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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