Physics Question...

[prohornblower]

I am wondering if you drop a sinkable 16 pound bowling ball into a deep sea, does it sink all the way to the Ocean floor? Or does the increased deep-down water pressure affect it to a point where the bowling ball will not sink any further?

In other words, would the bowling ball just get to a point of depth taht it would just float around? Or does it keep going to the floor?

Discussion amongst 2 co-workers. Looking for an answer quick.

Thanks!

[Borodog]

It sinks all the way down. Water is incompressible (more or less).

[prohornblower]

[ QUOTE ]
It sinks all the way down. Water is incompressible (more or less).

[/ QUOTE ]

Ok, so the water below the bowling ball will take the path of least resistance as the ball is sinking, so that water will just wrap itself around the ball, rather than support the ball. That makes sense.

Path of least resistance, dur!

[MelchyBeau]

Boro is right and wrong at the same time.

Yes water does not compress to any significant degree, however sea water is not pure H2O. This happens because the salinity affects the density of the liquid, and the increase in pressure along with the decrease in temperature has an effect on the density at deep ocean depths.

HOWEVER, it still won't be enough to prevent a bowling ball from falling to the bottom. though some lighter material that may sink in purely H2O could be suspended between 2 different density gradiants.

Melch

[Borodog]

That's why I said "more or less".

[ApeAttack]

http://www.windows.ucar.edu/tour/link=/earth/Water/density.html

[prohornblower]

[ QUOTE ]
http://www.windows.ucar.edu/tour/link=/earth/Water/density.html

[/ QUOTE ]

Interesting, it says that water circulates laterally deep in the Ocean.

[Borodog]

[ QUOTE ]
http://www.windows.ucar.edu/tour/link=/earth/Water/density.html

[/ QUOTE ]

Thanks. Less than 3 parts in 1000. Like I said, water is incompressible (more or less).

[phiphika1453]

No one has provided ANY evidence in this argument.

I am asuming the rate of acceleration of gravity in sea water is still -9.8m/s2. Obviously the water is exerting an upward force on the bowling ball at some point due to the ball not increaseing in speed as it falls towards the bottom.

The evidence we need is a measurement of upward force exerted on an object with the surface area of a bowling ball at or near the ocean floor.

[Borodog]

[ QUOTE ]
No one has provided ANY evidence in this argument.

I am asuming the rate of acceleration of gravity in sea water is still -9.8m/s2. Obviously the water is exerting an upward force on the bowling ball at some point due to the ball not increaseing in speed as it falls towards the bottom.

The evidence we need is a measurement of upward force exerted on an object with the surface area of a bowling ball at or near the ocean floor.

[/ QUOTE ]

I thought I made it clear. The only thing that matters is the buoyancy of the bowling ball as a function of depth. Since the OP stated that the bowling ball sinks, it is clearly more dense than water, and since the density of water changes by less than three parts in a thousand throughout the entire ocean column (since, as I have repeatedly stated, water is incompressible, more or less), the ball will sink to the bottom.

The only reason the ball ceases to accelerate downward at some point is that it reaches its terminal velocity because of drag caused by the ball moving through the water.

[ApeAttack]

[ QUOTE ]
No one has provided ANY evidence in this argument.

I am asuming the rate of acceleration of gravity in sea water is still -9.8m/s2. Obviously the water is exerting an upward force on the bowling ball at some point due to the ball not increaseing in speed as it falls towards the bottom.

The evidence we need is a measurement of upward force exerted on an object with the surface area of a bowling ball at or near the ocean floor.

[/ QUOTE ]


YOU WANT EVIDENCE?!?! HERE IS YOUR EVIDENCE!!!!

There are three competing forces on the ball.
1.) Gravity.
2.) drag forces. This will cause the ball to reach a terminal velocity.
3.) Buoyancy forces. This force is equal to the density of the fluid times the volume of the part of the object that is submerged times 9.8m/s^2.
Law of Archimedes:
The buoyant force is equal to the weight of the replaced liquid or gas.


Mathematically,
d = density
V = volume
g = 9.8 m/s^2


I am assuming a positive force means the ball will rise, a negative force means the ball will sink. The negative sign in 'g' is taken into account in that definition of my system ( I subtract the gravitational force below ).

The net force on the ball at any given time is:

F(net) = F(buoy) - F(grav) - F(drag)
= d(fluid)*V(submerged part of ball)*g - d(ball)*V(ball)*g - F(drag)

if the ball is completely submerged,
V(submerged part of ball) = V(ball)

and,
F(net) = V(ball)*g*( d(fluid) - d(ball) ) - F(drag)

F(drag) is proportional to the velocity of the ball falling and also proportional to the surface area of the ball. This force will only reduce the velocity of the ball, but will not change the direction it is falling, unlike the other two forces.

So, the important term (for this discussion) in F(net) is,
V(ball)*g*( d(fluid) - d(ball) )

As long as the density of the fluid is less than the density of the ball, the force will be downward (negative) and the ball will sink. The density of water changes by only a small fraction from the surface to the bottom of the ocean (less than 1%). The ball will sink all the way to the bottom IF the density of the ball is greater than the density of the water at the bottom of the ocean.

Ahhh... I see four years of physics didn't go to waste.