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Source: http://www.cbc.ca/news/canada/new-brunswick/friday-flood-new-brunswick-2018-1.4647979

Is there a way to calculate how much weight a vessel can carry (in fresh water) before it submerges?

Assumptions:

  • The water is not disturbed (no waves or wind) and the cargo does not move
  • The vessel weighs 200 lbs when empty
  • The vessel's volume is 60 cubic ft
User1974
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  • Feel free to add hypothetical variables. I don't have a background in engineering; I've just guessed at the variables that apply. – User1974 May 04 '18 at 14:13
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    You need the density of the liquid, the volume displaced, and the acceleration due to gravity. Research buoyancy on google and you should see how to set up the equation. – Dando18 May 04 '18 at 14:14

3 Answers3

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If the weight of an object is less than the weight of an amount of water of the same volume, it will float. The weight of water per cubic foot is about $62.5$ pounds. $60$ of these weigh about $3745$ pounds. The weight of the boat is $200$ lbs. That leaves an extra $3545$ pounds for water. That takes up about $56$ cubic feet.

William Grannis
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    Well argued. But please don't report all those nonsense significant digits. The only reasonable answer is "about 56 cubic feet". And you could see that from the start: it's the volume of the boat less the volume of water that would account for the 200 pound weight of the boat - about $4$ cubic feet. – Ethan Bolker May 04 '18 at 16:09
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    In agreement with Ethan, considering the starting values we only have $2$ significant figures. So $56 \textrm{ ft}^3$ is the appropriate answer with error of $\pm 0.5 \textrm{ ft}^3$. – Dando18 May 04 '18 at 18:40
  • I'm more of a pure math guy. I understand that one should round, but the idea of actually doing it brings me pain. – William Grannis May 05 '18 at 19:39
  • Nevertheless, I shall have my friend correct my post for me. – William Grannis May 05 '18 at 19:39
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We have the formula,

$$ F_{\textrm{buoyancy}} = \rho_{\textrm{fluid}}V_{\textrm{disp}}g, $$ where $g$ is the acceleration due to gravity. This will give us the force pushed upwards on the boat when it is in the water.

Now for an ideal floating scenario the boat is in static equilibrium, meaning acceleration is $0$, which shows $a=0 \implies \sum F = 0$. The boat should only have two forces acting on it: gravity and buoyancy.

$$ \begin{align*} \sum F &= W-F_{\textrm{buoyancy}} \\ 0 &= m_{\textrm{net}} g - \rho_{\textrm{fluid}}V_{\textrm{disp}}g \\ m_{\textrm{boat}} + m_{\textrm{contents}} &= \rho_{\textrm{fluid}}V_{\textrm{disp}} \\ m_{\textrm{contents}} &= \rho_{\textrm{fluid}}V_{\textrm{disp}} - m_{\textrm{boat}} \\ \end{align*} $$

This presents us with an equation for the mass of the contents. Note how you're given the weight (lbs), not the mass (slugs), so multiply both sides by $g \approx 32.2 \frac{\textrm{ft}}{\textrm{s}^2}$ to obtain the equation in terms of weight.

Note: This gives us the exact weight that will result in equilibrium, any more and the boat will begin to sink so it is beneficial to include a safety factor (i.e. replace $V_{\textrm{disp}}$ with $0.6V_{\textrm{disp}}$ or something similar).

Dando18
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Directly apply Archimedes principle which states that weight of boat with its contents equals weight of water ( density $\gamma$ ) that it displaces.

$$ ( W_{boat}+W_{loaded \,stuff} ) = \gamma_{water} \cdot V_{immersed} $$

$$(200+W_{loded stuff}) = 62.5 \, (60) \rightarrow \, W_{loded stuff}= 175\, lbs $$

not much.. just about a person's weight.

Narasimham
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