Fluid pressure increases with depth, so the force on a submerged object is not usually found by multiplying one pressure by one area. Calculus lets us add up many thin horizontal strips, each with its own pressure. This is important for designing dams, aquarium walls, submarine windows, and storage tanks.
The main idea is that pressure depends on depth while area depends on the shape of the plate.
Understanding Calculus: Fluid Force and Pressure
A pressure value tells how hard the fluid pushes on each small unit of area. Its units are newtons per square metre, called pascals. A deeper strip has more water above it, so it receives a larger push than a strip near the surface.
The force on one thin strip is found by multiplying the pressure at that depth by the area of that strip. Calculus is needed because a real plate contains infinitely many strips, each with a slightly different force. The integral combines those small forces into one total force.
The first careful decision is the coordinate system. Many problems measure y downward from the water surface. Then y directly represents depth, which makes the pressure expression simple.
Other problems place the origin at the centre or bottom of the plate. In that case, depth is not just y. It must be written as a water surface height minus or plus the chosen coordinate.
Drawing the waterline, the plate, and one representative strip prevents most setup errors. Label the strip thickness as a tiny change in y. Its length comes from the plate shape, while its width into the page is included if the problem gives one.
The function for strip width often causes more trouble than the integration itself. A rectangle has constant width. A triangle has a width that changes linearly with height.
A circular window has strips that are shortest near the top and bottom and widest through the centre. For a circle, the width comes from the circle equation after solving for the horizontal distance at a chosen height.
If the plate is only partly underwater, use only the portion below the waterline. The upper and lower integration bounds describe actual depths in the fluid, not merely convenient points on a graph.
The total force gives the size of the push, but engineers often need its location too. This location is called the centre of pressure. It lies below the geometric centre of a vertical submerged plate because deeper regions receive greater pressure.
A force applied near the lower part of a dam wall can create a strong turning effect about its base. Hinges, bolts, and supports must resist both the total force and this turning effect. Students should check dimensions at every stage.
Pressure times area must end in newtons. A result in newtons per metre usually means a missing width. They should also keep density, gravity, lengths, and areas in one consistent unit system before calculating.
Key Facts
- Fluid pressure at depth h is P = rho g h.
- Hydrostatic force on a flat surface is F = integral P dA.
- For a vertical plate, use a thin horizontal strip with dA = w(y) dy.
- If y measures depth below the water surface, then F = integral rho g y w(y) dy.
- The integration limits must match the top and bottom depths of the submerged plate.
- Use rho = 1000 kg/m^3 and g = 9.8 m/s^2 for water near Earth unless stated otherwise.
Vocabulary
- Hydrostatic pressure
- Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above a point.
- Hydrostatic force
- Hydrostatic force is the total force a fluid at rest exerts on a submerged surface.
- Depth
- Depth is the vertical distance below the fluid surface, and it determines the pressure in a stationary fluid.
- Density
- Density is mass per unit volume and is represented by rho in fluid pressure formulas.
- Differential area
- A differential area is a very small piece of a surface, such as a thin horizontal strip with area dA = w(y) dy.
Common Mistakes to Avoid
- Using the area of the whole plate with the pressure at the top, which is wrong because pressure changes with depth and must be integrated over the plate.
- Measuring y upward from the bottom but using P = rho g y as if y were depth, which gives incorrect pressure values unless the coordinate system is converted correctly.
- Forgetting the strip width w(y), which is wrong because the area of each horizontal strip depends on the shape and width of the plate at that depth.
- Using inconsistent units, which is wrong because rho, g, depth, width, and force must be in compatible units such as kg/m^3, m/s^2, m, and newtons.
Practice Questions
- 1 A vertical rectangular plate is 2 m wide and 3 m tall, with its top edge at the water surface. Using rho = 1000 kg/m^3 and g = 9.8 m/s^2, find the hydrostatic force on one side of the plate.
- 2 A vertical rectangular plate is 4 m wide and 2 m tall, with its top edge 1 m below the water surface. Set up and evaluate the integral for the hydrostatic force on one side of the plate using rho = 1000 kg/m^3 and g = 9.8 m/s^2.
- 3 A triangular plate and a rectangular plate have the same area and extend from the water surface to the same maximum depth. Explain why their hydrostatic forces may be different.