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Wize University Calculus 1 Textbook > Applications of Integration for Physical Science

Pressure and Force

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Pressure and Force

When objects are submerged in fluid, Pressure (hydrostatic pressure) is exerted on the object. This pressure is determined by the density of the fluid and how far the object is submerged.
Pressure=weight density ×depth\text{Pressure}=\text{weight density }\times \text{depth}
For an object submerged in a fluid with density ρ\rho at depth hh, the pressure PPon the the object is given by
P=ρ×g×h\boxed{P= \rho \times g \times h}
where g≈9.8g \approx 9.8 is the force of gravity.

Note: ρ×g\rho \times g is often called weight-density , ww,and reduces our formula

P=w×h\boxed{P=w\times h}

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Fluid Force

The force (hydrostatic force) FF from a constant pressure PP applied to plate of area AA submerged horizontally is
F=P×A\boxed{F=P \times A}

Fluid Force Integrals

For variable depth, the force FF applied to an object from depth aa to bbis given by
F=∫abw⋅h(y)⋅L(y)dy\boxed{F=\displaystyle \int_a^b w \cdot h(y)\cdot L(y)dy}
where h(y)h(y)is the depth of the object and L(y)L(y)is the cross sectional length of the object over [a,b][a,b].


Wize Tip
Using symmetry or similar triangles often helps in determining L(y)L(y).

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Example: Pressure and Force

A right circular cylindrical tank lying on its side and 10 ft.10 \ \text{ft}. in diameter contains a fluid with weight constant density 19.6 lbs/ft319.6 \ \text{lbs}/\text{ft}^3. How much force does the fluid exert on the sides of the tank when the tank is half full?


F=∫abw⋅h(y)⋅L(y)dy\boxed{F=\displaystyle \int_a^b w \cdot h(y)\cdot L(y)dy}

w=19.6w=19.6

Viewing our tank as a cross section the fluid fills the tank from y=−5 to y=0y=-5 \text{ to } y=0. (here the x-axis represents 0 depth since the tank is half full.)



The depth varies linearly below the x-axis:h(−5)=5,h(0)=0h(-5)=5, h(0)=0

⇒h(y)=−y\Rightarrow h(y)=-y

We can use symmetry to determine L(y)L(y).

y=25−x2  ⟺  x=25−y2y=\sqrt{25-x^2}\iff x=\sqrt{25-y^2}

The cross sectional length varies from −x-x to xx but is the same length on each side of the y-axis.

L(y)=2x=225−y2L(y)=2x=2\sqrt{25-y^2}

Thus

F=∫−5019.6⋅−y⋅225−y2dyF=\displaystyle \int_{-5}^0 19.6 \cdot -y\cdot2 \sqrt{25-y^2}dy

=−19.6×2∫−50y⋅25−y2dy=-19.6\times 2\displaystyle \int_{-5}^0 y\cdot \sqrt{25-y^2}dy

u=25−y2 u=25-y^2
du=−2ydydu=-2ydy
dy=du−2y\displaystyle dy=\frac{du}{-2y}
y=−5→u=0,y=0→u=25y=-5 \rightarrow u=0, y=0 \rightarrow u=25

=−19.6×2∫025yudu−2y=-19.6\times 2 \displaystyle \int_{0}^{25} y \sqrt{u}\frac{du}{-2y}

=19.6∫025u du=19.6∫025u12 du=19.6\displaystyle \int_{0}^{25} \sqrt{u} \ du=19.6\displaystyle \int_{0}^{25} u^{\frac{1}{2}} \ du

=19.6×2u323∣025\displaystyle =19.6 \times \frac{2u^\frac{3}{2}}{3} \bigg|_{0}^{25}

=19.6×2[(25)32−(0)32]3\displaystyle =19.6 \times \frac{2[(25)^\frac{3}{2}-(0)^\frac{3}{2}]}{3}

≈1633.33 lbs\approx1633.33\text{ lbs}

A triangular trough contains a certain oil with weight density w=55 lbs/ft3w=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the tank when the tank is full?





Extra Practice
Applications of Integration: Pressure and Force
A triangular trough contains a certain oil with weight density w=55 lbs/ft3w=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the tank when the tank is full?
Practice: Pressure and Force
A triangular trough contains a certain oil with weight density δ=55 lbs/ft3\delta=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the side of the tank when the tank is full? Note that the sides of the tank are given by the equations y=5y=5, y=xy=x, and y=−xy=-x.
(a) 125δ125\delta (b) 25δ25\delta (c) 1253δ\frac{125}{3}\delta (d) 253δ\frac{25}{3}\delta (e) 250δ250\delta
Practice: Pressure and Force
A triangular trough contains a certain oil with weight density δ=55 lbs/ft3\delta=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the side of the tank when the tank is full? Note that the sides of the tank are given by the equations y=5y=5, y=xy=x, and y=−xy=-x.
Practice: Pressure and Force
A triangular trough contains a certain oil with weight density δ=55 lbs/ft3\delta=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the side of the tank when the tank is full? Note that the sides of the tank are given by the equations y=5y=5, y=xy=x, and y=−xy=-x.
Applications of Integration: Pressure and Force
A triangular trough contains a certain oil with weight density w=55 lbs/ft3w=55 \ \text{lbs}/\text{ft}^3. The cross section of the tank is shown below. How much force does the fluid exert on the tank when the tank is full?
New Practice Question
An engineer plans to design a levitating piston which can support loads up to 500kg. He decides to make a cylindrical chamber with radius of 0.5m with a moving piston which weighs 20kg. The chamber is filled with 10 moles of a diatomic ideal gas which is initially kept at 200K. By adding loads on the piston, the gas underneath is heated such that the height of the piston remains unchanged. (Assume that heating process is happening extremely fast such that the height of the piston remains constant while the gas is heated)
a) When there is no load on the piston, what is the pressure of the gas inside the chamber?
b) What is the temperature of the gas when maximum load is added to the piston?
Applications of Integration: Pressure and Force
The change in momentum, called impulse (J), is calculated mathematically as J=∫t0tfF(t)  ⁣dt\displaystyle J=\int_{t_0}^{t_f}F(t)\de{t} where F(t)F(t) is the force exerted on an object as a function of time over the interval from initial time t0t_0 to final time tft_f. Determine the impulse within 1 second using the force F(t)=∣t−0.9∣F(t)=|t-0.9|.