An open rectangular tank is fixed on a truck…

An open rectangular tank is fixed on a truck. The tank is $4.0\ \text{m}$ long (along the road), $1.5\ \text{m}$ wide, and its walls are $1.5\ \text{m}$ high. With the truck at rest, it holds water to a depth of $1.2\ \text{m}$ , as shown in the figure. The truck then moves off along a straight, level road with a steady acceleration of $1.0\ \text{m/s}^2$ , and the water settles into a new shape (it moves with the tank). Take $g = 10\ \text{m/s}^2$ and the density of water as $1000\ \text{kg/m}^3$ . Give pressures as gauge pressures (above the air's pressure).

Side view of an open rectangular tank on a truck, at rest. The tank is 4.0 m long and its walls are 1.5 m high; it holds water 1.2 m deep with a flat surface. An arrow shows that the truck will accelerate forwards, to the right, at a = 1.0 m/s squared. The rear wall is on the left and the front wall on the right.

What angle does the water surface make with the horizontal?
What is the depth of the water at the rear wall, and at the front wall?
Find the gauge pressure at the bottom of the tank, at the rear wall and at the front wall.
Find the total force of the water on the rear wall and on the front wall.
What is the greatest steady acceleration the truck can have without water spilling over the rear wall?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Work in the truck's frame, where the water is at rest under its weight and a pseudo force. Find the tilt of the surface, then use the depth at each wall.

Step 1. The tilt (a). Each bit of water of mass $dm$ feels $g\,dm$ downwards and $a\,dm$ backwards. The free surface is at right angles to the combined pull, so it makes an angle $\theta$ with the horizontal where

\displaystyle \tan\theta = \frac{a}{g} = \frac{1.0}{10} = 0.1, \qquad \theta = 5.71^\circ

It slopes up towards the rear of the truck.

Step 2. Depths (b). The amount of water is fixed, so the surface turns about the middle of the tank. Over half the length, $2.0\ \text{m}$ , it rises by $2.0\ \text{m} \times 0.1 = 0.2\ \text{m}$ . So the depth is $1.4\ \text{m}$ at the rear and $1.0\ \text{m}$ at the front.

Step 3. Pressures (c). Vertically nothing has changed: the pressure grows by $\rho g$ per metre of depth below the surface above that point. At the rear, $1000 \times 10 \times 1.4\ \text{Pa} = 14.0\ \text{kPa}$ ; at the front, $10.0\ \text{kPa}$ .

Step 4. Forces on the end walls (d). On a vertical wall wetted to a depth $h$ , the pressure grows steadily from zero at the top to $\rho g h$ at the bottom, so the force is $\tfrac12\rho g h^2$ times the width:

\displaystyle F_{\text{rear}} = \tfrac12 \times 1000 \times 10 \times 1.96 \times 1.5\ \text{N} = 14700\ \text{N}, \qquad F_{\text{front}} = \tfrac12 \times 1000 \times 10 \times 1.00 \times 1.5\ \text{N} = 7500\ \text{N}

Step 5. The limit (e). Water spills when the rear depth reaches the top of the wall, a rise of $0.3\ \text{m}$ over half the length: $\dfrac{a}{g} = \dfrac{0.3}{2.0}$ , so $a = 1.50\ \text{m/s}^2$ .

Answer: (a) $5.71^\circ$ . (b) $1.4\ \text{m}$ at the rear, $1.0\ \text{m}$ at the front. (c) $14.0\ \text{kPa}$ and $10.0\ \text{kPa}$ . (d) $14700\ \text{N}$ on the rear wall, $7500\ \text{N}$ on the front wall. (e) $1.50\ \text{m/s}^2$ .

Check. The tank's walls must push the water forwards to give it its acceleration. The water's mass is $1000 \times 4.0 \times 1.5 \times 1.2\ \text{kg} = 7200\ \text{kg}$ , so the net push needed is $7200\ \text{kg} \times 1.0\ \text{m/s}^2 = 7200\ \text{N}$ . The end walls give $14700\ \text{N} - 7500\ \text{N} = 7200\ \text{N}$ . They agree.

Watch out. The pressure at the middle of the bottom stays the same as at rest, $12.0\ \text{kPa}$ . The pressure grows towards the rear and falls towards the front, because the depth changes, not because the rule $p = \rho g h$ fails.

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