any different, but never mind that.

Assuming that the mass is at the midpoint of the wire and that the length of one side of the formed triangle is $a$, these are the equations I got:
\begin{cases} a^2=y^2+l_o^2/4 \\ 2a=l_o+\Delta l \end{cases}
Solving for $y$ I get:
$$y=0.022$$
Not quite the correct answer.

As mentioned before, $x$ in the pictures is not related to Hook's law - just a poor choice of variables.

Those equations are now correct, but you have not explained how you get the value of y.
In post #1 you wrote T=mg. That is not correct. In your corrected diagram, what are the forces acting at the point where the mass is suspended from the wire?

 

[kaspis245]

Assuming that the tension in each piece of the wire is $Tcosθ$, then the equilibrium equation at the point where the mass is suspended is:
$$2Tcosθ=mg$$$$T=\frac{mga}{2y}$$
when I take $y=l_o/20$ and solve for $T$, I get the correct answer $T=5mg$. The only thing left is to find out how to arrive at $y=l_o/20$.

 

[BvU]

Assuming that the tension in each piece of the wire is $T\cos\theta$

What is the basis for such an assumption ? Make a free body diagram and clearly indicate what forces play a role.

 

[haruspex]

Assuming that the tension in each piece of the wire is $Tcosθ$,

I assume you mean the vertical component of the tension.
You know what the extension is when the mass hangs from the wire vertically. What equation does that give you for the elastic constant, k?

 

[kaspis245]

Putting everything together:
$$\frac{mga}{2y}=\frac{mg}{\Delta l}z⇒a\Delta l=2yz$$
Then using $a=l_o/2+z$ and $2a=l_o+\Delta l$, I get that $y=0.5005l_o$.

 

[haruspex]

View attachment 244579
Putting everything together:
$$\frac{mga}{2y}=\frac{mg}{\Delta l}z⇒a\Delta l=2yz$$
Then using $a=l_o/2+z$ and $2a=l_o+\Delta l$, I get that $y=0.5005l_o$.

There are two different $\Delta l$s, one when the wire is vertical and one when as in the diagram. You are given the value of one, and from that you can get an equation involving k.

 

[kaspis245]

I forgot that. Then I can't use $2a=l_o+\Delta l$ in the above equations. I can use $a^2=y^2+l_o^2/4$ instead, but once I plug it into $a\Delta l=2yz$, the equation becomes unsolvable.

 

[haruspex]

I forgot that. Then I can't use $2a=l_o+\Delta l$ in the above equations. I can use $a^2=y^2+l_o^2/4$ instead,

No, you can still use both those equations, where Δl is defined as the extension in the diagram position.
Now think about the vertical position. You have a different extension δl=10^-3l₀. If the "spring constant " of the wire is k, write an equation involving k and δl and another involving k and Δl.

 

[kaspis245]

Starting from the beginning.

Step 1
Initially, when the mass is hung at the end of the string vertically, we have
$$mg=kx$$
Since the string extends by $\Delta l=10^{-3}l_o$, we have
$$mg=k\Delta l\\⇒k=\frac{mg}{\Delta l}$$
Step 2
Next, from the diagrams above I can deduce, that the horizontal forces $Tsinθ$ of the system cancel out, while for the vertical ones we have
$$2Tcosθ=mg\\⇒2T\frac{y}{a}=mg$$
Step 3
Examining a single right triangle with a hypotenuse $a$, I say that the tension $T$ in that part of the string causes an extension $δ$. So the string of length $a$ extends by $δ$
$$T=kδ.$$
Step 4
Lastly, I relate the original length of the string with the newly stretched one with the following equation
$$2a=l_o+2δ$$
Solution
Putting everything together I get a system of equations
\begin{cases} k=\frac{mg}{\Delta l} \\ 2T\frac{y}{a}=mg \\ T=kδ \\ 2a=l_o+2δ \end{cases}
which don't give me a solution. I could also put an additional equation using Pythagoras theorem $a^2=y^2+\frac{l_o^2}{4}$, but that only complicates things. Could someone point out in which step did I make a mistake?

 

[haruspex]

Putting everything together I get a system of equations
\begin{cases} k=\frac{mg}{\Delta l} \\ 2T\frac{y}{a}=mg \\ T=kδ \\ 2a=l_o+2δ \end{cases}
which don't give me a solution. I could also put an additional equation using Pythagoras theorem $a^2=y^2+\frac{l_o^2}{4}$, but that only complicates things. Could someone point out in which step did I make a mistake?

