Forces on a rope when catching a free falling weight

[Zeusex]

Homework Statement

A rope 3mm dia is tied to a 10kg mass and other side is attached to a spool. At a length of 20m, spool starts rotating uncontrollably at about 2m/s initially. After about 2s, brake is applied. How much force will this rope absorb due to sudden braking?

Relevant Equations

F = 2mgh/s

If I am not wrong,
F = 2*10*9.81*4/0.2 = 3924N

(0.2m considering 5% stretch in 4m long rope)

 

[LastScattered1090]

@Zeusex This seems like a poorly-written problem with insufficient information to me. There's no mention of 5% stretch in the problem, yet you assume this in your solution. Are you sure this is the full text of the problem?

What is the significance of "after 20 m"? Rotating uncontrollably suggests that perhaps we are supposed to assume this mass is in freefall (which doesn't really make sense when attached to a spool). But even if assuming freefall, you can't just assume the height change is 4 m after 2 seconds, because you can't assume the velocity is constant at 2 m/s. Freefall means acceleration. So I don't really know what this problem is driving at.

Finally, in your equation, it seems that "W" is supposed to represent weight, because you clearly substitute in mg for it in your solution. But if that's the case, then why is there a factor of 2 in the equation? The energy needing to be absorbed should just be equal to the gravitational potential energy lost, which is mgh over the height of the fall. If you want to convert that into elastic potential energy, then it needs to be something like mgh = (1/2ks^2), but that doesn't work because we don't even know what k is here. I'd say go back to the teacher and clarify what is being asked.

 

[Zeusex]

@Zeusex This seems like a poorly-written problem with insufficient information to me. There's no mention of 5% stretch in the problem, yet you assume this in your solution. Are you sure this is the full text of the problem? What is the significance of "after 20 m"? Rotating uncontrollably suggests that perhaps we are supposed to assume this mass is in freefall (which doesn't really make sense when attached to a spool). But even if assuming freefall, you can't just assume the height change is 4 m after 2 seconds, because you can't assume the velocity is constant at 2 m/s. Freefall means acceleration. So I don't really know what this problem is driving at. Finally, in your equation, it seems that "W" is supposed to represent weight, because you clearly substitute in mg for it in your solution. But if that's the case, then why is there a factor of 2 in the equation? The energy needing to be absorbed should just be equal to the gravitational potential energy lost, which is mgh over the height of the fall. If you want to convert that into elastic potential energy, then it needs to be something like mgh = (1/2ks^2), but that doesn't work because we don't even know what k is here. I'd say go back to the teacher and clarify what is being asked.

This is not a question from an assignment. Yes there are many unknowns I understand and trying to gather as much as information possible. This exercise is to give me an estimate. To do so, currenlt I assumed 5% stretch. What I meant by uncontrolled rotation is could be due to some mechanical issues, such as failure of coupling connecting motor to the spool.
Factor of 2 is because of the work done by the rope = Fs/2 where F is the maximum force and s is the amount it stretches.
Considering acceleration, F = 2*10*9.81*23.6/1.18 = 3924N. does this look correct to you?


k can be calculated assuming 5% of static stretch (x). k = mg/(xh) = 83.13 N/m

 

[haruspex]

F = 2*10*9.81*23.6/1.18 = 3924N

Hard to follow your calculation when it's just numbers.
Please define variables to represent all the aspects and work algebraically.

 

[Zeusex]

Hard to follow your calculation when it's just numbers.
Please define variables to represent all the aspects and work algebraically.

Thanks for replying.
I tried to explain below:

If I ignore the mass of the rope and assume it stretches a small amount compared with the distance the mass falls, and assume the rope remains within its elastic limit, force should be found by calculating the work done to stretch the rope when it stops the mass.

