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# Pressure in Fluids

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Practice Pressure in Fluids
Progress
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Pressure in Fluids

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

### Key Equations

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

Guidance
• 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):

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

(b):

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

(c):

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

### Time for Practice

1. A $1500 \;\mathrm{kg}$ car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is a pipe with radius $24 \;\mathrm{cm}$ .
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 $2.00\;\mathrm{cm}$ . How much force must be exerted at this end?
3. To generate an upward acceleration for the car of $1.0 \;\mathrm{m/s}^2$ , 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. $15 \;\mathrm{m}.$
2. $50 \;\mathrm{m}.$
3. $100 \;\mathrm{m}.$
3. Ouch! You stepped on my foot! That is, you put a force of $550 \;\mathrm{N}$ in an area of $9 \;\mathrm{cm}^2$ 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?

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