Pressure exerted by water on Sphere partially exposed at end

• iitjee10
In summary, the objective of this problem is to determine the force exerted by water on a sphere of greater density that is submerged in a cylindrical container filled with water to a certain height. Two methods are proposed: one using Archimedes' Principle and the other using pressure integration. Both methods involve calculating the volume of water displaced by the sphere. However, the second method integrates the hydrostatic pressure over the wetted surface of the sphere and may require a correction for the distance from the water surface to the center of the sphere. Both methods are conceptually correct and should give the same result.
iitjee10

Homework Statement

There is a cylindrical container that has a small hole of radius ##a ( < R)## at the bottom. A sphere of radius ##R## and density ##\rho_{s} > \rho_{w}## is placed in the cylinder such that it completely covers the hole (a part of it sticks out as in attached figure). The container is now filled with water (##\rho_{w}##) to a height ##h##.

The objective is to determine the force exerted by the water on the sphere. (specifically on its dependence with the height of water)

Homework Equations

I have two methods in mind, but not sure which is correct.
- Using Archimedes Principle
- Using pressure integration

The Attempt at a Solution

Method 1:

If we assume that the hole along with the portion that remains outside the container did not exist .. and the resulting contact surface between the flat portion of the sphere and the bottom of the cylinder was not perfectly smooth, then the force by the liquid would be equal to the weight of the liquid displaced (##V \rho_{w}g##).

This force would be comprised of the force on the bottom surface and the force on the remaining cylindrical surface. Now, the force on the bottom surface is ##\rho_{w}hg \pi a^{2}##.

So the force on the cylindrical section is ## V \rho_{w} g - \rho_{w} h g \pi a^{2}##.

Since this is the force we want for the original question, this should be the answer after calculating ## V ## appropriately.

Method 2.

This involves integration.

Consider the diagram.

The pressure at a circular (shell) element that makes an angle ##\theta## is given by ##P = (h - R \cos \alpha - R \cos \theta) \rho_{w}g ##. The elemental area ## A = 2 \pi R \sin \theta R d \theta##.

So the net vertical force on the element (downward) is ## F_{v} = PA \cos \theta##

Integrating this from ## \theta = 0 ## to ## \theta = \pi - \alpha ## should give the answer.

However, evaluating this integral does not seem to match the answer for the above method.

Is one of the two methods conceptually wrong?
Any help would be highly appreciated.

iitjee10 said:

Homework Statement

There is a cylindrical container that has a small hole of radius ##a ( < R)## at the bottom. A sphere of radius ##R## and density ##\rho_{s} > \rho_{w}## is placed in the cylinder such that it completely covers the hole (a part of it sticks out as in attached figure). The container is now filled with water (##\rho_{w}##) to a height ##h##.

The objective is to determine the force exerted by the water on the sphere. (specifically on its dependence with the height of water)

Homework Equations

I have two methods in mind, but not sure which is correct.
- Using Archimedes Principle
- Using pressure integration

The Attempt at a Solution

Method 1:

If we assume that the hole along with the portion that remains outside the container did not exist .. and the resulting contact surface between the flat portion of the sphere and the bottom of the cylinder was not perfectly smooth, then the force by the liquid would be equal to the weight of the liquid displaced (##V \rho_{w}g##).

This force would be comprised of the force on the bottom surface and the force on the remaining cylindrical surface. Now, the force on the bottom surface is ##\rho_{w}hg \pi a^{2}##.

So the force on the cylindrical section is ## V \rho_{w} g - \rho_{w} h g \pi a^{2}##.

Since this is the force we want for the original question, this should be the answer after calculating ## V ## appropriately.

I'm not sure I follow your reasoning here. It seems much too complicated since you are applying Archimedes' Principle, i.e. the force applied by the water = the weight of water actually displaced by the sphere. It seems to me, the trick is to calculate the volume of the spherical cap which emerges from the hole in the bottom of the cylinder and subtract this volume from that of the whole sphere. This gives you the volume of displaced water.

Fortunately, the formula for calculating the volume of a spherical cap can be found quite readily:

https://en.wikipedia.org/wiki/Spherical_cap

Method 2.
This involves integration.

Consider the diagram.

The pressure at a circular (shell) element that makes an angle ##\theta## is given by ##P = (h - R \cos \alpha - R \cos \theta) \rho_{w}g ##. The elemental area ## A = 2 \pi R \sin \theta R d \theta##.

So the net vertical force on the element (downward) is ## F_{v} = PA \cos \theta##

Integrating this from ## \theta = 0 ## to ## \theta = \pi - \alpha ## should give the answer.

However, evaluating this integral does not seem to match the answer for the above method.

Is one of the two methods conceptually wrong?
Any help would be highly appreciated.

I think, looking at your diagram for the integration of hydrostatic pressure over the wetted surface of the sphere, it would be better to take h as the distance from the surface of the water in the cylinder measured to the center of the sphere. Then, you can use regular spherical coordinates to set up the integration.

It's not clear if you are using vector calculus in your integration, because, while the hydrostatic pressure acts normal to the surface of the sphere, you are interested only in the net vertical force acting on the sphere due to this pressure.

iitjee10 said:
Is one of the two methods conceptually wrong?
Any help would be highly appreciated.

I believe your two methods are correct. I get the same result using both methods.

