Hydrostatic force: practice problems with step-by-step solutions

Free Hydrostatic force practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: vertical wall, submerged gate, centre of pressure.

Free practice problemEasy
h = 3 m

Water stands 3 m deep against a vertical wall. Find the height ȳ above the base at which the resultant force acts.

h = 3 mb = 4 mρ = 1000 kg/m³g = 9.81 m/s²
m
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What you practise

  1. Easyvertical wall
  2. Mediumsubmerged gate
  3. Hardcentre of pressure

Key formulas

Pressure at depth
p = ρgh
Force on a vertical wall
F = ρg·hc·A
Centre of pressure
yp = yc + Ic/(ycA)
Rectangle
Ic = bH³/12

Learn Hydrostatic force from scratch

1What hydrostatic force is

Water at rest pushes on every surface it touches: the wall of a tank, a dam, a lock gate. That push is the hydrostatic force.

The deeper you go, the harder the water pushes. You feel this in your ears when you dive to the bottom of a pool.

The pressure at a depth h below the free surface is the weight of the water column above one square metre:

p = ρ · g · h

ρ is the density (1000 kg/m³ for fresh water) and g = 9.81 m/s². With h in metres, p comes out in Pa (N/m²). Divide by 1000 for kPa.

Pressure acts at right angles to the surface. It does not depend on the shape or width of the tank, only on the depth.

Self-check What is the water pressure 5 m below the surface of a lake? (ρ = 1000 kg/m³, g = 9.81 m/s²)
  1. 24.5 kPa
  2. 49.05 kPa
  3. 5 kPa
  4. 98.1 kPa

Answer: 49.05 kPa. p = 1000 · 9.81 · 5 = 49 050 Pa = 49.05 kPa. Leaving out g gives 5 kPa, which is a mass per area, not a pressure.

2Force on a vertical wall

On a vertical wall the pressure grows in a straight line from zero at the surface to ρgh at the bottom. The load on the wall is a triangle.

h = 3 mF

The total force is the area of that triangle times the width b of the wall:

F = ½ · ρg·h · h · b = ½ · ρ · g · h² · b

The force acts through the centroid of the triangle, one third of the height above the base:

ȳ = h / 3 (above the base)

So the force acts lower than the middle of the wall. This makes sense: the bottom part is pushed hardest.

Self-check Water 2 m deep stands against a vertical wall 1 m wide. What is the resultant force? (ρg = 9810 N/m³)
  1. 19.62 kN
  2. 39.24 kN
  3. 4 kN
  4. 9.81 kN

Answer: 19.62 kN. F = ½ · 9810 · 2² · 1 = 19 620 N = 19.62 kN. Using the base pressure over the whole wall (39.24 kN) forgets the ½: the pressure is zero at the top.

3Submerged gates and the centre of pressure

A gate deep in the wall of a tank does not start at the surface. For any flat surface, the force equals the pressure at its centroid times its area:

F = ρ · g · yc · A

yc is the depth of the centroid. For a rectangular gate of height H whose top edge is d below the surface, yc = d + H/2 and A = H·b.

The force does not act at the centroid. It acts a little lower, at the centre of pressure, because the lower part of the gate is pushed harder:

ycp = yc + Ic / (yc · A), Ic = b·H³/12

Ic is the second moment of area of the gate about its own centroid. The deeper the gate, the closer ycp comes to yc.

For a wall that starts at the surface (d = 0) this gives ycp = 2h/3 below the surface, which is the same as h/3 above the base.

4The method step by step

  1. Draw the surface and mark its depth below the free surface: top edge d and height H (or depth h for a wall).
  2. Find the depth of the centroid yc and the area A.
  3. Use the density of the liquid in the question: water 1000, seawater 1025, oil 880 kg/m³.
  4. Force: F = ρ·g·yc·A. Divide by 1000 for kN.
  5. If the line of action is asked: ycp = yc + Ic/(yc·A), measured down from the free surface.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A vertical rectangular gate 2 m wide and 2 m high sits in the wall of a water tank, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.

1

Look at the gate

d = 1 mH = 2 m

The gate runs from 1 m to 3 m below the surface. Its width into the page is b = 2 m.

2

Centroid depth

yc = d + H/2 = 1 + 2/2 = 2 m
3

Area

A = H · b = 2 · 2 = 4 m²
4

Pressure at the centroid

pc = ρ·g·yc = 1000 · 9.81 · 2 = 19 620 Pa = 19.62 kPa
5

Resultant force

F = pc · A = 19.62 · 4 = 78.48 kN
6

Check the answer

d = 1 mH = 2 mF

The pressure is 9.81 kPa at the top edge and 29.43 kPa at the bottom edge. Their average is 19.62 kPa, so the force is 19.62 · 4 = 78.48 kN again.

