homework and exercises - If I immerse a coin in a cup of water, help me define the forces acting on it

So the coin is hanging in an equilibrium position midway in the water filled cup, concerning the force acting on the upper face of the coin, I define it as F=mg where m is the weight of the water directly above the upper face of the coin and g is the present gravitational field, if I divide and multiply by V and then divide by A to get an expression for the total pressure affecting the upper face of the coin, I get P=Dgh, where V is the volume of water directly above the coin and A is the surface area of the coin (not the cup) and h is the height of water affecting the coin and D is the density of water. What am I struggling with : -The two horizontal forces affecting the lateral surface of the coin, they aren't any related to P=Dgh as the gravitational field is vertical, however my book disagrees with me.

Answer

In an ordinary cup, the weight of liquid directly above the top face of the coin does indeed equal the downward force on the top (horizontal) face of the coin. But this approach masks much of what is really going on, and doesn't help in more general cases, as you've seen for yourself when trying to calculate the pressure on the (vertical) edge of the coin.

Liquid at any given level (L, say) is slightly compressed by the liquid above it. Because of the ability of liquid molecules to slide over each other, the compressed liquid pushes on all surfaces in contact with it, including vertical surfaces. We can show that (if the liquid is stationary)

• the force is normal to any surface in contact with it

• the pressure (normal force per unit area) is equal on all surfaces at the same level, no matter what their orientation.

• the pressure at a depth $h$ in a liquid of density $\rho$ is given by $$p=h \rho g + \text{pressure}, p_s, \text{on exposed liquid surface.}$$

[An easy and elegant way to show these things equates the work done, against the liquid pressure, in pushing a piston immersed in the liquid to the gravitational PE gained by the displaced liquid.]

You'll note that this equation doesn't involve the shape of the container. So even in a tapering container, getting less wide as we go upwards, in which the exposed water surface is of smaller area than the area of your horizontal coin surface, the pressure all over that coin surface will still be given by $h \rho g +p_s .$ Your argument based on the liquid column directly above the coin would let you down!

If the last paragraph seems crazy, there's a simple explanation. The liquid above the coin exerts normal forces on the sides of the container. Because of the taper, these forces have an upward component. So the container sides exert an equal and opposite downward force component on the liquid. These downward force components compensate for the weight of liquid missing on account of the taper!