# Why is Buoyancy Force = to the density of the liquid * Vobj(submerg)*g

1. Aug 13, 2014

### shangriphysics

Why is the weight of the liquid displaced = buoyancy force and not the weight of the submerged volume of the object?

Does it have something to do with normal force of the surface of the liquid on the object or pressure?

2. Aug 13, 2014

### Matterwave

Think about your proposal. If the bouyancy force was equal to the submerged weight of the volume of the object itself...then nothing would ever sink would it? Everything would float.

3. Aug 13, 2014

### Staff: Mentor

For a floating object, the buoyant force happens to equal the weight of the entire object, not just the submerged part.

4. Aug 13, 2014

### Staff: Mentor

When an object is submerged in liquid, the surrounding liquid does not know that it has been replaced by a solid object. As far as it knows, there is still liquid there. So it exerts a force on the object equal to the weight of the liquid that would have filled that volume.

Chet

5. Aug 13, 2014

### shangriphysics

Hi Chester,
I am still a bit confused on this concept of Buoyancy force. Why does the liquid believe that the liquid is still there when it is displaced?

Also, in the situation where buoyancy force is less than the weight of the object, the object sinks and the buoyancy force is the density of the liquid times the volume of the object times gravity.

I am confused why the buoyancy force uses the liquids density and why, when displaced, the liquid believes that the liquid is still there when the object has a different density?

Thank you everyone for helping me! I am truly grateful for this community!

6. Aug 13, 2014

### Staff: Mentor

If the object were held stationary, the surrounding liquid would be exerting pressure forces on the object that are the same as if the object had not replaced the liquid. The surrounding liquid would not know what is there. If you integrate the pressure distribution of the surrounding liquid acting on the object, it will be equal to the weight of the liquid that filled the space before it was replaced by the object.

Hope this helps.

Chet

7. Aug 13, 2014

### rcgldr

Assume that the liquid is not accelerating in any direction and is not moving relative to the liquids center of mass (to keep things simple). Note that pressure increases with depth. Now consider the forces involved with any volume of liquid within the liquid. Gravity exerts a downwards force on that volume of liquid equal to the weight of that volume of liquid, the pressure differential versus depth of the surrounding fluid exerts an equal and opposing upwards force on that volume of liquid, so there is no net force on the volume of liquid and it does not move or accelerate.

Replace that volume of liquid with an object with the same shape and size of the volume of liquid. The pressure differential versus depth of the surrounding fluid still exerts the same upwards force that it did before (the force is equal in magnitude but opposing the weight of such a volume filled with the liquid). If this upwards force is greater than the weight of the object, the object initially accelerates upwards (until it reaches terminal velocity due to drag). If the upwards force is less than the weight of the object, the object initially accelerates downwards. If the objects net density is the same as the surrounding liquid, the object hovers.

8. Aug 14, 2014

### ZealScience

Well, if you haven't replaced the liquid, the surrounding liquid would exert the force that equals to the weight of the liquid; by replacing the liquid with a solid object, there is no reason that the property of the liquid should change anyhow...And thus the force exerted should remain the same.

Actually, I think your argument is valid. After all, buoyancy is certain force exerted by the surrounding liquid, and the only force there is when the liquid is stationary is that caused by the pressure.

9. Aug 14, 2014

### vanhees71

It's all about the pressure difference on a body within a fluid due to gravity. We can easily derive it for a cubic box without much of tensor calculus. Of course, it holds true for bodies of any shape, but for this we'd need some tensor calculus.

So let's first look on the pressure of a fluid at rest. To that end put a sheet parallel to the earth under water at a depth $h$. The force excerted on the upper surface is given by gravity due to the water column over this surface. Now the mass of this column is $m=A h \rho$. The force thus is
$$\vec{F}=-m g \vec{e}_z=-A h \rho g \vec{e}_z.$$
The pressure is then given by the negative normal component of this force to the surface, taken per area:
$$P=-\frac{1}{A} \vec{e}_z \vec{F}=\rho g h.$$
Now put a box with two of its faces parallel to the earth. Now the pressure exerts a force pointing downward on the upper of these parallel surfaces and one upward on the lower. Let the height of the box be $c$ the surface parallel to the earth has area $A=a b$. Since the forces excerted by the pressure on the surfaces perpendicular to the cancel each other, the net force points to the $z$ direction. Let the upper surface be on depth $h$. Then the lower surface is at depth $h+c$. Thus we get
$$\vec{F}=\vec{e}_z=a b[(h+c) \rho g-h \rho g]=abc \rho g \vec{e}_z.$$
Now $m_{\text{disp fluid}}=abc \rho$ is the mass of the fluid displaced by the body. As you see the net force is pointing upwards, because the downward force on the upper surface is smaller in magnitude than the upward force on the lower surface. So the buoyant force is upwards and as large as the weight of the displaced fluid, which is Archimedes's principle.

The vector-calculus argument for a general body goes as follows. In a fluid the stress tensor is isotropic and thus given by the pressure via
$$\hat{\sigma}(\vec{x})=-P(\vec{x}) \hat{1}.$$
The force on an arbitrary body is
$$\vec{F}=\int_{\partial V} \vec{d}^2 \vec{f} \cdot \hat{\sigma}=-\int_{\partial V} \vec{d}^2 \vec{f} P(\vec{x}).$$
Now let $\vec{e}_j$ be the unit basis vectors of a Cartesian coordinate system. Then
$$F_j=\vec{e}_j \cdot \vec{F}=-\int_{\partial V} \vec{d}^2 f \cdot \vec{e}_j P(\vec{x})=-\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot (\vec{e}_j P(\vec{x}).$$
Now in our case, as shown above, we have
$$P(\vec{x})=-\rho g z$$
and thus
$$\vec{\nabla} \cdot [\vec{e}_j P(\vec{x})]=\partial_j P(\vec{x})=-\rho g \delta_{j3}.$$
Thus the only non-vanishing force component is
$$F_3=\int_{V} \mathrm{d}^3 \vec{x} \rho g=V \rho g=m_{\text{displaced fluid}} g>0.$$
Again we find Archimedes's principle for a body of arbitrary shape :-).

10. Aug 14, 2014

### shangriphysics

I get it now! The pressure at that depth will be the same, and since the water is static, the forces are canceling each other out. Thank you everyone for the help!