# Determine equivalent Young's Modulus

• vcsharp2003
In summary: From there, the equivalent wire can be used to calculate the strain and tension in the system. Therefore, in summary, the conversation discusses finding a single wire to support the load, determining the equivalent stiffness of the system, and calculating the strain and tensions in the system using the equivalent wire.
vcsharp2003
Homework Statement
What will be the equivalent Young's Modulus of the system shown in diagram below.
Relevant Equations
Y= stress/strain
We need to find a single wire supporting the load. But, when I think about it there can so many different types of materials we could use for the single wire and each of these single wires would have a different Young's Modulus. So, I don't get what equivalent means here?

Anyways, I get the following equations assuming each wire will extend by a different amount and also tensions in each wire are taken as different. I am not sure if this approach is correct i.e. whether wires would have different tensions and extend by different amounts.

It looks like the wires are equidistant from the ends.
Taking moments about the centre of the mass, what do you deduce about the tensions?

Given that the extensions will be different, what would be a reasonable way to define the strain of the system?

vcsharp2003
haruspex said:
It looks like the wires are equidistant from the ends.
Taking moments about the centre of the mass, what do you deduce about the tensions?
That would make the tensions equal in both wires.
haruspex said:
Given that the extensions will be different, what would be a reasonable way to define the strain of the system?
Not sure. My guess is that we assume equal extensions of each wire since the large mass appears horizontal. Then extension of the system could be either of the equal extensions. Then tensions wouldn't be equal in the two wires.

Last edited:
Is option b correct?

#### Attachments

• soln.pdf
826.1 KB · Views: 92
vcsharp2003
Curiosity_0 said:
Is option b correct?
Yes

Curiosity_0 said:
Is option b correct?
I need to go through it and understand.

vcsharp2003 said:
My guess is that we assume equal extensions of each wire
but that is not possible because you already deduced:
vcsharp2003 said:
the tensions equal in both wires.
If you were to replace the two wires by a single wire, where would it be attached? What would its extension be for the mass to be in the same position?

haruspex said:
but that is not possible because you already deduced
If extensions are not equal then how can the block remain horizontal? If they are unequal then the block will be tilted i.e. inclined to the horizontal and also then the moment arms of the two tension forced would be unequal.
haruspex said:
If you were to replace the two wires by a single wire, where would it be attached?
Then the line from the wire would need to pass through the center of gravity of the block.

haruspex said:
What would its extension be for the mass to be in the same position?
Extension would be such that mg = stress in wire x area of cross-section of wire

In my judgment, you are supposed to assume that the extension in the wires are equal and that the tensions are not. This would be possible if the ratio of the distances of the wires from the center of mass of the bar were such that their moments about the center of mass of the bar were zero.

vcsharp2003
Curiosity_0 said:
Is option b correct?
For information, at the end of your Post #4 attachment (soln.pdf) you wrote $$y_3 = \frac {y_1 + y_2}{y_3}$$This may not be what you intended to write!

Curiosity_0 and vcsharp2003
When I looked at the solution in the book, as given in below screenshot, it doesn't seem logical to me. For example, I don't get why the solution assumed that the equivalent wire would have a cross sectional area of 2A. It's like combining the two wires of cross sectional area A into one along their length.

vcsharp2003 said:
When I looked at the solution in the book, as given in below screenshot, it doesn't seem logical to me. For example, I don't get why the solution assumed that the equivalent wire would have a cross sectional area of 2A. It's like combining the two wires of cross sectional area A into one along their length.

View attachment 317106
IMO it's simply a badly written question. The question does not adequately define what 'equivalent' means.

vcsharp2003 and Chestermiller
Steve4Physics said:
IMO it's simply a badly written question. The question does not adequately define what 'equivalent' means.
I think it may make sense, but only if one looks at it from the perspective as mentioned in a Wikipedia article. There is a table given in above article about how forces and elongations compare when two springs are in series and in parallel.

https://en.wikipedia.org/wiki/Series_and_parallel_springs#:~:text=The following table gives formula for the spring

We know that an elastic wire obeys the equation F = kx where x is extension in wire due to an applied force F and k is like a spring constant for the elastic wire. So, we could look at this problem as two springs connected in parallel and use the formula mentioned in the table.

