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I’m having difficulty getting my head around this scenario (we assume incompressibility):

We have a boat on a river moving at speed $u$ which in turn causes a current of speed $v$. The area of the base of the boat is $A_B$.

If we take a streamline from the base of the boat to the surface of the water, then how do we apply Bernoulli’s equation in such circumstance?

My guess is that the speed of the river at the surface is $u$ with atmospheric pressure. And my guess is that the speed on the base of the boat is $v$ but how is the pressure on the water defined at this point - is it just the buoyancy force / $A_B$?

I am also not sure whether the fluid under the boat has any potential energy (well of course if the boat were not there then the fluid would be higher - so is this negative potential?)

I would be grateful if anyone could help clear up my understanding, thank you.

maths54321
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1 Answers1

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Check out this link on the bernoulli principle.

The Bernoulli Principle:

$$P_1+\frac{1}{2}\rho v_1^2+ \rho g h_1=P_2+\frac{1}{2}\rho v_2^2+ \rho g h_2$$

The principle of buoyancy:

Take the velocity at all points to be equivalent ($v_1 = v_2$).

Thus, the equation becomes:

$$P_1=P_2+ \rho g (h_2-h_1) $$ $$ \implies P_1=P_2+ \rho g \Delta h $$ $$ \implies P_1-P_2 = \rho g \Delta h $$

i.e., the difference in pressure between any two points is proportional to the density of the fluid and the difference in height.

For a rectangular boat that displaces a volume $V$ of water, the buoyant force

$$F_B = \int_A {(P_1-P_2)}= \rho g V $$

Thus, we have used the bernoulli principle to derive the buoyant force.

The equation extends to non-rectangular boats, with a little effort.

enter image description here

bittahProfessional
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