Linearly accelerating hydrostatic fluid

[wahaj]

Homework Statement

A cart is acclerating to the right with $a=3m/s^2$. Fluid is in hydrostatic state. Find the force on the back wall. Cart goes .8m into the page. In the image dotted line is free surface when cart is stationary.

Homework Equations

$$\vec{\nabla P}=\rho (\vec{g} - \vec{a} ) \\<br /> dP = d \vec{R} \bullet \vec{\nabla P} \\<br /> F = \iint P \hat{n} dA$$

The Attempt at a Solution

$$\int_1^2 dP = \int_1^2 d \vec{R} \bullet \vec{\nabla P} \\ <br /> \int_1^2 dP = \int_1^2 (dx \hat{i} + dz \hat{k}) \bullet \rho (-g \hat{k} - a \hat{i} ) \\ <br /> P_2 - P_1 = - \rho g(z_2 - z_1) - \rho a (x_2 - x_1) \\ <br /> P_2 - P_1 = 0-0 = 0 \\ <br /> g(z_2 - z_1) = a (x_2 - x_1) \\ <br /> x_2 = 2 \ ; \ x_1 = 0 \ ; \ z_2 = 0.5+h \ ; \ z_1 = 0.5-h \\<br /> g(2h) = a(2) \\ <br /> h = 3/9.81 = 0.3058 m \\ <br /> <br /> F = \iint P \hat{n} dA \\ <br /> P = -\rho g z + C \\ <br /> P= \rho g \ when \ z = 0 \\ <br /> \rho g = C \\ <br /> P = \rho g (1-z) \\ <br /> F = \int_0^.8 \int_0^.8058 \rho g (1-z) \hat{i} dzdy \\<br /> F = 3.776 kN$$
The actual answer is 2.55 kN. I think my mistake lies in me not including the pressure due to the cart accelerating to the right. How would I account for that acceleration?

 

[wahaj]

I uploaded the image. Sorry about that.

 

[Chestermiller]

What is the volume of water being accelerated (it is water, correct?)? What is the mass of water being accelerated? By how much does the force exerted by the left wall have to exceed the force exerted by the right wall in order to accelerate the water at 3 m/s²? What is the force exerted by the left wall when the fluid is not being accelerated? What is the force exerted by the right wall when the fluid is not being accelerated?

Chet