A water tanker on a busy road brakes hard at a signal. Inside the tank the water surges forward and piles up against the front wall, and the whole truck lurches. On a bus, a glass of tea on the dashboard does the same thing in miniature. While the bus brakes steadily, the tea stays tilted, higher at the front of the glass.
A slab of liquid that accelerates
In a liquid at rest, the pressure grows downward at the rate \(\rho g\) and does not change along a level line. Now let the vessel accelerate with a steady horizontal acceleration \(a\). After a moment the liquid settles and moves with the vessel, so every part of it has the same acceleration \(a\). A horizontal slab of liquid then needs a net horizontal force. Only a difference of pressure between its ends can supply it.
\[ m = \rho A\,\Delta x \]
Why. a horizontal cylinder of liquid of end area \(A\) and length \(\Delta x\), lying along the acceleration.
\[ (P_1 - P_2)A = \rho A\,\Delta x\; a \]
Why. Newton's second law. \(P_1\) pushes on the rear face, \(P_2\) on the front face, and the net force gives the mass its acceleration \(a\).
\[ P_2 - P_1 = -\rho a\,\Delta x \]
Why. divide by \(A\) and rearrange. The front is at the lower pressure.
\[ \dfrac{\Delta P}{\Delta x} = -\rho a \]
Read it. the pressure falls along the direction of the acceleration, at the rate \(\rho a\).
Vertically nothing has changed: the liquid does not accelerate up or down. So the pressure still grows downward at \(\rho g\), exactly as at rest. The pressure now changes in two directions. It grows downward at \(\rho g\) and backward at \(\rho a\), against the acceleration.
The effective gravity
The two rates can be put in one vector. The pressure gradient \(\nabla P\) points in the direction in which the pressure grows fastest, and its size is that rate. At rest \(\nabla P = \rho \vec g\), straight down. In the accelerating vessel, \(\nabla P = \rho(\vec g - \vec a)\). The liquid behaves exactly as if gravity were \(\vec g - \vec a\). This is the effective gravity.
The same result comes from the vessel's frame. In the chapter on Accelerated Objects, a body in a frame with acceleration \(\vec a\) felt a pseudo force \(-m\vec a\). In the vessel's frame, each bit of liquid feels its weight \(m\vec g\) and the pseudo force \(-m\vec a\). Together they are \(m(\vec g - \vec a)\), a weight in the effective gravity. The liquid is at rest in this frame, so every rule of the last lessons holds with \(\vec g_{\text{eff}}\) in place of \(\vec g\).
A liquid that moves with a vessel of acceleration \(\vec a\) behaves as a liquid at rest in a gravity \(\vec g - \vec a\). Its pressure grows along \(\vec g - \vec a\) at the rate \(\rho|\vec g - \vec a|\). Its surfaces of equal pressure, the free surface among them, are at right angles to \(\vec g - \vec a\).
The tilted surface
The free surface is at one pressure, the atmosphere's. So it must be at right angles to \(\vec g - \vec a\). For a horizontal acceleration \(a\), the effective gravity points down and backward, at an angle \(\theta\) to the vertical with \(\tan\theta = a/g\). The surface tilts by the same angle \(\theta\) from the horizontal. It rises towards the rear.
The same angle comes from pressure alone. Take two points at the same level near the base, a distance \(L\) apart: A at the rear and B at the front. Horizontally, \(P_A - P_B = \rho a L\), from the derivation above. Vertically, each point is at \(P_0\) plus \(\rho g\) times the depth of water above it. So \(\rho g (h_A - h_B) = \rho a L\), and \((h_A - h_B)/L = a/g\). The difference in depth over the length is the slope, \(\tan\theta\).
A tank \(2\,\text{m}\) long on a cart holds water \(1\,\text{m}\) deep. The cart accelerates at \(5\,\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\). Find the angle of the surface, the depth at each end, and the difference in pressure between the ends at the level of the base.
The surface is at right angles to \(\vec g - \vec a\), so \(\tan\theta = a/g\). The water's volume is fixed, so the surface turns about its middle, which stays \(1\,\text{m}\) deep.
\(\tan\theta = a/g\). Over half the length, the level changes by \((L/2)\tan\theta\).
\(\tan\theta = 5/10 = 0.5\), so \(\theta \approx 26.6°\). Each end changes by \(1 \times 0.5 = 0.5\,\text{m}\): the rear is \(1.5\,\text{m}\) deep, the front \(0.5\,\text{m}\). At the base, \(\Delta P = \rho g \times 1.0 = 10^4\,\text{Pa}\).
