- #36

- 33

- 3

Negativeerobz said:and what is its sign ( positive or negative)?

mgH + mgx = 1/2kx^2

=> kx^2-2mgx-2mgH = 0

from #18 post, H=23.6 m & from #16, k = 490.5 N/m (assuming linear spring)

490.5x^2 - 196.2x - 4630.32 = 0

x=3.27896

x=−2.87896

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the conversation discusses a problem involving a falling mass attached to a spool and the need to calculate the force needed to stop the mass. However, there is insufficient information provided in the problem and assumptions have to be made. The conversation also mentions the concept of elastic potential energy and the calculation of force using variables and algebraic equations. The final solution provides an estimate of the force needed to stop the falling mass.

- #36

- 33

- 3

Negativeerobz said:and what is its sign ( positive or negative)?

mgH + mgx = 1/2kx^2

=> kx^2-2mgx-2mgH = 0

from #18 post, H=23.6 m & from #16, k = 490.5 N/m (assuming linear spring)

490.5x^2 - 196.2x - 4630.32 = 0

x=3.27896

x=−2.87896

Physics news on Phys.org

- #37

Homework Helper

Gold Member

- 3,315

- 1,352

Ok. now slow down on the ##k## value...Zeusex said:Negative

mgH + mgx = 1/2kx^2

=> kx^2-2mgx-2mgH = 0

from #18 post, H=23.6 m & from #16, k = 490.5 N/m (assuming linear spring)

490.5x^2 - 196.2x - 4630.32 = 0

x=3.27896

x=−2.87896

The Bad news. I feel the manufacturer has not provided you with enough information to determine the ##k## value.

They have given you a force of ##70 \rm{N}## and a percent elongation ##0.7## %. They need to supply us with the length of the test specimen for this to be useful as far as I can tell.

- #38

- 33

- 3

Please check this.erobz said:Ok. now slow down on the ##k## value...

The Bad news. I feel the manufacturer has not provided you with enough information to determine the ##k## value.

They have given you a force of ##70 \rm{N}## and a percent elongation ##0.7 \percent ##. They need to supply us with the length of the test specimen for this to be useful as far as I can tell.

Can't I write k = mg/sH where s is 0.7%

```
Below is our answer.
Spring rate: 70 N is required for 0.7% elongation.
Youngs modulus: 50GPa
```

- #39

- 33

- 3

True but 0.7% of any length, no?erobz said:They have given you a force of ##70 \rm{N}## and a percent elongation ##0.7## %. They need to supply us with the length of the test specimen for this to be useful as far as I can tell.

- #40

- 33

- 3

In parallel, let's just assume k = 490.5 N/mZeusex said:True but 0.7% of any length, no?

F = |kx| = 490.5*3.28 = 1608.8N = 164 Kgf.

Max mass for this rope is 100kg so that means it would break.

If I take the other root, F = 143.5 Kgf.

So in both case rope is going to break.

Which root is relevant?

- #41

Homework Helper

Gold Member

- 3,315

- 1,352

to calculate ##k## we need the deflection ##\delta## under some load ##F = 70 \, \rm{N}##. We do not have the deflection. We have the strain ##\epsilon = 0.007## and the Modulus of Elasticity ##E = 50 \, \rm{GPa}##. Using stress- strainZeusex said:True but 0.7% of any length, no?

$$\sigma = E \epsilon \implies F = A E \epsilon \implies F = A E \frac{\delta}{L} \implies \delta = \frac{F}{AE}L $$The deflection ##\delta## the rope experiences under tensile load ##F## depends on the initial length of the rope ##L##.

Furthermore if:

$$ F = k \delta$$

Then we have that:

$$ \delta = \frac{F}{AE}L = \frac{k \delta }{AE}L \implies k = \frac{AE}{L} $$

The dependency on ##L## remains. We do not (yet) have enough information to determine ##k##.

