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

• gnits
In summary, Mitch thinks that if he uses ##h## instead of ##x## in his equations, the book will be correct. But he also agrees with the majority that x=1/4 and the book is correct.
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.

gnits said:
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?

Last edited:
PeroK said:
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)

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.

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

gnits said:
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! )

hutchphd said:
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.

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

Then
8gx=2g
x=1/4​

What are you looking at?

hutchphd said:

( 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

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

haruspex said:
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.

gleem said:
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.

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

gleem said:
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.

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.

gleem said:
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.

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.

gleem said:
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
gleem said:
It stretches by 1/8
I.e. lengthens by 12.5%. You meant it stretches by 1/8 m.

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.

Last edited:
gleem said:
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.

## 1. What is an equilibrium position?

An equilibrium position is the position where the forces acting on an object are balanced, resulting in no net movement or acceleration.

## 2. How is the equilibrium position of a mass on an elastic string determined?

The equilibrium position of a mass on an elastic string can be determined by finding the position where the forces of gravity and the elastic force of the string are equal and opposite.

## 3. What factors affect the equilibrium position of a mass on an elastic string?

The equilibrium position is affected by the mass of the object, the stiffness of the string, and the gravitational force acting on the object.

## 4. How can the equilibrium position be changed?

The equilibrium position can be changed by adjusting the mass of the object, the stiffness of the string, or the gravitational force acting on the object.

## 5. Why is finding the equilibrium position important in scientific research?

Finding the equilibrium position is important in scientific research because it allows us to understand the balance of forces in a system and predict the behavior of the object. It is also a key concept in many areas of physics, such as mechanics and thermodynamics.

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