How To Calculate The Force Of Water Exerted On A Full Tank

  • Thread starter tomtomtom1
  • Start date
  • #1
160
8
Homework Statement:
To Calculate Force of Water Acting On A Full Tank Of Water
Relevant Equations:
Pressure = Density * Gravity * Depth Below Water Surface
Pressure = Force / Area
Hello Community

I was hoping someone could me with the following question:-

A rectangular tank contains water.
It is 12m long, 4m wide and has a height of 3m.
If the tank is full determine:-
1) The Force on the long side
2) The Force on the short side
3) The moment at the bottom of the long side and its centre.

I am just tying to figure out part 1 for now.

Trying to visualise the problem I would say that the area of the long side of the tank wall is 12m * 3m = 36m^2, and the question is asking how much force is the water exerting on this wall.

I know/worked out the following:-
- Water Density = 1000kg/m^3
- Pressure = Density * Gravity * Depth Below Water Surface
- Pressure = Force / Area
- The total volume of the water in the tank is 12 * 4 * 3 = 144.
- Using the equation Density = Mass/Volume, I can say that 1000 = Mass/144, by rearranging I get Mass to equal 144000kg.

I don't really know where to go from here, I have been told that the answer is 529.2kN - but I want to know how to get to the answer.
I am struggling because I know pressure increases with depth so my gut feeling is that some form of integration is needed but I have been told that integration can be used but is not needed.

Can someone help?

Thank you.
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,363
3,460
You need to do an integral because the pressure (force per unit area) depends on the distance from the surface. Go with your gut feeling, get the answer, then see if you can get the same answer without integration.
 
  • #3
160
8
You need to do an integral because the pressure (force per unit area) depends on the distance from the surface. Go with your gut feeling, get the answer, then see if you can get the same answer without integration.
I would even know what to integrate???
 
  • #4
160
8
Apologies, was meant to say I wouldn't even know what to integrate?
 
  • #5
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,363
3,460
Consider a horizontal strip at depth ##y## of width ##dy##.
1. What is the contribution ##dF## of that strip to the total force? (Pretend that ##dy## is small enough so that the pressure is the same over the strip.)
2. Add all such contributions to find the total force on the side.
 
  • #6
256bits
Gold Member
3,428
1,446
Would you be able to find the force on the bottom surface?
The pressure should be constant there.

Then contemplate, say for the long side, where the force varies with depth.
What is the pressure at the top surface?
What is the pressure at the bottom surface?
( You have already said that Pressure = ρgh so that should be easy )
Then, how should the force on the surface vary on an incremental area as one moves down the surface?
Then how can I add up all these forces to obtain the whole force on the surface.

And you have heard correctly.
For this problem to find the forces on the sides, integration is not needed.
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,386
6,930
I have been told that integration can be used but is not needed.
If you draw a graph of pressure against depth, what shape do you get? The total force will be the area under it,
 
  • #8
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,363
3,460
If you draw a graph of pressure against depth, what shape do you get? The total force will be the area under it,
In a graph of pressure (expressed in N/m2) against depth (expressed in m), the area under the curve comes out in units of N/m, not N. Setting aside the dimensional mismatch, what non-calculus argument identifies the "area under the curve" to the quantity sought without using the idea of adding infinitesimal contributions to obtain the whole? It seems to me that "because the integral can be easily found as the area of a simple geometrical shape" is insufficient justification to say that no integration is needed.
 

Related Threads on How To Calculate The Force Of Water Exerted On A Full Tank

  • Last Post
Replies
1
Views
5K
Replies
3
Views
15K
Replies
15
Views
1K
  • Last Post
Replies
6
Views
6K
Replies
1
Views
9K
  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
9
Views
5K
Replies
1
Views
12K
Replies
12
Views
1K
Top