Intro to Pressure: Videos & Practice Problems

Pressure is defined as force divided by area, measured in pascals (Pa). Atmospheric pressure, the weight of air above, is approximately 1.01 x 10⁵ Pa at sea level. As altitude increases, atmospheric pressure decreases due to less air above. In liquids, pressure increases with depth, calculated using the equation: P_bottom = P_top + ρgh. Understanding these concepts is crucial for applications in fluid dynamics and atmospheric science.

Pressure and Atmospheric Pressure

Pressure is a fundamental concept defined as the force applied per unit area, expressed mathematically as:

\[ P = \frac{F}{A} \]

where \( P \) is pressure, \( F \) is force, and \( A \) is area. The standard unit of pressure is the Pascal (Pa), which is equivalent to one Newton per square meter (N/m²). Understanding pressure is crucial, as it describes how force is distributed over a surface.

To illustrate this, consider two identical wooden blocks with a density of 800 kg/m³ and dimensions of 0.2 m x 0.2 m x 1 m. When these blocks are placed on a horizontal surface, they exert pressure on that surface due to their weight. The weight of each block can be calculated using the formula:

\[ F = mg \]

where \( m \) is mass and \( g \) is the acceleration due to gravity (approximately 10 m/s² for simplification). The mass can be derived from the density and volume:

\[ m = \rho V \]

where \( \rho \) is density and \( V \) is volume. The volume of the block is calculated as:

\[ V = 0.2 \times 0.2 \times 1 = 0.04 \, \text{m}^3 \]

Thus, the mass of the block is:

\[ m = 800 \, \text{kg/m}^3 \times 0.04 \, \text{m}^3 = 32 \, \text{kg} \]

Substituting this mass into the weight formula gives:

\[ F = 32 \, \text{kg} \times 10 \, \text{m/s}^2 = 320 \, \text{N} \]

Now, to find the pressure exerted by the block on the surface, we calculate the area of contact:

\[ A = 0.2 \times 0.2 = 0.04 \, \text{m}^2 \]

Thus, the pressure is:

\[ P = \frac{F}{A} = \frac{320 \, \text{N}}{0.04 \, \text{m}^2} = 8000 \, \text{Pa} \]

In a different orientation, where the block has a larger area of contact, the pressure will be lower due to the same force being distributed over a larger area. This illustrates that pressure is inversely related to area when force remains constant.

Another important concept is atmospheric pressure, which is the pressure exerted by the weight of air molecules surrounding us. At sea level, the standard atmospheric pressure is approximately:

\[ P_{\text{atm}} = 1.01 \times 10^5 \, \text{Pa} \]

This value can also be expressed as 1 atmosphere (1 ATM) for convenience. Atmospheric pressure can vary with altitude; it decreases as one ascends to higher elevations.

To calculate the force exerted by atmospheric pressure on the top surface of the blocks, we can use the relationship:

\[ F = P \times A \]

For the first block, with an area of 0.04 m², the force is:

\[ F = 1.01 \times 10^5 \, \text{Pa} \times 0.04 \, \text{m}^2 = 4040 \, \text{N} \]

This force represents the weight of the air above the block, which is substantial, equivalent to the weight of approximately 400 kg.

For the second block, with a larger area of 0.2 m², the force becomes:

\[ F = 1.01 \times 10^5 \, \text{Pa} \times 0.2 \, \text{m}^2 = 20200 \, \text{N} \]

This demonstrates that a larger surface area results in a greater force due to the increased weight of the air above it.

In summary, understanding pressure, its calculation, and the effects of atmospheric pressure are essential in various scientific and engineering applications. The relationship between force, area, and pressure is foundational in physics, influencing how we interact with the physical world.

A large warehouse is 100 m wide, 100 m deep, 10 m high:
a) What is the total weight of the air inside the warehouse?
b) How much pressure does the weight of the air apply on the floor?

