It's a Drag

[Gareth56]

If a boat is moving with constant speed through the water are there any drag forces acting on it?

[YT2095]

yes, look up Laminar Flow and Reynolds Number(s).

[Klaynos]

Yes, if you turn off the engine (or drop the sail) the boat will slow and stop.

[swansont]

Yes, if you turn off the engine (or drop the sail) the boat will slow and stop.

 

Put another way, with either the engine or sail, you have a force. Why is it moving at constant velocity?

[Gareth56]

Put another way, with either the engine or sail, you have a force. Why is it moving at constant velocity?

 

Because it's being pulled along by a horse (it's a barge really in the book) and the book only gives the force the horse provides to the barge. Sadly this is an even numbered question in my physics book

[swansont]

Because it's being pulled along by a horse (it's a barge really in the book) and the book only gives the force the horse provides to the barge. Sadly this is an even numbered question in my physics book

 

So the horse exerts a force. Now, apply F = ma to the problem.

[Gareth56]

So the horse exerts a force. Now, apply F = ma to the problem.

 

I did that as Fx = 2000N x cos25deg but the question asked what was the drag force of the water on the boat.

[Klaynos]

If the boat is moving at a constant velocity is it accelerating?

 

F=ma=?

[h4tt3n]

Put another way, with either the engine or sail, you have a force. Why is it moving at constant velocity?

 

Because viscous drag is velocity dependant. Sooner or later the boat will reach its so called terminal velocity where friction force and engine force "cancels out" and accelleration becomes zero.

 

Here are a few equations on air / fluid friction:

 

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri.html

[MrMongoose]

yes, look up Laminar Flow and Reynolds Number(s).

 

How's that going to help?

[Klaynos]

Because viscous drag is velocity dependant. Sooner or later the boat will reach its so called terminal velocity where friction force and engine force "cancels out" and accelleration becomes zero.

 

Here are a few equations on air / fluid friction:

 

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri.html

 

I think this is too complicated for this question, it's moving at a constant velocity so terminal velocity has already been reached.

[Gareth56]

I think this is too complicated for this question, it's moving at a constant velocity so terminal velocity has already been reached.

 

So presumably the drag force of the water on the barge is Zero?

[swansont]

So presumably the drag force of the water on the barge is Zero?

 

We've established that the net force on the barge must be zero. Next step is to identify all of the forces, and set up the equations so they add to zero.

 

What are the forces acting on the barge?

[tvp45]

Why did you use cos(25 degrees)?

[Gareth56]

Why did you use cos(25 degrees)?

 

This was the angle the force was making with the barge.

[tvp45]

This was the angle the force was making with the barge.

 

So, what keeps the barge from slamming into the bank?

[thedarkshade]

yes, look up Laminar Flow and Reynolds Number(s).

Laminar flow:

[math]\frac{\Delta v}{\Delta t}=0[/math] ; [math]\frac{\Delta p}{\Delta t}=0[/math] ; [math]\frac{\Delta \rho}{\Delta t}=0[/math]

 

Reynolds Number: [math]Re=\frac{dv\rho}{\eta}[/math]

[Gareth56]

So, what keeps the barge from slamming into the bank?

 

Because the horizontal component of the force (F x costheta) is greater than the vertical(towards the bank) component (F x sintheta)of the force. That is Fx > Fy

[Klaynos]

That would make no difference....

 

So what do we know:

 

There is no acceleration.

Therefore no net force....

But there is a finite and known forward force.

An unknown possibly 0 drag force, what does this force have to = for the first 3 statements to be true?

[MrMongoose]

Laminar flow:

[math]\frac{\Delta v}{\Delta t}=0[/math] ; [math]\frac{\Delta p}{\Delta t}=0[/math] ; [math]\frac{\Delta \rho}{\Delta t}=0[/math]

 

Reynolds Number: [math]Re=\frac{dv\rho}{\eta}[/math]

 

Those top three equaions look more like steady flow than laminar flow. I still don't see how laminar to turbulent transition is going to help with gaining a fundamental insight into the problem. It seems like trying to teach someone binary to help them realise thier pc isnt turning on because its not plugged in.

 

Even looking at simple skin friction drag and form drag equations would seem excessive when the OP is simply trying to grasp Newtons 1st Law.

[Gareth56]

That would make no difference....

 

So what do we know:

 

There is no acceleration.

Therefore no net force....

But there is a finite and known forward force.

An unknown possibly 0 drag force, what does this force have to = for the first 3 statements to be true?

 

 

I'm sorry but I think it would make a difference. If the vertical component of the force is greater than the horizontal component then the barge will be pulled more towards the bank. Try changing the angle of pull to 60 deg and see what happens.

 

With respect to the second part above I suggest it's equal to the x component of the force.

[swansont]

I'm sorry but I think it would make a difference. If the vertical component of the force is greater than the horizontal component then the barge will be pulled more towards the bank. Try changing the angle of pull to 60 deg and see what happens.

 

The x and y components are independent; an x-direction force can only influence motion in the x-direction. That's why you break forces down into orthogonal components — it's a powerful tool.

 

Since the velocity is zero, you know that both the x- and y-component forces are zero.

[Skye]

So, what keeps the barge from slamming into the bank?

In practice, a rudder. It's not important to the question though, since you know what the velocity is anyway. You could break it down into it's component forces like swansont says and work out just how drag would be required though, if that's your thing.

[Gareth56]

The x and y components are independent; an x-direction force can only influence motion in the x-direction. That's why you break forces down into orthogonal components — it's a powerful tool.

 

Since the velocity is zero, you know that both the x- and y-component forces are zero.

 

Exactly that's why (in this example) if Fy > Fx then the barge will tend to go towards the y direction ie the bank. As for the velocity being zero, it's being pulled at constant speed which isn't zero.

[swansont]

Oops. That should be "velocity is constant"