4.2 Laws for Fluids at Rest
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fluid under pressure. Wherever a local change of pressure occurs in the fluid, the
nearby fluid must respond by moving, and each subsequent more distant layer of
fluid must move by the extra pressure of the prior layer. If the fluid returns to static
conditions, there will be new pressure forces on each element of fluid, but with no
shearing forces.
Suppose a small triangular object, with sides of length a, b, and c, and some
thickness d, were placed in the fluid. Under static conditions in a fluid, the forces on
any surface of the object can only be pressure forces perpendicular to the object’s
surfaces. The three pressure forces on the sides must add to zero, so they also form
a triangle, with the same angles as those of the physical triangle. The sides of these
two similar triangles must be proportional: F a /a = F b /b = F c /c. Since F a =
p a a d, F b = p b b d, and F c = p c c d, the pressure p a = p b = p c is the same
on all sides of the triangle, no matter its orientation. At a given depth of water, the
pressure on any side of a fish will be nearly constant. However, the ‘high’ side of
the fish will have less pressure than the ‘low’ side, being at a shallower depth. The
result is buoyancy.
A balancing of forces on each fluid element throughout the fluid leads to Pascal’s
principle: If A and B are two locations where a change in pressure has occurred,
then, after static conditions have returned,
p A = p B .
(4.1)
For a fluid which returns to rest, any change in pressure at one location will be
reflected by the same pressure change elsewhere in that fluid.
For a fluid at rest held by gravity, such as our atmosphere and oceans, the pressure
increases with depth. This is seen by balancing the forces on an element of the fluid
at a given height. Taking a rectangular volume of height dz and area A, the pressure
on the bottom of the volume must be greater than that on the top because of the
weight of the material in the volume. 2
Suppose the fluid has a density ρ at a given height z. Balancing forces on the
fluid element gives
p(z)A − p(z + dz)A = ρgdzA ,
(4.2)
or
−
dp
dz
= ρg .
(4.3)
2 Gas does have weight, because the Earth’s gravity accelerates gas molecules downward, and
decelerates them upward. This makes the impulse of the molecules on the top of the volume less
than that on the bottom. If the gas were held in a container and weighed, the extra impulse of the
gas between the top and bottom surfaces causes the container to press further on the scales.
79
fluid under pressure. Wherever a local change of pressure occurs in the fluid, the
nearby fluid must respond by moving, and each subsequent more distant layer of
fluid must move by the extra pressure of the prior layer. If the fluid returns to static
conditions, there will be new pressure forces on each element of fluid, but with no
shearing forces.
Suppose a small triangular object, with sides of length a, b, and c, and some
thickness d, were placed in the fluid. Under static conditions in a fluid, the forces on
any surface of the object can only be pressure forces perpendicular to the object’s
surfaces. The three pressure forces on the sides must add to zero, so they also form
a triangle, with the same angles as those of the physical triangle. The sides of these
two similar triangles must be proportional: F a /a = F b /b = F c /c. Since F a =
p a a d, F b = p b b d, and F c = p c c d, the pressure p a = p b = p c is the same
on all sides of the triangle, no matter its orientation. At a given depth of water, the
pressure on any side of a fish will be nearly constant. However, the ‘high’ side of
the fish will have less pressure than the ‘low’ side, being at a shallower depth. The
result is buoyancy.
A balancing of forces on each fluid element throughout the fluid leads to Pascal’s
principle: If A and B are two locations where a change in pressure has occurred,
then, after static conditions have returned,
p A = p B .
(4.1)
For a fluid which returns to rest, any change in pressure at one location will be
reflected by the same pressure change elsewhere in that fluid.
For a fluid at rest held by gravity, such as our atmosphere and oceans, the pressure
increases with depth. This is seen by balancing the forces on an element of the fluid
at a given height. Taking a rectangular volume of height dz and area A, the pressure
on the bottom of the volume must be greater than that on the top because of the
weight of the material in the volume. 2
Suppose the fluid has a density ρ at a given height z. Balancing forces on the
fluid element gives
p(z)A − p(z + dz)A = ρgdzA ,
(4.2)
or
−
dp
dz
= ρg .
(4.3)
2 Gas does have weight, because the Earth’s gravity accelerates gas molecules downward, and
decelerates them upward. This makes the impulse of the molecules on the top of the volume less
than that on the bottom. If the gas were held in a container and weighed, the extra impulse of the
gas between the top and bottom surfaces causes the container to press further on the scales.