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Recall and use the equation for the change in pressure beneath the surface of a liquid: $\Delta p = \rho g \Delta h$
Pressure is defined as the force exerted per unit area. In liquids, pressure acts equally in all directions.
The pressure at a certain depth in a liquid depends on the density of the liquid, the acceleration due to gravity, and the depth.
Where:
This equation states that the pressure at a depth is directly proportional to the density of the liquid and the depth.
Consider a liquid with a constant surface area. The pressure at a certain depth is caused by the weight of the liquid above that point. The weight of the liquid is given by mass x gravity. Mass is related to density and volume. Volume is related to area and depth. Combining these relationships leads to the pressure formula.
Example 1: Calculate the change in pressure at a depth of 2 meters in water, given that the density of water is 1000 kg/m3.
Using the formula: $\Delta p = \rho g \Delta h$
$\Delta p = (1000 \, \text{kg/m}^3) \times (9.8 \, \text{m/s}^2) \times (2 \, \text{m})$
$\Delta p = 19600 \, \text{Pa}$
Example 2: A scuba diver is at a depth of 10 meters in seawater. The density of seawater is approximately 1030 kg/m3. Calculate the pressure on the diver.
Using the formula: $\Delta p = \rho g \Delta h$
$\Delta p = (1030 \, \text{kg/m}^3) \times (9.8 \, \text{m/s}^2) \times (10 \, \text{m})$
$\Delta p = 99940 \, \text{Pa}$
Note: The pressure on the diver is the sum of atmospheric pressure and the pressure due to the water above.
Density ($\rho$) | Gravity ($g$) | Depth ($\Delta h$) | Change in Pressure ($\Delta p$) |
---|---|---|---|
1000 kg/m3 | 9.8 m/s2 | 1 m | 9800 Pa |
1000 kg/m3 | 9.8 m/s2 | 10 m | 98000 Pa |
1030 kg/m3 | 9.8 m/s2 | 1 m | 10094 Pa |
The pressure at a point in a liquid is always in the same direction as the force acting on a small horizontal area at that point.
The pressure is cumulative. The total pressure at a depth is the sum of the pressure due to all the layers of liquid above.