Answer:
[tex]h_{max} = 51.8 cm[/tex]
Explanation:
given data:
height of tank = 60cm
diameter of tank =40cm
accelration = 4 m/s2
suppose x- axis - direction of motion
z -axis - vertical direction
[tex]\theta[/tex] = water surface angle with horizontal surface
[tex]a_x = [/tex]accelration in x direction
[tex]a_z = [/tex]accelration in z direction
slope in xz plane is
[tex] tan\theta = \frac{a_x}{g +a_z}[/tex]
[tex]tan\theta = \frac{4}{9.81+0}[/tex]
[tex]tan\theta =0.4077[/tex]
the maximum height of water surface at mid of inclination is
[tex]\Delta h = \frac{d}{2} tan\theta[/tex]
[tex] =\frac{0.4}{2}0.4077[/tex]
[tex] \Delta h 0.082 cm[/tex]
the maximu height of wwater to avoid spilling is
[tex]h_{max} = h_{tank} -\Delta h[/tex]
= 60 - 8.2
[tex]h_{max} = 51.8 cm[/tex]
the height requird if no spill water is [tex]h_{max} = 51.8 cm[/tex]
