Coefficient of friction in sea water

[daraphy]

TL;DR

What is the average coefficient of friction in sea water?

Is there any friction coefficient in the sea as there is in ground? I've tried searching but I see results between 0.4 and 25.7 (?). If there's no such thing as a coefficient of friction to water, how do I calculate the amount of force to start moving a X mass object in water?

 

[Orodruin]

Water is a fluid. It has viscosity, not a friction coefficient.

 

[phinds]

how something moves in water (sea or fresh) depends on its shape, mass, direction, and buoyancy as much as the force driving it. How it displaces water is the equivalent of "friction" and as @Orodruin just said, that's called viscosity.

Salt water is just slightly more buoyant and slightly more viscous than freshwater.

 

[erobz]

TL;DR Summary: What is the average coefficient of friction in sea water?

Is there any friction coefficient in the sea as there is in ground? I've tried searching but I see results between 0.4 and 25.7 (?). If there's no such thing as a coefficient of friction to water, how do I calculate the amount of force to start moving a X mass object in water?

When a body is moving through a fluid, the property of the fluid know as its viscosity and shape of the body geometry combine together in a very non-linear way to make a "drag force". It's not a trivial problem to tackle. For a dimensionless measure called the "Reynolds Number" less than 2300 ( laminar flow characteristics), the drag force is usually proportional to the velocity $v$. For greater than 2300 its usually proportional to the square of velocity $v^2$. I'm not certain about the linearity of the proportionality in laminar flow, but I know that in turbulent flow the proportionality is non-linear.

You can assume a basic model where the drag coefficient $C_d$ is constant like:

$$F_d = \frac{1}{2} \rho A C_d v^2$$

Where $F_d$ is the drag force
$\rho$ is the fluid density
$A$ is the projected area of the body

but it's not likely to hold in general, because $C_d$ is a function of the Reynolds number, which is a function of fluid viscosity, flow/body geometry, and the flow velocity itself.

 

[erobz]

how do I calculate the amount of force to start moving a X mass object in water?

Also...its not exactly clear what you mean by this. To start anything moving you must accelerate it. I suppose at the instant you start to move something the force required depends on how much you try to accelerate it and what mass of fluid/solid body is accelerating at (also other forces acting on it). I believe "drag" is zero the instant you try to accelerate it.

If you are talking about how the velocity varies as you apply some known force to a body in a fluid(which I think is more likely what you intend), then you are in the realm of drag force.

There are professionals around so I'm prepared to be beaten mercilessly( if need be) about making this point...

 

[daraphy]

Also...its not exactly clear what you mean by this. To start anything moving you must accelerate it. I suppose at the instant you start to move something the force required depends on how much you try to accelerate it and what mass of fluid/solid body is accelerating at (also other forces acting on it). I believe "drag" is zero the instant you try to accelerate it.

If you are talking about how the velocity varies as you apply some known force to a body in a fluid(which I think is more likely what you intend), then you are in the realm of drag force.

There are professionals around so I'm prepared to be beaten mercilessly( if need be) about making this point...

What I know is that in ground, when trying to move an object, another force appears that you have to beat in order to move the object, so for example you'll need more than 98N to start moving a 10kg object in a frictionless environment, I think im right. So I would like to know how that scenario applies in sea water. For example the minimum force needed to move a 100kg boat without a lot of technical things, if that's how it works... Anyway first time I see the drag force thing, I'll investigate and see if that's what I'm looking for

 

[erobz]

What I know is that in ground, when trying to move an object, another force appears that you have to beat in order to move the object,

Thats sounds right, in reality, friction is present.

so for example you'll need more than 98N to start moving a 10kg object in a frictionless environment, I think im right.

No, in a frictionless environment without any other forces acting on the body no force is required to move the block. It could be the empire state building, if its sitting on a frictionless surface, and I'm not... Then I can move it. I won't be able to accelerate it much (change its velocity at a great rate )because of its large mass and my limited physiology, but none the less a non-zero net force means (center of mass) motion. This is by Newtons Second Law.

