To find the equilibrium position of a mass attached to an elastic string

[gnits]

Homework Statement

To find the equilibrium position of a mass attached to an elastic string

Relevant Equations

F=ma

Hi, can anyone see if I have made an error in answering this simple question, my answer is not the one given in the textbook.

Here's my diagram of the system in equilibrium. So the mass has dropped a distance x below the midpoint of PQ:

So the forces balance as:

T1 = g + T2

So, using Hooke's law T = Yx/a (where 'Y' is the modulus of elasticity of the string and 'a' the natural length of the string) we have:

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)

Because both parts of the string above and below the mass have a natural length of 1/2 and the part above is of length 1 + x and so has an extention of (1/2) + x and that below the mass is of length 1 - x and so has anextention of (1/2) - x

This solves to give x = 1/16 and so the height above Q would be 15/16.

The book answer if 3/4 = 12 / 16.

Thanks for any help,
Mitch.

 

[PeroK]

Homework Statement: To find the equilibrium position of a mass attached to an elastic string
Homework Equations: F=ma

Hi, can anyone see if I have made an error in answering this simple question, my answer is not the one given in the textbook.

View attachment 252832

Here's my diagram of the system in equilibrium. So the mass has dropped a distance x below the midpoint of PQ:

View attachment 252835
So the forces balance as:

T1 = g + T2

So, using Hooke's law T = Yx/a (where 'Y' is the modulus of elasticity of the string and 'a' the natural length of the string) we have:

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)

Because both parts of the string above and below the mass have a natural length of 1/2 and the part above is of length 1 + x and so has an extention of (1/2) + x and that below the mass is of length 1 - x and so has anextention of (1/2) - x

This solves to give x = 1/16 and so the height above Q would be 15/16.

The book answer if 3/4 = 12 / 16.

Thanks for any help,
Mitch.

Why not try again using $h$?

I'm really confused by your equations involving $x$.

Okay. I see what you are doing with $x$. It looks correct from what I can see.

Why introduce $x$, when it's $h$ you want?

 

[gnits]

Why not try again using $h$?

I'm really confused by your equations involving $x$.

Okay. I see what you are doing with $x$. It looks correct from what I can see.

Why introduce $x$, when it's $h$ you want?

I can repeat with equations in terms of h. This gives:

4g*(2-(1/2)-h)/(1/2) = g + 4g(h-(1/2)/(1/2)

This leads directly to the same answer of h = 15/16

But thanks for saying that you believe I am correct. The book I am using is rarely wrong. Having other people tell me that they agree with me gives confidence that the book may indeed be wrong in this case.

Thanks,
Mitch.

 

[gleem]

Are you using Tension = Y*elongation/original length?

 

[hutchphd]

Homework Statement: To find the equilibrium position of a mass attached to an elastic string
Homework Equations: F=ma

This solves to give x = 1/16 and so the height above Q would be 15/16.

If I understand what you are doing this solves to x=1/4 and the book is correct. ( your answer is fine but your arithmetic is even worse than mine! )

 

[haruspex]

If I understand what you are doing this solves to x=1/4 and the book is correct. ( your answer is fine but your arithmetic is even worse than mine! )

No, I think you lose the arithmetic contest, on consensus at least. I also get 1-x=15/16.

Another flaw in the question is stating the elasticity as "4gN". g has dimension; the "N" is superfluous.

 

[hutchphd]

If I start with this (from #1)

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)​

Then

8gx=2g
x=1/4​

What are you looking at?

 

[haruspex]

If I start with this (from #1)

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)​

Then

8gx=2g
x=1/4​

What are you looking at?

( 4g( (1/2) + x) ) / (1/2) = g + ( 4g( (1/2) - x) ) / (1/2)
2( 4g( (1/2) + x) ) = g + 2( 4g( (1/2) - x) )
( 4g( 1 + 2x) ) = g + ( 4g( 1 - 2x) )
4 ( 1 + 2x) = 1 + 4( 1 - 2x)
4 (1+2x-1+2x) = 1
16x=1

 

[gleem]

Start with T = Y*(Δl ± x)/L where Δl = 1 , L=2

Then 4g(1+x)/2 = g +4g(1-x)/2

x= 1/4

 

[hutchphd]

16x=1

You know I should never do arithmetic (or anything for that matter at my age) after 10pm. I worked the problem through today and concur with majority. Apologies.

 

[haruspex]

Start with T = Y*(Δl ± x)/L where Δl = 1 , L=2

By the definition of Y, T=Y(L+ΔL)/L, where, in the present case, L=1 and ΔL is represented by x.

 

[gleem]

This has been a bad week ! right L =1 I also misread the problem.

 

[haruspex]

This has been a bad week ! right L =1 I also misread the problem.

On the plus side, you and @hutchphd have shown how the problem setter may have arrived at that answer.

 

[gleem]

OK I've been pulling my hair out making silly mistakes but let me give you my approach.

Step one.

Add 1 kg to the string at the 4/8 m point. It stretches by 1/8 to a length of 5/8 m. The weight is 11/8 above the lower attachment point and the end of the lower part of the string is 7/8 above the lower attachment point.

Step two.

Pull the lower string and fix it to the lower attachment point. You increase the length of the upper string by X. so its elongation is 1/8 +X. the end of the lower string is moved 7/8 down and attached. Its elongation is therefore 7/8 -X.

This gives the books answer.

 

[haruspex]

It stretches by 1/8 to a length of 5/8 m.

The relaxed string is 1m in total. The unstretched length above the half way point is 0.5m. Extending by 1/8 means an extension of 1/16 m.

 

[gleem]

I read the problem as the weight is attached to the 1/2 m point along the string before stretching.

e =T*l/Y l=1/2 m, T =1g , y =4g , e=1/8 m

I looked at the problem as involving two strings one 1/2 m (relaxed) with a 1 kg weight attached and the other 1/2 m (relaxed) string attached to the weight with its free end pulled 7/8 m.

 

[haruspex]

I read the problem as the weight is attached to the 1/2 m point along the string before stretching.

e =T*l/Y l=1/2 m, T =1g , y =4g , e=1/8 m

I looked at the problem as involving two strings one 1/2 m (relaxed) with a 1 kg weight attached and the other 1/2 m (relaxed) string attached to the weight with its free end pulled 7/8 m.

Ah, you misled me by writing

It stretches by 1/8

I.e. lengthens by 12.5%. You meant it stretches by 1/8 m.
Ok, but please post the rest of your working.

Or we can work it backwards..
With the book answer, the upper section extends from 0.5m to 1.25m. The tension should be 4g(1.25-0.5)/0.5=6g kg.
The lower section extends from 0.5m to 0.75m. The tension should be 4g(0.75-0.5)/0.5=2g kg.
Net elastic force, 6g-2g = 4g kg upwards.
But the gravitational force is only 1g kg.

 

[hutchphd]

Pull the lower string and fix it to the lower attachment point. You increase the length of the upper string by X. so its elongation is 1/8 +X. the end of the lower string is moved 7/8 down and attached. Its elongation is therefore 7/8 -X.

But this defines the new zero and thereby gravity is completely included. When the lower end is attached its pull is balanced by the additional extension X and so

X=(7/8)-X
X=7/16​

and this new equilibrium is therefore 1/16 below the center line.