Lesson Explainer: Pressure Produced by Fluids Physics • 9th Grade

In this explainer, we will learn how to use the formula 𝑝=πœŒπ‘”β„Ž to calculate the pressure produced at different depths by different fluids that gravity acts on.

Pressure is a property of fluids. All liquids and gases produce pressure because they are made of matter that has weight and can flow to fill a volume.

One way to experience pressure from a fluid is to submerge ourselves in a body of water. By varying our depth below the surface of the water, we can feel changes in pressure. The deeper we go, the more pressure the water exerts.

Depth is one factor that affects the pressure a fluid produces. Another factor is the fluid’s density, symbolized using the Greek letter rho (𝜌).

The more dense a fluid is, the greater the pressure it produces. This explains the difference we feel between the pressure created by several metres of air and several metres of water.

Each fluid produces pressure, but because water is more dense than air, its pressure at the same depth is greater.

The last factor that affects the pressure produced by a fluid is the acceleration due to gravity (𝑔). We know that a difference in gravitational field strength, and therefore in acceleration due to gravity, will cause two objects of the same mass to have different weights.

A person standing on Earth for example will have a greater weight than if they stood on the Moon. This difference carries over to a difference in pressure produced by a fluid. A given fluid pressure on Earth will be different from the pressure of that same fluid at the same depth if it is located in a region of different gravitational field strength.

The factors of fluid depth (β„Ž), density (𝜌), and acceleration due to gravity (𝑔) can be combined to give a mathematical relationship for the pressure produced by a fluid.

Formula: Pressure Produced by a Fluid

If the pressure produced by a fluid of density 𝜌 at a depth β„Ž in a gravitational field of strength 𝑔 is 𝑝, then 𝑝=πœŒπ‘”β„Ž.

The SI standard unit of pressure is the pascal (Pa), which is equal to a newton of force per square metre of area (N/m2). If a fluid’s density is given in kilograms per cubic metre (kg/m3), its depth is given in metres (m), and the surrounding acceleration due to gravity is given in units of metres per second squared (m/s2), then the fluid pressure 𝑝 will have units of pascals.

One noteworthy aspect of the pressure produced by a fluid is that it has no associated direction. If we imagine a horizontal surface submerged in a fluid, intuitively we understand that the fluid pressure acts down on the surface.

Perhaps surprisingly, it also acts up on the surface with the same strength.

Indeed, at any given point within a fluid, the pressure acts equally in all directions. This is why a submerged object will not be moved in any direction, except as its weight and buoyancy forces may cause it to sink or float. The pressure on the object will equalize in all directions and have no influence on object motion.

Example 1: Calculating Water Pressure on a Diver

A scuba diver dives to a depth of 1.25 m below the surface of the sea. The sea water has a density of 1β€Žβ€‰β€Ž025 kg/m3. What pressure does the water exert on the diver, to the nearest pascal?

Answer

To calculate the pressure exerted by the water on the diver, we need to know the diver’s depth, the water’s density, and the acceleration due to gravity.

The diver is at a depth of 1.25 m, while the water density is given as 1β€Žβ€‰β€Ž025 kg/m3. We can take the acceleration due to gravity to be 9.8 m/s2.

The pressure on the diver equals the product of these three values: 𝑝=ο€Ή1025/×9.8/×(1.25)=12556.25.kgmmsmPa

Rounding this result to the nearest pascal, our answer is 12β€Žβ€‰β€Ž556 Pa.

Example 2: Determining Acceleration due to Gravity from Fluid Pressure

A pool containing a liquid with a density of 1β€Žβ€‰β€Ž000 kg/m3 at the surface of an unknown planet produces a pressure of 8β€Žβ€‰β€Ž400 Pa at a depth of 2.4 m. What is the acceleration due to gravity at the surface of this planet?

Answer

Fluid pressure (𝑝), depth (β„Ž), density (𝜌), and the surrounding acceleration due to gravity (𝑔) are related to one another by the equation 𝑝=πœŒπ‘”β„Ž.

In this example, we want to solve not for 𝑝 but for 𝑔, so we divide both sides of the equation by 𝜌 times β„Ž: π‘πœŒΓ—β„Ž=𝑔.

Switching the left- and right-hand sides, 𝑔=π‘πœŒΓ—β„Ž.

Here, 𝑔 is the acceleration due to gravity on the surface of the unknown planet. We solve for 𝑔 by substituting in the given values for fluid pressure, depth, and density: 𝑔=84001000/Γ—2.4=3.5/.Pakgmmms

The acceleration due to gravity on the unknown planet is 3.5 m/s2.

Example 3: Calculating Pressure in Liquids of Different Densities

Two identical steel ball bearings are dropped into two different liquids 𝐴 and 𝐡. Liquid 𝐴 has a density of 1β€Žβ€‰β€Ž200 kg/m3 and liquid 𝐡 has a density of 1β€Žβ€‰β€Ž500 kg/m3. How many times deeper does the ball bearing have to fall into liquid 𝐴 to be subject to the same pressure as the ball bearing in liquid 𝐡?

Answer

Remembering that fluid pressure is directly proportional to fluid density, we know that the ball bearing in this scenario will need to fall to a greater depth in liquid 𝐴 than in liquid 𝐡 to experience the same pressure in each case.

The general mathematical relationship between fluid pressure 𝑝, depth β„Ž, density 𝜌, and environmental acceleration due to gravity 𝑔 is as follows: 𝑝=πœŒπ‘”β„Ž.

Since we are considering two different fluids, liquid 𝐴 and liquid 𝐡, we can write expressions for the fluid pressure in each one: 𝑝=πœŒπ‘”β„Ž,𝑝=πœŒπ‘”β„Ž.

We are requiring that these two pressures be equal; that is, 𝑝=𝑝.