@SteamKing

For the first method, I can calculate the volume. What I intend to find out is the force exerted by the liquid on the curved section of the hypothetical ball (which will be the answer as needed for the original question). Since the buoyant force force for the hypothetical case is equal to the force acting on the bottom + the force on the curved surface, it should be easy to find out

For the second method, I ignored the horizontal component of the forces, since they would negate out to zero. The change of measurement of h seems like good advice, probably can reduce calculations a bit.

@TSny

It's a bit late here. I'll post my complete calculations in roughly around 8 hours. I was getting conflicting results. Hope you can review my calculations and help me out if needed.

SteamKing said:
I'm not sure I follow your reasoning here. It seems much too complicated since you are applying Archimedes' Principle, i.e. the force applied by the water = the weight of water actually displaced by the sphere. It seems to me, the trick is to calculate the volume of the spherical cap which emerges from the hole in the bottom of the cylinder and subtract this volume from that of the whole sphere. This gives you the volume of displaced water.
That seems to be addressing how to find V in litjee's post, so it is complementary to litjee's work, but does not replace it. Litjee is correct that the pressure the water would exert over the area of the hole (were the sphere replaced by water) has to be subtracted.
Note in particular that the weight of water displaced does not depend on depth, but the subtracted term does. This explains why a sufficient depth of water will hold the sphere in place, while shallow water might allow it to float up.

Last edited:
Thanks everyone.

I went through the calculations again .. checking each and every line and voila the culprit was that I missed a factor of 2.

Just for completeness, I get the force by the liquid (curved area) to be ##\displaystyle \pi \rho _{w} g \left [ a^{2} h - \frac{1}{3} \left ( a^{2} \sqrt{R^{2} - a^{2}} + 2R^{3} + 2R^{2} \sqrt{R^{2} - a^{2}}\right ) \right ]##

Did I get it right?

Last edited:
iitjee10 said:
Thanks everyone.

I went through the calculations again .. checking each and every line and voila the culprit was that I missed a factor of 2.

Just for completeness, I get the force by the liquid (curved area) to be ##\displaystyle \pi \rho _{w} g \left [ a^{2} h - \frac{1}{3} \left ( a^{2} \sqrt{R^{2} - a^{2}} + 2R^{3} + 2R^{2} \sqrt{R^{2} + a^{2}}\right ) \right ]##

Did I get it right?
Try a couple of sanity checks: a=R, a=0. Do those answers look right?

Woopsie, there was a typo in the earlier post.

##a = 0## gives ##- \pi \rho_{w} g \cdot \frac{4}{3}R^{3}## which seems right.

## a = R ## gives ## \pi \rho_{w} g R^{2} h - \frac{2}{3} \pi R^{3} \rho_{w} g ## which also seems right
Positive indicates a downward force.

Yay! sanity checks are holding.. Thanks! I'm reasonably certain these are correct now.
It's strange how you forget doing basic things like sanity checks, checking for miscalculations when you've been out of touch from actually going to paper pencil calculations and not relying on a piece of software.

Last edited:
iitjee10 said:
Woopsie, there was a typo in the earlier post.

##a = 0## gives ##\pi \rho_{w} g \cdot \frac{4}{3}R^{3}## which seems right.

## a = R ## gives ## \pi \rho_{w} g R^{2} h - \frac{2}{3} \pi R^{3} \rho_{w} g ## which also seems right

Yay! sanity checks are holding.. Thanks! I'm reasonably certain these are correct now.
Yes, except that you are being inconsistent wrt sign. You ought to state which direction is positive for your answer.

True, will edit the relevant post.

Thanks! :

iitjee10 said:
I get the force by the liquid (curved area) to be ##\displaystyle \pi \rho _{w} g \left [ a^{2} h - \frac{1}{3} \left ( a^{2} \sqrt{R^{2} - a^{2}} + 2R^{3} + 2R^{2} \sqrt{R^{2} - a^{2}}\right ) \right ]##

Did I get it right?

For what it's worth, I get the same result.

1. What is the formula for calculating the pressure exerted by water on a partially exposed sphere?

The formula for calculating pressure exerted by water on a sphere is P = ρgh, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid.

2. How does the pressure exerted by water on a partially exposed sphere change as the depth of the water increases?

The pressure exerted by water on a partially exposed sphere increases with the depth of the water. This is because the weight of the water above the sphere increases with depth, resulting in a higher pressure being exerted on the sphere.

3. Are there any other factors that affect the pressure exerted by water on a partially exposed sphere?

Yes, there are other factors that can affect the pressure exerted by water on a partially exposed sphere. These include the size and shape of the sphere, the density of the fluid, and the acceleration due to gravity.

4. How can the pressure exerted by water on a partially exposed sphere be measured?

The pressure exerted by water on a partially exposed sphere can be measured using a pressure gauge or a manometer. These devices measure the force per unit area exerted by the water on the surface of the sphere.

5. What is the significance of understanding the pressure exerted by water on a partially exposed sphere?

Understanding the pressure exerted by water on a partially exposed sphere is important in various fields of science and engineering, such as fluid mechanics, oceanography, and marine engineering. It can also help in understanding the behavior of objects submerged in water and designing structures that can withstand underwater pressure.

• Introductory Physics Homework Help
Replies
18
Views
779
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
364
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
860
• Introductory Physics Homework Help
Replies
19
Views
419
• Introductory Physics Homework Help
Replies
63
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
998