Stretching of a wire due to it's weight

In summary, the conversation discusses the calculation of the longest length of steel wire that can hang vertically without breaking, using the breaking stress and density of steel. It also delves into the concept of stress and strain, with the first question solving for stress and the second question considering strain. The expert explains that the maximum tension occurs at the top of the wire, and that the wire will break at this point when the external force (weight) is equal to the breaking stress. In the second case, the wire will stretch and decrease in cross sectional area, leading to an increase in stress.
  • #1
palkia
52
2

Homework Statement



1) calculate the longest
length of steel wire that can hang vertically without breaking
.breaking stress for steel=7.982*10^8N/m^2 and density of
steel(d)=8.1*10^3kg/m^3.2)A bar of mass m and length l is hanging from a point A as shown in the figure.Find the increase in it's length due to it's own weight.The Young's modulus of eleasticity of the wire is Y and area of cross section of the wire is A.

Homework Equations



Y=ΔL/L, Stress=F/A

The Attempt at a Solution



The first questions solves using equation of stress

The second question solves on the basis of strain

I am confused on which formulae to apply... I don't know when to apply which one so can you tell me when to use when and what exactly is the factor deciding to use them.

Can we integrate in the second question for all points in the bar...
[/B]
 
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  • #2
For the first part, you have noted the correct equation for stress. The max tension in the wire for a given length must be equal to ? from Newton's first law? Where does it occur?

For part 2, you could integrate , but it is easier to use the average tension in the bar, due to the linear variation of the load, but your equation is for constant strain, and not equal to Y, and not equal to the deflection,
 
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  • #3
Since the stress is magnitude of the internal restoring force per unit which in this case is the tension then it will be greatest at the top
How exactly did you know to apply the formulae for strain in the second case...I struggled a lot for that

My definitions prior to solving a question for stress and strain is:

StressThe magnitude of the internal force per unit area that occurs due to external force which tends to restore the body to it's original stateStrain:When the size gets changed of the body due to external forcesIn the first case,is that it is asking about the greatest length so as not to undergo any change in it's dimensions( Does breaking means being stretched)

In the second case,is it because it is said it is stretching that we have to use the definition of strain?Am I thinking right ?
 
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  • #4
Well, not quite. Let's look at part a first. The max tension occurs at the top, correct. What is that tension equal to ? Note that stress is force per unit area, leave out that part about 'restoring to its original length'. Tensile stress under axial loading is F/A, period. Where F in this case is T. And T at the top must be equal to the ___?___ of the wire, using equilibrium law.
Also, breaking stress is a property of the material, and is the stress at which the wire will break. It has to elongate before it breaks, all materials deform under stress.
 
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  • #5
Since at the top,the maximum tension acting on is the weight of the wire mg (Why are we calculating this for the upprmost point tho)If at breaking stress,it has to elongate before it breaks then can we use the formulae of strain here maybe
 
  • #6
palkia said:
Why are we calculating this for the upprmost point tho
It is the same radius all the way down, so if it is going to break, where will that be?
palkia said:
If at breaking stress,it has to elongate before it breaks
True, but how is that relevant in part 1? Will elongation at a point somehow stop it breaking there? The tension remains the same.
 
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  • #7
haruspex said:
It is the same radius all the way down, so if it is going to break, where will that be?

Makes sense then.So the uppermost point is most probable to break first
haruspex said:
True, but how is that relevant in part 1? Will elongation at a point somehow stop it breaking there? The tension remains the same.

So the strain will not tell us the condition for breaking but the stress will,right?
If the breaking stress is equal to the external force (weight) at the topmost point then beyond this the wire breaks,right?What does breaking mean here by the way...the wire gets detached
 
  • #8
palkia said:
So the strain will not tell us the condition for breaking but the stress will,right?
If the breaking stress is equal to the external force (weight) at the topmost point then beyond this the wire breaks,right
Right. In practice, it will stretch first, but that reduces the cross sectional area, so increases the stress. Unless by some means the stretch also increases the modulus, the stretching will accelerate.
palkia said:
What does breaking mean here by the way...the wire gets detached
Yes.
 
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  • #9
In the second case,the wire gets elongated but the restoring force that is tension remains constant but it;'s width get elongated so will stress also act in that case?
 