Kinetic energy of the mass = Wh where W is the weight and h is the distance it falls
Work done by the rope = Fs/2 where F is the maximum force and s is the amount it stretches
so F = 2Wh/s

Considering 2 seconds of fall at an initial speed of 2 m/s:
h = v₀t + (1/2)gt² = 23.6 m
Fall velocity: vf = v₀ + gt = 21.6 m/s

Assumed static elongation is 5% therefore s should be 5% of h = 1.18m

Now, putting above values to equation
F = 2*9.81*10*23.6/1.18 = 3924N

 

[Zeusex]

But, at the moment of braking momentum is m*v = 10 kg*21.6 m/s = 216 kg-m/s.
Impulse equals the change in momentum, so the impulse would be 216 kg-m/s. Impulse is also equal to the force*time. Therefore change in momentum = force*time.
force = (216 kg-m/s)*time

If the mass comes to a stop in 1 second, the force would be 216 N.

I am confused.

 

[LastScattered1090]

Doesn't that suggest that, given your assumed stiffness/spring constant, the mass will come to a stop in much less than one second? Can you think of ways to apply the equations for a spring to figure out what the stopping time would be?

 

[haruspex]

Assumed static elongation is 5%

This seems backwards. For a given elastic constant, the elongation will depend on the force, which is what you are trying to determine. Seems more natural to pick an elastic constant and determine the maximum force from that.

change in momentum = force*time

= average force * time, Ft/2 here.

If the mass comes to a stop in 1 second

You can determine the time from your other assumptions,

 

[Zeusex]

Please explain.

You can determine the time from your other assumptions,

This seems backwards. For a given elastic constant, the elongation will depend on the force, which is what you are trying to determine. Seems more natural to pick an elastic constant and determine the maximum force from that.

I agree but there are two unknowns in the Hook's equation: force and elongation.

 

[Zeusex]

Doesn't that suggest that, given your assumed stiffness/spring constant, the mass will come to a stop in much less than one second? Can you think of ways to apply the equations for a spring to figure out what the stopping time would be?

Do you mean decay time for spring oscillation?

 

[haruspex]

Please explain.I agree but there are two unknowns in the Hook's equation: force and elongation.

Assume the elastic constant is k. The energy to be absorbed tells you the elongation (not forgetting the additional gravitational energy from the descent during the elongation), and that tells you the max force, F. But you cannot assume the average force is F/2. Use SHM equation to find the time taken.

 

[Zeusex]

Assume the elastic constant is k. The energy to be absorbed tells you the elongation (not forgetting the additional gravitational energy from the descent during the elongation), and that tells you the max force, F. But you cannot assume the average force is F/2. Use SHM equation to find the time taken.

Elastic constant for PE is around 4 GPa = 4*10^9 N/m²
Absorbed energy Es = 1/2kx^2 = 0.5*4x10^9*1.18 = 2.36E+09 joules? Thats huge. If I add mgh (gravitational potential energy, this would be higher)
Potential energy lost by mass due to fall = mgh = 10*9.81*23.6 = 2315.6 joules.
Does that mean the rope won't tear?
Fmax = Es/h = 1E+08 N?

 

[Zeusex]

Absorbed energy Es = 1/2kx^2 = 0.5*4x10^9*1.18 = 2.36E+09 joules? Thats huge. If I add mgh (gravitational potential energy, this would be higher)

units don't match.
Spring constant I calculated earlier
k can be calculated assuming 5% of static stretch (x). k = mg/(xh) = 83.13 N/m
gives Es = 57.8 joules.

 

[haruspex]

Thats huge

yes, because you insist on taking the stretch to be 5% despite the high k value.
Don't keep plugging in numbers, just work algebraically until you have all the equations.
Find the extension in terms of k, m, g and the drop height.

 

[erobz]

Apply Conservation of Energy to solve for the desired spring constant $k$ which assumes an elongation fraction $f$.

Then Apply Newtons Second Law to determine the position of the mass $x(t)$ with initial conditions $x(0) = 0, \dot x(0) = v_o$.

