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Fluid at Rest in an Accelerating Container

Relative equilibrium means the liquid has no motion relative to its container, but the whole liquid mass may be accelerating. Replace gravity with an effective acceleration field.

$$\nabla p = \rho(\vec{g}-\vec{a})$$$$\tan\theta = \frac{a_x}{g}$$

The free surface is perpendicular to the resultant effective gravity.

Vertical Acceleration Pressure

When the tank accelerates vertically, use an effective unit weight. Upward acceleration increases pressure; downward acceleration reduces it.

$$p = \rho(g+a)h \quad \text{upward}$$$$p = \rho(g-a)h \quad \text{downward}$$

Horizontal Acceleration Surface Rise

An open tank 2 m long accelerates horizontally at $2.5\text{ m/s}^2$. Find the difference in liquid level between the ends.

$$\tan\theta=\frac{a}{g}=\frac{2.5}{9.81}=0.255$$$$\Delta h = L\tan\theta=2(0.255)=0.51\text{ m}$$

Answer: The liquid level differs by about 0.51 m, rising opposite the direction of acceleration.

Relative Equilibrium Exam Reminders

Sketch the pseudo-force direction before writing equations. For open tanks, conserve volume when finding spill or final depth; for closed tanks, include trapped air pressure when given.

Problem: Water Surface Slope in an Accelerating Truck

An open rectangular tank 3.0 m long (in direction of motion) and 1.5 m wide is mounted on a truck. The tank is filled with water to a depth of 0.90 m. The truck accelerates forward at 3.0 m/s². Determine: (a) the slope of the water surface, (b) the rise of water at the rear wall and the drop at the front wall, and (c) whether any water spills.

$$\tan\theta = \frac{a}{g} = \frac{3.0}{9.81} = 0.3058$$ $$\Delta h = L \tan\theta = 3.0(0.3058) = 0.917 \text{ m total difference}$$ $$\text{Rise at rear} = \frac{\Delta h}{2} = 0.459 \text{ m}, \quad \text{Drop at front} = 0.459 \text{ m}$$ $$h_{rear} = 0.90 + 0.459 = 1.359 \text{ m}$$

Tank depth is not given, but if the tank walls are at least 1.36 m tall, no spill occurs. If tank wall height is 1.20 m (common), the rear overflows.

Answer: The water surface tilts at $\tan\theta = 0.306$. Water rises 0.459 m at the rear and drops 0.459 m at the front. The rear level reaches 1.359 m — water would spill if the tank walls are less than 1.36 m tall.

Problem: Pressure at the Bottom of a Tank Accelerating Vertically

A closed tank 1.5 m tall is completely filled with water and is accelerating upward at 4.0 m/s². Determine the pressure in kPa at the bottom of the tank if the pressure at the top is zero gage. Compare this to the static case.

$$p_{bottom} = \rho(g + a)h = 1000(9.81 + 4.0)(1.5) = 1000(13.81)(1.5)$$ $$p_{bottom} = 20{,}715 \text{ Pa} = 20.72 \text{ kPa}$$

For static case (a = 0):

$$p_{static} = \rho g h = 1000(9.81)(1.5) = 14.72 \text{ kPa}$$ $$\text{Increase} = 20.72 - 14.72 = 6.0 \text{ kPa} = \rho a h = 1000(4.0)(1.5)$$

Answer: Bottom pressure under upward acceleration is 20.72 kPa vs. 14.72 kPa static — an increase of 6.0 kPa. Downward acceleration at 4 m/s² would give only 8.72 kPa, confirming that upward acceleration amplifies hydrostatic pressure.

Problem: Volume Spilled from Open Tank Under Horizontal Acceleration

A rectangular open tank is 4.0 m long, 2.0 m wide, and has walls 1.20 m tall. It is initially filled to a depth of 1.00 m and accelerates horizontally at 5.0 m/s². Determine whether water spills and, if so, compute the volume of water lost.

$$\tan\theta = \frac{a}{g} = \frac{5.0}{9.81} = 0.510$$ $$\Delta h = L\tan\theta = 4.0(0.510) = 2.04 \text{ m}$$ $$h_{rear,no-spill} = h_0 + \frac{\Delta h}{2} = 1.00 + 1.02 = 2.02 \text{ m}$$

Since $h_{rear} = 2.02 \text{ m} > h_{wall} = 1.20 \text{ m}$, spill occurs at the rear.

