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68. Problem 17.21(HR) Consider the horizontal acceleration of a mass of liquid in
an open tank. Acceleration of this kind causes the liquid surface to drop at the
front and to rise at the rear. (a) We have to show that the liquid
surface slopes at and angle
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with
the horizontal, where
, a
being the horizontal acceleration. (b) How does the pressure with depth?|
Solution: Click For PDF Version Concepts that we will use in solving this problem are that
pressure force is perpendicular to fluid surface and that hydrostatic gauge
pressure varies with depth by the standard relation
We consider an element of the fluid that is moving horizontally with acceleration. Relevant data concerning the motion of this element have been shown in the following diagram.
Let L be the length and W be the width of this element of
fluid, and let
Horizontal forces acting on this element are the force due to pressure acting
on its vertical face, height
from 0 to
Horizontal component of the force due to atmospheric pressure
The net horizontal force on the fluid element will be
Mass of the fluid element is
As it is moving with acceleration a, from the Newton’s second law of motion we have
or
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.
If a fluid element is under acceleration a then the net force on it due
to gauge pressure has to be equal to its mass times the acceleration a.
be the
angle that its surface makes with the horizontal when it is moving horizontally
with acceleration a. Let the atmospheric pressure acting on this element
of fluid be
.
and width
W , and the horizontal component of the force due to atmospheric pressure
and width
W. Force on the vertical face can be calculated by integrating
. We get
,
that is
.
.
,