Last edited:

- #42

- 33

- 3

Ok thanks. I will ask the test specimen length when they found spring rate.erobz said:to calculate ##k## we need the deflection ##\delta## under some load ##F##. We do not have the deflection, we are given the strain ##\epsilon##. We also have the Modulus of Elasticity ##E##. Using stress- strain

$$\sigma = E \epsilon \implies F = A E \epsilon \implies F = A E \frac{\delta}{L} \implies \delta = \frac{F}{AE}L $$The deflection ##\delta## the rope experiences under tensile load ##F## depends on the initial length of the rope ##L##.

Furthermore if:

$$ F = k \delta$$

Then we have that:

$$ \delta = \frac{F}{AE}L = \frac{k \delta }{AE}L \implies k = \frac{AE}{L} $$

- #43

- 33

- 3

Please comment.Zeusex said:In parallel, let's just assume k = 490.5 N/m

F = |kx| = 490.5*3.28 = 1608.8N = 164 Kgf.

Max mass for this rope is 100kg so that means it would break.

If I take the other root, F = 143.5 Kgf.

So in both case rope is going to break.

Which root is relevant?

- #44

Homework Helper

Gold Member

- 3,315

- 1,352

The negative root ( ##x## above PE = 0 ) is an artifact of us modeling this as anZeusex said:Please comment.

Last edited:

- #45

- 33

- 3

Thank you for your patience and help. I will share further information from manufacturer.erobz said:The negative root ( ##x## above PE = 0 ) is an artifact of us modeling this as anidealspring. The rope however does not care about our model of it over a particular range. It simply does not apply resistive force under compression... Its rope. It just moves out of the way on the way up after PE = 0 datum.

- #46

Homework Helper

Gold Member

- 3,315

- 1,352

Assume you have some spring with unstretched length ##l_o##. You pull it to some length ##l## and you measure the Force ##F##.

We have that:

$$ F = k \left( l - l_o \right)$$

Now, someone has given you that the percent elongation ##EL##% accompanying that force.

$$ EL \% = \frac{l -l_o}{l_o} \cdot 100 = 100 \epsilon $$

We have that:

$$l = l_o\left( 1 + \epsilon \right)$$

Plugging that back into ##F##:

$$ F = k \left( l - l_o \right) = k \left( l_o\left( 1 + \epsilon \right) - l_o \right) = k l_o \left( 1+\epsilon - 1\right) = k l_o \epsilon $$

We are one variable shy of solving for ##k## , namely ##l_o##.

- #47

- 33

- 3

Thanks. If manufacturer can't provide this data then I can build a small setup to measure the initial and stretched length using a known load.erobz said:

Assume you have some spring with unstretched length ##l_o##. You pull it to some length ##l## and you measure the Force ##F##.

We have that:

$$ F = k \left( l - l_o \right)$$

Now, someone has given you that the percent elongation ##EL##% accompanying that force.

$$ EL \% = \frac{l -l_o}{l_o} \cdot 100 = 100 \epsilon $$

We have that:

$$l = l_o\left( 1 + \epsilon \right)$$

Plugging that back into ##F##:

$$ F = k \left( l - l_o \right) = k \left( l_o\left( 1 + \epsilon \right) - l_o \right) = k l_o \left( 1+\epsilon - 1\right) = k l_o \epsilon $$

We are one variable shy of solving for ##k## , namely ##l_o##.

- #48

Homework Helper

Gold Member

- 3,315

- 1,352

Is the rope really ##3 \, \rm{mm}## diameter? That is tiny rope?Zeusex said:Thanks. If manufacturer can't provide this data then I can build a small setup to measure the initial and stretched length using a known load.

- #49

- 33

- 3

Yeserobz said:Is the rope really ##3 \, \rm{mm}## diameter? That is tiny rope?

- #50

- 33

- 3

Answer from manufacturer:Zeusex said:Yes

```
The tensile force is measured by pulling at a speed of 300 mm per minute at intervals of 250 mm and measuring the force at rupture.
We are sorry, but we do not disclose detailed data to consumers.
```

- #51

Homework Helper

Gold Member

- 3,315

- 1,352

Thats weird. I don't know what to tell you other than you might be stuck experimenting yourself to estimate ##k##.Zeusex said:Answer from manufacturer:

`The tensile force is measured by pulling at a speed of 300 mm per minute at intervals of 250 mm and measuring the force at rupture. We are sorry, but we do not disclose detailed data to consumers.`

- #52

- 33

- 3

Yes they weren't cooperating. I will update as soon as I get results from my experiments.erobz said:Thats weird. I don't know what to tell you other than you might be stuck experimenting yourself to estimate ##k##.

