浮力和阿基米德原理

Suppose that we have an incompressible static fluid of density $\rho$. In this fluid we will immerse a right solid of base area $A$ and height $h$.

Recall from an earlier topic that the pressure at a depth $d$ is $\rho gd$ (this is a very good approximation in almost all cases).

Obviously, the forces on all of the sides of the box cancel out. Thus, we only need to consider the forces on the tip and the bottom.

The top of the box is at a depth $d$, and the bottom of the box is at a depth $d+h$.

We can calculate the downward force on the top of the box: $$F_{down}=P_{top}A$$ $$F_{down}=A\rho gd$$ Similarly, we can calculate the upward force on the bottom of the box: $$F_{up}=P_{bottom}A$$ $$F_{up}=A\rho g(d+h)$$ Calculating the net upward force on the box, $$F_{net}=F_{up}-F_{down}$$ $$F_{net}=A\rho g(d+h)-A\rho gd$$ $$F_{net}=A\rho gh$$ But wait! Since $F_{net}$ is positive, the fluid exerts a net upward force on the box! This force is called the buoyant force.

Now, we make some key observations: $$F_{net}=A\rho gh$$ Notice that $Ah$ is the volume of the fluid displaced by the box! $$F_{net}=V\rho g$$ We may simplify again, since the volume times the density of the fluid is simply the fluid's mass! $$F_{net}=mg$$ Now, the mass times the gravitational force is simply the weight! $$F_{net}=W_{fluid}$$ This is the famous Archimedes's Principle, which states that the buoyant force acting on an object is equal to the weight of the fluid displaced by the object.

Objects in a fluid seem to weigh less than they do because of a concept called buoyancy. Buoyancy occurs because the pressure of a fluid increases with depth. When an object is in a fluid, the force pushing up on the bottom of the object is larger than the force pushing down from the top. The buoyant force can be calculated by finding the difference between the force pushing on the bottom of the object (F2) and the force pushing on the top of the object (F1):FB = F2 – F1