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vcsharp2003 said:
I think it may make sense, but only if one looks at it from the perspective as mentioned in a Wikipedia article. There is a table given in above article about how forces and elongations compare when two springs are in series and in parallel.

https://en.wikipedia.org/wiki/Series_and_parallel_springs#:~:text=The following table gives formula for the spring

View attachment 317107
View attachment 317108
We know that an elastic wire obeys the equation F = kx where x is extension in wire due to an applied force F and k is like a spring constant for the elastic wire. So, we could look at this problem as two springs connected in parallel and use the formula mentioned in the table.
But if they start with the same free length that must have different extensions, and the bar must be at an angle such that the torques balance, or they are not positioned symmetrically about the CM of the bar?

erobz said:
But if they start with the same free length that must have different extensions, and the bar must be at an angle such that the torques balance, or they are not positioned symmetrically about the CM of the bar?
I think it's reasonable to assume that extensions are equal, but their tensions are not. The block still remains horizontal since the distances of each tension from the center of mass of block are not equal.

vcsharp2003 said:
I think it's reasonable to assume that extensions are equal, but their tensions are not. The block still remains horizontal since the distances of each tension from the center of mass of block are not equal.
We are dealing with wires, not with springs.
The differential length for any wire should be imperceptible to the naked eye, and we will still see the horizontal mass in the same position.

Making both stresses equal simplifies the problem, keeping the strain of each wire as the variable.

Note that the solution also keeps both cross-section areas equal; hence, having equal tensions seems more reasonable to me.

The equivalent wire of double cross-section area will have double strees and average stretching.

vcsharp2003 and erobz
vcsharp2003 said:
I think it may make sense, but only if one looks at it from the perspective as mentioned in a Wikipedia article. There is a table given in above article about how forces and elongations compare when two springs are in series and in parallel.

https://en.wikipedia.org/wiki/Series_and_parallel_springs#:~:text=The following table gives formula for the spring

View attachment 317107
View attachment 317108
We know that an elastic wire obeys the equation F = kx where x is extension in wire due to an applied force F and k is like a spring constant for the elastic wire. So, we could look at this problem as two springs connected in parallel and use the formula mentioned in the table.
The parallel springs formula assumes the springs are adjacent, and on the same assumption works perfectly well for wires.
Lnewqban said:
Making both stresses equal simplifies the problem, keeping the strain of each wire as the variable.
No, equal strains is simpler.

The difficulty in the posed question is that the wires are clearly not adjacent.
If we assume the attachments are symmetric and the extensions are different, it gets a bit messy because of the angle of the bar and its thickness, but to a first approximation it shouldn’t change the answer much.
If we assume the attachments are asymmetric so that the extensions are the same, it seems to produce exactly the same answer as for adjacent wires.

vcsharp2003
Steve4Physics said:
For information, at the end of your Post #4 attachment (soln.pdf) you wrote $$y_3 = \frac {y_1 + y_2}{y_3}$$This may not be what you intended to write!
I wanted to write y3=(y1+y2)/2

vcsharp2003

## 1. What is Young's Modulus?

Young's Modulus, also known as the elastic modulus or modulus of elasticity, is a measure of the stiffness of a material. It describes how much a material will deform when subjected to a given amount of stress.

## 2. How is Young's Modulus determined?

Young's Modulus is typically determined through a tensile test, where a sample of the material is subjected to a known amount of stress and its resulting strain is measured. The ratio of stress to strain is then calculated to determine the Young's Modulus.

## 3. What is the unit of measurement for Young's Modulus?

The unit of measurement for Young's Modulus is typically expressed in Pascals (Pa) or megapascals (MPa). However, it can also be expressed in other units such as pounds per square inch (psi) or gigapascals (GPa).

## 4. How does temperature affect Young's Modulus?

Young's Modulus is affected by temperature, as changes in temperature can cause changes in the internal structure of a material. In general, as temperature increases, the modulus decreases, making the material more flexible. However, this relationship can vary depending on the material and its composition.

## 5. What are some common materials and their Young's Modulus?

The Young's Modulus of a material can vary greatly depending on its composition and structure. Some common values for Young's Modulus include 190-210 GPa for steel, 69 GPa for aluminum, and 2-4 GPa for rubber. Other materials, such as glass, wood, and plastics, can have a wide range of Young's Modulus values depending on their specific properties.

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