Therefore, πœŒπ‘”β„Ž=πœŒπ‘”β„Ž.

Note that the acceleration due to gravity 𝑔 is common to both sides of the equation and can be divided out πœŒβ„Ž=πœŒβ„Ž.

Our question asks how many times deeper a ball bearing needs to fall into liquid 𝐴 to experience the same fluid pressure as in liquid 𝐡. This value can be expressed as the ratio β„Žβ„ŽοŒ οŒ‘. In our equation above, we can solve for this fraction by dividing the entire equation by β„ŽοŒ‘ and by 𝜌: β„Žβ„Ž=𝜌𝜌.

Our answer to this question then equals the ratio of the density of liquid 𝐡 to that of liquid 𝐴. Substituting in for these values, β„Žβ„Ž=1500/1200/=1.25.kgmkgm

The ball bearing needs to fall 1.25 times farther down into liquid 𝐴 than into liquid 𝐡 to experience an equal fluid pressure.

Example 4: Determining Fluid Density from Pressure and Depth

The pressure exerted by a liquid at a depth of 2.5 m is 36β€Žβ€‰β€Ž750 Pa. What is the density of the liquid, to the nearest kilogram per cubic metre?

Answer

Fluid density (𝜌), pressure (𝑝), depth (β„Ž), and the surrounding acceleration due to gravity (𝑔) are related through the equation 𝑝=πœŒπ‘”β„Ž.

In this example, we want to solve for the unknown liquid’s density. To begin doing this, we can divide both sides of the equation above by 𝑔 times β„Ž: π‘π‘”Γ—β„Ž=𝜌.

Switching the left and right sides, 𝜌=π‘π‘”Γ—β„Ž.

We are given the pressure 𝑝 and fluid depth β„Ž, and we can recall that the acceleration due to gravity near Earth’s surface is 9.8 m/s2.

Considering the pressure unit of pascals (Pa), we remember that a pascal is equivalent to a newton per square metre (N/m2). A newton, moreover, equals a kilogram-metre per second squared (kgβ‹…m/s2).

Plugging the three known values into our equation for density and using SI base units in each case, we find that 𝜌=36750β‹…//9.8/Γ—2.5=1500/.kgmsmmsmkgm

The unknown liquid density is exactly 1β€Žβ€‰β€Ž500 kg/m3.

Example 5: Calculating Total Force Exerted due to Pressure

The hull of a sunken boat, laying on the seabed, is 12 m below the surface of the sea, where the sea water has an average density of 1β€Žβ€‰β€Ž025 kg/m3. The hull’s surface area is 15 m2. What is the total force exerted by seawater on the hull?

Answer

To solve for the total force on the hull, we must know the pressure the hull experiences as well as its surface area. This is because, in general, pressure is equal to an amount of force spread out over a certain area: 𝑝=𝐹𝐴.

Since, in this case, we want to solve for the force 𝐹, we can rearrange the expression above and find 𝐹=𝑝×𝐴.

That is, the total force exerted on the hull equals the fluid pressure on it multiplied by its area.

The pressure on the hull from the sea water can be written using an equation combining pressure 𝑝, fluid density 𝜌, fluid depth β„Ž, and acceleration due to gravity 𝑔, as follows: 𝑝=πœŒπ‘”β„Ž.

Substituting this expression for pressure into the equation for force 𝐹, 𝐹=(πœŒπ‘”β„Ž)×𝐴.

In our question, we are told that the sea water density is 1β€Žβ€‰β€Ž025 kg/m3, that the hull is a depth of 12 m below the water’s surface, and that it has a surface area of 15 m2. We can also recall that the acceleration due to gravity near Earth’s surface is 9.8 m/s2. Inserting these values into our equation for force, we get 𝐹=ο€Ή1025/×9.8/×(12)Γ—ο€Ή15=1808100.kgmmsmmN

The total force on the hull of the sunken boat is 1β€Žβ€‰β€Ž808β€Žβ€‰β€Ž100 N.

Example 6: Determining Pressure Difference on Different Parts of a Diver

A diver swims in water of density 1β€Žβ€‰β€Ž015 kg/m3, as shown in the diagram. What is the difference between the water pressure at the diver’s head and at his feet? Answer to the nearest pascal.

Answer

The diagram shows us that the diver’s head is 1.2 m and his feet are 1.8 m below the surface of the water.

We recall that the pressure 𝑝 produced by a fluid is given by the following expression: 𝑝=πœŒπ‘”β„Ž, where 𝜌 is the fluid’s density, 𝑔 is the acceleration due to gravity, and β„Ž is the depth of the fluid.

Since pressure varies directly with depth, we know the pressure at the diver’s feet will be greater than that at his head.

If we call the difference in pressure between these two depths Δ𝑝, then we can write Δ𝑝=πœŒπ‘”Ξ”β„Ž, where Ξ”β„Ž is the difference in depth between the diver’s head and feet. That difference is 1.8 m minus 1.2 m, so Ξ”β„Ž=0.6.m

Since the density of water is 1β€Žβ€‰β€Ž015 kg/m3 and the acceleration due to gravity is 9.8 m/s2, Δ𝑝=ο€Ή1015/×9.8/×(0.6)=5968.2.kgmmsmPa

Rounding this result to the nearest pascal, our answer is that the difference in pressure between the diver’s head and feet is 5β€Žβ€‰β€Ž968 Pa.

Key Points

  • All fluidsβ€”liquids and gasesβ€”produce pressure.
  • Pressure equals force divided by area.
  • Given a fluid of density 𝜌 and depth β„Ž experiencing an acceleration due to gravity 𝑔, the pressure 𝑝 produced by the fluid is 𝑝=πœŒπ‘”β„Ž.

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