# Magnitude of a Force on a hatch in a Vessel.

• Basch
In summary, the question involves finding the force on a hatch with a surface area of 1.00m^2 when the height of the liquid in the neck is 0 and the pressure difference between the inside and outside due to the 5.4 cm wall thickness is not taken into account. The vessel is made of acryllic with a density of 1185kg/m^3 and the liquid in the vessel (LAB) has a density of 860 kg/m^3. The water on the outside has a density of 1000 kg/m^3 and the wall thickness of the vessel is 5.4 cm. The formula to use is Pressure = Force/Area, and the attempt at solving the question involved
Basch

## Homework Statement

We're told the density of ever material involved here, the liquid in the vessel, called "LAB", the water on the outside, and the acryllic that the vessed is made of.
LAB: 860 kg/m^3
Water: 1000 kg/m^3
Acryllic: 1185kg/m^3
Also, the wall thickness of the vessel is 5.4 cm. The height of the neck is labeled delta h, we're also told to not take the pressure difference between the inside and the outside, due to the 5.4cm wall.

This is a three part question, but I'm only looking for help on the first part, which asks us to find the force on the hatch, which is 1.00m^2 big, when delta h = 0.

## Homework Equations

We need to use Pressure = Force/Area for this, solving for force gives us
Force = Pressure*Area
density * gravity * height = pressure

## The Attempt at a Solution

My attempt to solve the question, was to solve for the value of pressure of the inside of the system, and then use that to solve for force. Since the height of liquid in the neck is 0, we have no liquid in the neck, so we only have to consider what's inside the circular area, which has a height of 12m, density of 860kg/m^3, and gravity is 9.81 m/s^2, so which we have to add the pressure of 1 atm due to the air at the top, 1.01e-5 Pa.

Which gives us: 101239 Pa * 1.00 m^2, so the answer is 1.01e5 N?

Edit: 1 atm is 101kPa, making it 202564 Pa * 1.00 m^2, to be 2.03e5 N.

Last edited:
1 atm due to the air at the top, 1.01e-5 Pa
Isn't air pressure about 100 kilopascals?

The wording of the question is bothering me. Are you sure you are to ignore the force of the water pressure on the outside?

Delphi51 said:
Isn't air pressure about 100 kilopascals?

The wording of the question is bothering me. Are you sure you are to ignore the force of the water pressure on the outside?

I'm not 100% sure about the air pressure, so you're probably right, as for the second part, i'll quote it exactly from the question.

"treat the vessel as a thin shell, ie, do not take the pressure difference between the inside and outside due to the 5.4 cm wall thickness into account"

"treat the vessel as a thin shell, ie, do not take the pressure difference between the inside and outside due to the 5.4 cm wall thickness into account"
I think this just means that you take the radius to be 6 m whether you are doing the inside or the outside.

What exactly is the question? If it says find the total or net force on the hatch, you would take the difference between inside and outside forces.

We're looking to find the magnitude of the force on the hatch. so looks like i should throw the pressure of water into the mix, too.

I think so. That is the force you would be interested in if you were thinking of opening the hatch!

Alright, thanks a lot delphi :D

## 1. What is the magnitude of a force on a hatch in a vessel?

The magnitude of a force on a hatch in a vessel refers to the strength or size of the force that is acting on the hatch. It is typically measured in units of Newtons (N) in the metric system or pounds (lbs) in the imperial system.

## 2. How is the magnitude of a force on a hatch in a vessel calculated?

The magnitude of a force on a hatch in a vessel can be calculated using Newton's Second Law of Motion, which states that force equals mass multiplied by acceleration (F=ma). Alternatively, it can also be calculated using the formula F=P/A, where F is the force, P is the pressure, and A is the area of the hatch.

## 3. What factors affect the magnitude of a force on a hatch in a vessel?

The magnitude of a force on a hatch in a vessel can be affected by various factors, such as the weight of the hatch, the pressure inside the vessel, the area of the hatch, and the direction and intensity of external forces acting on the vessel.

## 4. Why is it important to consider the magnitude of a force on a hatch in a vessel?

It is important to consider the magnitude of a force on a hatch in a vessel because it can help determine the strength and stability of the hatch, as well as the overall structural integrity of the vessel. Excessive or insufficient force can lead to damage or failure of the hatch, which can have serious consequences for the vessel and its occupants.

## 5. How can the magnitude of a force on a hatch in a vessel be controlled or adjusted?

The magnitude of a force on a hatch in a vessel can be controlled or adjusted by changing the variables that affect it, such as the weight of the hatch, the pressure inside the vessel, and the external forces acting on the vessel. This can be achieved through various methods, such as adjusting the contents of the vessel, modifying the design of the hatch, or using external supports or reinforcements.

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