# A sphere of radius R, made from material of specific gravity, SG is submerged in a tank of water

... The sphere is placed over a hole, of radius a, in the tank bottom. Develop a general expression for the range of specific gravities for which the sphere will float to the surface.

I've attached the diagram for the problem.

As an attempt at the solution, I've summed the forces on the sphere, or the buoyant force $$F_b$$ - the weight of the sphere. Buoyant force is equal to the weight of the displaced water by the sphere, $$\frac{4}{3}\pi R^3\rho_{water}\times32\frac{ft}{s^2}$$. The weight of the sphere is $$\frac{4}{3}\pi R^3\rho_{sphere}\times32\frac{ft}{s^2}$$. Based on force equation, in order for the ball to float the buoyant force would have to be greater than the weight of the sphere, or $$F_b-W>0$$. After some algebra you can say that the specific gravity of the sphere has to be $$<1$$, which seems too simple to be the actual answer.

I'm sure I'd have to take the hole that the sphere is resting atop into consideration, but I'm not sure how to go about that. My only ideas on how to include that would be to include the force from the ambient pressure on the bottom of the ball pushing upward in the force equation, but even if that effects the specific gravity at which the ball will float, once it started to float and that force was no longer there, the ball would simply sink again, unless $$S.G.<1$$. My other thought was that I would have to consider the volume of the sphere not submerged due to the hole and thus not contributing to the buoyant force, but I have no idea how to calculate what that volume would be.

Is my first solution correct, that $$S.G.<1$$ would cause the ball to float? That again feels too simple for the given parameters.

Any direction would be appreciated.

## 2 Answers

The net force on the sphere due to the water around it is obtained by integrating the pressure as a vector at each point around the area of the sphere.

$$F_w = \iint \vec{p}(\theta, \phi)\bullet\vec{n}\ d\theta\ d\phi$$

Allow that water is incompressible and the pressure at the top of the tank is $$p_o$$. The magnitude of the pressure depends on distance from the bottom of the tank $$d_b$$ as

$$p = p_o + \rho_w g \left(H - d_b\right)$$

The relationship between $$d_b$$ and $$\theta, \phi, R$$ can be developed by geometry. The vector relationship for $$p$$ can be developed based on position $$(r, \theta, \phi)$$. Assume the pressure is otherwise isotropic (it only depends on $$d_b$$ not on $$\theta$$ or $$\phi$$. For a free-floating sphere, the double integral becomes $$4\pi R^2$$. But ... the segment of the sphere at the bottom where no water exists must be dropped from the integral in this system. Alternatively, the accurate approach is to replace the pressure over only that segment with an integral where $$p$$ is constant (at $$p_o$$). The double integral does not easily become just $$4\pi R^2$$ because of the change in limits on $$\phi$$ and $$\theta$$.

The request was for a hint at the right direction, not for a complete answer. In due respect to this (and indeed in great appreciation that you want to do the work yourself once you know how to start), I leave the grunt work, as one might say, as an exercise for the reader.

There are some points and some assumptions we need to make.

• If the ball floats the water will drain from the hole.

• If we assume there is a fill valve or somehow we close he hole just after the ball starts to float, your assumption is correct, the ball has to be lighter than water..

If these are taken care of then by doing a bit of geometry we see by your sketch the hole is very small compared to the sphere, we can assume small angles, 9 degrees.

$$sin(2/20)\pi/2 = (1/10)\cdot \pi/2$$

and subtract the surface area of the hole multiplied by the water pressure, as opposed to calculating the sphere sectional surface and pertinent water pressure on it, at the bottom of the tank $$F_{floating}= \frac{4}{3}\pi R^3\rho_{water} - \pi*4\cdot(o.8*o.1kg/m height) \\ = \frac{4}{3}\pi (1/5)^3*1kg/liter \\ -\pi*4mm*(1)liter/(1000)milL \cdot(o.8*o.1kg/m height)$$ assuming the atmosphere pressure constant for the top and bottom of the tank. I did a rough calculation and the effect of the hole is minimal as with your dimensions less than 3%. This is within 4 decimal point rounding.