# Drop a spring into the ocean... and other stuff

by D9 XTC
Tags: ocean, spring
P: 1,985
 Quote by JHamm My calculation took a single, infinitesimal strip, maybe I'm not understanding you properly but I don't see how a strip of zero height could have different densities at top and bottom. If each strip has forces balancing out then no matter how many strips I add I'll still have no net force.
I am not sure about the geometry you consider either. Or what orientation of the hemispheres in respect to the water level.
The formulas that you wrote are confusing as the units do not match.
What you call a "force" is not, according to the formulas. Pressure times radius is not a force.
To find a force you need to consider and infinitesimal area and multiply by the pressure on that area. Then you may calculate the component of the force that is of interest.
Now, when you integrate to find the total force over the surface, you need to consider the variation of pressure from element to element.

You may start with a simple case, like the hemisphere being oriented with the plane surface horizontal.
Anyway, this is just interesting as an exercise.
Archimedes' law tells you the force on the hemisphere. You may just want to see that is comes from adding up the pressure forces.
P: 1,985
 Quote by Naty1 Good analogy: I was thinking this way: Drop a hemisphere into the ocean....if there were more pressure on the spherical portion, it would rocket around underwater forever...free power!!!!
If the hemisphere is upside-down (flat surface on top) the upward force on the spherical portion is larger than the downward force on the flat portion. It doe not mean that what you assume above will happen.
If the difference is larger than the weight of the hemisphere, it will jump towards the surface. It's called buoyancy.
If the difference is less it will just sink to the bottom.
The same behavior will be observed no matter the orientation of the hemisphere. The resultant of the pressure forces will be always oriented the same way.
 P: 390 No, what he is saying is that if there is a net force due to the geometry it will move forever, in which case he is right. And your complaint that my units do not match is silly, I did not multiply by a radius, I multiplied by a length which corresponded to a radius, I can consider the pressure constant since I am working at constant depth. Additionally the claim by hellboy4444 that the spring will compress under high pressure is slightly misleading I think due to conflicting definitions of compress. To clarify the spring WILL shrink because the cross sectional area of the wire will decrease, however this has nothing to do with the winding of the wire into a spring.
P: 15,319
 Quote by nasu If the hemisphere is upside-down (flat surface on top) the upward force on the spherical portion is larger than the downward force on the flat portion. It doe not mean that what you assume above will happen. If the difference is larger than the weight of the hemisphere, it will jump towards the surface. It's called buoyancy.
No.

Bouyancy is a property of the volume of an object and its weight. Shape is utterly irrelevant.
 Mentor P: 22,313 Nitpick edit: Bouyancy is a property of the volume of an object and therefore the displaced water's weight. Shape is utterly irrelevant. Whether an object floats or not is not strictly buoyancy, it is due to a resultant force consisting of buoyancy minus the weight of the object.
P: 15,319
 Quote by russ_watters Nitpick edit: Bouyancy is a property of the volume of an object and therefore the displaced water's weight. Shape is utterly irrelevant. Whether an object floats or not is not strictly buoyancy, it is due to a resultant force consisting of buoyancy minus the weight of the object.
After I went to bed I started thinking I should have added "volume of the displaced water", just to be thorough.
P: 1,985
 Quote by JHamm No, what he is saying is that if there is a net force due to the geometry it will move forever, in which case he is right. And your complaint that my units do not match is silly, I did not multiply by a radius, I multiplied by a length which corresponded to a radius, I can consider the pressure constant since I am working at constant depth.