Thanks.

- #53

Homework Helper

Gold Member

- 3,315

- 1,352

Do you have an experimental setup in mind, and do you know the material?Zeusex said:Yes they weren't cooperating. I will update as soon as I get results from my experiments.

Thanks.

- #54

- 33

- 3

Yes, linear stage, linear encoder and a load cellerobz said:Do you have an experimental setup in mind, and do you know the material?

- #55

Homework Helper

Gold Member

- 3,315

- 1,352

If it's a plastic material just be mindful of "Creep". If you put some load on it and it keeps slowly elongating over time at that load the material is experiencing "Creep". From memory I think plastics are prone to it.Zeusex said:Yes, linear stage, linear encoder and a load cell

- #56

- 33

- 3

I am planning to first measure the length L0 when load cells shows 0 N. Then start moving the linear strange until the thread breaks. Let's say the thread breaks at 1 KN. I can then plot F vs L and would know strain: (L-L0)/L0 for the linear regime.erobz said:If it's a plastic material just be mindful of "Creep". If you put some load on it and it keeps slowly elongating over time at that load the material is experiencing "Creep". From memory I think plastics are prone to it.

E - young's modulus I know from manufacturer 50 GPa.

k = AE/L0, where A is the crossectional area of thread

- #57

Homework Helper

Gold Member

- 3,315

- 1,352

That can be used as a check, but you are measuring Force (with a load cell) and Length (with the linear encoder). The slope of that graph is the spring rate ##k## in the Hookean regime. Don't over complicate it with stress-strain. Put some length of rope under some small initial load. Measure the length of the rope. Then increase the load making successive length measurements at each load. Make a plot of Force vs length.Zeusex said:I am planning to first measure the length L0 when load cells shows 0 N. Then start moving the linear strange until the thread breaks. Let's say the thread breaks at 1 KN. I can then plot F vs L and would know strain: (L-L0)/L0 for the linear regime.

E - young's modulus I know from manufacturer 50 GPa.

k = E*A/L, where A is the crossectional area of thread

- #58

- 33

- 3

Sure. I will do that and share the plot. Thanks.erobz said:That can be used as a check, but you are measuring Force (with a load cell) and Length (with the linear encoder). The slope of that graph is the spring rate ##k## in the Hookean regime. Don't over complicate it. Put some length of rope under some small initial load. Measure the length of the rope. Then increase the load making successive length measurements at each load. Make a plot of Force vs length.

- #59

Homework Helper

Gold Member

- 3,315

- 1,352

Most importantly. If you are determined to test to failure (I think it is unnecessary because it's beyond the scope of the Hookean model we are proposing) it should go without saying, but do this safely with PPE, and shield between you and the rope.Zeusex said:Sure. I will do that and share the plot. Thanks.

Probably best to keep the rope reasonably short as well. You don't want to max out the encoder stretching it.

- #60

- 33

- 3

Sure, I will do that :)erobz said:Most importantly. If you are determined to test to failure (I think it is unnecessary because it's beyond the scope of the Hookean model we are proposing) it should go without saying, but do this safely with PPE, and shield between you and the rope.

Probably best to keep the rope reasonably short as well. You don't want to max out the encoder stretching it.

Share:

- Replies
- 7

- Views
- 315

- Replies
- 7

- Views
- 103

- Replies
- 11

- Views
- 341

- Replies
- 1

- Views
- 97

- Replies
- 3

- Views
- 383

- Replies
- 3

- Views
- 125

- Replies
- 9

- Views
- 822

- Replies
- 3

- Views
- 203

- Replies
- 21

- Views
- 616

- Replies
- 6

- Views
- 2K