A ship stops suddenly - Forces acting on the cargo?

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  • #1
praame
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A ship is traveling at constant speed, i.e. the engine exerts a force equal to the drag force acting on the ship (right?). A cargo box is standing on the deck of the ship and is secured with a rope to the mast (see pic). When the ship kills the engine, the drag will create a sudden change in velocity, which (given that the ship's initial velocity is high enough) causes the box to continue moving relative to the ship and there will be a tug on the rope. How do I calculate the force acting on the rope? I figure that I can't use the same calculation as with a collision, since the deceleration is created by the ship's drag, which means that the deceleration won't be uniform. I already know the drag of the ship, it's speed and the mass of the box.
 

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  • #2
jbriggs444
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the drag will create a sudden change in velocity
Can you be more quantitative than that? How rapidly will the ships velocity be changing? What things would you want to know to calculate the rate of change? Is there a word for the rate of change of velocity over time?
 
  • #3
scottdave
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What is the formula for drag? That should give you a clue how to proceed. Also, the velocity doesn't "suddenly" change. Before, the ship had constant speed (no acceleration), now it is decelerating.
 
  • #4
anorlunda
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I assume you're looking for the maximum force exerted by the rope, not the force as a function of time. If that's so, then the maximum deceleration should come at the first instant after engine stop. I also assume that the sea conditions are not heavy.

The ship's drag through the water at a given speed should be proportional to engine power at that speed. Do you have that information, power versus speed?
 
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  • #5
praame
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the velocity doesn't "suddenly" change
Well, if it is a vessel with high drag (say a barge, rather than a kayak), then when the engine stops, there will usually be an initial jerk (that sends the cargo flying), after which the deceleration smooths out, right?

Can you be more quantitative than that?
I guess that's what I am struggling with here. I can't figure out how to calculate the rate of change, which I fell is probably necessary to find the force on the rope. But since I know the drag at any given velocity, it should be possible to figure it out, I figure.

What is the formula for drag?
I have used the following formula to calculate drag: F = 0.5 * C * ρ * A * V2
F: Drag force
C: Drag coefficient (dependent on the shape and proportion of the hull https://en.wikipedia.org/wiki/Drag_coefficient)
ρ: Density of the fluid
A: Cross-sectional area of the body traveling through the fluid
V: Velocity

I assume you're looking for the maximum force exerted by the rope
Exactly.

I also assume that the sea conditions are not heavy
Yes, let's assume an ideal, stationary body of water.

The ship's drag through the water at a given speed should be proportional to engine power at that speed. Do you have that information, power versus speed?

Well, let's make some assumptions (and do let me know if I make a mistake here);
While the ship is traveling at constant speed, the drag force is equal to the force exerted by the engine.
Let's assume that the hull is a streamlined half-body, which gives us C=0.09.
ρ is 1.000 kg/m3 for water.
Let's say that the cross-section of the submerged part is 5m2.
Finally, let's say that the ship is traveling at 15 m/s.

So the drag Force is 0,5 * 0,09 * 1.000 * 5 * 152 = 50.625 N

Hence, while the ship is traveling at constant speed, the engine is exerting 50.625 N to overcome the drag. If the engine cuts, there will be 50.625 N of force acting against the ship, forcing it to decelerate. BUT, this is just in the first instance, since the change in velocity leads to a change in drag force.
 
  • #6
anorlunda
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BUT, this is just in the first instance, since the change in velocity leads to a change in drag force.
The first instance is the same time as maximum force.

Do you have the total displacement of ship+cargo?
 
  • #7
anorlunda
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Well, if it is a vessel with high drag (say a barge, rather than a kayak), then when the engine stops, there will usually be an initial jerk (that sends the cargo flying), after which the deceleration smooths out, right?
I should have mentioned before, if the line holding the cargo is not taut and has slack, then it will produce a jerk as it pulls taut. But there will be no jerk in the ship's speed, it should start decelerating right away with max acceleration in the first instant. (forget jerk defined as rate of change of acceleration, that's not what is relevant here.)
 