• A. (a) 1.2×10⁴ N;
  (b) 1200 Pa
• B. (a) 1.2×10⁵ N;
  (b) 120 Pa
• C. (a) 1.2×10⁵ N;
  (b) 1200 Pa
• D. (a) 1.2×10⁶ N;
  (b) 120 Pa

Answer: (D)

Pressure In Air and In Liquids

Understanding pressure is crucial in both atmospheric and liquid contexts. At sea level, the standard atmospheric pressure is approximately 1 ATM, which is equivalent to 101,325 Pascals. As you ascend in altitude, the air pressure decreases because there is less air above you exerting weight. This decrease in air pressure is accompanied by a reduction in air density, as the molecules become less compressed at higher elevations. For instance, at 100 meters above sea level, the change in pressure is negligible, and it can be approximated as 1 ATM.

In contrast, when submerged in a liquid, such as water, the pressure increases with depth due to the higher density of liquids compared to air. The pressure in a liquid can be calculated using the equation:

P = P₀ + ρgh

In this equation, P represents the pressure at a certain depth, P₀ is the pressure at the surface, ρ is the density of the liquid, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the depth of the liquid. The pressure at the bottom of a liquid column is referred to as absolute pressure, while the pressure at the top is known as relative pressure. The difference between these two is termed gauge pressure.

For example, if a person is 1.8 meters tall and their heart is located 1.4 meters from their feet, the height difference from the heart to the top of the head is 0.4 meters. If the blood pressure near the heart is 1.3 × 10⁴ Pascals, we can calculate the blood pressure at the top of the head using the aforementioned equation. By rearranging the equation, we find:

P_top = P_bottom - ρgh

Substituting the known values, we can determine the pressure at the top of the head. Similarly, to find the blood pressure at the feet, we would set up the equation with the heart as the top reference point and the feet as the bottom, adjusting the height accordingly.

As a general rule, pressure increases with depth in liquids, while it decreases with altitude in gases. This principle is essential for understanding various phenomena in physics and engineering, particularly in fields related to fluid dynamics and atmospheric science.

Calculating Pressure in Liquids

Understanding pressure in liquids is crucial, especially as it varies with depth. The relationship can be expressed with the equation:

$$P_{\text{bottom}} = P_{\text{top}} + \rho g h$$

In this equation, P represents pressure, ρ is the density of the liquid, g is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and h is the depth measured from the top of the liquid down to the point of interest. It is essential to note that h should be treated as a depth rather than a height, which is a common misconception.

When dealing with multiple layers of liquids, the points where two different fluids meet are referred to as interfaces. At these interfaces, the pressure is equal for both fluids, meaning there are no abrupt changes in pressure across the boundary. For example, the pressure at the topmost layer of liquid in contact with air is equal to the atmospheric pressure, which is approximately \(101,000 \, \text{Pa}\) (Pascals).

To illustrate these concepts, consider a scenario with two liquids in a beaker: a blue liquid with a density of \(1200 \, \text{kg/m}^3\) and a yellow liquid with a density of \(800 \, \text{kg/m}^3\). If the blue liquid has a height of \(0.06 \, \text{m}\) and the yellow liquid has a height of \(0.04 \, \text{m}\), the total height of the beaker is \(0.12 \, \text{m}\). The gap at the bottom of the beaker is \(0.02 \, \text{m}\).

To calculate the absolute pressure at the blue-yellow interface, we can use the previously mentioned equation. The pressure at this interface can be calculated by considering the pressure at the top (atmospheric pressure) and adding the pressure contribution from the yellow liquid:

$$P_{\text{blue-yellow}} = P_{\text{air}} + \rho_{\text{yellow}} g h_{\text{yellow}}$$

Substituting the known values:

$$P_{\text{blue-yellow}} = 101,000 \, \text{Pa} + (800 \, \text{kg/m}^3)(9.81 \, \text{m/s}^2)(0.04 \, \text{m})$$

This results in an absolute pressure of approximately \(101,320 \, \text{Pa}\).

Next, gauge pressure, which is the pressure relative to atmospheric pressure, can be calculated as:

$$P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{air}}$$

In this case, the gauge pressure at the blue-yellow interface is simply the pressure contribution from the yellow liquid, which is \(320 \, \text{Pa}\).

For further calculations, such as finding the absolute pressure at the bottom of the blue liquid, the same principles apply. The pressure at the bottom can be calculated using the pressure at the blue-yellow interface and adding the contribution from the blue liquid:

$$P_{\text{blue}} = P_{\text{blue-yellow}} + \rho_{\text{blue}} g h_{\text{blue}}$$

By substituting the known values, we find:

$$P_{\text{blue}} = 101,320 \, \text{Pa} + (1200 \, \text{kg/m}^3)(9.81 \, \text{m/s}^2)(0.06 \, \text{m})$$

This results in an absolute pressure of approximately \(102,040 \, \text{Pa}\).

