Resultant force and its point of action

Click For Summary

Discussion Overview

The discussion revolves around calculating the resultant force acting on an inclined surface of a storage tank and determining its point of action. Participants explore various methods and equations related to fluid pressure, geometry, and moments of inertia, with a focus on both theoretical understanding and practical application.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents an initial calculation for the resultant force and point of action, using pressure equations and trigonometric relationships.
  • Another participant questions the necessity of finding the angle and suggests using Pythagorean theorem directly, while also raising concerns about the dimensional consistency of the equations used for point of action.
  • A subsequent post indicates a second attempt at the calculations, adjusting the pressure and resultant force values, but still faces challenges with the point of action.
  • Further clarification is sought regarding the shape of the area in question, with differing interpretations of the cross-section being discussed.
  • Another participant emphasizes the importance of understanding the point of application and line of action in vector analysis.
  • In later attempts, participants refine their calculations for pressure, resultant force, and moment of inertia, but discrepancies in values and methods persist.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate methods for calculating the resultant force and point of action, with no consensus reached on the correct approach or final values. The discussion remains unresolved with multiple competing interpretations and calculations presented.

Contextual Notes

Participants highlight potential limitations in their calculations, including assumptions about the shape of the area and the application of various equations. There are also unresolved questions regarding the dimensional consistency of certain terms used in the calculations.

Who May Find This Useful

This discussion may be useful for students and practitioners in engineering and physics who are dealing with fluid mechanics, particularly in the context of calculating forces on inclined surfaces and understanding the implications of geometry in such scenarios.

MCTachyon
Messages
51
Reaction score
3

Homework Statement


A storage tank has the cross-sectional shape shown attached and is of 1m breadth. Calculate the resultant force acting on the inclined surface AB and its point of action.

The density of the liquid is 900kgm–3 and take g = 9.81ms–2.

hahahahahahahahahahahahah.png


Homework Equations


P = ρgh
F = ρA
Point of action = hc + (Ig Sin(θ)2 / A hc)

The Attempt at a Solution


Length of AB:

Tanθ = (O/A)
θ = Tan-1 (1.5/1)
θ = 56.31°

Cos(56.31) = 1/AB
AB = 1/Cos(56.31)
AB = 1/0.5547
AB = 1.8m

Pressure at centroid of AB:

P = ρgh
P = 900 x 9.81 x (4 + (1.8/2))
P = 900 x 9.81 x 4.9
P = 432621Pa

Resultant force at AB:

F = ρA
F = 432621 x (1.8 x 1)
F = 77818N
F = 77.82kN

This is bit when I get a bit lost.

Point of Action:

= hc + (Ig Sin(θ)2 / A hc)

Where Ig = db/12 = 1.8/12 = 0.15

= 4.9 + (0.15 x Sin(90-56.31)2 / 1.8 x 1 x 4.9)

= 4.9 + 0.00523

= 4.90523m

Therefore force of 77.82kN is acting 4.90523m from top of storage tank.
------------------------------------------------------------------------------------------------------

Am I on the right track with my method? Pretty confident till working out the Point of Action.
 

Attachments

  • hahahahahahahahahahahahah.png
    hahahahahahahahahahahahah.png
    2 KB · Views: 789
Physics news on Phys.org
MCTachyon said:
θ = 56.31°
It is rarely necessary to find the angle. You can go straight from one trig ratio to another. E.g. use Pythagoras.
MCTachyon said:
P = 900 x 9.81 x (4 + (1.8/2))
Pressure varies as depth. The 1.8m is not vertical.
MCTachyon said:
Where Ig = db/12 = 1.8/12 = 0.15
This doesn't seem right. Shouldn't there be a quadratic term? Certainly it seems like the resulting equation for point of action is dimensionally inconsistent...

... Or maybe I am misreading it. Please make it clearer by use of parentheses and define all the variables.
 
Last edited:
  • Like
Likes   Reactions: MCTachyon
Thank you.

Lots of food for thought there. I will go back and retry.

Thanks again.
 
