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Drop a spring into the ocean and other stuff

  1. Nov 8, 2011 #1
    I was taking with my dad and we got on the subject of what would happen if you dropped a spring into the ocean. Well, I said nothing it would sink to the bottom and not compress because pressure is pushing on it equally from all directions.

    He agreed.

    Then I said what would happen if you slapped a hemisphere on each end of the spring (imagine cutting a ball bearing in half and welding it to the spring... cut sides of the bearing being welded to the spring). Basically my argument is that the spring will compress as the ocean pressure rises because there is more surface area on the outsides of the hemispheres than there is on the insides (more surface area = more pressure pushing in on each side of the spring than pushing out).

    My dad says no that won't happen it will do nothing as the pressure increases just like the spring with nothing on it.

    Who is right?
  2. jcsd
  3. Nov 8, 2011 #2


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    The spring will not compress as the pressure on all sides is equal. I'm not sure on the why though. I was thinking because even if there is more surface area on the outside, the inside is flat and so all the pressure is against compression, while the outside is round and so the pressure is not solely in one direction but spread out, causing everything to equal out. Not sure though, anyone know?
  4. Nov 8, 2011 #3
    If placing objects at the ends of a spring would cause the spring to compress then we can see violations of the conservation laws:
    Take a spring with these hemispheres on each end and drop it into a fluid so that two small balls press up against the ends, then reach in and slide off these end pieces (since the only effect they produce is net force inwards the work done removing them is negligible); the effect will be the expansion of the spring and the balls will begin to move giving you free energy.

    A second possibility is to take one of these hemisphere, if it would cause a compression of the spring then by itself the imbalances of forces on either side should cause it to accelerate and you would end up with perpetual motion.
  5. Nov 8, 2011 #4


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    That isn't right, because pressure at each point on the curved surface acts perpendicular to the surface, not all in one direction. When you add up all the presure force, the sum is equal and opposite to the force on the flat side.

    Think about the "opposite" situation. Suppose you have a hollow half-sphere (not under water!) that contains gas at high pressure. Will it spontaneously move because there is "more gas pressure" on the curved surface than the flat surface? I don't think so.
  6. Nov 8, 2011 #5
    The resultant of all forces (from water pressure) on the hemisphere (or any submerged object) will be equal to the buoyant force. It will be directed upwards and have the same value, no matter what is the orientation of the object. So the forces on the two hemisphere will have same magnitude and direction so it should not be any tension on the spring from this.
  7. Nov 8, 2011 #6
    Mathematically consider the forces on each side (the rounded side of the hemisphere or the flat side you weld onto), since the pressure is equal in all directions I presume we can take the forces to be perpendicular to the surface at every point, we can then set up an expression for the net force to compress the spring.

    First the rounded edge, I will consider an infintesimal slice of the half sphere as a half circle with radius R and compare this to the force on the corresponding line on the circular side since the two scale proportionally. I will also take only the components of force directed into the spring since this shape is symmetric about the other axis.

    For a force directed towards the spring at a single point X with [itex]\rho[/itex] being the pressure and [itex]\theta[/itex] being the angle of the applied force (directed inwards not to the spring) with the axis perpendicular to that of the spring
    F = \rho \sin \theta
    Where (given y is the equation of the semicircle)
    \theta = \tan ^{-1} (-\frac{dx}{dy})
    \frac{dy}{dx} = \frac{-x}{\sqrt{R^2 - x^2}}
    Then the sums of the forces directed inwards is the line integral over the semicircle of these values where
    ds = \sqrt{1 + \frac{dy/dx} ^2}dx = \sqrt{1 + \frac{x^2}{R^2 - x^2}}dx
    So our expression is
    F_{net} = \int_{-R}^{R} \rho \sqrt{1 - \frac{x^2}{R^2}}\sqrt{1 + \frac{x^2}{R^2 - x^2}}dx
    Which when I evaluate in Mathematica gives [itex]2\rho R[/itex]

