A rectangular tank contains water to a depth of 1.5 m. Find the horizontal acceleration in the direction of its length so that the spilling of water is just on the verge of taking place. Also calculate the total forces on each end of the tank in each case and verify the results.

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Accelerating Tank - Spilling and Exposure Cases

Problem Statement

A rectangular tank (L=6m, W=2.5m, D=2m) contains water to a depth of 1.5 m. Find the horizontal acceleration in the direction of its length so that (i) the spilling of water is just on the verge of taking place, (ii) the front bottom corner is just exposed, (iii) the bottom is exposed upto its mid-point. Also calculate the total forces on each end of the tank in each case and verify the results.

Given Data

  • Length of tank, \( L = 6 \, \text{m} \)
  • Width of tank, \( b = 2.5 \, \text{m} \)
  • Total depth of tank = 2 m
  • Initial depth of water, \( h_{initial} = 1.5 \, \text{m} \)

Solution

(i) Case 1: Spilling is on the Verge

For spilling to be on the verge of taking place, the water level must rise to the top rear corner of the tank. The available vertical distance is \(2 - 1.5 = 0.5\) m, over half the tank length (\(L/2 = 3\) m).

Diagram for Case 1: Spilling on the verge

(a) Horizontal Acceleration

$$ \tan \theta = \frac{\text{rise}}{\text{run}} = \frac{2 - 1.5}{L/2} = \frac{0.5}{3} \approx 0.1667 $$ $$ a = g \tan \theta = 9.81 \times 0.1667 \approx 1.635 \, \text{m/s}^2 $$

(b) Forces on Each End

Depth at rear end, \(h_{rear} = 2\) m. Depth at front end, \(h_{front} = 1.5 - 0.5 = 1\) m.

$$ F_{rear} = \rho g A_{rear} \bar{h}_{rear} $$ $$ = 1000 \times 9.81 \times (2 \times 2.5) \times (2/2) = 49050 \, \text{N} $$
$$ F_{front} = \rho g A_{front} \bar{h}_{front} $$ $$ = 1000 \times 9.81 \times (1 \times 2.5) \times (1/2) = 12262.5 \, \text{N} $$

(c) Verification

$$ F_{net} = F_{rear} - F_{front} = 49050 - 12262.5 = 36787.5 \, \text{N} $$ $$ \text{Volume} = L \times b \times h_{initial} = 6 \times 2.5 \times 1.5 = 22.5 \, \text{m}^3 $$ $$ \text{Mass} = \rho \times V = 1000 \times 22.5 = 22500 \, \text{kg} $$ $$ M \times a = 22500 \times 1.635 = 36787.5 \, \text{N} $$

The net force matches the force required for acceleration.

(ii) Case 2: Front Bottom Corner Just Exposed

In this case, the water surface slopes from the top rear corner (or some point on the rear wall) down to the front bottom corner.

Diagram for Case 2: Front corner exposed

(a) Horizontal Acceleration

The total depth at the rear must be 2m to maintain the original volume. The slope runs over the full length of the tank, L=6m.

$$ \tan \theta = \frac{\text{total depth}}{\text{total length}} = \frac{2}{6} = \frac{1}{3} \approx 0.3333 $$ $$ a = g \tan \theta = 9.81 \times \frac{1}{3} \approx 3.27 \, \text{m/s}^2 $$

(b) Forces on Each End

$$ F_{rear} = \rho g A_{rear} \bar{h}_{rear} $$ $$ = 1000 \times 9.81 \times (2 \times 2.5) \times (2/2) = 49050 \, \text{N} $$ $$ F_{front} = 0 \, \text{N (since the corner is exposed)} $$

(c) Verification

$$ F_{net} = 49050 - 0 = 49050 \, \text{N} $$ $$ \text{Volume} = \frac{1}{2} \times L \times h_{rear} \times b = 0.5 \times 6 \times 2 \times 2.5 = 15 \, \text{m}^3 $$ $$ \text{Mass} = 1000 \times 15 = 15000 \, \text{kg} $$ $$ M \times a = 15000 \times 3.27 = 49050 \, \text{N} $$

The net force again matches the force required for acceleration.

(iii) Case 3: Bottom Exposed to Mid-Point

The water surface now slopes from the top rear corner to the midpoint of the tank bottom.

Diagram for Case 3: Bottom exposed to midpoint

(a) Horizontal Acceleration

The slope now covers a horizontal distance of 3 m.

$$ \tan \theta = \frac{\text{total depth}}{\text{half length}} = \frac{2}{3} \approx 0.6667 $$ $$ a = g \tan \theta = 9.81 \times \frac{2}{3} = 6.54 \, \text{m/s}^2 $$

(b) Forces on Each End

$$ F_{rear} = \rho g A_{rear} \bar{h}_{rear} $$ $$ = 1000 \times 9.81 \times (2 \times 2.5) \times (2/2) = 49050 \, \text{N} $$ $$ F_{front} = 0 \, \text{N (since the front end is dry)} $$

(c) Verification

$$ F_{net} = 49050 - 0 = 49050 \, \text{N} $$ $$ \text{Volume} = \frac{1}{2} \times (L/2) \times h_{rear} \times b = 0.5 \times 3 \times 2 \times 2.5 = 7.5 \, \text{m}^3 $$ $$ \text{Mass} = 1000 \times 7.5 = 7500 \, \text{kg} $$ $$ M \times a = 7500 \times 6.54 = 49050 \, \text{N} $$

The results are verified for the final case.

Final Results Summary:

Case (i) - Verge of Spilling: \( a \approx 1.635 \, \text{m/s}^2 \), \( F_{rear} = 49050 \, \text{N} \), \( F_{front} = 12262.5 \, \text{N} \).

Case (ii) - Front Corner Exposed: \( a = 3.27 \, \text{m/s}^2 \), \( F_{rear} = 49050 \, \text{N} \), \( F_{front} = 0 \, \text{N} \).

Case (iii) - Bottom Exposed to Midpoint: \( a = 6.54 \, \text{m/s}^2 \), \( F_{rear} = 49050 \, \text{N} \), \( F_{front} = 0 \, \text{N} \).

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