In summary, understanding how to calculate pressures in fluids involves recognizing the relationship between depth, density, and gravitational force, as well as the significance of interfaces between different fluids. This knowledge is essential for solving various problems in fluid mechanics.

The deepest known point on Earth is called the Mariana Trench, at ~11,000 m (~36,000 ft). If the surface area of the average human ear is 20 cm2, how much average force would be exerted on your ear at that depth?

• A. 2.22×10⁵ N
• B. 2.22×10⁷ N
• C. 1.11×10⁸ N
• D. 2.22×10⁹ N

Answer: (A)

A tall cylindrical beaker 10 cm in radius is placed on a picnic table outside. You pour 5 L of an 8,000 kg/m³ liquid and 10 L of a 6,000 kg/m³ liquid into. Calculate the total pressure at the bottom of the beaker. (Use g=10 m/s².)

• A. 3.20×10⁴
• B. 3.52×10⁴
• C. 1.33×10⁵
• D. 1.36×10⁵

Answer: (C)

A wooden cube, 1 m on all sides and having density 800 kg/m³ , is held under water in a large container by a string, as shown below. The top of the cube is exactly 2 m below the water line. Calculate the difference between the force applied by water to the top and to the bottom faces of the cube (Hint:calculate the two forces, then subtract. Use g=10 m/s².)

• A. F_bot − F_top = 800 N
• B. F_bot − F_top = 1000 N
• C. F_bot − F_top = 8,000 N
• D. F_bot − F_top = 10,000 N

Answer: (D)

Here’s what students ask on this topic:

What is the definition of pressure in physics?

Pressure in physics is defined as the force exerted per unit area. Mathematically, it is expressed as:

$P=\frac{F}{A}$

where $P$ is the pressure, $F$ is the force, and $A$ is the area over which the force is distributed. The SI unit of pressure is the pascal (Pa), which is equivalent to one newton per square meter (N/m²).

How does atmospheric pressure change with altitude?

Atmospheric pressure decreases with increasing altitude. This is because as you ascend, there is less air above you, resulting in a lower weight of the air column pressing down. At sea level, the standard atmospheric pressure is approximately 1.01 x 10⁵ pascals (Pa). As you go higher, the air becomes less dense, and the pressure drops. This relationship is crucial for understanding phenomena in meteorology and aviation.

What is the equation for pressure in a liquid and how is it used?

The pressure in a liquid increases with depth and can be calculated using the equation:

$P_{\mathrm{bottom}}=P_{\mathrm{top}}+\rho gh$

where $P_{\mathrm{bottom}}$ is the pressure at the bottom, $P_{\mathrm{top}}$ is the pressure at the top, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the height (or depth) of the liquid column. This equation is essential for solving problems related to fluid statics and understanding pressure variations in liquids.

What is gauge pressure and how is it different from absolute pressure?

Gauge pressure is the pressure relative to the atmospheric pressure. It is calculated by subtracting the atmospheric pressure from the absolute pressure. Mathematically, it is expressed as:

$P{\mathrm{gauge}}_{}=P{\mathrm{absolute}}_{}-P{\mathrm{atmospheric}}_{}$

where $P{\mathrm{gauge}}_{}$ is the gauge pressure, $P{\mathrm{absolute}}_{}$ is the absolute pressure, and $P{\mathrm{atmospheric}}_{}$ is the atmospheric pressure. Gauge pressure is often used in practical applications, such as tire pressure measurements, where the atmospheric pressure is not included in the reading.

How do you calculate the force applied by atmospheric pressure on a surface?

To calculate the force applied by atmospheric pressure on a surface, you can use the equation:

$F=PA$

where $F$ is the force, $P$ is the atmospheric pressure, and $A$ is the area of the surface. For example, at sea level, the atmospheric pressure is approximately 1.01 x 10⁵ pascals (Pa). If you have a surface area of 1 square meter, the force applied by the atmospheric pressure would be:

$F=1.01\times {10}^5\times 1$

This results in a force of 101,000 newtons (N).