So I would like to know how that scenario applies in sea water. For example the minimum force needed to move a 100kg boat without a lot of technical things, if that's how it works...

You're not going to find something that general.

 

[hutchphd]

What I know is that in ground, when trying to move an object, another force appears that you have to beat in order to move the object, so for example you'll need more than 98N to start moving a 10kg object in a frictionless environment,

You need to define "frictionless environment".
The general issue of "instantaneous rates of change" is at the conceptual heart of calculus (Dr. Newton didn't talk of fluxions and fluents for nothing) The "rate of change of the rate of change of position " can be exactly mg in free fall.
In the real world there is something called "stiction" which is the tendency for objects to develop a static adhesion. This is a real problem in vacuo, where surfaces remain pristine.

 

[Dullard]

You're describing 'static friction.' There is no fluid equivalent.

 

[daraphy]

Thats sounds right, in reality, friction is present.

No, in a frictionless environment without any other forces acting on the body no force is required to move the block. It could be the empire state building, if its sitting on a frictionless surface, and I'm not... Then I can move it. I won't be able to accelerate it much (change its velocity at a great rate )because of its large mass and my limited physiology, but none the less a non-zero net force means (center of mass) motion. This is by Newtons Second Law.


You're not going to find something that general.

So if I want to know the amount of force a motor has to do in order to move a boat, there's no easy to calc answer?

 

[hutchphd]

What does "move a boat" mean? It takes no force to maintain the existing state of motion. The details depend upon, well, the details.....

 

[daraphy]

What does "move a boat" mean? It takes no force to maintain the existing state of motion. The details depend upon, well, the details.....

Make a rectangular shape boat move at, for example, 1 m/s constant speed

 

[daraphy]

Make a rectangular shape boat move at, for example, 1 m/s constant speed

straight forward

 

[erobz]

straight forward

There’s is drag on the hull from
The water and from the air. There is also drag on the propeller. It’s complex to be anywhere near precise.

 

[daraphy]

There’s is drag on the hull from
The water and from the air. There is also drag on the propeller. It’s complex to be anywhere near precise.

Then I'll just test until it moves

 

[erobz]

Then I'll just test until it moves

A light boat will move with ease from rest in the water. That’s different from pushing it at 1m/s steadily. There are many variables to specify.

 

[Orodruin]

So I would like to know how that scenario applies in sea water

Simply put: It does not.

 

[Baluncore]

There is no fluid equivalent to static friction, there is only surface tension, that holds very small things in place.

The fluid equivalent, to sliding friction of a solid block on a solid surface, is the fluid drag on the wetted surface area of the hull.

For solids, the work done to keep a friction surface sliding, is proportional to velocity, force and the friction coefficient between the solids.

The work done to overcome surface drag, of the wetted area of a solid hull and a fluid, is proportional to velocity squared, wetted area, and a surface drag coefficient.

The wave making properties of a moving hull, is an additional complexity, that is determined by hull profile.

 

[Arjan82]

A fluid like water is defined by its inability to sustain tangential stresses (i.e. frictional forces), or rather, a fluid will deform under even the slightest tangential stress. Thus in principle, when there is no wind or current, you can propel a boat forward with a gram of force, or even a milligram, or microgram... you get the point. What will change is the speed forward. A gram of force will not give you a very high velocity.

For ships however, friction is only one of the resistive forces. Another important one is the wave resistance. How important that is depends on the velocity and hull form (length of the hull is very important). Air resistance generally is not so important.

One way to compute the ship speed for a given power is the Holtrop-mennen method. It contains a lot of formulas. But here there is some online tool (seems ok, haven't tried it myself)

 

[sophiecentaur]

TL;DR Summary: What is the average coefficient of friction in sea water?