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  • #8
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if the line holding the cargo is not taught
And thus remains ignorant... :oldbiggrin:
 
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  • #9
anorlunda
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And thus remains ignorant... :oldbiggrin:
Ah, the spell checker did me in again. I changed it, thanks.

The spell checker got me again this afternoon. I typed "toastmaster" and it wanted to change it to something so pornographic that I can't repeat it here. I'm sure we all have a love-hate relationship with spell checkers.
 
  • #10
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I typed "toastmaster" and it wanted to change it to something so pornographic that I can't repeat it here.
The mind boggles at trying to come up with what the spellchecker might have come up with...
 
  • #11
praame
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The first instance is the same time as maximum force.

Do you have the total displacement of ship+cargo?
No, I don’t have that I’m afraid. Is it relevant, though? Shouldn’t the drag force be sufficient information?
 
  • #12
praame
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I should have mentioned before, if the line holding the cargo is not taut and has slack, then it will produce a jerk as it pulls taut. But there will be no jerk in the ship's speed, it should start decelerating right away with max acceleration in the first instant. (forget jerk defined as rate of change of acceleration, that's not what is relevant here.)
OK, I’ll buy that. Having thought about this some more, I am wondering about the inertia of the cargo. Could it be like this;
We assume that the cargo is standing on a frictionless surface. For the rope to become taut, the drag force on the ship when it stops, must be greater than the inertial force of the cargo. So, if the drag force was infinite, the ship would stop instantly and the force on the rope could be calculated if I knew its elasticity (which would create a uniform deceleration of the cargo and give me the stopping distance/speed of the cargo relative to the now stationary mast). However, since the drag force is not infinite, there will be added stopping distance as the ship slows, which lessens the force in the rope. But I don’t know the ships rate of deceleration, so I’m back to square one.
 
  • #13
jbriggs444
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OK, I’ll buy that. Having thought about this some more, I am wondering about the inertia of the cargo. Could it be like this;
We assume that the cargo is standing on a frictionless surface. For the rope to become taut, the drag force on the ship when it stops, must be greater than the inertial force of the cargo.
You should start thinking more like a physicist. The two forces you are talking about are not comparable.

Let us begin with Newton's second law applied to the ship, ##F=ma##. Here ##F## is the total fore and aft force on the ship. The overwhelming bulk of this is the drag force. ##m## is the ships mass. ##a## is the ships fore and aft acceleration.

The ship is also subject to forward force from the partially restrained cargo. But for the sake of simplicity, let us ignore that.

You speak about the "inertial force of the cargo". This would presumably mean the fictitious force that appears to move the cargo forward relative to the rearward-accelerating ship. This force is given by ##F=ma##, but this time ##F## is the fictitious forward force on the cargo. ##m## is the mass of the cargo and ##a## is the ship's actual rearward acceleration.

You spoke of comparing the drag force on the ship with the inertial force on the cargo. As you can see above, the two are not the same. The accelerations are the same but the masses are vastly different. So the forces cannot be the same.

You want to compare the fictitious forward force on the cargo with the rearward force of friction and of partially taut straps. You have to figure out how the cargo shifts as the ship decelerates.

If this is a real world problem, you should not be calculating this stuff from first principles. You should be taking measurements. If this is a toy problem, you need to come up with a mathematical model for how the cargo box shifts. For instance, how much slack is there in the rope initially? What is the coefficient of static friction of cargo box on deck? What is the coefficient of kinetic friction of cargo box on deck? How elastic are the ropes? Can we assume that they obey Hooke's law when going from slack to its breaking strength? Can we assume that they all become taut and reach their breaking strength at once? Can the contents of the cargo box shift relative to the box?

Personally, I would simplify with...

Coefficient of static friction is irrelevant. Assume that the cargo box actually moves. Otherwise the problem does not exist in the first place.

