Hydrostatic Force Ratio

Consider just the top half of the sphere (with a solid bottom). Suppose that the sphere is just submeged. The vertical (z) component of the hydrostatic force is then

$\int \int _{\text{hemisphere}}\rho g z \mathbf{ \hat{z}\cdot dA}$ and the divergence theorem transforms this to

$\int \int \int_{\text{hemisphere}}\rho g dV=\frac{2}{3}\pi R^{3}\rho g$

(Alternatively, if you prefer Archimedes to Gauss, the buoyancy force equals the weight of the displaced water!)

We want the hydrostatic force acting on the domed upper surface of the hemisphere, so we must subtract the force that is acting on the flat disk at a depth of R, namely

$\text{pressure} \times \text{area}=\rho g R \times \pi R^{2}=\pi R^{3} \rho g$

The hydrostatic force on the upper domed surface is thus $\frac{2}{3}\pi R^{3}\rho g-\pi R^{3} \rho g=-\frac{1}{3}\pi R^{3} \rho g\dots \text{equation 1}$

Slightly adjusting the argument to suit the bottom hemisphere gives the force on the bottom dome as

$\frac{2}{3}\pi R^{3}\rho g+\pi R^{3} \rho g=\frac{5}{3}\pi R^{3} \rho g\dots \text{equation 2}$

Comparing the two equations shows that the required ratio is $\boxed{5}$

(strictly speaking it is -5)

The hydrostatic force is the weight of water above, the volume of a sphere inscribed in a cylinder is 2/3 the volume of the cylinder, so the volume of water above the sphere is 1/6 the volume of the cylinder, and the volume of water above the bottom half surface of the sphere is 5/6. Then the ratio (5/6)/(1/6)=5.

This could be done in spherical coordinates, but I'll do it in Cartesian coordinates, because that way is more elementary. Imagine the sphere as being built out of infinitesimally thin disks stacked up on top of each other. Only a tiny bit of area on each disk is exposed to the vertical pressure force, because the rest of the area is covered by the (very slightly smaller) disk on top. Suppose that $x$ is the horizontal direction, and $y$ is the vertical direction (in which there is a pressure gradient).

$x^2 + y^2 = R^2 \\ 2x + 2y \frac{dy}{dx} = 0 \\ x \, dx = -y \, dy$

Define the infinitesimal area:

$dA = \pi (x^2 - (x - dx)^2) = 2 \pi x \, dx = -2 \pi y \, dy$

Define the pressure (assume pressure $P_0$ at the sphere center $(y=0)$ and water density $\rho$ ):

$P = P_0 - \rho g y$

Differential force:

$dF = P \, dA = (P_0 - \rho g y)(-2 \pi y \, dy) = -2 \pi (P_0 y - \rho g y^2) \, dy$

Find the force on the lower half:

$F_L = -2\pi \int_{-R}^0 (P_0 y - \rho g y^2) \, dy \\ = -2\pi (-P_0 \frac{R^2}{2} - \rho g \frac{R^3}{3}) \\ = 2\pi (P_0 \frac{R^2}{2} + \rho g \frac{R^3}{3})$

Find the force on the upper half:

$F_U = -2\pi \int_0^R (P_0 y - \rho g y^2) \, dy \\ = -2\pi (P_0 \frac{R^2}{2} - \rho g \frac{R^3}{3}) \\ = 2\pi (-P_0 \frac{R^2}{2} + \rho g \frac{R^3}{3})$

As a check, find the total force, which we know must be the weight of the displaced water.