After spilling, the rear water surface is at the wall top (1.20 m). Let $x$ = depth at front wall:

$$\text{Slope} = \frac{1.20 - x}{4.0} = \tan\theta = 0.510 \Rightarrow x = 1.20 - 2.04 = -0.84 \text{ m}$$

Since $x < 0$, the front is dry — the water surface intersects the bottom. The water surface is a triangle in the rear portion. Let $L_w$ = length from rear where water contacts floor:

$$L_w = \frac{1.20}{\tan\theta} = \frac{1.20}{0.510} = 2.353 \text{ m}$$ $$V_{remaining} = \frac{1}{2}(1.20)(2.353)(2.0) = 2.824 \text{ m}^3$$ $$V_{original} = 4.0(2.0)(1.00) = 8.0 \text{ m}^3$$ $$V_{spilled} = 8.0 - 2.824 = 5.176 \text{ m}^3$$

Answer: Water spills. The volume lost is approximately 5.18 m³ — the remaining water occupies only a triangular wedge against the rear wall during acceleration.

Problem: Acceleration at an Angle — Surface Slope and Pressure

A closed rectangular tank 2.0 m long and 1.0 m tall is completely filled with water and accelerates to the right along an incline at 30° above horizontal. The magnitude of acceleration is 5.0 m/s². Determine: (a) the slope of the pressure isobars (equipressure planes) inside the tank, and (b) the gage pressure at the rear-bottom corner if the front-top corner has gage pressure of 10 kPa.

Resolve the acceleration: $a_x = 5\cos30° = 4.33 \text{ m/s}^2$ (horizontal), $a_z = 5\sin30° = 2.50 \text{ m/s}^2$ (upward).

$$\tan\alpha = \frac{a_x}{g + a_z} = \frac{4.33}{9.81 + 2.50} = \frac{4.33}{12.31} = 0.352$$ $$\alpha = 19.4° \text{ (slope of isobars from horizontal)}$$

Pressure gradient: along horizontal (x-direction): $\frac{\partial p}{\partial x} = -\rho a_x = -1000(4.33) = -4330 \text{ Pa/m}$. Along vertical (z-direction): $\frac{\partial p}{\partial z} = -\rho(g + a_z) = -1000(12.31) = -12{,}310 \text{ Pa/m}$.

$$\Delta p = -(-4330)(2.0) - (-12{,}310)(1.0) = 8660 + 12{,}310 = 20{,}970 \text{ Pa}$$ $$p_{rear-bottom} = 10 + 20.97 = 30.97 \text{ kPa}$$

Answer: Pressure isobars slope at 19.4° from horizontal. The rear-bottom corner has gage pressure 30.97 kPa.

Problem: Downward Acceleration — Reduced Pressure and Cavitation Risk

A pump draws water from a tank through a 100 mm pipe. The pump is 3.5 m above the water surface in the tank. The pump-suction assembly accelerates downward (momentarily) at 3.0 m/s² due to vibration. The velocity in the pipe is 2.0 m/s and there are no friction losses during this instant. Determine the absolute pressure at the pump inlet. Does cavitation risk increase? Atmospheric pressure = 101.3 kPa, vapor pressure of water = 2.34 kPa absolute.

Applying Bernoulli with effective gravity $g_{eff} = g - a = 9.81 - 3.0 = 6.81 \text{ m/s}^2$ from the free surface (point 1) to pump inlet (point 2):

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + z_2$$ $$\frac{101.3}{9.81} + 0 + 0 = \frac{p_2}{9.81} + \frac{(2.0)^2}{2(9.81)} + 3.5$$ $$10.326 = \frac{p_2}{9.81} + 0.204 + 3.5$$ $$\frac{p_2}{9.81} = 6.622 \Rightarrow p_2 = 64.97 \text{ kPa (abs)}$$

With downward acceleration — the effective head available is reduced further. Using effective weight $\gamma_{eff} = \rho(g-a)$, the static head contribution decreases.

Answer: Pump inlet absolute pressure drops to about 64.97 kPa, well above vapor pressure of 2.34 kPa in this case. However, during downward acceleration, effective gravity is reduced, making suction head more critical. In deep-suction pumps, vibration-induced downward acceleration can trigger cavitation.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q369

HGE - Hydraulics / Relative Equilibrium of Fluids / Engr. Janclyde Espinosa (Clidez)

An open rectangular tank mounted on a moving vehicle 6m long, 2m wide and 3m high is filled with water to a depth of 2.5 m.

What is the maximum horizontal acceleration in m/sec2 that can be imposed on the tank without spilling any water?

