Wind exerts a constant pressure p (force per unit area)

Wind exerts a constant pressure $p$ (force per unit area) on a rectangular door of width $b$ and height $h$, hinged along one vertical edge. Find the moment of the wind-pressure force about the hinge line.

Step 1: Set up coordinates

Let $x$ run from $0$ (the hinge) to $b$ (the free edge), and $y$ run over the door's height $h$.

Step 2: Moment of a thin strip

A vertical strip at distance $x$ from the hinge, of width $dx$ and full height $h$, has area $h\,dx$, force $dF=p\cdot h\,dx$, and moment arm $x$ about the hinge, so $dM=x\cdot p h\,dx$.

Step 3: Integrate over the width

\[M=\int_0^bx\cdot ph\,dx=ph\left[\frac{x^2}{2}\right]_0^b=\frac{pb^2h}{2}.\]

\[\boxed{M=\frac{pb^2h}{2}}\]

Step 4: Cross-check via the resultant force at the centroid

Since the pressure is uniform, the resultant force $F=p\cdot bh$ acts at the door's centroid, a distance $b/2$ from the hinge, giving $M=F\cdot(b/2)=pbh\cdot(b/2)=pb^2h/2$, matching Step 3.

Step 5: Physical interpretation

The moment grows with the square of the width $b$ (since both the total force and its moment arm scale with $b$) but only linearly with the height $h$. This is why wider doors need disproportionately stronger hinges and closers as width increases.