$F_T = F_L + F_U = 2\pi (P_0 \frac{R^2}{2} + \rho g \frac{R^3}{3}) + 2\pi (-P_0 \frac{R^2}{2} + \rho g \frac{R^3}{3}) = \frac{4}{3} \pi R^3 \rho g$

We know that the sphere is just barely submerged. Therefore, the pressure at the sphere center is as follows:

$P_0 = \rho g R$

Substitute into upper and lower force equations:

$F_L = 2\pi \Big(\rho g R \frac{R^2}{2} + \rho g \frac{R^3}{3}\Big) = 2 \pi \rho g \Big(\frac{5}{6} R^3\Big) \\ F_U = 2\pi \Big(-\rho g R \frac{R^2}{2} + \rho g \frac{R^3}{3}\Big) = 2 \pi \rho g \Big(-\frac{1}{6} R^3\Big) \\ \frac{|F_L|}{|F_U|} = 5$

No need of integrals. Imagine cutting the ball into an upper and a lower hemisphere (both solid independent bodies) and move them apart horizontally. Now for the upper half, its buoyant force can be calculated with F = ρgV = ρg · ( $\frac{2}{3}$ πr³) = $\frac{2}{3}$ πρgr³. This force is, on the other hand, the resultant of the hydrostatic forces on the curved top ( $F_{1}$ ) and the flat bottom ( $F_{2}$ ) . $F_{2}$ = p·S = (ρgr) · (πr²) = πρgr³. Thus by subtraction, $F_{1}$ equals $\frac{1}{3}$ πρgr³. This is identical to the hydrostatic force on the upper hemisphere of the original body. Analogous for the lower half, we get $\frac{5}{3}$ πρgr³. So the answer is 5.

Suppose that the centre of the sphere is a distance $kR$ below the surface, where $R$ is the radius of the sphere. Thus $k \ge 1$ . Consider splitting the sphere into two hemispherical halves by making a horizontal cut through its centre.

The upper hemisphere is acted upon by the hydrostatic force $F_T$ on the top half of the sphere, plus a hydrostatic force $kR \rho g \times \pi R^2 = k\pi \rho g R^3$ on the circular base. The difference between these forces is the buoyancy force on the hemisphere, and so $k\pi \rho g R^3 - F_T \; = \; \tfrac23\pi \rho g R^3$ and hence $F_T \; = \; \tfrac13\pi \rho g(3k-2)$ Similarly, the lower hemisphere is held in place by the hydrostatic force $F_B$ on the bottom of the sphere and a hydrostatic force $k\pi\rho g R^3$ on the circular top of the hemisphere. Thus $F_B - k\pi\rho g R^3 \; = \; \tfrac23\pi \rho g R^3$ and hence $F_B \; = \; \tfrac13\pi \rho g(3k+2)$ Thus the ratio of the hydrostatic forces on the upper and lower hemispheres is $r \; = \; \frac{F_B}{F_T} \; = \; \frac{3k+2}{3k-2}$ The question does not specify the depth of the sphere , but the fact that $k \ge 1$ tells us that $1 < r \le 5$ . Given the answer options for this question, we must have $k=1$ and $r=\boxed{5}$ . The sphere must be just below the surface of the water.

The pressure experienced by the upper sphere can be written as

$p = \rho g h = \rho g R (1-cos\phi)$

If we want to calculate the force, we might use $F = pdA$ but it is slightly different,

It should be $F = pdA cos\phi$ because we added up the force altogether to form a force which straightly downward.

$F = pdA cos\phi = \int \rho g R (cos\phi - cos^2 \phi )dA$

Set surface vector $r = <Rsin\phi cos\theta, Rsin\phi sin\theta, Rcos\phi >$

The mini area $dA$ calculated as $| r_\phi \times r_\theta |d\phi d\theta$ as well.

We get $dA = R^2 sin\phi d\phi d\theta$ .

So we can calculate the force act on the upper sphere $F_{upper}$ and substitute $V = \frac{4}{3}\pi R^3$ :

$F_{upper} = \int pdAcos\phi \\ = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \rho g R (cos\phi - cos^2 \phi) R^2 sin\phi d\phi d\theta \\ = \frac{1}{4}\rho g V$

And this is important part, we know that because the sphere is submerge in the water, so the total hydrostatic force act on it will be

$F = mg = \rho g V$

So , the bottom force minus the upper force will be $\rho g V$

$F_{bottom} = \frac{5}{4}\rho g V$

So the ratio is $\boxed{5}$

I think it should be mentioned that the sphere is JUST below the surface..it is not that obvious from the wordings or the picture