Fluid Mechanics with teacher's notes

SECTION 9-2
Visual Strategy
Figure 9-8
Point out that the fluid exerts
pressure in all directions. Pressure
against the sides of the box
also increases with depth.
Suppose the box in the dia-
Q gram is immersed in water
and L = 2.0 m. How does pressure
on the box’s sides 1.0 m
from its top compare with pressure
on the sides near the top and
near the bottom?
∆P = rg∆h = (1.00 ×
A
328
10 3 kg/m 3 )(9.81 m/s 2 )
(1.0 m) = 9800 Pa. This means
that the pressure on the middle
area of the sides of the box is
9800 Pa higher than the pressure
near the top and 9800 Pa lower
than the pressure near the bottom
of the box.
ANSWERS TO
Conceptual Challenge
1. The pressure inside the
building is approximately
equal to the pressure outside
the building.
2. The small piston must be
pushed a long distance to
move the object a short
distance.
3. With snowshoes, her weight
is applied over a larger area,
so the pressure is small.
Without snowshoes, the force
is applied over a smaller area,
so the pressure is large.
L
328
Chapter 9
P0
P0 + ρgh1
P0 + ρgh2
Figure 9-8
The fluid pressure at the bottom of
the box is greater than the fluid
pressure at the top of the box.
Pressure varies with depth in a fluid
1. Atmospheric pressure Why doesn’t the
roof of a building collapse under the tremendous
pressure exerted by our atmosphere?
2. Force and work In a hydraulic lift, which of
the two pistons (large or small) moves through a
longer distance while the hydraulic lift is lifting an
object? (Hint: Remember that for an ideal machine,
the input work equals the output work.)
As a submarine dives deeper in the water, the pressure of the water against the
hull of the submarine increases, and the resistance of the hull must be strong
enough to withstand large pressures. Water pressure increases with depth
because the water at a given depth must support the weight of the water above it.
Imagine a small area on the hull of a submarine. The weight of the entire
column of water above that area exerts a force on the area. The column of
water has a volume equal to Ah, where A is the cross-sectional area of the column
and h is its height. Hence the mass of this column of water is m = rV =
rAh. Using the definitions of density and pressure, the pressure at this depth
due to the weight of the column of water can be calculated as follows:
F mg
P = ⎯⎯ = ⎯⎯ = ⎯
A A
rVg
rAhg
⎯ = ⎯⎯ = rhg
A A
Note that this equation is valid only if the density is the same throughout the
fluid.
The pressure in the equation above is referred to as gauge pressure. It is not
the total pressure at this depth because the atmosphere itself also exerts a pressure
at the surface. Thus, the gauge pressure is actually the total pressure
minus the atmospheric pressure. By using the symbol P 0 for the atmospheric
pressure at the surface, we can express the total pressure, or absolute pressure,
at a given depth in a fluid of uniform density r as follows:
FLUID PRESSURE AS A FUNCTION OF DEPTH
P = P 0 + rgh
absolute pressure =
atmospheric pressure + (density × free-fall acceleration × depth)
This expression for pressure in a fluid can be used to help understand
buoyant forces. Consider a rectangular box submerged in a container of water.
The water pressure pushing down on the top of the box is −(P 0 + rgh 1), and
3. Snowshoes A woman
wearing snowshoes stands
safely in the snow. If she
removes her snowshoes,
she quickly begins to
sink. Explain what happens
in terms of force
and pressure.
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