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Demonstration.- Use a solution of salt, which is heavier than water, and it will be found to require less height for the same pres


From this we see that the pressure on the bottom of a vessel depends also upon the density of the liquid.

140. Vertical Upward Pressure. It has been shown (§ 138) that the upward pressure at any point below the surface of a liquid is equal to the downward pressure.

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Demonstrations. Select a glass tube, like a lamp chimney, and a glass plate or disk just large enough to cover the end of the tube. Grind the end of this tube, and the glass plate, to a water-tight joint with emery. Fasten three cords to the plate or disk, and to a single cord, as in Fig. 116. Hold the disk in place over the bottom of the tube with the cord, and push the tube down into the water in a jar. The upward pressure will hold the disk in place without the cord. Pour water into the tube until the disk falls off; then the weight of the water poured in, added to the weight of the disk, will measure the upward pressure.


FIG. 116

117, with the

Place in the bottom of a shallow pan a piece of smooth glass, and stand upon it a heavy lamp chimney of the form shown in Fig. ends ground smooth. Pour water into the chimney, and observe that when it rises to some definite point, A, it will not rise any farther because it runs out at the bottom. If the chimney is held down by placing a finger on the top and it is then filled with water, it will remain full as long as it is held down, but when the finger is removed, the water suddenly drops to A and


FIG. 117

stops there. What is the effect of the upward pressure of the water on the collar of the chimney in this demonstration?

141. Pressure on the Side of a Vessel. When a liquid is contained in a vessel with vertical sides, the pressure at any point of a side depends upon its distance from the surface of the liquid. The total pressure on the sides of the vessel is the sum of all these pressures, which vary from zero at the surface to a maximum at the bottom.

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The pressure of a liquid upon any submerged surface is equal to the weight of a column of the liquid having the area of the surface for its base, and the depth of the center of gravity 1 of the given surface below the surface of the liquid for its height.


in which H is the height of the surface of the liquid above the center of gravity of the submerged surface, a is the area of the submerged surface, and W is the weight of a unit volume of the liquid.

This rule applies to all submerged surfaces, whether vertical, horizontal, or inclined, plane or curved, If the surface is the horizontal base of the vessel, the height of the column will be the total depth of the liquid.

The law may be expressed in a formula as follows:

Pressure =


FIG. 118


1 See § 83. In plane surfaces the center of gravity is the center of area. The center of gravity of a triangle, for instance, is a point two thirds of the distance from any angle to the mid-point of the opposite side.

A cubic foot of water weighs about 62.5 lb., or 1000 oz.

Example. The pressure of water on any submerged body, as in the figure, is found as follows:

Pressure on top (HaW) = 2 × (2 × 3) × 62.5
Pressure on bottom
× (2 × 3) × 62.5
Pressure on ends 2 × 2.5 × (2 × 1) × 62.5
Pressure on sides 2 × 2.5 × (3 × 1) × 62.5
Total pressure


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142. Center of Pressure. The center of pressure on a submerged surface is the point of application of the resultant of all the forces acting upon it, due to the pressure of the water. If we have a rectangular side to a vessel containing water, since the pressure increases from the top to the bottom, it is evident that this point must be below the middle of the side. Calculation and experiment show that it is two thirds of the distance from top to bottom.

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FIG. 119



A convenient way to determine the position of this point is as follows. Lay off a line CB (Fig. 119) perpendicular to the side AB

to represent the pressure at B. Draw the line CA. Any lines, as ED and GF, drawn perpendicular to AB, will represent the pressure at the points D and F respectively. Why?



750 lb.

1125 lb.
625 lb.

The area of the triangle ABC will represent the entire pressure upon AB, and the center of pressure will be at

the point H where the perpendicular from the center of area of the triangle meets the side AB. It is evident that this point will be in such a position that A H = AB, since the center of area is at O, a point such that AOAK.

If the side AB is movable, a support at H will prevent either the top or the bottom from being pushed out.

937.5 lb.

3437.5 lb.

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143. The Surface of a Liquid at Rest. We have already seen that when the resultant of all the forces that act upon

any point in a liquid is zero, there will be a condition of equilibrium, and the liquid will be at rest (§ 137).

In order that the surface of a liquid may be at rest, it must be horizontal. Suppose that the surface is not horizontal, as in Fig. 121.

The force of gravity AB, which acts upon any molecule upon the surface, as A, may be resolved into two forces, one of which, AC, is perpendicular, and the other, AD, parallel to the surface. Now the first, AC, will be opposed by the resistance of the liquid, while the other, AD, will move the molecule to a lower level. When the surface is horizontal, the action of gravity is perpendicular to it, and if we try to resolve this force into two components as before, we find that the component perpendicular to the surface is equal to the force, and the horizontal component is zero. Hence no movement will take place, and the liquid will remain at rest.



FIG. 121


144. Equilibrium in Communicating Vessels. Whenever a number of vessels are connected, and water is poured into one of them, it will, when

it comes to rest, stand at the same level in all. This is in direct accordance with the transfer of liquid pressure stated

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