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Applications of Integration

7.7 Fluid Pressure and Fluid Force

Focus on hydrostatic pressure, submerged plates, force from horizontal strips.

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Ideas in this lesson

01Hydrostatic pressureWorked steps · five questions02Submerged platesWorked steps · five questions03Force from horizontal stripsWorked steps · five questions
Idea 1 of 3

Hydrostatic pressure

Pressure equals fluid weight density times depth below the fluid surface. It increases linearly with depth.

Worked example 1

A vertical rectangular plate is 2m2 m wide and 1m1 m high, with its top at the water surface. Use weight density 9800Nm39800 \dfrac{N}{m^{3}}. Find the hydrostatic force.
  1. p(y)=9800yp(y) = 9800y; dA=2dydA = 2 dy
  2. F=∫0119600y dyF = \int_{0}^{1} 19600y\,dy
  3. F=9800(2)(1)22=9800F = \dfrac{9800(2)(1)^{2}}{2} = 9800

Answer: 98009800

Worked example 2

A vertical rectangular plate is 3m3 m wide and 2m2 m high, with its top at the water surface. Use weight density 9800Nm39800 \dfrac{N}{m^{3}}. Find the hydrostatic force.
  1. p(y)=9800yp(y) = 9800y; dA=3dydA = 3 dy
  2. F=∫0229400y dyF = \int_{0}^{2} 29400y\,dy
  3. F=9800(3)(2)22=58800F = \dfrac{9800(3)(2)^{2}}{2} = 58800

Answer: 5880058800

Idea 2 of 3

Submerged plates

Integrate pressure times strip width over the plate’ss depth interval, not automatically from zero.

Worked example 1

A vertical plate is 2m2 m wide and 1m1 m high, starting 1m1 m below the water surface. Use 9800Nm39800 \dfrac{N}{m^{3}}. Find the fluid force.
  1. dF=9800y(2)dydF = 9800y(2)dy; 1≤y≤21 \le y \le 2
  2. F=∫1219600y dyF = \int_{1}^{2} 19600y\,dy
  3. F=196002[(2)2−(1)2]=29400F = \dfrac{19600}{2}[(2)^{2} - (1)^{2}] = 29400

Answer: 2940029400

Worked example 2

A vertical plate is 3m3 m wide and 2m2 m high, starting 2m2 m below the water surface. Use 9800Nm39800 \dfrac{N}{m^{3}}. Find the fluid force.
  1. dF=9800y(3)dydF = 9800y(3)dy; 2≤y≤42 \le y \le 4
  2. F=∫2429400y dyF = \int_{2}^{4} 29400y\,dy
  3. F=294002[(4)2−(2)2]=176400F = \dfrac{29400}{2}[(4)^{2} - (2)^{2}] = 176400

Answer: 176400176400

Idea 3 of 3

Force from horizontal strips

A thin strip has nearly constant pressure. Summing strip forces gives the force integral.

Worked example 1

A vertical rectangular plate is 2m2 m wide and 1m1 m high, with its top at the water surface. Use weight density 9800Nm39800 \dfrac{N}{m^{3}}. Find the hydrostatic force.
  1. p(y)=9800yp(y) = 9800y; dA=2dydA = 2 dy
  2. F=∫0119600y dyF = \int_{0}^{1} 19600y\,dy
  3. F=9800(2)(1)22=9800F = \dfrac{9800(2)(1)^{2}}{2} = 9800

Answer: 98009800

Worked example 2

A vertical rectangular plate is 3m3 m wide and 2m2 m high, with its top at the water surface. Use weight density 9800Nm39800 \dfrac{N}{m^{3}}. Find the hydrostatic force.
  1. p(y)=9800yp(y) = 9800y; dA=3dydA = 3 dy
  2. F=∫0229400y dyF = \int_{0}^{2} 29400y\,dy
  3. F=9800(3)(2)22=58800F = \dfrac{9800(3)(2)^{2}}{2} = 58800

Answer: 5880058800

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.7,p7.7, p. 497497. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.