(^724) A Textbook of Engineering Mechanics
Now from the geometry of the figure, we find that
AD : AC =x 1 : x 2 =
612
ABD : ABD
PP
wa wa
= 1: 2 Ans.
35.9.TOTAL PRESSURE ON AN INCLINED IMMERSED SURFACE
Fig. 35.7. Total pressure on an inclined surface.
Consider a plane inclined surface, immersed in a liquid as shown in Fig. 35.7.
Let w = Specific weight of the
liquid,
A = Area of the immersed surface,
x = Depth of centre of gravity of the immersed surface from the
liquid surface, and
θ = Angle at which the immersed surface is inclined with the
liquid surface.
Divide the whole surface into a no. of small parallel strips as shown in the figure. Let us
consider a strip of thickness dx, width b and at a distance l from O (the point, on the liquid surface,
where the immersed surface will meet, if produced).
We know that the intensity of pressure on the strip
=wl sin θ
and area of the strip =b dx
∴ Pressure on the strip = Intensity of pressure × Area
=wl sin θ b dx
Now* total pressure on the surface,
P =∫wlsinθb dx
=wlbdxsinθ∫
- The total pressure may also be found out by dividing the whole surface into a no. of small strips.
Let a 1 , a 2 , a 3 ... = Areas of the strips,
l 1 , l 2 , l 3 ... = Distance of the corresponding strips from 0.
∴Pressure on the first strip =wa 1 l 2 sin θ
Similarly, pressure on the second atrip
=wa 2 l 2 sin θ
and pressure on the third strip =wa 3 l 3 sin θ
∴Total pressure on the surface,
P=wa 1 l 1 sin θ + wa 2 l 2 sin θ + wa 3 l 3 sin θ ...
=w sin θ (a 1 l 1 + a 2 l 2 + a 3 l 3 + ...)
=wAl sin θ
= wAx (Q l sin θ = x)