Density and Pressure — Question 5

Question 5

A spatially varying pressure shifts the resultant

A rectangular panel has width $b$ and length $L$. A prescribed net normal pressure varies along the coordinate $x$ as $q(x)=q_0(1+x/L)$, $0\le x\le L$. It is uniform across the width. The same normal direction applies everywhere; the pressure profile is given, not derived from depth.

1. Find the resultant force and its point of application $x_R$.

2. Evaluate for $q_0=2.0\,\mathrm{kPa}$, $b=0.40\,\mathrm{m}$, $L=0.60\,\mathrm{m}$. Why does using the midpoint pressure find the force but not its moment?

Question 5 - Solution

A spatially varying pressure shifts the resultant

(a) Derivation and calculation

$$dF=bq_0(1+x/L)\,dx.$$$$F=bq_0\int_0^L(1+x/L)\,dx=\boxed{\tfrac32bq_0L}.$$$$Fx_R=bq_0\int_0^L(x+x^2/L)\,dx=\tfrac56bq_0L^2.$$$$\boxed{x_R=5L/9}.$$

(b) Result and interpretation

$$\boxed{F=\tfrac32(.40)(2000)(.60)=720\,\mathrm N},\quad\boxed{x_R=.33333\,\mathrm m}.$$$$M_0=Fx_R=240\,\mathrm{N\,m},\quad F(L/2)=216\,\mathrm{N\,m}.$$Midpoint pressure equals the mean for a linear profile, but placing the resultant at the midpoint loses the correct moment.

Physical check

The resultant lies toward the higher-pressure end: $5L/9>L/2$.