How do you calculate forces needed to move a floating object against a flow of water?

Question

A while ago I came across the problem of having to work out how to remove a narrowboat (long steel boat) from a river weir. The boat was 6 foot, 10 inches wide, 60 foot long and about 22 tonnes in weight. The boat was stuck against a weir by the pressure of a rivers flow of water, going at about 3.5 meters per second.

Trying to figure out the amount of force required to pull the boat off of the weir really stumped me and we eventually had to just call out a tug to help. I was wondering if anyone could explain how I could have mathematically calculated this problem?

Trying to come up with a solution in the pub people kept saying I'd need over 22 ton of force because the boat weighed 22 ton, this is of course stupid as the boats weight is negated by its buoyancy. Its the force of the flow of the river that we needed to negate but how could that be calculated?

A weir is like a rope across the path of a river with lots of floats spread across a river to stop boats going down dangerous rapids, the boat was roughly at right angles to the flow and I'd estimate the draft to be between 2.5 and 3 foot.

Follow this link to see a map of the area with the weir. http://maps.google.co.uk/?ie=UTF8&ll=52....

It would have damaged the boat to drag the boat along the floats. The only way would have been to pull the boat away at right angles. I was wondering if a half ton winch tied to a tree on the bank opposite the weir would have enough power to pull the boat via a rope across the river.

Despite much searching for fluid dynamics, hydrodynamics and the Bernoulli equation i'm no closer to figuring this one out.

Answer 1

Your reasoning is correct that 22 tons of weight is no equal to 22 tons of pull. Let's look at the  forces needed to overcome:

1. the friction along the weir

2. moving the boat against the current.

Let's consider a a hydrodynamic problem of a force the flowing water exerts on the boat Fw.

Fw= 0.5 p A V^2 (Yes it is a quite a force see ref 1 and keep in mind that Power is a product of force and velocity P = F x V)

p- density of water

A effective area of water hitting the boat (assuming laminar flow...)

V - velocity of water

The boat returns that favor by exerting the force on the weir. The force of friction between the boat and the weir is

f = uN

u -coefficient of friction

N - normal to the weir component of  Fw

N= Fw sin(A)

A -the angle Fw makes with the weir.

Here I have introduced the  basic components.

To estimate the actual tug force you must consider a geometry of were the tug force is applied relative to Fw.

If properly executed first the tug force must overcome the f and Fw and then just Fw.

I hope it helps.

Did you say 1/2 ton winch?

Lets compute the force the current of (3.5 m/ s = 11.5 ft/s) exerts on the boat

Lets assume the worst case scenario that your boat was pinned by a normal to the hull current.

A= 60 feet x 3 feet

p = 62 lb/ ft^3

we have

Fw= 0.5 x  62 x ( 60 x 3) x (11.5 )^2

Fw= 738,000 lb  or

Fw= 370 tons  ( No kidding!)  in reality due to effective area of the hull is actually about 50 % less we would need  at least

185 tons just to start.

OMG I think I need a drink....

Before I have a drink ...

I see that the current is 45 degrees relative the weir and that reduces your force even more

F= Fw cos(45)= 185 x .707= 130 tons.

____________More____

Thank you, I have tried.

Also the current will tilt the boat further reducing the effective area and lift the boat out of the water. There we must consider Coandă's  effect.  How much in all? To give a crack at it I I would need boat hull geometry and its center of mass. Please note that these figures are a grotesque since we are not considering the hydrodynamic aspects of the boat. It would produce these forces if the boat turned into a sheet of 60 x 3 feet ^2 area.  

My calculations showed tat this area  can get reduced (hydrodynamically) maximum by a factor of 10!