Yes, you will need the Pythagoras equation too.
You can easily eliminate k, T, m and g from your equations. What three equations does that leave you with?
You will probably need to make an approximation on the basis that the extension is much smaller than the overall length.

 

[kaspis245]

If I'm including the Pythagoras equation then the full system of equations is
\begin{cases} k=\frac{mg}{\Delta l} \\ 2T\frac{y}{a}=mg \\ T=kδ \\ 2a=l_o+2δ \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Eliminating $k$ and $T$ reduces the system to
\begin{cases} 2yδ=a\Delta l \\ 2a=l_o+2δ \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Next I eliminate δ
\begin{cases} (2a-l_o)y=a\Delta l \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Finally I try to eliminate $a$ and get the following expression
$$\left(2\sqrt{y^2+\frac{l_o^2}{4}}-l_o\right) y=\sqrt{y^2+\frac{l_o^2}{4}}\Delta l$$
As I mentioned previously, I see no clear way to solve that for $y$. If I attempt to make an approximation that $y^2<<\frac{l_o^2}{4}$, this gives zero on one side of the equation:
$$\left(2\frac{l_o}{2}-l_o\right)y=\frac{l_o}{2}\Delta l$$

Any other ideas?

 

[haruspex]

If I'm including the Pythagoras equation then the full system of equations is
\begin{cases} k=\frac{mg}{\Delta l} \\ 2T\frac{y}{a}=mg \\ T=kδ \\ 2a=l_o+2δ \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Eliminating $k$ and $T$ reduces the system to
\begin{cases} 2yδ=a\Delta l \\ 2a=l_o+2δ \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Next I eliminate δ
\begin{cases} (2a-l_o)y=a\Delta l \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Finally I try to eliminate $a$ and get the following expression
$$\left(2\sqrt{y^2+\frac{l_o^2}{4}}-l_o\right) y=\sqrt{y^2+\frac{l_o^2}{4}}\Delta l$$
As I mentioned previously, I see no clear way to solve that for $y$. If I attempt to make an approximation that $y^2<<\frac{l_o^2}{4}$, this gives zero on one side of the equation:
$$\left(2\frac{l_o}{2}-l_o\right)y=\frac{l_o}{2}\Delta l$$

Any other ideas?

You have made it a bit complicated by the way you eliminated a.
After eliminating $\delta$, use the linear equation to get an expression for a and substitute that in the quadratic equation.
The approximation to use later is for small $\Delta l$, initially. Maybe for small y afterwards.

 

[kaspis245]

Ok, so once again our final system of equations is
\begin{cases} (2a-l_o)y=a\Delta l \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Expressing $a$ from the first equation
$$a=\frac{l_oy}{2y-\Delta l}$$
Putting that into the second one
$$\frac{l_o^2y^2}{(2y-\Delta l)^2}=y^2+\frac{l_o^2}{4}$$
$$l_o^2y^2=\left(y^2+\frac{l_o^2}{4}\right)(4y^2-4y\Delta l+(\Delta l)^2)$$
At this point I don't know how to move forward - taking $\Delta l≈0$ or $(\Delta l)^2≈0$ doesn't get me anywhere.

 

[haruspex]

Ok, so once again our final system of equations is
\begin{cases} (2a-l_o)y=a\Delta l \\ a^2=y^2+\frac{l_o^2}{4}\end{cases}
Expressing $a$ from the first equation
$$a=\frac{l_oy}{2y-\Delta l}$$
Putting that into the second one
$$\frac{l_o^2y^2}{(2y-\Delta l)^2}=y^2+\frac{l_o^2}{4}$$
$$l_o^2y^2=\left(y^2+\frac{l_o^2}{4}\right)(4y^2-4y\Delta l+(\Delta l)^2)$$
At this point I don't know how to move forward - taking $\Delta l≈0$ or $(\Delta l)^2≈0$ doesn't get me anywhere.

Certainly discard the $(\Delta l)^2$ term.
Cancelling a factor y, you should have a y² term, a y³ term, and and a y⁰ term. From this, you can see that y is somewhat small compared with l₀, allowing you to try discarding the y² term. We can check later whether that is valid.
(Alternatively, use the formula for solving a cubic.)

 

[kaspis245]

Discarding the $(\Delta l)^2$ term and expanding the right side of the equation
$$l_o^2y^2=4y^2-4y^3\Delta l+y^2l_o^2-yl_o^2\Delta l$$$$4y^3-4y^2\Delta l-l_o^2\Delta l=0$$
I don't see why I should discard the $y^2$ term and leave $y^3$.