The surface slopes at about \(26.6°\), \(1.5\,\text{m}\) deep at the rear and \(0.5\,\text{m}\) at the front. The rear of the base is at \(10^4\,\text{Pa}\) more pressure, which pushes the water forward.
Now the braking tanker of the opening. Braking is an acceleration backward, so the "rear" is now the front of the truck. Suppose the tanker brakes at \(2\,\text{m/s}^2\), with water \(1\,\text{m}\) deep in a tank \(6\,\text{m}\) long. Then \(\tan\theta = 0.2\), about \(11°\). The water rises \(3 \times 0.2 = 0.6\,\text{m}\) at the front, to \(1.6\,\text{m}\), and falls to \(0.4\,\text{m}\) at the back.
An open tank \(2\,\text{m}\) long and \(1.4\,\text{m}\) tall holds water \(1\,\text{m}\) deep. Take \(g = 10\,\text{m/s}^2\). What is the largest horizontal acceleration, in m/s², before the water spills over the rear wall?
A closed box, partly filled with water, slides freely down a smooth incline of \(30°\). How does the water's surface lie while the box slides?
A bucket in a lift
Let the acceleration be vertical, as in a lift. Take upward as positive. For an upward acceleration \(a\), \(\vec g - \vec a\) points down with size \(g + a\). So the pressure at a depth \(h\) in a bucket of water is \(P_0 + \rho (g + a) h\). For a downward acceleration, \(a\) is negative, and the pressure is smaller. The surface stays level, since \(\vec g - \vec a\) is still vertical.
A tub holds water \(1\,\text{m}\) deep in a lift. At rest, the gauge pressure at its bottom is \(10\,000\,\text{Pa}\). The cable breaks, and the lift falls freely. What is the gauge pressure at the bottom now?
The chips of the figure above step through the cases for water \(1\,\text{m}\) deep. With the lift accelerating up at \(2\,\text{m/s}^2\), \(g_{\text{eff}} = 12\,\text{m/s}^2\) and the gauge pressure is \(12\,000\,\text{Pa}\). Accelerating down at \(2\,\text{m/s}^2\), it is \(8000\,\text{Pa}\). In free fall it is zero, exactly as a weighing scale in a falling lift reads zero.
A lift accelerates upward at \(5\,\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\). What is the gauge pressure \(40\,\text{cm}\) below the surface of the water in a bucket in the lift, in Pa?
A U-tube that accelerates
Fix a U-tube of water on a trolley, with its vertical arms a distance \(L\) apart along the direction of motion. When the trolley accelerates at \(a\), the liquid in the horizontal bottom limb must accelerate with it. That needs a pressure \(\rho a L\) larger at the rear end of the limb than at the front. Only a taller column in the rear arm can supply it. So \(\rho g\,\Delta h = \rho a L\), and \(\Delta h = aL/g\).
The two surfaces in Fig 5.1 are two points of one tilted surface. The line joining them slopes at \(\tan\theta = a/g\), exactly as in the tank. This makes the U-tube an accelerometer: measure \(\Delta h\) and \(L\), and \(a = g\,\Delta h/L\). For arms \(30\,\text{cm}\) apart at \(2\,\text{m/s}^2\), \(\Delta h = 2 \times 30/10 = 6\,\text{cm}\).
A U-tube of water with vertical arms \(40\,\text{cm}\) apart is fixed in a car, along its length. The car accelerates at \(2.5\,\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\). What is the difference between the water levels, in cm?
The U-tube of Fig 5.1 is on a trolley moving to the right at a steady \(3\,\text{m/s}\). Which arm stands higher?
The pendulum in an accelerating car, from the chapter on Accelerated Objects, hung back at \(\tan\theta = a/g\). The liquid surface tilts by the same angle for the same reason. The string of the pendulum lines up with \(\vec g - \vec a\), and the liquid's surface lies at right angles to it.
A tanker carrying a liquid is divided inside by walls with holes, called baffles. Without them, the whole load surges forward when the driver brakes and back when he speeds up. The surge pushes the truck and can tip it on a turn. The baffles split the liquid into short compartments, so each surge is small.
- Letting the surface rise at the front of a tank that speeds up.The water stays behind as the tank speeds up. It rises at the rear, at right angles to \(\vec g - \vec a\).
- Using the velocity in place of the acceleration.Only acceleration tilts a surface. At steady velocity the liquid is as at rest.
- Writing \(g - a\) for a lift accelerating upward.With upward positive, \(g_{\text{eff}} = g + a\). Upward acceleration raises the pressure; downward lowers it.
- Using the whole length for the rise at one end of a tank.The surface turns about its middle, so each end changes by \((L/2)\tan\theta\).