Take the coefficient of kinetic friction as zero something small enough that the cargo box actually moves on the deck. For a typical ships deceleration under drag, this will be something pretty small, even unrealistically small. If this is a real problem, this fact is a hint that our toy mathematical model is inadequate. Hence, it is time for real measurements rather than armchair assumptions. Call the coefficient of friction ##\mu## and throw it into the model.

Assume that the ropes all become taut together and have slack range ##s## before they begin resisting. Assume that Hooke's law applies and let the rope's elasticity be given by a spring constant ##k##.

Assume that the box and contents are effectively a single rigid and non-rotating body.

Can you write down an equation for the acceleration of the cargo box relative to the ship prior to the ropes becoming taut?
 
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  • #14
praame
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Personally, I would simplify with...
I would actually prefer to simplify further. The box is rigid and unifrom (no space or contents inside), there is no friction whatsoever and also, assume that the rope is an ideal rope which is already taut and has zero elasticity. It could just as will be a rigid connection between the box and the mast.

The elasticity of the rope (or some other part that deforms in the model) would have been crucial to calculate the force in a collision. But here, there is no firm stop and uniform deceleration (like there is in a collision) but rather a slowing down, due to drag. Which, by the way, maybe never reaches zero?

It is indeed a toy problem (although I am hoping it will help me understand another, real world problem).
 
  • #16
jbriggs444
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I would actually prefer to simplify further. The box is rigid and unifrom (no space or contents inside), there is no friction whatsoever and also, assume that the rope is an ideal rope which is already taut and has zero elasticity. It could just as will be a rigid connection between the box and the mast.

The elasticity of the rope (or some other part that deforms in the model) would have been crucial to calculate the force in a collision. But here, there is no firm stop and uniform deceleration (like there is in a collision) but rather a slowing down, due to drag. Which, by the way, maybe never reaches zero?

It is indeed a toy problem (although I am hoping it will help me understand another, real world problem).
Very well. That particular problem in various guises is a staple of first year physics courses.

We have two blocks sitting atop one another on a frictionless table. The upper block has mass m. The lower block has mass M. A massless rope is tied to the lower block and pulls rightward with force F.

Assume that the force of static friction is great enough that the two blocks move as one.

a. What is the acceleration of the lower block?
b. What is the acceleration of the upper block?
c. What is the magnitude of the horizontal force of the lower block on the upper block? Does it act rightward or leftward?


1586950491091.png
 
  • #17
jbriggs444
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I tried making sense of the information here in the hopes that it would help me, but I couldn't wrap my head around it: https://physics.stackexchange.com/q...lculate-the-distance-a-ship-will-take-to-stop
This problem has more meat to it. The solutions offered up involved differential equations and an assumption of quadratic drag.

Can you help us out. Is your problem following the differential equations? Or did phrases like "kinematic viscosity (momentum diffusion constant)" put you off?

[That phrase also puts me off, but I recognize that the writer is talking about how one would distinguish between a regime where quadratic drag is a good model and a regime where linear drag is a better model. As long as we are confident in the drag formula, we can ignore the fancy words and proceed]
 
  • #18
anorlunda
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Can you help us out. Is your problem following the differential equations?
In #5, he confirmed that he is looking only for a single number, maximum force, rather than force as a function of time.



@praame , You seem determined to reason this out with intuition. But engineers work with theory and numbers. The applicable theory in this case is Newton's Second Law F=ma. You say you don't know the mass m (called displacement for a ship). We are saying that you are doomed to failure until you learn the math, and then find the critical data (m) that are key to the question.

So if you are ready to learn how the physics really works, try starting here:
https://www.khanacademy.org/science...second-law-ap/v/Newton-s-second-law-of-motion
 
  • #19
praame
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Very well. That particular problem in various guises is a staple of first year physics courses.

We have two blocks sitting atop one another on a frictionless table. The upper block has mass m. The lower block has mass M. A massless rope is tied to the lower block and pulls rightward with force F.

