Question 1.5.1: A rectangular tank 3 m long, 2 m wide and 2 m deep contains ......
A rectangular tank 3 m long, 2 m wide and 2 m deep contains water to a depth of 1.25 m. If it is accelerated horizontally at 3 m/s² in the direction of its length, find (a) (i) the inclination of the water surface with the horizontal, (ii) depth of water at the two ends, (iii) total force on the sides at the two ends of the tank and (iv) net horizontal force on the fluid mass.
(b) What maximum acceleration is permissible in this case?
(c) How much water would spill out if the tank is accelerated at 8 m/s²?
(a) Equation 1.5.3 gives the slope of the free surface; the angle θ is \tan^{–1} (3/9.81) = 17°.
\mathrm{z/y=\tan \theta =-a_y/(a_z+g)} (1.5.3)
∴ AA′ = BB′ = (3/2) tan θ = 0.4587 m.
Depth of water at the rear end = 1.5 + 0.4587 = 1.7087 m and Hydrostatic force on rear side of the tank = w A z = 9810 × (2 × 1.7087) × 1.7087/2 = 28.642 kN
Depth of water at the front end = 1.5 – 0.4587 = 0.7913 m
Hydrostatic force on front side of the tank = 9810 × (2 × 0.7913) × 0.7913 /2 = 6.143 kN.
∴ Net horizontal force = 28.642 – 6.143 = 22.499 kN, opposite to the direction of \mathrm{α_y}.
(b) The tank can be accelerated further till the water level at the rear side just touches the top of the tank. (Further increase in acceleration will spill out the water from the tank.) For this condition, the slope of the free surface is tan θ′ = (2 – 1.25)/1.5 = ½. This gives a_y = g × tan θ′ = 4.905 m/s².
(c) In this case tan θ ′′ = 8/9.81, and the free surface will be CD. As seen in the Fig., there is no water on the front end. From the triangle OCD, OD = OC/tan θ” = 2.4525 m.
∴ Volume of water in the tank = ½ × OD × OC × width of tank = ½ × 2.4525 × 2 × 2 = 4.905 m³.
Volume of water originally present = 3 × 1.25 × 2 = 7.5 m³.
→ Volume of water spilled out from the tank = 7.5 – 4.905 = 2.595 m³.