# Hydrostatic Pressure at Base of Floating Object in Liquid with Greater Density

Let's say there's an object floating in liquid. The depth of the object below the surface is $$d$$ and its height above the water is $$h$$. The density of the ball is $$p_b$$ and the density of the liquid is $$p_{L}$$. The pressure above the water is $$P_{atm}$$.

I understand this situation when only the density of the ball matters; you integrate to get P(z) = $$p_bgz$$ + C, and given the initial condition P(d) = $$P_{atm}$$. (pressure at water-air interface is atmospheric), C = $$P_{atm}$$. - $$p_bgd$$ , so P(z) = $$p_bgz$$ + $$P_{atm}$$ - $$p_bgd$$ .

What I am confused about is how to find the pressure given the influence of the density of the liquid. I don't really understand why or how the density of the liquid affects the equation for pressure at all, besides that in order for the object to be floating, $$p_b$$ < $$p_L$$. It seems like since you're only interested in the ball, so the only distance that matters is the distance of the ball under the water, and the only density relevant there would be the one of the ball. Or do you take the ratio of the densities? Is it their difference? Why does the liquid matter at all?

• What is the shape of the object you are interested in? What question are you trying to answer? Please be more specific and try to ask only one question per post. Commented Apr 29, 2022 at 16:05
• A floating object displaces its weight of the displaced fluid. You may have to do an integral to determine the maximum depth that the sphere sinks to. Commented Apr 29, 2022 at 18:34

First of all, hydrostatic pressure is only applicable in a fluid, not a solid, so $$P=\rho_b g z + C$$ doesn't work - the mechanical forces within a solid give a pressure distribution that isn't hydrostatic, normally governed by elastic stresses and strains and structural mechanics

So instead you have to apply the hydrostatic balance inside the liquid, so $$P(-d)=P_{atm}+\rho_L g d$$

$$\rho_b$$ only affects the depth the ball floats at:

For the ball to float, the overall upward force has to be equal to it's weight, so $$F=\rho_b g \frac{4}{3} \pi r^3$$

The total buoyancy force is equal to the weight of fluid displaced (because if the original fluid was there, it wouldn't be moving so needs a buoyancy force that balances it's weight - the surface of the object below the fluid is the same size and shape so has the same pressure and force)

The weight of fluid displaced is $$\rho_L g$$ times the volume, and the volume of a spherical cap from wikipedia is $$V=\frac{\pi d^2}{3}(3r - d)$$, where $$d$$ is the submerged depth

So setting $$\rho_L \frac{\pi d^2}{3}(3r - d) g = \rho_b g \frac{4}{3} \pi r^3$$ gives a cubic ($$(\frac{d}{r})^3-3(\frac{d}{r})^2+4\frac{\rho_b}{\rho_L}=0$$) you can solve for $$d$$ (only one solution will be physically realistic, with $$\frac{d}{r}$$ real and $$0<\frac{d}{r}<2$$) and then plug into $$P(-d)=P_{atm}+\rho_L g d$$ to get the pressure