    Comparing this to the corresponding force on the line of the circle with length 2R at that point we obtain [itex] 2\rho R [/itex] So we see that the forces will balance out in this case.
    Last edited: Nov 8, 2011
  8. Nov 8, 2011 #7
    You assume that the pressure is constant when it is actually a function of depth. This is why you get the two forces to balance. In reality they don't.
  9. Nov 9, 2011 #8
    Unless I make the assumption that the spring lies parallel to the water surface. The argument is about whether shape causes force imbalances in a fluid, the assumption that there is equal pressure over the entire surface is valid.
  10. Nov 9, 2011 #9
    It does not matter how the spring lies. As long as the hemispheres are completely submerged there is a force imbalance on each one of them (called buoyant force). This again does not depend on the orientation of the hemispheres. And the assumption that the pressure is the same over the entire surface would be valid if the entire surface would be at the same depth in the liquid. Which is not the case for a 3D object. If your assumption were true there would be no buoyant force.
  11. Nov 9, 2011 #10
    But arranging the spring level with the water surface creates no net force from the objects at the ends of the spring directed at the center due to buoyancy, it's the components of force acting to compress the spring that we are concerned about. Additionally I might take the limiting case where the height of the spring tends to zero and so the differences in pressure from top to bottom become negligible.

    Edit: On further thought I have realised that my calculations depend only on having a constant pressure at constant depths (assuming the spring is sitting parallel with the water surface) since I calculated only an infintesimal strip parallel to the axis of the spring as having a constant density, I showed that the corresponding strip on the curved surface provided equal force to that on the flat surface. Even if pressure changed with depth enough to make a difference this is accounted for as each new slice would have a new [itex]\rho[/itex] and the forces would still equal out.
    Last edited: Nov 9, 2011
  12. Nov 9, 2011 #11
    I was referring to your calculation for a single hemisphere and not to the force on the spring.
    The pressure changes with depth (from the top to the bottom of the object) enough to produce a net force equal to the weight of the water displaced by the object. It does not balance out, no matter how you take it.See Archimedes' law.

    For a full hemisphere of radius R, the net force will be (2/3 πR^3)*ρ*g where ρ is the density of water and g is the gravitational acceleration.
  13. Nov 10, 2011 #12
    Last edited by a moderator: May 5, 2017
  14. Nov 10, 2011 #13
    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.
  15. Nov 10, 2011 #14
    First, neglect pressure changes over the entire area of your device(this is very realistic). But don't be afraid!
    Now, the wire your spring is made of shall definitely have a finite cross-sectional area, say "a". The component of the pressure force acting on the curved surface of one hemisphere along the axis, inwards of your assemble (of spring and hemisphere) is precisely equal to P times ∏xRxR. (P = pressure, R= radius). But, the pressure force on the flat surface along the axis but away from the assembly is P times (∏xRxR - a).
    same from the other hemisphere. Net inwards force is hence = 2 times 'P' times 'a'.
    This causes spring to compress to a length at which the reaction from spring equals this cause.

    So go tell your dad, the spring will compress. Ask him this:
    Take a very soft(easily compressible) but solid rubber ball. Put it in water. Does it compress. If that doesn't impress him, ask him to try the swimming pool and feel for himself, if his feet are compressed when under water.
  16. Nov 10, 2011 #15
    I'm not sure the analogs of a ball or feet apply in this case since those things compress because the pressure inside is constant whereas the pressure outside is dependent on the environment, dropping them to the bottom of a pool would cause them to compress because of this discrepancy. However with a spring anything that pushes it in will also have a counterpart pushing it out, for instance I doubt a spring would go straight if placed in a vacuum while a rubber ball would expand greatly.
  17. Nov 10, 2011 #16
    Hi JHamm!! I know you're not sure; try this:

    I am sorry but would you believe if I said that the radius of the hemisphere will also decrease when the assembly is plunged into water.
    If you want more, I shall say that the area of cross-section of the wire of the spring will also increase!!!
    Never thought of it right?
    Actually, any closed finite volume when put inside a region of higher pressure(than before) shall become stressed. Hence the strain, or, the change in dimensions.
    The OP wanted to know about the axial changes in dimension only(actually because it will be highly pronounced, due to the highly elastic, deformable spring). Tell you what, even if it were a rod of steel, the length would reduce, again due to the stress inwards of the assembly.

    In perfect vaccum, won't the particles(atoms or molecules) rip apart and the spring wire become a gas? Think!
  18. Nov 10, 2011 #17
    I agree that an increase in pressure will cause the spring to shrink, but this isn't affected by the objects we place on the ends, interesting though.
  19. Nov 10, 2011 #18
    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!!!!
  20. Nov 10, 2011 #19
    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.
  21. Nov 10, 2011 #20
    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.
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