How much weight can a floating vessel carry before submerging?

Question

^{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

Answer 1

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.

Answer 2

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).

Answer 3

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.