If there's no such thing as a coefficient of friction to water, how do I calculate the amount of force to start moving a X mass object in water?

A light boat will move with ease from rest in the water.

And so will a massive boat (of many tons). At near-zero speeds, things follow Newton's Second Law of Motion (F=Ma) but in real life there are always other (man size) forces around, such a water flow, wind and even things like surface tension.
I think the OP needs to come up with a realistic, practical problem to solve, rather than a theoretical, idealised one. Then there will be a solution (with a whole list of possible errors - as in real life).

 

[erobz]

And so will a massive boat (of many tons). At near-zero speeds, things follow Newton's Second Law of Motion (F=Ma).

I think you know this, but when you say "Newtons Second Law" with (F = Ma) in parentheses as If I'm a complete noob, it's kind of insulting...

I get it... I'm not as intelligent or educated as you...must you go out of your way to remind me when you address me (or my content)?

In order to change its velocity, you must accelerate the mass of the vessel and accelerate some mass of fluid in contact with the hull, the more in contact... the more that "effective mass" is. Per unit force applied the light "bass boat", accelerates easier than the aircraft carrier. Even in a perfectly ideal world without any other external forces on each vessel (even inviscid fluid), because it the carrier has significantly has greater mass. We measure "effort" (the ease) to do something in terms of force (or work) it would take to do something.

I think the OP needs to come up with a realistic, practical problem to solve, rather than a theoretical, idealised one. Then there will be a solution (with a whole list of possible errors - as in real life).

What makes you think I believe otherwise?

 

[sophiecentaur]

as If I'm a complete noob, it's kind of insulting

That's a disappointing response. Did you consider the actual 'noobs' who may be reading this thread? In your opening question you actually give the impression that you are not specifically aware of F=Ma because there is no definition of "starting moving"; the acceleration is a relevant quantity.

What makes you think I believe otherwise?

So, do you have a practical problem and a required solution? Ask away.

 

[erobz]

It’s not insulting if one IS a “noob” ( new beginner). I’m no doctor, but I’m not completely uninformed.

In your opening question you actually give the impression that you are not specifically aware of F=Ma because there is no definition of "starting moving".

What opening question? I’m not the OP…

 

[Orodruin]

In fact, OP has not been seen for more than two weeks ... This was light-necroed by post #19.

 

[sophiecentaur]

What opening question? I’m not the OP…

Owch. Sorry about that. But how did you imagine my F=Ma was aimed only at you? It was a general comment, hoping to add clarification.

in contact with the hull, the more in contact... the more that "effective mass" is

You have turned the problem into a vast, unsolvable one here. Obvs, with an enormous scoop at the front of the hull, the effective mass would be greater. The original discussion was surely about an idealised boat - same as an idealised block being dragged across the ground. The initial effect when pushing a real boat forward is dominated by the mass of the boat. Any change of lateral momentum of water at the(pointed) bow would be equal and opposite to the net change of momentum of the water coming together at the stern. After that,as speed becomes relevant the fluid behaviour of the water becomes relevant. Is that ok with you?

 

[Orodruin]

But how did you imagine my F=Ma was aimed only at you? It was a general comment, hoping to add clarification.

I would assume it is difficult not to read it like an answer to something @erobz said when you quote their post like that …

Not imposing possibly unwanted meaning in a forum post is an art form in itself.

 

[erobz]

You have turned the problem into a vast, unsolvable one here. Obvs, with an enormous scoop at the front of the hull, the effective mass would be greater. The original discussion was surely about an idealised boat - same as an idealised block being dragged across the ground. The initial effect when pushing a real boat forward is dominated by the mass of the boat. Any change of lateral momentum of water at the(pointed) bow would be equal and opposite to the net change of momentum of the water coming together at the stern. After that,as speed becomes relevant the fluid behaviour of the water becomes relevant. Is that ok with you?

Yea, I believe you know what you are talking about. Sounds reasonable to me.