Archimedes’s Principle for Floating Objects:

So hears the theory:

Archimedes principle says that the magnitude of the buoyant
force always equals the weight of the fluid displaced by the object.
This buoyant force always acts upward through the point that was the center
of gravity of the displaced fluid. In the case of floating objects
the buoyant force is equal to the force of gravity on the object. Knowing
that the change in pressure is equal to the Buoyant force per unit area
(ΔP = B/A) we see that B = (ΔP)A and ΔP = ρgH where ρ is the density of
the fluid g is the acceleration due to gravity and H is the height of the
fluid displaced. Therefore B= ρgH(A) = ρgV where V is the volume of
the fluid displaced by the object. If we look at the sum of forces
acting in the vertical direction for a floating object we get the equation

where M is the mass of the object displacing the fluid. By rewriting the equation to include what we know about the buoyant force and then solving for the height H we have what where looking for.

The equation gives us the height H of the fluid that is displaced by the object which is equivalent to the height from the bottom of our floating object to the water level on the object.

Because my boat is primarily used in saltwater I run the numbers (for dimensions click here) using the density of saltwater ρ = 1.03x103kg/m^{3}. I plug all my values
in to the equation and I’ve got it.

**H = Mg/ρgA** = M/ρA = (141 + 141 + 239)kg / (1.03x10^{3}kg/m^{3}
x 5.55m^{2}) = 9.11cm = 3.6 inches

By building the second deck greater than a height of 3.6 inches above the bottom of the boat I could ensure even with a full tank of gas the saltwater would not run back in to the boat. The height I got from the local boat builder was a safe 5.75 inches. So approximately**how much extra
mass can my boat hold before the plugs start letting in water**?
By plugging the value of H into the equation we get the following:

**My boat can hold approximately 283lbs of extra weight** (in its center)
before I need to put the plugs in the drain holes.

ΣF = B – Mg = 0

where M is the mass of the object displacing the fluid. By rewriting the equation to include what we know about the buoyant force and then solving for the height H we have what where looking for.

B = ρVg = Mg

ρgH(A) = Mg

ρgH(A) = Mg

The equation gives us the height H of the fluid that is displaced by the object which is equivalent to the height from the bottom of our floating object to the water level on the object.

Because my boat is primarily used in saltwater I run the numbers (for dimensions click here) using the density of saltwater ρ = 1.03x103kg/m

By building the second deck greater than a height of 3.6 inches above the bottom of the boat I could ensure even with a full tank of gas the saltwater would not run back in to the boat. The height I got from the local boat builder was a safe 5.75 inches. So approximately

5.75in=.0226m

ρgH(A) = Mg

M = ρHA = 1.03x10^{3} x 0.226m x 5.55m^{2} = 129.19kg ≈
283 lbs

ρgH(A) = Mg

M = ρHA = 1.03x10

The largest approximation I use in this is in assuming that the sides of my boat are actually straight up and down as well as the transom. Because my sides are at 26