# Fluids - buoyancy

1. Jul 1, 2008

### c.j.

Hi - I'm having trouble with the bouyancy force. I understand that in static equilibrium a force from below a submerged object (not resting on the floor of a container but in the liquid) equals the overlying liquid plus the weight of the object. So that the bouyancy force is the difference of the force below minus the force above and thus equals the weight of the object. But I don't seem to have a sense as to why this equals the weight of the displaced fluid. Maybe something about the behavior of liquids?

Thanks

2. Jul 1, 2008

### Staff: Mentor

This page describes a simple "thought experiment" that may help convince you that the buoyant force must equal the weight of the displaced fluid: "http://hyperphysics.phy-astr.gsu.edu/Hbase/pbuoy.html" [Broken]

Last edited by a moderator: Apr 23, 2017 at 1:55 PM
3. Jul 1, 2008

### c.j.

Thank you! I'm off to class and will give a more focussed read after class. Looks like this will help!

4. Jul 1, 2008

### c.j.

I read the materials and the link is great. I tend to come from a more chemistry perspective so I think in terms of bonds, etc. So basically, the fluid's bonds, energy, etc. will allow it to "support" a certain amount of matter. If there was a tank of liquid that was in equilibrium so that all the forces in the liquid were balanced, the max the liquid could "support" would be it's own density. Anymore and it would overcome the liquid's bonds and sink and less it would not sink into the liquid. And, if the force equation is written in terms of press x area, the equations use the density and volume for mass. I guess the hydrostatic press. difference (if something goes deeper vs. a shallow submerged object) gets eliminated since the buoyancy force is the difference between the force below and the force above?

5. Jul 2, 2008

### bilgealp

I may suggest you think about a submerged rectangle first then a submerged prism and then convince yourself with a general shape.

First Rule: In a fluid at rest, pressure at a point is equal to the height of the fluid body(h) * unit weight of the fluid, namely at point A, fluid pressure is equal to

$$p_A=\gamma h$$

Pressure distribution on a rectangle: (Unrealistic but helps) Think of the rectangle's top and bottom surfaces are perfectly parallel to the still fluid surface. Top surface (edge) of the rectangle is h unit-lengths below the surface. Dimensions of the rectangle is axb unit-length.

Top surface (line in this case) : Equal pressure distribution with magnitude

$$p=\gamma h$$

and the total force acting on this surface

$$F_{top}=\gamma h a$$

where a is the width of the rectangle.

Side Surfaces(Edges): Pressure forces acting on the side surface are equal and opposite. So we don't need to take them into account.

Bottom Surface(Edge): Equal pressure distribution with magnitude

$$p=\gamma (h+b)$$

and the total force acting on this surface

$$F_{bottom}=\gamma (h+b) a$$

where b is the height of the rectangle

Force Balance: For equilibrium net body and external forces must vanish.

$$\sum_{i}F_{i}=0$$

$$F_{bottom}-F_{top}-W=0$$

where W is the weight of the body. Hence

$$W=\gamma a b$$

in which a b is the total area of the rectangle. Weight of the body is equal to the weight of the fluid of the same area.

Prism: Prism will have a third dimension, say c. Then, similar to the previous discussion we can obtain the weight of the submerged prism.

$$W=\gamma abc$$

Weight of the body is equal to the fluid body of the identical volume.

The pressure difference here is in fact the buoyant force.

Now one can conclude with the well known principle :

Buoyant forces acting on a body is equal to the weight of the fluid displaced by the body.

This is valid for a general shape of object which don't need to submerge. I mean, it can also float.

Now, to carry this topic further, my question at this point to the audience is what is the relationship between the unit weight of the body and the fluid. What can you say about the unit weight of a floating object?