I have a cross section of a half-tube with a pressure gradient across it causing a force outwards. I am attempting to extract the vertical component (in relation to diagram) of the force on this semi-circular cross-section. The semi-circle has radius r and the tube an internal pressure P1 higher than the external P2.

Diagram of the case

Let Δp = P1 - P2, r be the radius of the semi-circle and l be the length of the tube.

The force perpendicular to a small area of the tube surface is dF = Δp dA. So across a small length of the circumference of the tube is dF = Δp l dx.

From reading I know that the sum of the vertical components of these forces is is 2rlΔp as apparently you can just multiply the horizontal projection of the curved surface by the length and the pressure difference.

However I thought that you would need to integrate the individual forces over the circumference of the semi-circle, πr, and multiply by $ \int_0^\pi \mathrm{sin}(\theta) \, \mathrm{d}\theta$ in order to extract the vertical component of each small force. This gives 2rlπΔp instead of the correct answer above.

Can anybody explain the flaw in my reasoning to me?

  • $\begingroup$ Knowing that the force at every point is constant, you can realize that a gross over estimate of the downward force would be to simply add up the magnitudes, which would give: $F = \Delta p * \Delta A$ where $\Delta A = \pi rL$. It should definitely be less than this amount since the vertical components must be less than the magnitudes. $\endgroup$
    – Jared
    May 2, 2015 at 13:24

1 Answer 1


The differential amount of force is given by:

$$ dF = \Delta p* dA $$

You do need to add up the vertical components, which are given by:

$$ dF_\downarrow(\theta) = \Delta p * dA * \sin(\theta) $$

This goes from $0 \leq \theta \leq \pi$. At each value of $\theta$ the differential area is the same:

$$ dA = rd\theta * dx $$

This gives that:

$$ dF_\downarrow(x, \theta) = r\Delta p\sin(\theta) * d\theta * dx $$

Integrating from $0$ to $\pi$ and from $0 \leq x \leq L$ gives:

$$ F_\downarrow = r\Delta p \int\limits_0^L dx \int\limits_0^\pi\sin(\theta)d\theta = 2rL\Delta p $$


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