Gauge pressure due to a floating body

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SUMMARY

The discussion centers on the gauge pressure due to a floating body, specifically a block of wood. The conclusion is that the pressures at two points, A and B, are equal, represented mathematically as P_A = P_B. This equality is derived from Archimedes' principle, which states that a floating object displaces a volume of liquid equal to its weight. The analysis involves calculating pressures from both the water column and the block, confirming that the pressures balance out under the conditions described.

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brotherbobby
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Homework Statement
Both liquid columns stand to a height ##h## in the diagram below, but the second has a block of wood floating in it. How does the gauge pressure at the point marked ##\mathbf A## differ from the gauge pressure at ##\mathbf B##?
Relevant Equations
Gauge pressure at a depth ##h## inside a liquid is given by : ##P_G = \rho_L gh##, where ##\rho_L## is the density of the liquid.
water and wood.png


My answer : Both pressures are equal, i.e. ##\boxed{P_A = P_B}##.

Reason : (1) The block of wood displaces an amount (mass) of liquid equal to its weight (archimedes' principle for floating bodies, or law of floatation). Hence we can imagine removing the block in the second case and filling it up by water equal to the block's weight. Water would stand to the same height in both the liquid columns and the pressures at A and B would be the same.

The same can be shown in a different way, keeping the block in place.(2)
1580629360834.png
Let us divide the total height in the second case into two parts : ##h_1 \rightarrow## the height of the water column only and ##h_2 \rightarrow## the height of the block of wood. The pressure due to water column ##P_1 = \rho_W gh_1##. The pressure due to block ##P_2 = \rho_B g h_2##. [Pressure = (Force) / (area) = (m_B g) / (width_B) = (density_B volume_B g) / (width_B) = density_B h_2 g]. By archimedes' principle, ##P_2 = \rho_W g h'_2## where ##h'_2## is the height of water from the top to the base of the block (clearly ##h'_2 < h_2##). Hence the pressure ##P_B = \rho_W g h_1+ \rho_W g h'_2 = \rho g (h_1 + h'_2) = \rho g h = P_A##.

Is my answer correct?
 
Last edited:
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Can you convince yourself ?
That's more important than an approval stamp from PF (we're not in that business, anyway)
 
yes I am convinced I am right.

But as you know, it's easy to go wrong and not know about it. Sometimes it's carelessness. At other times, there are things we simply didn't know.
 
Stick to your conviction. You read the problem statement correctly. That's the most important part. The math and the physics is straightforward.
 
brotherbobby said:
But as you know, it's easy to go wrong and not know about it. Sometimes it's carelessness. At other times, there are things we simply didn't know.
When one has doubt there are good ways to allay it. In this case can you provide for yourself another good plausability argument (The one you use is good enough for me...). Conversely you could assume that the alternative answer is correct and show it leads to nonsense like perpetual motion.
But in life the answers are not in the back of the book and sometimes lives depend upon getting it right, so it is necessary to this skill as desribed by @BvU in #2. But always ask when necessary!!!
 

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