  • #10
palkia said:
In the second case,the wire gets elongated but the restoring force that is tension remains constant?
Yes.
I had not looked at part 2. I am dubious of this:
palkia said:
due to its own weight.
Is that really what the question says? That would mean due to the weight of the wire, but surely it is the weight of the bar that is to be used?
 
  • #11
Yep.It's the weight of the bar...I kinda mixed the wire from the first case

In the second case,some stress will be acting because the width gets elongated...I have heard that young's modulus makes sense when stress and strain are proportional.

In the second case,the wire is acting along the length of the wire which is elongating the wire and which varies with the distance along the bar so we have to integrate it
 
  • #12
palkia said:
Yep.It's the weight of the bar...I kinda mixed the wire from the first case

In the second case,some stress will be acting because the width gets elongated...I have heard that young's modulus makes sense when stress and strain are proportional.

In the second case,the wire is acting along the length of the wire which is elongating the wire and which varies with the distance along the bar so we have to integrate it
Is this also wrong:
palkia said:
A bar of mass m and length l
It's a bar of mass m and a wire of length l, right?

There's no need for integration. This is not the same wire as in the first part, so we are not given a mass, and we can assume it is to be ignored.
Thus the tension is the same throughout, and the percentage extension likewise.

palkia said:
Y=ΔL/L
That is incorrect.
ΔL/L is the strain (ε)
F/A is the stress (σ)
Young's modulus (E or Y) = σ/ε
 
  • #13
I think the book considers the bar and wire to be same because they have done the integration using the weight per unit length
 
  • #14
palkia said:
I think the book considers the bar and wire to be same because they have done the integration using the weight per unit length
OK, that makes sense - no wire, just a bar.

Problems like this can be a bit tricky because you have two measures of the position of an element of the bar: its distance from one end when there is no tension and its position when there is tension.
I recommend you to consider an element of thickness dx, and distance x from (what will be) the bottom, when the bar is horizontal, so relaxed.
When suspended, what is tension on the element? What is the elongation of the element?
 
  • #15
In the second question, the problem is asking for the elongation, or change on length, of a bar of mass m and length l hanging from the ceiling, but it then used the word wire when giving properties. I think it is an error. A figure would help.
 
  • #16
Palkia
Please read again my response in post 2. You must have missed it.
 
  • #17
The tension will be - (mg/L)*x where x is the distance from the bottom

Elongation...it should be the force causing the elongation ...in this case the weight so it is (W/AY)dx where dx is the length of the segment
 
  • #18
PhanthomJay said:
Palkia
Please read again my response in post 2. You must have missed it.
I have read it sir...I understood that part about the maximum stress
 
  • #19
palkia said:
The tension will be - (mg/L)*x where x is the distance from the bottom

Elongation...it should be the force causing the elongation ...in this case the weight so it is (W/AY)dx where dx is the length of the segment
Right, so substitute the tension you found for W and integrate.
 
  • #20
Got that
 

Related to Stretching of a wire due to it's weight

1. How does the weight of a wire affect its length?

The weight of a wire causes it to stretch and increase in length due to the force of gravity pulling down on it. This stretching is known as "sag" and is more pronounced in longer wires with heavier weights.

2. What factors impact the amount of stretch in a wire due to its weight?

The amount of stretch in a wire due to its weight depends on several factors, including the length and weight of the wire, the material it is made of, and the temperature of the wire.

3. Can the stretching of a wire due to its weight be prevented?

While it is not possible to completely prevent the stretching of a wire due to its weight, it can be minimized by using a stronger or thicker wire, reducing the weight hanging from the wire, and properly supporting the wire at regular intervals.

4. How can the stretching of a wire due to its weight be measured?

The stretching of a wire can be measured by using a ruler or measuring tape to determine the change in length of the wire when a weight is added to it. This can be repeated at different weights to determine the relationship between weight and stretch.

5. How does the stretching of a wire affect its strength and durability?

The stretching of a wire can decrease its strength and durability over time, as it may become more susceptible to breaking or snapping under the weight it is supporting. Regular maintenance and proper support can help maintain the strength and durability of a wire.

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