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/m^{3} 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?
Solution
For all parts of these problems, we'll be using the equation for pressure given above where the atmospheric pressure at sea level is P_{o}.
(a):
\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*}
(b):
\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*}
(c):
\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
- 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*}.
- If there is no net force on the car, calculate the pressure in the pipe.
- 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?
- 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?
- A SCUBA diver descends deep into the ocean. Calculate the water pressure at each of the following depths.
- \begin{align*}15 \;\mathrm{m}.\end{align*}
- \begin{align*}50 \;\mathrm{m}.\end{align*}
- \begin{align*}100 \;\mathrm{m}.\end{align*}
- 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!
- What was the pressure on my feet?
- What is the ratio of this pressure to atmospheric pressure?
Answers to Selected Problems
- 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*}
- a. \begin{align*}248 \;\mathrm{kPa}\end{align*} b. \begin{align*}591 \;\mathrm{kPa}\end{align*} c. \begin{align*}1081 \;\mathrm{kPa}\end{align*}
- a. \begin{align*}611 \;\mathrm{kPa}\end{align*} b. \begin{align*}6 \;\mathrm{atm}\end{align*}