# Fluid Dynamics - Hydrostatic Pressure #2

• Lukas_RSA
In summary: Lukas van RooyenIn summary, the sliding bolt is pushing against a force of 24.525kN from the water. The hinges resist this force and keep the door in place.
Lukas_RSA

## Homework Statement

a Flat rectangular door in a mine is submerged from one side in water. The door dimensions are 2m high, 1m wide and the water level is 1.5m higher than the top of the door. The door has two hinges on the vertical edge, 160mm from each corner and a sliding bolt on the other side in the middle. Calculate the forces on the hinges and sliding bolt.
(Hint: consider the door from a side view and from a plan view respectively and take moments about a point each time.)

## The Attempt at a Solution

i am able to calculate the pressure at the sliding bolt but i don't know how to approach the problem of the hinges, please help me to understand this concept.Sliding bolt:
Fsb = p(water density) g 'y A(door) / 2 'y = 2/2 + 1.5
= (1000)(9.81)(2.5)(2x1) / 2 'y = 2.5
= 24.525 kN

now i thought that to get the answer to the hinges you simply just change 'y to the desired distance, 'y = 1.5+ 0.160
and for the second hinge 'y = 1.5 + ( 2 - 0.160 )

the ansers i get is not correct, the textbook has these questions as excercises after each module.

i hope to hear from you soon.

regards
Lukas van Rooyen

Lukas_RSA said:

## Homework Statement

a Flat rectangular door in a mine is submerged from one side in water. The door dimensions are 2m high, 1m wide and the water level is 1.5m higher than the top of the door. The door has two hinges on the vertical edge, 160mm from each corner and a sliding bolt on the other side in the middle. Calculate the forces on the hinges and sliding bolt.
(Hint: consider the door from a side view and from a plan view respectively and take moments about a point each time.)

## The Attempt at a Solution

i am able to calculate the pressure at the sliding bolt but i don't know how to approach the problem of the hinges, please help me to understand this concept.Sliding bolt:
Fsb = p(water density) g 'y A(door) / 2 'y = 2/2 + 1.5
= (1000)(9.81)(2.5)(2x1) / 2 'y = 2.5
= 24.525 kN

now i thought that to get the answer to the hinges you simply just change 'y to the desired distance, 'y = 1.5+ 0.160
and for the second hinge 'y = 1.5 + ( 2 - 0.160 )
The hinges and the sliding bolt keep the water pressure from pushing the door our of its frame. The force on the hinges or the sliding bolt is not just ρgh * Adoor

Remember to use the Hint given in the Problem Statement.

thanks for the feedback

i tried all possible angles to tackles this problem but i cant, so if someone can help me and show me step by step with notes i would really appreciate it.

regards
Lukas

Sorry, but the Rules of PF state that you must do your own work.

If you care to share your best attempt at solving this problem, perhaps you might get a helpful suggestion which will lead you to a solution.

Realy

i am not asking for the answer i have the answer, i am asking to explain how to get to the answer i don't grasp the concept, shees what a hasstle for some help on this site, luckily i found CHEGG.com there i even speak to a tutor who explains to me the concept.

but yey well thanks for wasting my time, will not use this website again.

cheers

## 1. What is the difference between hydrostatic pressure and dynamic pressure in fluid dynamics?

Hydrostatic pressure is the pressure exerted by a fluid at rest, whereas dynamic pressure is the pressure exerted by a fluid in motion. Hydrostatic pressure is dependent on the depth and density of the fluid, while dynamic pressure is dependent on the velocity and density of the fluid.

## 2. How is hydrostatic pressure calculated in a fluid system?

Hydrostatic pressure can be calculated by multiplying the density of the fluid by the acceleration due to gravity and the depth of the fluid. This is known as the hydrostatic equation: P = ρgh, where P is the hydrostatic pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid.

## 3. How does hydrostatic pressure affect objects submerged in a fluid?

Objects submerged in a fluid experience hydrostatic pressure acting on all sides. This pressure increases with depth and can cause objects to float, sink, or be compressed depending on their density and the density of the fluid.

## 4. What is Pascal's principle in fluid dynamics?

Pascal's principle states that pressure applied to a confined fluid is transmitted equally in all directions. This principle is the basis for hydraulic systems, where a small amount of force applied to a small piston can generate a larger force on a larger piston.

## 5. How does hydrostatic pressure vary with depth in a fluid?

Hydrostatic pressure increases with depth in a fluid due to the weight of the fluid above it. This is because the weight of the fluid creates a force that is transmitted equally in all directions, resulting in an increase in pressure with depth.

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Thermodynamics
Replies
11
Views
405
• Introductory Physics Homework Help
Replies
11
Views
6K
• Mechanics
Replies
40
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
975
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
9K
• Engineering and Comp Sci Homework Help
Replies
22
Views
2K