Second attempt:

Length of AB:

Pythagorean Theorem:

A2 + B2 = C2

C = √ (12 + 1.52)
C = √3.25
C = 1.8m

Pressure at centroid of AB:

P = ρgh

P = 900 x 9.81 x (4 + (1/2))
P = 39731 Pa
P = 39.73 kPa

Resultant force at AB:

F = ρA
A = (h x b) / 2
F = 39731 x ((1 x1.5) / 2)
F = 39731 x 0.75
F = 29798 N
F = 29.80 kN

Point of Action:

Area Moment of Inertia:

Iy = IR + Ar2

Where:

IR = (d x b3) / 36

IR = (1 x 1.53) / 36
IR = 3/32 = 0.09375m4

Therefore:

Iy = 3/32 + ((1 x 1.5) / 2) x 4.52
Iy = 15.28m4

Radius of Gyration:

Ry = √(I / A)
Ry = √ (15.28 / 0.75)
Ry = √ 20.37
Ry = 4.513m

Therefore force of 29.80 kN is acting 4.513m from top of storage tank.

-----------------------------------------------------------------------------------------------------------

Any closer to sussing out the method?

Thanks for your help with this.
 
MCTachyon said:
F = ρA
PA not ρA.
MCTachyon said:
A = (h x b) / 2
MCTachyon said:
(d x b3) / 36
What shape do you think the area is?
 
  • Like
Likes   Reactions: MCTachyon
haruspex said:
What shape do you think the area is?

ABC, so a right angled triangle.
 
MCTachyon said:
ABC, so a right angled triangle.
That's not how I interpret the diagram.
C seems to be just a reference point for showing the width and height of the sloped section. The text says it is a cross-section, implying the "breadth 1m" is into the page. The area of interest is therefore a rectangle, seen side-on as AB.
 
Right then, this is starting to make some sense now.

It is late where I am, I will come back tomorrow with some new numbers and see if this info has put me on the right track.

Thanks again for your help.
 
With vectors you have a point of application, and a line of action.
 
  • #10
Third attempt:

Pressure at centroid of AB:

P = ρgh

P = 900 x 9.81 x (4 + (1/2))
P = 39731 Pa
P = 39.73 kPa

Resultant force at AB:

F = PA
A = (h x b)
F = 39731 x (1 x1.5)
F = 39731 x 1.5
F = 59597 N
F = 59.60 kN

Point of Action:

Area Moment of Inertia:

Iy = IR + Ar2

Where:

IR = (d x b3) / 12

IR = (1 x 1.53) / 12
IR = 9/32 = 0.28125m4

Therefore:

Iy = 9/32 + (1 x 1.5) x 4.52
Iy = 30.66m4

Radius of Gyration:

Ry = √(I / A)
Ry = √ (30.66 / 1.5)
Ry = √ 20.44
Ry = 4.52m

Therefore force of 59.60 kN is acting 4.52m from top of storage tank.

------------------------------------------------------------------------------------------------------------------

Getting closer?

Thanks again for all your help.
 

Similar threads

Book plots position & force on the same graph, still gets right answer
  • · Replies 8 ·
Replies
8
Views
1K
How Much Weight Is Needed to Keep the Water Gate Closed?
  • · Replies 1 ·
Replies
1
Views
3K
To find the resultant of forces on a lamina
  • · Replies 5 ·
Replies
5
Views
3K
To find the magnitude of the resultant of forces around a triangle
  • · Replies 4 ·
Replies
4
Views
2K
Fluid mechanics civil engineering
  • · Replies 1 ·
Replies
1
Views
2K
Fluid Mechanics: Net Force & Point of Action Calculation
  • · Replies 2 ·
Replies
2
Views
1K
Find moment arm of resultant force of two forces applied at two points
  • · Replies 3 ·
Replies
3
Views
1K
Hydrostatics Pressure - Finding thrust and resultants
  • · Replies 1 ·
Replies
1
Views
2K
Heat transfer and Fluid Flow Struggling Problems
  • · Replies 25 ·
Replies
25
Views
5K
Finding 2 of 3 Forces Given 1 and a Resultant
  • · Replies 9 ·
Replies
9
Views
4K