Assume that the force of static friction is great enough that the two blocks move as one.

a. What is the acceleration of the lower block?
b. What is the acceleration of the upper block?
c. What is the magnitude of the horizontal force of the lower block on the upper block? Does it act rightward or leftward?


View attachment 260685

a. a = F/(M+m)?
b. a = F/(M+m)?
c. m * a? Rightward?

Is this correct?

Let's modify this example to get closer to my initial question:
1586962821905.png


Instead of a sitting on frictionless table, the blocks are now floating (as in, being buoyant) in a fluid. Instead of a massless rope, the blocks are now being propelled by an engine through force Fp. The drag of the fluid is creating a force, Fd, resisting the rightward motion of the blocks. Assume that the blocks are moving with a greater than 0 velocity (v) that is constant, which means that Fp and Fd are equal.

Now, Fp becomes 0 (loss of propulsion), leading to a deceleration of the blocks caused by the drag force, Fd.

a. What is the deceleration of the lower block?
b. What is the deceleration of the upper block?
c. What is the magnitude of the horizontal force of the lower block on the upper block? Does it act rightward or leftward?
 
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  • #20
jbriggs444
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a. a = F/(M+m)?
b. a = F/(M+m)?
c. m * a? Rightward?
Yes, indeed.

Now if we have the ship (##M+m##) going left, drag (##F_d##) pulling right and we turn off the engines, the answers are still as above.

I am not sure what the cargo box of mass m is doing to make the problem more understandable.
 
  • #21
praame
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You seem determined to reason this out with intuition
On the contrary. I want to find the correct math to get away from intuition and assumptions. Since I am no physicist or physics student, but just a layman, I am turning to you guys for help with this.

The applicable theory in this case is Newton's Second Law F=ma
I don't see how that applies here. If the acceleration would be constant (and the velocity thus changing linearly), then yes, it would be this straightforward. However, the formula for the drag force contains velocity (squared), which ought to lead to acceleration being non-constant and the drag force (and thereby the forces acting on the cargo of the ship) changing in a non-linear way.

You say you don't know the mass m (called displacement for a ship)
Sorry, I confused displacement with volume. You are right that it is mass, so then yes. I know the mass.
 
  • #22
jbriggs444
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However, the formula for the drag force contains velocity (squared), which ought to lead to acceleration being non-constant and the drag force (and thereby the forces acting on the cargo of the ship) changing in a non-linear way.
Newton's second law is strictly and exactly true at all times and in all circumstances. ##F=ma## always. [In relativity we would rephrase it slightly]

If the force changes then the acceleration changes accordingly. Yes, this means that the acceleration will change in a non-linear way over time (for instance, perhaps, as a decaying exponential) but it will still change in a way that is directly proportional to force.

It would help if you had some background with calculus. Do you know anything about "derivatives", "integrals" or "differential equations"?
 
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  • #23
gmax137
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I don't see how that applies here. If the acceleration would be constant (and the velocity thus changing linearly), then yes, it would be this straightforward.
Newton's second, F=Ma, does not mean acceleration "a" is constant. It can be a function of time (or position). But that doesn't really matter in this case, as @anorlunda mentioned, because you said you were interested in the maximum force on the cargo. And that will occur at the time the engine is cut (because the velocity is highest then, which means the drag force is also highest).

Sorry, I confused displacement with volume. You are right that it is mass, so then yes. I know the mass.
For something floating (like a ship), "displacement," volume, and mass are all related because the volume below the waterline (the "displaced" water) weighs the same as the entire ship. But that's another subject...
 
  • #24
praame
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Newton's second law is strictly and exactly true at all times and in all circumstances. ##F=ma## always. [In relativity we would rephrase it slightly]

If the force changes then the acceleration changes accordingly. Yes, this means that the acceleration will change in a non-linear way over time (for instance, perhaps, as a decaying exponential) but it will still change in a way that is directly proportional to force.

