The pressure loading on the plate is described by the function $$\rho = 10 \left[ \frac{6}{x+1} + 8 \right] \; \mathrm{lb}/\mathrm{ft}^2.$$ Determine the magnitude of the resultant force and the coordinates $(\overline{x}, \overline{y})$ of the point where the line of action of the force intersects the plate.
I was able to solve this question using a simple integration. However, I tried to solve it as a co-planar distributed load as the width of the surface is constant and the loading varies only by x. When i tried this, the answers I got were wrong. Could anyone explain to me If this can really be expressed as a co-planar load and what the conditions are for a general distributed load to be represented as a co-planar load.
No magic, just grind:
$F= \int_{x=0}^2 \int_{y=0}^3 p(x,y) dx dy = 3 \int_0^2 10({6 \over x+1}+8)dx= 60 (3 \ln 3 + 8)$.
$\bar{x} = {1 \over F} \int_{x=0}^2 \int_{y=0}^3 xp(x,y) dx dy = {1 \over F} 60(14-3 \ln 3)$
$\bar{y} = {1 \over F} \int_{x=0}^2 \int_{y=0}^3 y p(x,y) dx dy = {1 \over F} \int_0^3 y dy \int_0^2 10({6 \over x+1}+8)dx = {3 \over 2}$