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Al P
Lv 7
Al P 發問於 Science & MathematicsPhysics · 10 年前

A ball floating in the ocean : reinstate?

A ball is floating in the ocean at latitudes south

of 60 degree south latitude. What is the density of the

ball? calculus or geometry derivation ok.

Thank you.

更新:

no formula

更新 2:

Density ocean water = 1027 kg/m^3

更新 3:

solid ball uniform density

更新 4:

@ Koshka

The original question teaches an important lesson.

This question will go to vote or BA.

Thanks to bill for answering.

更新 5:

@Anyone

There is one numerical answer. Pick a coordinate

system (or geometry) and derive away!

更新 6:

Koshka is absolutely on the right

track with the formula version.

更新 7:

Koshka my hero. This is what Scythian did:

The bottom 30 degrees of the ball are below water:

pf = 1027 kg/m^3

a = pi / 3

r = 1

h = r - r * Sin(a)

vc = (1 / 3) * pi * h ^ 2 * (3 * r - h)

vs = (4 / 3) * pi * r ^ 3

p_ball = pf*(vc/vs)

更新 8:

Thanks to all. Lesson: Confucius say:

Sometimes it's better to keep your head

out of the ocean, maybe ;)

I was trying to discourage the use of the

cap formula, without derivation, with my

"no formula" comment. I should have clarified this.

Here's the way I originally did this: I like

nuts and bolts, but to each their own.

The volume below water: "volume cap"

d (rcosθ)/dθ = -rsinθ

Vc = ∫ dvol = ∫ π(rsinθ)^2 * -rsinθdθ

Vc = -πr^3 ∫ sinθ^3dθ [π/6,0]

Vc = -πr^3 [cos3θ - 9cosθ]/12 |π/6,0

Vc = 0.0538708182 cubic units

Volume of the ball:

Vs = (4/3)πr^3

In equilivrium we have

ρ => density of ball

ρf => density of ocean water

Fw = Fb

m g = ρf Vc g

ρ Vs g = ρf Vc g

ρ = ρf (Vc/Vs) = 13.2079497 kg/m^3

5 個解答

相關度
  • Koshka
    Lv 5
    10 年前
    最愛解答

    No formula?

    The part that is "immersed" is (pi*h^2)/3* (3a-h) where h represents the height of the cap and "a" the radius of the cap. The weight (mg) of the spherical cap is the same as the water displaced.

    Volume of the sphere minus volume of the spherical cap

    (4*pi*r^3)/3 - (pi*h^2)/3* (3a-h)

    Is this going somewhere?

    It is 11:43 PM here

    Perhaps someone might take it from there?

    -------------------------

    Édited: Scythian is my hero!

    At least someone understands the question!!

    Athough, I still can not figure out how he got the 1/2 * sqrt of 3 part.

    Goodby!

  • Scythian1950
    Lv 7
    10 年前

    Taking a cue from Koshka's interpretation of the problem, the ratio of the volumes of an unit spherical cap subtending 60° and an unit sphere is:

    (π/3)(3 - h²)h / (4/3)π

    which is equal to, for h = 1 - (1/2)√3

    (1/2) - (9/32)√3 = 0.0128607

    so that the density of the ball is 0.0128607 x 1027 = 13.2079 kg/m³

    Practically a beach ball.

    Edit: You should give Koshka the BA, since I merely provided the "mechanics", while she cracked the problem.

  • Fred
    Lv 7
    10 年前

    Is there Earth ocean that far south? Google Earth says, yes, and plenty of it.

    No calculus or geometry required. The answer is an inequality.

    0 < ρ[ball] < 1027 kg/m^3

    Important lesson? Am I that dense? [pun intended!]

    PS. Thanks, Er. H.

    I was excluding the neutral buoyancy, "submerged" case, because I don't consider that to be "floating."

  • Mike Richard
    Lv 4
    10 年前

    Yeah, Fred is right.

    If it is partially floating then density of ball would be <1027 kg/m^3

    and if it floating fully submerged then density would be =1027 kg/m^3

    Thanks

  • billrussell42
    Lv 7
    10 年前

    Impossible to answer. The density could be anywhere from slightly lower than salt water to zero.

    And the location doesn't matter.

    .

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