# Fluid Mechanics: resultant force of oil and water on a door

• Frankenstein19
In summary: Therefore, you can use the density of water and the atmospheric pressure to calculate the force exerted by the water. As for the Ixx, it does not matter how the door will open as long as you take into account the first moment of the force. You can find the average point at which the total force is being applied by dividing the total first moment by the total net force.
Frankenstein19

## The Attempt at a Solution

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My questions:

1. By resultant force, does that mean the force exerted on the side of the water + force exerted on side of the oil? Because I'd think that the force on the side of the water would be different than the force exerted on the door on the side of the oil.

2. When I find the force exerted by the oil, I multiple the SG of the oil by the density of water to get the specific weight of the oil, but in chegg solutions I've been seeing people use 31.75lb/ft^3 and saying that's the specific weight of AIR. Can I use the density of water?

3. When finding the force exerted by the water on the door, I should be considering the 10psig of the air, right?

4. When finding Ixx, I should consider the way the door will open, right? How do I know which way it'll open? I was taking b=12 for Fr so I carried that convention when doing Ixx and used 1/12ba^3 instead of 1/12b^3a which was what I was seeing in the chegg solutions.

Any help is greatly appreciated.

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Frankenstein19 said:
[...]
My questions:

1. By resultant force, does that mean the force exerted on the side of the water + force exerted on side of the oil? Because I'd think that the force on the side of the water would be different than the force exerted on the door on the side of the oil.
Yes. (as vectors or at least magnitude and indicated direction.) and yes, the two forces will be different in general. The door is locked in place and presumably if unlocked would open one way or the other depending on the net resultant force.

2. When I find the force exerted by the oil, I multiple the SG of the oil by the density of water to get the specific weight of the oil, but in chegg solutions I've been seeing people use 31.75lb/ft^3 and saying that's the specific weight of AIR. Can I use the density of water?
You should work out the force from each side independently and then (vectorally as in pay attention to direction) add them. The presence of the door isolates the two sides so nothing about the water side affects the oil side et vice versa.
3. When finding the force exerted by the water on the door, I should be considering the 10psig of the air, right?
Right. As also with the oil you have to consider atmospheric pressure. The only question, and an ambiguity I see in the problem is whether the 10psi is relative to vacuum or relative to atmospheric pressure. i.e. is the pressure above the water Atmospheric pressure plus 10psi or just 10psi. I would assume the simpler case (relative to atmospheric) so that you don't need to input the additional information of what 1 atm is defined to be in psi.
4. When finding Ixx, I should consider the way the door will open, right? How do I know which way it'll open? I was taking b=12 for Fr so I carried that convention when doing Ixx and used 1/12ba^3 instead of 1/12b^3a which was what I was seeing in the chegg solutions.
Any help is greatly appreciated.

I don't think how it will open will matter. The positioning of the force is at the center of force so that the force on the door plus torque around any pivot point will be the same as if all the force were applied at a single point. To find this you'll need the first moment of the force by integrating the differential force due to pressure times the position at which that pressure is being applied. Find the total first moment by adding (including sign of the normal force so really subtracting) the contributions from both water side and oil side. Then divide that moment by the total net force to find the average point at which this total force is being applied.

jambaugh said:
The only question, and an ambiguity I see in the problem is whether the 10psi is relative to vacuum or relative to atmospheric pressure. i.e. is the pressure above the water Atmospheric pressure plus 10psi or just 10psi. I would assume the simpler case (relative to atmospheric) so that you don't need to input the additional information of what 1 atm is defined to be in psi.
The diagram says 10 psig (i.e., gauge pressure)

## 1. What is fluid mechanics?

Fluid mechanics is the branch of physics that deals with the study of how fluids (liquids and gases) behave and interact with forces.

## 2. How is the resultant force of oil and water calculated?

The resultant force of oil and water on a door can be calculated by using the formula F = ρgV, where ρ is the density of the fluid, g is the acceleration due to gravity, and V is the volume of the fluid that is in contact with the door.

## 3. Why is the resultant force of oil and water important in fluid mechanics?

The resultant force of oil and water is important in fluid mechanics because it helps us understand how fluids interact with objects and how they can cause movement or deformation.

## 4. How does the density of the fluids affect the resultant force?

The density of the fluids plays a significant role in determining the resultant force. The greater the density, the greater the resultant force will be. This is because denser fluids have more mass and therefore exert more force on the object they are in contact with.

## 5. Can the resultant force of oil and water be negative?

Yes, the resultant force of oil and water can be negative. This means that the forces acting on the door are in opposite directions, causing a net force of zero or a negative value. This can happen when the densities of the oil and water are similar, resulting in a canceling out of forces.

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