Determine equivalent Young's Modulus

[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.

 

[haruspex]

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]

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.

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.

 

[Curiosity_0]

Is option b correct?

 

[vcsharp2003]

Is option b correct?

Yes

 

[vcsharp2003]

Is option b correct?

I need to go through it and understand.

 

[haruspex]

My guess is that we assume equal extensions of each wire

but that is not possible because you already deduced:

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?

 

[vcsharp2003]

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.

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.

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

 

[Chestermiller]

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.

 

[Steve4Physics]

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!

 

[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.

 

[Steve4Physics]

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]

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.

 

[erobz]

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?

 

[vcsharp2003]

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.

 

[Lnewqban]

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.

 

[haruspex]

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.

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.

 

[Curiosity_0]

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