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Question

The gas tank in a sports car is a cylinder lying on its side. If the diameter of the tank is 0.42 m and if the tank is filled with gasoline to within 0.21 m of the top, find the force on one end of the tank. The density of gasoline is 735 kg/m3. Use 𝑔=9.8 m/s2.

(Use decimal notation. Give your answer to one decimal place.)

Answer 1

First notice that the tank is filled exactly to the middle.

Let

d = 0.42 m be the diameter of the tank,

r = 0.21 m be the radius of the tank,

ρ = 735 kg/m³ be the density of gasoline,

g = 9.8 m/s² be gravitational acceleration.

Let me first explain the physics of the situation.

At some depth h from the surface, the gasoline has pressure

p = ρgh

This pressure is equal in all directions, including horizontally against the sidewall.

In general, a uniform pressure p against a surface of size S creates force

F = pS

Now let's go back to calculus.

Imagine a very thin horizontal layer of thickness dh and depth h from the surface of the gasoline.

Please draw a circle representing the tank's side with a horizontal line distance h from the center.

By Pythagorean theorem that line segment inside the circle has length

2√(r²-h²)

That layer of gasoline then projects on the tank's side a surface of area

S = 2√(r²-h²)dh

Therefore the gasoline is applying force

F = ρgh2√(r²-h²)dh

Now we need to add the forces by all such layers as h ranges from 0 to r.

That is the definite integral from 0 to r

∫ρgh2√(r²-h²)dh

First calculate the indefinite integral

G(h) = ∫ρgh2√(r²-h²)dh = 2ρg∫h√(r²-h²)dh = (-2/3)ρg(r²-h²)^3/2

The definite integral then evaluates to

G(r) - G(0) = (-2/3)ρg(r²-r²)^3/2 - (-2/3)ρg(r²-0²)^3/2 = (2/3)ρgr³

Substituting actual numbers, the total force is

(2/3)×735×9.8×0.21³ = 44.5 N