Answer

F = 78.48 kN, acting horizontally on the gate.

6Common mistakes

  • Putting the force at the centroid. It acts lower: at h/3 above the base for a wall, at ycp for a gate.
  • Measuring ȳ from the surface when the question asks for the height above the base, or the other way round.
  • Using the depth of the top edge, or of the bottom edge, instead of the centroid depth.
  • Forgetting the ½ for a wall that starts at the surface.
  • Leaving out g, or using water density for oil or seawater.
Self-check Water is 6 m deep against a vertical wall. Where does the resultant force act?
  1. 2 m above the base
  2. 2 m below the surface
  3. 3 m above the base
  4. 4 m above the base

Answer: 2 m above the base. The triangular load acts through its centroid, h/3 = 2 m above the base. 3 m is the middle of the wall, where the force would act only if the pressure were uniform.

7How exam questions look

  • Easy: water against a vertical wall from the surface down. Find F = ½ρgh²b or the height ȳ = h/3 above the base.
  • Medium: a submerged rectangular gate in water. Find F = ρg·yc·A, as in the example.
  • Hard: the same gate in oil, seawater or glycerin, and sometimes the depth of the centre of pressure ycp.

For the example gate: Ic = 2 · 2³/12 = 1.333 m⁴, so ycp = 2 + 1.333/(2 · 4) = 2.167 m below the surface. That is 0.167 m below the centroid.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
h = 4 m

Water stands 4 m deep against a vertical wall. Find the height ȳ above the base at which the resultant force acts.

h = 4 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

Find ȳ (m)

Show hints
  1. Pressure in a still liquid is p = ρ·g·z.
  2. The pressure diagram on the wall is a triangle. F is its area times the width; it acts through its centroid.
  3. The pressure at the base is 39.24 kPa.
Show the answer and the step-by-step solution

Hydrostatic force: on a plane surface, F = ρ·g·yc·A, where yc is the depth of the centroid. It acts below the centroid, at the centre of pressure.

h = 4 mF

Pressure varies linearly with depth

p = ρ·g·z, from zero at the surface to ρ·g·h at the base, so the load on the wall is a triangle.

pbase = ρgh = 1000·9.81·4 = 39.24 kPa

Line of action

A triangular load acts at its centroid, one third of the height from the base
ȳ = h/3 = 4/3 = 1.333 m above the base

Answer

ȳ = 1.333 m
Example 2Medium
gated = 4 mH = 1.5 m

A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 4 mH = 1.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

Find F (kN)

Show hints
  1. For a plane surface, F = ρ·g·yc·A, with yc the depth of the gate’s centroid.
  2. The centroid of a rectangle is at mid-height: yc = d + H/2.
  3. yc = 4.75 m.
Show the answer and the step-by-step solution

Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.

Centroid depth and area

yc = d + H/2 = 4 + 1.5/2 = 4.75 m
A = H·b = 1.5·3 = 4.5 m²
gated = 4 mH = 1.5 mF

Resultant force

F = ρ·g·yc·A = 1000·9.81·4.75·4.5 = 209.7 kN

Answer

F = 209.7 kN
Example 3Hard
gated = 2 mH = 2 m

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.

d = 2 mH = 2 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²

Find ycp (m)

Show hints
  1. For a plane surface, F = ρ·g·yc·A, with yc the depth of the gate’s centroid.
  2. Use ycp = yc + Ic/(yc·A) with Ic = bH³/12.
  3. yc = 3 m.
Show the answer and the step-by-step solution

Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.

Centroid depth and area

yc = d + H/2 = 2 + 2/2 = 3 m
A = H·b = 2·1.5 = 3 m²
gated = 2 mH = 2 mF

Resultant force

F = ρ·g·yc·A = 1025·9.81·3·3 = 90.5 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

Ic = b·H³/12 = 1.5·2³/12 = 1 m⁴
ycp = yc + Ic/(yc·A) = 3 + 1/(3·3) = 3.111 m

Answer

ycp = 3.111 m

Questions and answers

What is Hydrostatic force?

Water at rest pushes on every surface it touches: the wall of a tank, a dam, a lock gate. That push is the hydrostatic force.

The deeper you go, the harder the water pushes. You feel this in your ears when you dive to the bottom of a pool.

How do I solve Hydrostatic force problems step by step?

Hydrostatic force: on a plane surface, F = ρ·g·yc·A, where yc is the depth of the centroid. It acts below the centroid, at the centre of pressure.

What do the Hydrostatic force exercises cover?
  • Easy: vertical wall
  • Medium: submerged gate
  • Hard: centre of pressure

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Hydrostatic force worksheet with answers
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