Calculate the average force on the side walls of a container

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 :
$$
F = 1/2 \rho g h_{y} A_{y}
$$

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

haruspex said:

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.

patric44 said:

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.

haruspex said:

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.

patric44 said:

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.

patric44 said:

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.