Calculation of the force exerted by a liquid

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Discussion Overview

The discussion revolves around calculating the force exerted by a liquid on a slant wall of a vessel, considering the height, width, and density of the liquid. Participants explore the relevant formulas and the implications of pressure at different depths, focusing on the mathematical approach to derive the force in the x-direction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to solve for the force by integrating a differential slice of the liquid but struggles with the angle alpha's role in the problem.
  • Another participant points out the need to clarify whether the force in question is the normal force or the force in the x-direction.
  • It is confirmed that the force being calculated is in the x-direction.
  • Participants discuss the relationship between pressure, area, and force, with one noting that pressure depends on height, density, and gravity.
  • There is a clarification that pressure varies with depth, increasing as depth increases.
  • A suggestion is made to derive an equation for the force acting on a small strip of the target at varying depths, assuming pressure remains constant over the strip's height.
  • One participant proposes breaking down pressure into force and area, leading to an expression involving the area and height, but seeks clarification on how to express the width in terms of the vessel's dimensions.

Areas of Agreement / Disagreement

Participants generally agree on the principles of pressure variation with depth and the need for clarity in defining the force direction. However, there is no consensus on the specific expressions for area and width, indicating ongoing exploration and differing interpretations.

Contextual Notes

The discussion includes assumptions about the constancy of pressure over small differential heights and the need for clarity in defining variables such as width and area in relation to the vessel's dimensions.

Arka420
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Question: The height of the given vessel is h,and the width of the given vessel is b (as given in the diagram). The density of the liquid is r.Find the force exerted by the liquid on the slant wall.

Relevant formulae : P = F/A
F = Vdg

An attempt at the solution : I had tried taking a differential slice of the liquid and then integrating. But I am having trouble handling the alpha part. Where does the angle alpha fit into this question?
 

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Check the problem statement for any details regarding the force. What you've posted does NOT specify whether it's the normal force or force in the "x" direction.
 
Bystander said:
Check the problem statement for any details regarding the force. What you've posted does NOT specify whether it's the normal force or force in the "x" direction.
It's the force in the x-direction. Sorry for not mentioning earlier.
 
One step at a time: what does "P" depend upon?
 
P,or the pressure,depends upon the area A and the force F.
 
You're trying to find "F" by looking at "P." What else determines "P?"
 
Pressure depends upon the height from the free surface of the liquid h,the density of the liquid d and the acceleration due to gravity g.
 
Arka420 said:
Pressure depends upon the height from the free surface of the liquid h,the density of the liquid d and the acceleration due to gravity g.
Is P the same at all depths?

Chet
 
Nope,pressure isn't the same at all depths. As depth increases,so does pressure.
 
  • #10
Can you find an equation for the force (in the x direction) acting on small strip of the target between depth h and depth h+dh? Assume that dh is small enough so that pressure does not vary significantly from top to bottom of the strip.
 
  • #11
jbriggs444 said:
Can you find an equation for the force (in the x direction) acting on small strip of the target between depth h and depth h+dh? Assume that dh is small enough so that pressure does not vary significantly from top to bottom of the strip.
Yes,that was my idea. We can break P as F X A,which gives Vrg (r is the density,I'm not taking d as the density as it will be used to denote the differential height,as you said),which on further reduction gives Ahrg. I'm only having trouble finding A(or the area). If we assume the differential height is at a distance x from the free surface of the liquid, then x is the height. dx (the differential height,for ease of calculation,I am taking the variable as x) multiplied by the width gives the area A. My question is - what is the width? (It has to be an expression in b,or the width of vessel wall, I'm guessing)
 

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