The tension in the rope/spring will be given by:

$$ |T(t)|= |k \, x(t)|$$

Or

If you are just interested in the maximum tensile maximum force developed:

$$ | T_{max}|= |k \, f x_o|$$

 

[Zeusex]

Apply Conservation of Energy to solve for the desired spring constant $k$ which assumes an elongation fraction $f$.

Then Apply Newtons Second Law to determine the position of the mass $x(t)$ with initial conditions $x(0) = 0, \dot x(0) = v_o$.

The tension in the rope/spring will be given by:

$$ |T(t)|= |k \, x(t)|$$

Or

If you are just interested in the maximum tensile maximum force developed:

$$ | T_{max}|= |k \, f x_o|$$

Reply from manufacturer:
Spring rate: 70 N is required for 0.7% elongation.
Youngs modulus: 50GPa

So, the initial elongation in 20m long rope due to 10kg mass is 1%.
10*9.81 N = - k*0.2 m
=> k = 490.5 N/m

I am interested in knowing if the rope would break at the sudden braking and if so, at which fall height.

 

[erobz]

Reply from manufacturer:
Spring rate: 70 N is required for 0.7% elongation.
Youngs modulus: 50GPa

So, the initial elongation in 20m long rope due to 10kg mass is 1%.
10*9.81 N = - k*0.2 m
=> k = 490.5 N/m

I am interested in knowing if the rope would break at the sudden braking and if so, at which fall height.

Lets just first check to see if the maximum force would break the rope? Is the Yield Strength of the rope quoted anywhere in the literature - should be in units $\rm{GPa}$?

We are assuming the mass is in freefall before braking (from what I'm reading). There isn't initial elongation with that simplifying assumption (we are ignoring the inertia of the drum ).

We are pretending the brake immediatly halts the drum, and the inital length of unwound rope/spring is $H$ (how far it fell), and the rope is a linear spring (this is likely not the case - ask the manufacturer about this if you are already in contact).

Let's restart here:

What is the kinetic energy (ignoring drag and the mass of the rope) in terms of $H$ just before the spring begins to apply a force?

 

[Zeusex]

Lets just first check to see if the maximum force would break the rope? Is the Yield Strength of the rope quoted anywhere in the literature - should be in units $\rm{GPa}$?

We are assuming the mass is in freefall before braking (from what I'm reading). There isn't initial elongation with that simplifying assumption (we are ignoring the inertia of the drum ).

We are pretending the brake immediatly halts the drum, and the inital length of unwound rope/spring is $H$ (how far it fell), and the rope is a linear spring (this is likely not the case - ask the manufacturer about this if you are already in contact).

Let's restart here:

What is the kinetic energy (ignoring drag and the mass of the rope) in terms of $H$ just before the spring begins to apply a force?

Considering 2 seconds of fall at an initial speed of 2 m/s:
H = v₀t + (1/2)gt² = 23.6 m
Fall velocity: vf = v₀ + gt = 21.6 m/s

KE = 1/2 mv^2 = 0.5*10*21.6^2 = 2333 joules.

Max allowed mass to be lifted by this rope is 100 kg. For now we can assume that the rope behaves like a linear spring.

 

[erobz]

Considering 2 seconds of fall at an initial speed of 2 m/s:
H = v₀t + (1/2)gt² = 23.6 m
Fall velocity: vf = v₀ + gt = 21.6 m/s

KE = 1/2 mv^2 = 0.5*10*21.6^2 = 2333 joules.

Max allowed mass to be lifted by this rope is 100 kg.

The kinetic energy of the mass just before the rope applies any force is simply $mgH$.

Now reset your PE datum to this location. Using Conservation of Energy ( and a linear spring rate assumption - perhaps a bad assumption ) what is the relationship for $E(0) = E(x)$ when the force is maximized ( assuming it does not break before this point ). In other words, express the conservation of energy from this point forward for the mass/spring system when the velocity of the mass at $x$ i.e $v(x)$ is $0$.