  1. 1.64
  2. 1.46
  3. 1.38
  4. 1.83

Determine the accelerating force on the liquid mass.

  1. 49.2
  2. 49.6
  3. 48.4
  4. 48.7

If the acceleration is increased to 6m/sec2, how much water is spilled out?

  1. 15.3
  2. 16.2
  3. 14.8
  4. 13.6

Part 1.

For a horizontally accelerating tank, the free-surface difference is:
$\Delta h=\frac{aL}{g}$
The tank is 3 m high and initially filled to 2.5 m, so the allowable rise at one end is 0.5 m. The total difference is twice this, or 1.0 m:
$1.0=\frac{a(6)}{9.81}$
$a=1.635$ m/s2
$\boxed{a\approx1.64}$

Part 2.

Volume of water in the tank:
$V=6(2)(2.5)=30$ m3
Mass is $m=1000(30)=30{,}000$ kg. With $a=1.64$ m/s2:
$F=ma=30{,}000(1.64)=49{,}200$ N
$\boxed{F=49.2\text{ kN}}$

Part 3.

At $a=6$ m/s2, the free-surface slope gives a height difference:
$\Delta h=\frac{aL}{g}=\frac{6(6)}{9.81}=3.67$ m
Since the tank is only 3 m high, water spills until the high end is at the rim and the free surface meets the bottom at distance:
$x=\frac{Hg}{a}=\frac{3(9.81)}{6}=4.905$ m
Remaining volume is triangular:
$V_r=\frac{1}{2}(3)(4.905)(2)=14.715$ m3
Spilled volume:
$30-14.715=15.3$ m3
$\boxed{15.3}$

Question Bank: v1

HGE - Hydraulics / Fluid Statics / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A bucket with water 0.55 m deep is carried inside an elevator that accelerates at 3 m/s2. Evaluate the pressure in kPa exerted by the water at the base of the bucket if the elevator is travelling downward.

  1. 4.42 kPa
  2. 5.40 kPa
  3. 7.05 kPa
  4. 3.75 kPa

Inside a vertically accelerating container the water behaves with an effective unit weight $\gamma_{eff}=\gamma\left(1\pm\dfrac{a}{g}\right)$ — the sign is minus for downward acceleration. The pressure at the base is

$$p=\gamma_{eff}\,h=\gamma h\left(1-\frac{a}{g}\right).$$
Computed answer: 3.75 kPa

Question Bank: v21

HGE - Hydraulics / Relative Equilibrium / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

An open rectangular tank 1.6 m wide, 6 m long, and 2.2 m high is mounted on a truck. The still-water depth is 1.5 m.

Determine the maximum horizontal acceleration without spillage, in m/s2.

  1. 2.289 m/s2
  2. 2.220 m/s2
  3. 2.747 m/s2
  4. 2.976 m/s2

Using that maximum acceleration, determine the force required to accelerate the water mass, in kN.

  1. 28.68 kN
  2. 25.38 kN
  3. 36.59 kN
  4. 32.96 kN

If the acceleration is increased to 5.1 m/s2, how much water is spilled, in cubic meters?

  1. 8.95 m3
  2. 10.95 m3
  3. 6.95 m3
  4. 12.95 m3

Part 1 — Maximum acceleration. Under horizontal acceleration the free surface tilts at $\tan\theta=\dfrac{a}{g}$. Impending spillage occurs when the back rises by $H_{tank}-h_0$ over the half-length:

$$H_{tank}-h_0=\frac{L}{2}\cdot\frac{a}{g}\ \Rightarrow\ a=\frac{2g\,(H_{tank}-h_0)}{L}.$$

Part 2 — Accelerating force. With water volume $\mathcal V=W L h_0$ and mass $m=\rho\mathcal V$,

$$F=m\,a=\rho\,(WLh_0)\,a.$$

Part 3 — Spillage at $a_2$. The surface slope is $a_2/g$. After spilling, the water is a triangular wedge of base $x=\dfrac{H_{tank}}{a_2/g}$, so

$$V_{left}=\tfrac12 H_{tank}\,x\,W,\qquad V_{spill}=WLh_0-V_{left}.$$

Computed answers:
1. 2.289 m/s2
2. 32.96 kN
3. 6.95 m3

Question Bank: v60

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A container holds water 1.1 m deep in an elevator accelerating upward at 1.5 m/s2. Determine the base pressure in kPa.

