solution
diagram for this equation can be drawn like this
let us take the triangular plate ABC
let us assume a small element of length dx and width D
then force on this element can be given as
"dF=P dA"
P=pressure at element and dA is area of it
so
"dF=\\rho gDxdx"
from triangle ABC and AEF we can write
"\\frac{D}{w}=\\frac{x}{h}\\\\"
"D=\\frac{wx}{h}.....eq.2"
from eq.1 and eq.2
"dF=\\rho gx^2\\frac{w}{h}dx"
so by integrating we can find force on it
"\\int dF=\\int^h_0(\\frac{\\rho gw}{h}x^2dx)"
so
"F=\\frac{\\rho g wh^2}{3}"
above expression is for the force on triangular plate.
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