Force on side of tank filled with liquid of variable specific weight

[CheesyPeeps]

Homework Statement

An open tank with a rectangular side 1 m wide and 4 m high is filled with a liquid of variable specific weight, γ, with γ = 50 + 2y (N/m3), where y is measured vertically downward from the free surface. Find the magnitude of the force on the side of the tank.

Relevant Equations

F = ∫pdA
p = ϒh

Homework Statement: An open tank with a rectangular side 1 m wide and 4 m high is filled with a liquid of variable specific weight, γ, with γ = 50 + 2y (N/m3), where y is measured vertically downward from the free surface. Find the magnitude of the force on the side of the tank.
Homework Equations: F = ∫pdA
p = ϒh

Using F_r = ∫pdA and p = ϒh to make F_r = ∫ϒhdA = ∫yhA, then substituting in the equation for ϒ, I end up with F_r = 200h + 8h². Note that I have equated y and h, as I will be taking them both to be the height of the wall's centroid.

Substituting in 2 for h (as 2m is the height of the wall's centroid), I end up with F_r = 432N. The correct answer is 421.3N, so I'm not far off, but the presence of the decimal tells me I've missed something.

Any help is appreciated.

 

[TSny]

p = ϒh

This equation is not correct when ϒ varies with depth. However, for an infinitesimal change in depth dh, the pressure change will be dp = ϒdh.

Also, when the density varies with depth, the centroid can no longer be used to obtain the force on the wall.

 

[mfig]

I think you need to integrate instead of using the results of the integration from a simple constant-density case.

For example:

$F = \int_0^{height} \int_0^{width} \gamma(y)_{avg} \cdot y \;dx dy$

 

[TSny]

I think you need to integrate instead of using the results of the integration from a simple constant-density case.

For example:

$F = \int_0^4 \int_0^1 \gamma(y) \cdot y \;dx dy$

The suggestion of integrating to find the force is a good one. However, the integral shown is not going to give the correct answer. The pressure at a depth $y$ is not given by $\gamma(y) \cdot y$.

We try to give just enough hint or help to the original poster so that they can do as much of the work on their own as possible. So, we should probably wait for @CheesyPeeps to let us know if they have been able to make any further progress.

 

[mfig]

The suggestion of integrating to find the force is a good one. However, the integral shown is not going to give the correct answer. The pressure at a depth $y$ is not given by $\gamma(y) \cdot y$.

We try to give just enough hint or help to the original poster so that they can do as much of the work on their own as possible. So, we should probably wait for @CheesyPeeps to let us know if they have been able to make any further progress.

You replied before I caught my omission of a subscript. The expression I used does indeed give the correct answer as supplied by CheesyPeeps. I thought it was vague enough that CheesyPeeps would have to do some thinking and figuring, and yet it might help push the solution process forward.

 

[TSny]

You replied before I caught my omission of a subscript. The expression I used does indeed give the correct answer as supplied by CheesyPeeps.

Yes. With the subscript "avg", it's right. Good.