Calculation of the force exerted by a liquid

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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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  • #2
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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.
 
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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.
It's the force in the x-direction. Sorry for not mentioning earlier.
 
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One step at a time: what does "P" depend upon?
 
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P,or the pressure,depends upon the area A and the force F.
 
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You're trying to find "F" by looking at "P." What else determines "P?"
 
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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.
 
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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
 
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Nope,pressure isn't the same at all depths. As depth increases,so does pressure.
 
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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.
 
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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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