Components of the acceleration along the horizontal and vertical directions are

\mathrm{a_y = 5 \cos 30 = 4.33  and  a_z = 5 \sin 30 = 2.5  m/s^2.}

Slope of the water surface is then \mathrm{\tan θ = a_y /(a_z + g) = 0.3519}
→                                           θ = 19.38°.

Depth at b = 2 – (3/2) tan θ = 1.472 m.

The pressure at b now can be determined using the equation for static fluid. However, there is acceleration in the z-direction too that would add to the pressure. We may use Eq. 1.5.2 without considering the first term in y as effect of [latex]\mathrm{a_y} has already been accounted while using the depth calculated above. Moreover, while going down, z is negative. With these comments, we get

\mathrm{p=-\rho [\alpha _y y+(\alpha _z+g)z]+p_0,}                   (1.5.2)

\mathrm{pb/w = (1 + a_z/g) × 1.472 = 1.847  m.}

or,

\mathrm{pb = 1.847 × 9787 × 0.8 = 14.463  kN/m^2} (gauge).