## Physics Friday 117

What is the net force exerted by a uniform pressure P on the interior of a hemispherical surface of radius R?

One could find the force via the surface integral
$\mathbf{F}=\iint_{\Sigma}P\mathbf{\hat{n}}\,dA$; more specifically, one could use the rotational symmetry, which requires that the net force must be in the direction of the axis, to show one only needs the component along the axis:
$\mathbf{F}=P\mathbf{\hat{z}}\iint_{\Sigma}\cos\theta\,dA$ (where θ is the angle between the surface normal and the axial direction, so $\cos\theta=\mathbf{\hat{n}}\cdot\mathbf{\hat{z}}$).

However, a simpler method is simply to add the flat disk with the same edge as the hemisphere, to create a closed, hemispherical cavity. Now (since we are neglecting gravity), a gas contained within the cavity should exert no net force on its container. Thus, the force on the hemispherical surface must be equal in magnitude and opposite in direction to that on the flat surface. For the flat surface though, the force is simply the pressure times the area. As the surface is a disk of radius R, it has area $A=\pi{R^2}$, so the force on the flat face is $F=\pi{R^2}P$, and thus the force on the hemispherical surface is also of magnitude $F=\pi{R^2}P$ in the direction of the symmetry axis.

Note: we can confirm this by performing the integral outlined above:
$\begin{array}{rcl}\mathbf{F}&=&P\mathbf{\hat{z}}\iint_{\Sigma}\cos\theta\,dA\\&=&P\mathbf{\hat{z}}\int_{0}^{\pi/2}\int_0^{2\pi}\cos\theta\cdot{R^2}\sin\theta\,d\phi\,d\theta\\&=&R^2P\mathbf{\hat{z}}\int_{0}^{\pi/2}\int_0^{2\pi}\cos\theta\sin\theta\,d\phi\,d\theta\\&=&R^2P\mathbf{\hat{z}}\int_{0}^{\pi/2}2\pi\cos\theta\sin\theta\,d\theta\\&=&\pi{R^2}P\mathbf{\hat{z}}\int_{0}^{\pi/2}\sin(2\theta)\,d\theta\\&=&\pi{R^2}P\mathbf{\hat{z}}\left[-\frac{\cos(2\theta)}2\right]_{0}^{\pi/2}\\\mathbf{F}&=&\pi{R^2}P\mathbf{\hat{z}}\end{array}$.

### 2 Responses to “Physics Friday 117”

1. Physics Friday 144 « Twisted One 151's Weblog Says:

[…] inside the sphere is P, then the force that pressure exerts on the sphere is πr2P, as shown here. This force on the hemisphere must be countered by the material forces, and so we see that , where […]

2. Physics Friday 145 « Twisted One 151's Weblog Says:

[…] due to pressure must be the pressure times the cross sectional area (due to similar arguments as here and here), or . This must be balanced by the longitudinal stress force, which is the longitudinal […]