It would help if you had some background with calculus. Do you know anything about "derivatives", "integrals" or "differential equations"?

I most definitely believe that either speed or acceleration (I can’t wrap my head around which one of them ATM) will change as a decaying exponential and I believe that the initial vector of that is key to finding the force I am looking for.

Yes, I am familiar with derivatives, integrals and diff equations.
 
  • #25
praame
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you said you were interested in the maximum force on the cargo. And that will occur at the time the engine is cut (because the velocity is highest then, which means the drag force is also highest).

Agreed. Do you know how to calculate it, though?
 
  • #26
gmax137
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Well, up above you determined the drag force at time zero (when the engine is cut) as 50 thousand Newtons. If you know the mass of the ship, this allows you to determine the acceleration at that time.

If the rope is taut, the cargo box will have that same acceleration. So if you know the box mass, you can determine the force on the box. If the box is frictionless on the deck, the only force on it (in the horizontal direction) is due to the rope.
 
  • #27
jbriggs444
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I most definitely believe that either speed or acceleration (I can’t wrap my head around which one of them ATM) will change as a decaying exponential and I believe that the initial vector of that is key to finding the force I am looking for.
Since acceleration is proportional to drag which is proportional to [the square of] speed, both will be decaying. Without solving the differential equation it is not immediately clear whether either, neither or both will be decaying exponentials. Both should definitely have similar decaying behavior though.

From https://physics.stackexchange.com/q...lculate-the-distance-a-ship-will-take-to-stop we are given the solution and told that the velocity will decay to be inversely proportional to elapsed time.

This seems eminently reasonable since ##\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}##

The referenced article does not make it clear that after the transition from quadratic to linear drag one is going from ##\frac{dv}{dt}=-k_qv^2## to ##\frac{dv}{dt}=-k_lv## where ##k_q## is the constant associated with the quadratic drag regime and ##k_l## is the constant associated with the linear drag regime.

The truth is that drag is never truly proportional to the square of velocity. Nor is it ever truly directly proportional to velocity. Rather, it is (well approximated as) a sum of the two. i.e. drag = ##k_qv^2 + k_lv##. For high speeds, the quadratic term dominates. For low speeds, the ##v^2## term gets smaller faster than the ##v## term and the linear term dominates. To make for a differential equation that can be solved analytically, it is convenient to ignore the non-dominant term. Note that I am spit-balling here outside my area of expertise. I am not a fluid dynamics guy.

In the quadratic drag regime the distance traveled is nominally infinite. But the time taken gets exorbitant and the velocities get very low. Clearly one slows down and transitions to the linear regime before going exceptionally far.

In the linear drag regime the new differential equation results in exponential decay and a finite distance traveled even over infinite elapsed time.
Yes, I am familiar with derivatives, integrals and diff equations.
Then you should be able to write the differential equation: ##\frac{dv}{dt} = -kv^2## for some positive constant k derived from the coefficient of drag, the cross-sectional area of the ship and the mass of the ship.
 
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  • #28
gmax137
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But the time taken gets exorbitant and the velocities get very low.
This reminded me, when my buddy was teaching me how to handle small boats, his advice was "never approach a dock faster than you're willing to hit it." :smile:
 
  • #29
praame
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I am starting to think that taking the rate of deceleration into account and all that, is complicating things unnecessarily and that the F=ma approach might be valid and a close enough approximation for my application.

The fact that I am only interested in Fmax, which occurs at V0, which in turn is the only instance at which I know the acceleration, makes F=ma applicable I guess.

Would I have needed to know the force acting on the cargo over time, I would have had to dive into jbriggs444's equation and plot the force(s) resulting from this. @jbriggs444 now that you have dived into the world of quadratic vs linear drag and all that - do you agree that the F=ma approach is sufficient for me to calculate the force within an error margin of, say, a half order of magnitude?

Thanks for all the help guys!
 

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