I’ve done the derivation for the buoyant force using the difference in pressures and after everything is said and done the buoyant force of some object in a fluid is equal to the gravitational force F=mg of the fluid displaced where m is the mass of that fluid. Here m can be substituted with Vp where V is the volume displaced and p is the density of the fluid and that’s our normal buoyant force formula F=pVg

Why is this?


2 Answers 2


Look at the picture. The pressure in a fluid is same at same height and increases as go in depth. The pressure is equal to $p ×h×g$. To get the force on surface multiply this pressure with Area of surface. The force from sideways will cancel out. Now the important step to be taken here is calculating upward and downward force. There are two ways to find force. First one, Convert the shape into cuboid or cylindrical shape so that base and top becomes flat. Both areas would becomes equal. Second one, First find each perpendicular component of force to base and then find their component pependicular to ground.

Force at top $$F1 = A\rho h1g$$ Force at base $$F2= Aph2g $$ ($\rho $ remains same as fluid is incompressible)

Net force $$ F = A\rho hg(h2-h1) upward$$ = - Weight of fluid displaced

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We know that the volume $V$ of displaced fluid is also the volume of the object immersed. The quantity $m=\rho V$ is the mass of the displaced fluid, which therefore has a force on it given by $$F=mg=\rho Vg$$ since $$\rho=\frac{m}{V}\\ \rightarrow m=\rho V$$

So the first equation above is the force on the body of fluid that get's displaced by the object.

Consider when the actual object is immersed in the fluid:

Since it displaces a volume of fluid equal to the volume of the object, gravity tries to push the displaced fluid back to its original position. But since this is already occupied by the immersed object, this immersed object will therefore experience a force equal to the weight of the displaced fluid pressing on it.

Imagine that the object is not present in the fluid. The net upward force on this volume of fluid will be equal to its weight (since the fluid parcel is in equilibrium). But when the object is immersed in the fluid, this exact same force will be acting on the object pushing it upward.


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