 

[Zeusex]

The kinetic energy of the mass just before the rope applies any force is simply $mgH$.

Now reset your PE datum to this location. Using Conservation of Energy ( and a linear spring rate assumption - perhaps a bad assumption ) what is the relationship for $E(0) = E(x)$ when the force is maximized ( assuming it does not break before this point ). In other words, express the conservation of energy from this point forward for the mass/spring system when the velocity of the mass at $x$ i.e $v(x)$ is $0$.

When v(x) = 0, 1/2 kx^2 = mgH

 

[erobz]

When v(x) = 0, 1/2 kx^2 = mgH

Almost. What about the potential energy of the mass?

 

[Zeusex]

Almost. What about the potential energy of the mass?

Because at this point, it is like "new equilibrium"?

 

[erobz]

Because at this point, it is like "new equilibrium"?

The mass lost some gravitational energy on its drop w.r.t the PE=0 datum. That energy must be absorbed by the spring.

 

[Zeusex]

The mass lost some gravitational energy on its drop w.r.t the PE=0 datum. That energy must be absorbed by the spring.

So x = sqrt(mgH/k)
and absorbed energy, Es = 1/2kx^2 joules

 

[erobz]

So x = sqrt(mgH/k)
and absorbed energy, Es = 1/2kx^2 joules

No, I think you are getting further away.

$$mgH = \frac{1}{2}k x^2 \pm PE(x)$$

I'm letting you figure out what the sign should be and the relationship.

 

[Zeusex]

No, I think you are getting further away.

$$mgH = \frac{1}{2}k x^2 \pm PE(x)$$

I'm letting you figure out what the sign should be and the relationship.

Should be + as it is gaining PE
So how do I find x and tension?

 

[erobz]

Should be + as it is gaining PE
So how do I find x and tension?

Are you sure it's gaining potential energy as its falling? Move the term over to the LHS. Does it look like the spring has absorbed the initial kinetic energy of $mgH$ and the "would be" kinetic energy from the mass falling a further distance $x$?

 

[Zeusex]

Are you sure it's gaining potential energy as its falling? Move the term over to the LHS. Does it look like the spring has absorbed the initial kinetic energy of $mgH$ and the "would be" kinetic energy from the mass falling a further distance $x$?

Yes you are right. I keep on thinking that the thread is still in tension although falling and applying tension which is why 'spring' tends to return back to equilibrium.

 

[erobz]

Yes you are right. I keep on thinking that the thread is still in tension although falling and applying tension which is why 'spring' tends to return back to equilibrium.

So whenever you have got the relationship, please share.

 

[Zeusex]

So whenever you have got the relationship, please share.

mgH = 1/2kx^2 - kx
=> kx^2-2kx-2mgH = 0.

 

[erobz]

mgH = 1/2kx^2 - kx
=> kx^2-2kx-mgH = 0.

No. Think gravitational potential energy. The missing term should look a whole lot like the LHS of the equation I posted.

 

[Zeusex]

No. Think gravitational potential energy. The missing term should look a whole lot like the LHS of the equation I posted.

2mgH = 1/2kx^2?
=> x = sqrt(4mgH/k)
F = |kx|?

 

[erobz]

2mgH = 1/2kx^2?
=> x = sqrt(4mgH/k)
F = |kx|?

No, not quite. That is a good approximation is the stretch $x$ is very small. But I would prefer you figured out what you are missing.

What is the gravitational potential energy of the mass at distance $x$ below the PE = 0 datum?

 

[Zeusex]

No, not quite. That is a good approximation is the stretch $x$ is very small. But I would prefer you figured out what you are missing.

What is the gravitational potential energy of the mass at distance $x$ below the PE = 0 datum?

mgx

 

[erobz]

mgx

and what is its sign ( positive or negative)?

 

[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.