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Pressure in Fluids

Pressure is force per unit area and increases with depth in an incompressible fluid.

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Pressure in Fluids

Students will learn about pressure and solving pressure problems in the context of fluids.

Key Equations

\begin{align*} P = \frac{F}{A} && \text{Pressure is force per unit area}\\ P = P_o - \rho gh && \text{Pressure in an incompressible fluid as a function of depth}\\ \end{align*}

  • The pressure of a fluid is a measure of the forces exerted by a large number of molecules when they collide and bounce off its boundary. The unit of pressure is the Pascal (Pa).
  • In a fluid at rest, pressure increases linearly with depth – this is due to the weight of the water above it.
  • Pascal’s Principle reminds us that, for a fluid of uniform pressure, the force exerted on a small area in contact with the fluid will be smaller than the force exerted on a large area. Thus, a small force applied to a small area in a fluid can create a large force on a larger area. This is the principle behind hydraulic machinery.
  • Liquids obey a continuity equation which is based on the fact that liquids are very difficult to compress. This means that the total volume of a fluid will remain constant in most situations. Imagine trying to compress a filled water balloon!

Example 1

A weather balloon is ascending through the atmosphere. If the density of air is 1.2 kg/m3 and atmospheric pressure at sea level is 101.3 kPa, then what is the pressure on the balloon at (a) 100 m above the ground, (b) 500 m above the ground, and (c) 1000 m above the ground?


For all parts of these problems, we'll be using the equation for pressure given above where the atmospheric pressure at sea level is Po.


\begin{align*} P&=P_o-\rho_{air}gh\\ P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 100\;\text{m}\\ P&=100.12\;\text{kPa}\\ \end{align*}


\begin{align*} P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 500\;\text{m}\\ P&=95.4\;\text{kPa}\\ \end{align*}


\begin{align*} P&=101.3\;\text{kPa} - 1.2\;\text{kg/m}^3 * 9.8\;\text{m/s}^2 * 1000\;\text{m}\\ P&=89.5\;\text{kPa}\\ \end{align*}

Watch this Explanation

Time for Practice

  1. A \begin{align*}1500 \;\mathrm{kg}\end{align*} car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is a pipe with radius \begin{align*}24 \;\mathrm{cm}\end{align*}.
    1. If there is no net force on the car, calculate the pressure in the pipe.
    2. The other end of the pipe has a radius of \begin{align*}2.00\;\mathrm{cm}\end{align*}. How much force must be exerted at this end?
    3. To generate an upward acceleration for the car of \begin{align*}1.0 \;\mathrm{m/s}^2\end{align*}, how much force must be applied to the small end of the pipe?
  2. A SCUBA diver descends deep into the ocean. Calculate the water pressure at each of the following depths.
    1. \begin{align*}15 \;\mathrm{m}.\end{align*}
    2. \begin{align*}50 \;\mathrm{m}.\end{align*}
    3. \begin{align*}100 \;\mathrm{m}.\end{align*}
  3. Ouch! You stepped on my foot! That is, you put a force of \begin{align*}550 \;\mathrm{N}\end{align*} in an area of \begin{align*}9 \;\mathrm{cm}^2\end{align*}on the tops of my feet!
    1. What was the pressure on my feet?
    2. What is the ratio of this pressure to atmospheric pressure?

Answers to Selected Problems

  1. a. \begin{align*}83,000 \;\mathrm{Pa}\end{align*} b. \begin{align*}104 \;\mathrm{N}\end{align*} c. \begin{align*}110 \;\mathrm{N}\end{align*}
  2. a. \begin{align*}248 \;\mathrm{kPa}\end{align*} b. \begin{align*}591 \;\mathrm{kPa}\end{align*} c. \begin{align*}1081 \;\mathrm{kPa}\end{align*}
  3. a. \begin{align*}611 \;\mathrm{kPa}\end{align*} b. \begin{align*}6 \;\mathrm{atm}\end{align*}

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