A tank contains water up to a depth of 1.5 m. The length and width of the tank are 4 m and 2 m respectively. The tank is moving up an inclined plane with a constant acceleration of 4 m/s². The inclination of the plane with the horizontal is 30°. Find: (i) the angle made by the free surface of water with the horizontal, and (ii) the pressure at the bottom of the tank at the front and rear ends.

Tank Accelerating Up an Inclined Plane

Problem Statement

Given Data

Initial depth of water, $ h = 1.5 \, \text{m} $
Length of tank, $ L = 4 \, \text{m} $
Width of tank, $ b = 2 \, \text{m} $
Acceleration along incline, $ a = 4 \, \text{m/s}^2 $
Inclination of plane, $ \alpha = 30^\circ $

Diagram of Tank on Inclined Plane

Diagram of a tank accelerating up an inclined plane

Solution

The acceleration along the inclined plane is resolved into horizontal ($a_x$) and vertical ($a_y$) components.

$$a_x = a \cos \alpha = 4 \times \cos 30^\circ \approx 3.464 \, \text{m/s}^2$$ $$a_y = a \sin \alpha = 4 \times \sin 30^\circ = 2 \, \text{m/s}^2$$

(i) Angle of the Free Surface ($\theta$)

The angle of the free surface with the horizontal depends on both the horizontal and vertical components of acceleration.

$$\tan \theta = \frac{a_x}{g + a_y}$$ $$\tan \theta = \frac{3.464}{9.81 + 2} = \frac{3.464}{11.81} \approx 0.2933$$ $$\theta = \tan^{-1}(0.2933) \approx 16.35^\circ$$

(ii) Pressure at Front and Rear Ends

First, we find the change in depth at the ends of the tank, which occurs over half the length ($L/2 = 2$ m).

$$\text{Change in depth} = \frac{L}{2} \times \tan \theta = 2 \times 0.2933 = 0.5866 \, \text{m}$$

Now, calculate the new depths at the rear ($h_2$) and front ($h_1$) ends.

$$h_2 (\text{rear}) = h + \text{change} = 1.5 + 0.5866 = 2.0866 \, \text{m}$$ $$h_1 (\text{front}) = h - \text{change} = 1.5 - 0.5866 = 0.9134 \, \text{m}$$

The pressure at the bottom is affected by the vertical component of acceleration ($a_y$).

$$p_D (\text{rear}) = \rho (g + a_y) h_2$$ $$p_D = 1000 \times (9.81 + 2) \times 2.0866$$ $$p_D \approx 24642.7 \, \text{N/m}^2$$

$$p_A (\text{front}) = \rho (g + a_y) h_1$$ $$p_A = 1000 \times (9.81 + 2) \times 0.9134$$ $$p_A \approx 10787.2 \, \text{N/m}^2$$

Final Results:

(i) The angle of the free surface with the horizontal is $ \theta \approx 16.35^\circ $.

(ii) The pressure at the rear bottom is $ p_D \approx 24642.7 \, \text{N/m}^2 $.

(ii) The pressure at the front bottom is $ p_A \approx 10787.2 \, \text{N/m}^2 $.

Explanation of Concepts

When a tank of fluid accelerates up an inclined plane, the acceleration vector has both horizontal and vertical components.

The horizontal component ($a_x$) causes the free surface of the water to tilt, similar to a tank accelerating on a flat surface.
The vertical component ($a_y$) acts in the same direction as gravity, effectively increasing the gravitational force on the fluid. This is why the term $ (g + a_y) $ is used. It increases the pressure at any given depth and also influences the angle of the free surface.

The final tilted surface is a result of the balance between the inertial force from horizontal acceleration and the combined downward force from gravity and vertical acceleration.