# Force due to pressure over a surface

• tomwilliam2
In summary, Tom William 2 attempted to solve a homework problem involving Bernoulli's equation and pressure on a cylinder immersed in a fluid. He found an equation for p which takes the form $$p = p(\theta)$$ and got an equation for length-force which depends on the trig functions inside the integral. In order to find the total pressure force, he integrated over the surface. Thanks, that makes perfect sense.
tomwilliam2

## Homework Statement

I'm trying to find the Force per unit length due to a pressure function across a cylinder immersed in a fluid.

## Homework Equations

Bernouilli's equation,
$$\mathbf{F}_s = -\mathbf{n} \int_s p dA$$

## The Attempt at a Solution

I've got an expression for p which takes the form $$p = p(\theta)$$.
My expression for the unit vector is $$\mathbf{n}=\cos \theta \mathbf{i} + \sin \theta \mathbf{j}$$
I then get an equation:
$$\mathbf{F}_s/length = -(\cos \theta \mathbf{i} + \sin \theta \mathbf{j}) \int_s p(\theta) r d\theta$$

Now it's clear from the solution that I need to have the trig functions inside the integral...but is this mathematically justified? I know that the values of the unit vector change as we move around the surface, but does that mean I can just include them into the integral? Where the i-component for example would become:
$$\mathbf{F}_i / length = \int_s -p(\theta)r \cos \theta d\theta$$
Is this right?

The general form of the equation is:
$${\boldsymbol F}_\mathrm{s} = -\!\!\int\limits_s \!\!p\hat{\boldsymbol n}\mathrm{d}S$$
If the normal is independent of the variables of integration, then you can write
$${\boldsymbol F}_\mathrm{s} = -\hat{\boldsymbol n}\!\!\int\limits_s\!\! p\mathrm{d}S$$
You can see this if you look at the force on a differential fluid element:
$$\mathrm{d}{\boldsymbol F}_\mathrm{s} = -p\mathrm{d}{\boldsymbol S} = -p\hat{\boldsymbol n}\mathrm{d}S$$
which is integrated over the surface to find the total pressure force.

Thanks, that makes perfect sense.

jhae2.718 said:
If the normal is independent of the variables of integration, then you can write
$${\boldsymbol F}_\mathrm{s} = -\hat{\boldsymbol n}\!\!\int\limits_s\!\! p\mathrm{d}S$$
Yes, but that's only going to be true for a flat surface. For a curved surface the unit vector will change according to position on the surface.
tomwilliam2, I feel that you've missed some important facts in the statement of the problem. Please post it word for word.

It's part of a much longer problem...my mistake was applying the formula which applies only to a flat surface (where the unit vector can be taken outside the integral) instead of applying the general formula, with the unit vector containing elements dependent on the variable of integration.
All sorted now, and it works out as I was expecting.
Thanks again

jhae2.718 said:
The general form of the equation is:
$${\boldsymbol F}_\mathrm{s} = -\!\!\int\limits_s \!\!p\hat{\boldsymbol n}\mathrm{d}S$$
If the normal is independent of the variables of integration, then you can write
$${\boldsymbol F}_\mathrm{s} = -\hat{\boldsymbol n}\!\!\int\limits_s\!\! p\mathrm{d}S$$
You can see this if you look at the force on a differential fluid element:
$$\mathrm{d}{\boldsymbol F}_\mathrm{s} = -p\mathrm{d}{\boldsymbol S} = -p\hat{\boldsymbol n}\mathrm{d}S$$
which is integrated over the surface to find the total pressure force.
Where is that general equation from? It makes sense, but I haven't seen it in that form before.

## What is force due to pressure over a surface?

Force due to pressure over a surface is the amount of force exerted on a given area of a surface due to the pressure applied to it.

## How is force due to pressure calculated?

Force due to pressure can be calculated by multiplying the pressure applied to a surface by the area over which the pressure is applied. The formula for force due to pressure is F = P x A, where F is force, P is pressure, and A is area.

## What are some common units of measurement for force due to pressure?

The most common units of measurement for force due to pressure are newtons (N) and pounds (lbs). Other units include pascals (Pa), atmospheres (atm), and pounds per square inch (psi).

## How does force due to pressure affect objects?

Force due to pressure can cause objects to compress, deform, or move depending on the strength and direction of the force. For example, a high amount of pressure applied to a small area can cause an object to break or fracture.

## What are some real-world applications of force due to pressure?

Force due to pressure is an important concept in engineering and physics. It is used in calculating the structural integrity of buildings and bridges, determining the buoyancy of objects in fluids, and in many other practical applications such as scuba diving and hydraulic systems.

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