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Pressure in liquid inside a tank with a sloped wall
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l
w
hθ ∆h
Figure 1: Sloping tank with liquid.
Consider the tank shown in Figure 1.
The volume of an infinitesimal strip of thickness ∆h is
∆V =12
∆h w
(l + l +
∆htan θ
).
The mass of this volume of liquid is∆m = ρ ∆V
where ρ is the density of the liquid.
The force exerted by this mass of liquid due to gravity is
∆f = g ∆m = ρ g w ∆h
(l +
∆h2 tan θ
).
where g is the acceleration due to gravity.
Therefore
lim∆h→0
∆f∆h
≡df
dh= ρ g w l(h) .
To get the total force at a location h = h1 due to water between locations h = h1 and h = h2 (h2 > h1) we integrate, and get
f(h1) = ρ g w
∫ h2
h1
l(h) dh .
The corresponding pressure at h = h1 is
p(h1) =ρ g
l(h1)
∫ h2
h1
l(h) dh .
Note that if l(h1) = l0 = 0, an infinite pressure is obtained.
If, as in the figure,l(h) = l0 + h cot θ
then
p(h1) =ρ g
l0 + h1 cot θ
[l0(h2 − h1) +
cot θ2
(h22 − h2
1)
].
1
For a vertical wall, θ = π/2 and we havep(h1) = ρ g (h2 − h1) .
And, if h1 = 0 and h2 = h, the pressure at the bottom is ρ g h.
2