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.
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I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – Feb 06 '19 at 14:37
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Why would you, this is about me trying to understand the concept that's way before homework and is from my mind. – user597368 Feb 06 '19 at 14:39
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@user597368 The description of the homework-and-exercises tag: Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. – BioPhysicist Feb 06 '19 at 14:59
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I've taken the question as a conceptual one about pressure in a liquid – which, in my experience – is poorly understood by many. – Philip Wood Feb 06 '19 at 15:57
2 Answers
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!
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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,
It's important to remember that pressure is a scaler as discussed at this PSE question/answer and links therein. The gravitational field magnitude is used to calculate the pressure gradient in fluids, and from that we can calculate the pressure at a point in the fluid. The pressure in the fluid may involve other activities/processes. Consider a balloon filled with water. There is pressure due to the elastic skin of the balloon. If a person squeezes the balloon, the pressure will change. Or consider a fluid in a gravitational field with vacuum above it (and the fluid doesn't boil away---this is a thought experiment). The pressure at the top of the fluid is zero. The direction of the gravitational field tells us the direction of increasing pressure, not the direction of forces resulting from the pressure in the fluid.
Because pressure is a scaler, the forces associated with the pressure depend on the area interface (size and orientation) you choose to analyze. If the pressure at some point in the fluid near the edge of your coin is $p_1$, then on some small area $\delta A$ to the right of the point there is a force $p_1 \delta A$ to the right. On some $\delta A$ immediately above the point there is an upward force $p_1 \delta A$. On some $\delta A$ below the point there is a downward force of $p_1 \delta A$. The direction of the gravitational field does not determine the direction of the force.
So, on the top of the immersed coin, the pressure in the fluid produces a downward force on the coin. On the bottom of the coin, the fluid produces an upward force, slightly larger than the force on the top because the pressure in the fluid at the bottom area is larger by $Dg\Delta y$, where $\Delta y$ is the coin thickness. That difference in forces is what we call buoyant force. The force magnitudes on the edge of the coin are all equal (in a static fluid), but the vectors add in "opposite" directions, so the net sideways force is zero .
