Forces on a rope when catching a free falling weight

[Zeusex]

and what is its sign ( positive or negative)?

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

 

[erobz]

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

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 $0.7$ %. They need to supply us with the length of the test specimen for this to be useful as far as I can tell.

 

[Zeusex]

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.

Please check this.
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

 

[Zeusex]

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.

True but 0.7% of any length, no?

 

[Zeusex]

True but 0.7% of any length, no?

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?

 

[erobz]

True but 0.7% of any length, no?

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- 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:

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

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

 

[Zeusex]

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:

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

Ok thanks. I will ask the test specimen length when they found spring rate.

 

[Zeusex]

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?

Please comment.

 

[erobz]

Please comment.

The negative root ( $x$ above PE = 0 ) is an artifact of us modeling this as an ideal spring. 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 crossing the PE = 0 datum.

 

[Zeusex]

The negative root ( $x$ above PE = 0 ) is an artifact of us modeling this as an ideal spring. 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.

Thank you for your patience and help. I will share further information from manufacturer.

 

[erobz]

Another way to show this:

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 $l_o$.

 

[Zeusex]

Another way to show this:

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 $l_o$.

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]

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.

Is the rope really $3 \, \rm{mm}$ diameter? That is tiny rope?

 

[Zeusex]

Is the rope really $3 \, \rm{mm}$ diameter? That is tiny rope?

Yes

 

[Zeusex]

Yes

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.

 

[erobz]

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.

Thats weird. I don't know what to tell you other than you might be stuck experimenting yourself to estimate $k$.

 

[Zeusex]

Thats weird. I don't know what to tell you other than you might be stuck experimenting yourself to estimate $k$.

Yes they weren't cooperating. I will update as soon as I get results from my experiments.
Thanks.

 

[erobz]

Yes they weren't cooperating. I will update as soon as I get results from my experiments.
Thanks.

Do you have an experimental setup in mind, and do you know the material?

 

[Zeusex]

Do you have an experimental setup in mind, and do you know the material?

Yes, linear stage, linear encoder and a load cell

 

[erobz]

Yes, linear stage, linear encoder and a load cell

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]

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.

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

 

[erobz]

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

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]

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.

Sure. I will do that and share the plot. Thanks.

 

[erobz]

Sure. I will do that and share the plot. Thanks.

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.

 

[Zeusex]

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.

Sure, I will do that :)