Calculate the average force on the side walls of a container

[patric44]

Homework Statement

calculate the average force on the face "y"

Relevant Equations

F = 1/2 rho*g*h*A

Hi All, in the following problem:

the book solution

I don't understand why he added the term
$$
\rho g h_{z} A_{y}
$$
shouldn't it just be :
$$
F = 1/2 \rho g h_{y} A_{y}
$$

 

[haruspex]

What would you say the pressure is at height $h_y$ on the x side?

 

[patric44]

What would you say the pressure is at height $h_y$ on the x side?

I guess it will be $\rho g h_{y}$ but it will be a downward pressure.

 

[haruspex]

I guess it will be $\rho g h_{y}$ but it will be a downward pressure.

Pressure in a fluid has no direction, which is to say it acts equally in all directions.

 

[patric44]

Pressure in a fluid has no direction, which is to say it acts equally in all directions.

OK, so you mean the pressure due to the weight of water contained at the z part ($\rho g h_{z}$) will act also along the right side affecting the area $A_{y}$ hence its force = $\rho g h_{z} A_{y}$, am I correct in this interpretation.
the confusing thing also is, if I use integration it gives an equation that is the difference between two values but also outputs the correct answer.

 

[haruspex]

OK, so you mean the pressure due to the weight of water contained at the z part ($\rho g h_{z}$) will act also along the right side affecting the area $A_{y}$ hence its force = $\rho g h_{z} A_{y}$, am I correct in this interpretation.

Yes. The pressure at depth h below the surface is $\rho gh$, regardless of the route. So on y the pressure varies from $\rho gh_z$ at the top of y to $\rho gh_x$ at the bottom.

the confusing thing also is, if I use integration it gives an equation that is the difference between two values but also outputs the correct answer.

Would that be involving $h_x$ in the difference instead of $h_y$? If not, please post the details.