Question

A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated is 4 m/s^2. Determine the allowable initial water height in the tank if no water is to spill out during acceleration.

Answer 1

Answer:

$h_{max} = 51.8 cm$

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

$\theta$ = water surface angle with horizontal surface

$a_x =$accelration in x direction

$a_z =$accelration in z direction

slope in xz plane is

$tan\theta = \frac{a_x}{g +a_z}$

$tan\theta = \frac{4}{9.81+0}$

$tan\theta =0.4077$

the maximum height of water surface at mid of inclination is

$\Delta h = \frac{d}{2} tan\theta$

            $=\frac{0.4}{2}0.4077$

$\Delta h 0.082 cm$

the maximu height of wwater to avoid spilling is

$h_{max} = h_{tank} -\Delta h$

            = 60 - 8.2

$h_{max} = 51.8 cm$

the height requird if no spill water is $h_{max} = 51.8 cm$