I tried solving the cubic equation with Wolfram Alpha, there are no solutions that are even close to the correct one. I'll remind you that the solution given by the textbook is $y=\frac{l_o}{20}$.

 

[haruspex]

I don't see why I should discard the $y^2$ term and leave $y^3$.

Because the y² term has a factor $\Delta l$, which makes it the smallest of the three terms.

tried solving the cubic equation with Wolfram Alpha, there are no solutions that are even close to the correct one.

I don't get the given answer, but it isn't that far off... $\frac {l_0}{10*2^\frac 23}$.

 

[kaspis245]

If I eliminate the $y^2$ term, the equation $$4y^3-l_o^2\Delta l=0$$ is much closer to a solution of $y=\frac{l_o}{16}$ than to $y=\frac{l_o}{20}$ - that's a 20% difference. I don't see why they would choose $\frac{l_o}{20}$.

 

[haruspex]

If I eliminate the $y^2$ term, the equation $$4y^3-l_o^2\Delta l=0$$ is much closer to a solution of $y=\frac{l_o}{16}$ than to $y=\frac{l_o}{20}$ - that's a 20% difference. I don't see why they would choose $\frac{l_o}{20}$.

No, I don't know how they get 20. We can make a better approximation by plugging our value for y into the y² term, but that makes the answer 5% larger, not 20% smaller.

 

[ehild]

You droppwd a "2" somewhere. I got the given result, ΔL=L₀/20, with approximating L≈L₀ where it was possible (Lis the stretched length):
Pythagoras: (L/2)²-(L₀/2)²=y² ----> (L-L₀)(L+L₀)=4y²
using L≈L₀,
(L-L₀)=2y²/L₀. (1)
Given for spring constant: k=1000mg/L₀, (2)
From Hook's Law, and using (1) and (2):
k(L-L₀)=T, that is 2000mgy²/L₀²=T. (3)
From force balance: mg=2Tcos(θ). As cos(θ)=y/(L/2) mg=4Ty/L (4).
From (3) and (4), and taking L≈L₀, the result is y=L₀/20.

 

[haruspex]

You droppwd a "2" somewhere. I got the given result, ΔL=L₀/20, with approximating L≈L₀ where it was possible (Lis the stretched length):
Pythagoras: (L/2)²-(L₀/2)²=y² ----> (L-L₀)(L+L₀)=4y²
using L≈L₀,
(L-L₀)=2y²/L₀. (1)
Given for spring constant: k=1000mg/L₀, (2)
From Hook's Law, and using (1) and (2):
k(L-L₀)=T, that is 2000mgy²/L₀²=T. (3)
From force balance: mg=2Tcos(θ). As cos(θ)=y/(L/2) mg=4Ty/L (4).
From (3) and (4), and taking L≈L₀, the result is y=L₀/20.

Thanks, I've found it.
Post #22 fourth equation should read $2a=l_0+\delta$, not $2a=l_0+2\delta$.

 

[magic]

I'm also confused as to where the factor of two comes into play. For my equations, I have:

$$[1] \qquad F_{net,y} = mg = 2Tsin(\theta) \approx 2Ttan(\theta) \Rightarrow T = \frac{mg}{2tan(\theta)}$$ where $\theta$ is the angle between the horizontal and and the wire.
$$[2] \qquad tan(\theta) = \frac{y}{\frac{l_0}{2}} = \frac{2y}{l_0}$$
$$[3] \qquad l'^2 = y^2 + (\frac{l_0}{2})^2 \ \Rightarrow l' \approx \frac{l_0}{2} + \frac{y^2}{l_0}$$ where I used the binomial expansion to expand the root term above
$$[4] \qquad T = k(l' - \frac{l_0}{2}) = \frac{mgy^2}{\delta l_0}$$

combining it all, I end up with:

$$y^3 = \frac{l_0^3}{4000}$$

and the only way I end up with the correct solution of $y=\frac{l_0}{20}$ is if in the 4th equation that $T=2k(l' - \frac{l_0}{2})$. Is there any physical reason as to why that should be the case?

 

[haruspex]

the only way I end up with the correct solution of $y=\frac{l_0}{20}$ is if in the 4th equation that $T=2k(l' - \frac{l_0}{2})$. Is there any physical reason as to why that should be the case?

Indeed there is.
The whole wire has relaxed length $l_0$ and is extended to length $2l'$.
k is the constant for the wire as a whole. Half the wire would have twice the constant.

 

[magic]

Thank you haruspex! It completely slipped from me that $k$ depends just as much on the geometry of the wire as it does with the material.