  1. 8.21 kPa
  2. 13.19 kPa
  3. 12.44 kPa
  4. 9.46 kPa
Upward acceleration increases apparent unit weight: $$p=\gamma h\left(1+\frac ag\right)=9.81(1.1)\left(1+\frac{1.5}{9.81}\right)=12.441\text{ kPa}.$$
Computed answer: 12.44 kPa

Question Bank: v118

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE May 2022

Formula-mode item rendered with fixed values for lecture/PDF export.

A water container filled to a depth of 0.45 m is inside an elevator accelerating upward at 1 m/s2. Compute the gage pressure at the bottom of the container.

  1. 3.80 kPa
  2. 5.53 kPa
  3. 6.13 kPa
  4. 4.86 kPa
For upward acceleration, the effective acceleration is $g+a$. Therefore, $$p=\gamma_wh\left(1+\frac{a}{g}\right)=9.81(0.45)\left(1+\frac{1}{9.81}\right)=4.8645\text{ kPa}.$$
Computed answer: 4.86 kPa

Question Bank: v142

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

An open tank contains water to a depth of 0.9 m while an elevator carries it upward at the constant speed 1.5 m/s.

Compute the gage pressure at the tank bottom.

  1. 6.90 kPa
  2. 10.04 kPa
  3. 8.83 kPa
  4. 11.13 kPa
Constant velocity means zero acceleration, so the ordinary hydrostatic relation p = γh applies.
Computed answers: 8.829.

Question Bank: v157

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

An open rectangular tank is 4 m long and contains water to a uniform depth 1.8 m. A truck accelerates horizontally at 3 m/s2 without spilling.

Find the hydrostatic resultant per meter width on the deeper end wall.

  1. 22.28 kN/m
  2. 28.53 kN/m
  3. 32.44 kN/m
  4. 35.97 kN/m
The free surface tilts with slope a/g. Obtain the deeper-end depth and integrate the triangular pressure distribution.
Computed answers: 28.5270623853.

Question Bank: v180

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A square open tank 1.5 m on each side contains oil of specific gravity 0.75 to depth 1.4 m. The tank accelerates vertically upward at 3.5 m/s2.

Find the resultant force on one vertical wall.

  1. 11.46 kN
  2. 16.68 kN
  3. 14.67 kN
  4. 18.50 kN
Replace g by g + a in the hydrostatic pressure relation and integrate over the wall.
Computed answers: 14.674275.

Question Bank: v187

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

When an open vessel accelerates upward along an inclined plane, how does its free surface orient relative to the acceleration?

  1. It remains horizontal.
  2. It tilts in the direction of acceleration.
  3. It tilts opposite the acceleration direction.
  4. Its orientation is unpredictable.
The free surface is normal to the effective gravity vector, which combines gravity and the opposite of the vessel acceleration.

Question Bank: v205

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

A partly filled tank on a truck accelerates steadily forward. How does the liquid level change?

  1. It rises and falls alternately at the front.
  2. It falls at the rear.
  3. It remains level.
  4. It rises at the rear and falls at the front.
The free surface tilts opposite the direction of acceleration.

Question Bank: v224

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

An open water tank accelerates vertically upward. How does bottom pressure compare with static pressure?

  1. Greater
  2. Equal
  3. Less
  4. Zero
Upward acceleration increases effective gravity to g + a.

Question Bank: v255

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A partly filled tanker accelerates horizontally at 0.4g. Determine the free-surface angle from horizontal and vertical.

Find the acute angle from vertical.

  1. 69.99 deg
  2. 101.90 deg
  3. 113.01 deg
  4. 89.62 deg
The surface slope is a/g.
Computed answers: 89.6194936229.

Question Bank: v257

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A rectangular tank 1.6 m long contains water depth 0.75 m and accelerates horizontally under force 11500 N. Tank-plus-water mass is 3800 kg.

Find hydrostatic force per meter width on the deeper end.

  1. 3806 N/m
  2. 5541 N/m
  3. 4874 N/m
  4. 6146 N/m
Find acceleration, tilted-surface rise, and the triangular end pressure resultant.
Computed answers: 4873.60303611.

Question Bank: v265

HGE - Hydraulics / Relative Equilibrium of Fluids / HGE Refresher Series

Formula-mode item rendered with fixed values for lecture/PDF export.

A passenger stands in an elevator accelerating downward at 3 m/s2 while holding a glass with water depth 0.275 m.

Find the bottom gage pressure.

  1. 1.87 kPa
  2. 1.46 kPa
  3. 2.13 kPa
  4. 2.36 kPa
Downward acceleration reduces effective gravity to g - a.
Computed answers: 1.87275.