6.4. PHYSICS APPLICATIONS: WORK, FORCE, AND PRESSURE
373
where P represents pressure, F represents force, and A the area of the region being
considered. Of course, in the equation F = P A, we assume that the pressure is constant
over the entire region A.
Most people know from experience that the deeper one dives underwater while
swimming, the greater the pressure that is exerted by the water. This is due to the fact
that the deeper one dives, the more water there is right on top of the swimmer: it is the
force that “column” of water exerts that determines the pressure the swimmer experiences.
To get water pressure measured in its standard units (pounds per square foot), we say that
the total water pressure is found by computing the total weight of the column of water that
lies above a region of area 1 square foot at a fixed depth. Such a rectangular column with
a 1 × 1 base and a depth of d feet has volume V = 1 · 1 · d ft 3 , and thus the corresponding
weight of the water overhead is 62.4d. Since this is also the amount of force being exerted
on a 1 square foot region at a depth d feet underwater, we see that P = 62.4d (lbs/ft 2 ) is
the pressure exerted by water at depth d.
The understanding that P = 62.4d will tell us the pressure exerted by water at a depth
of d, along with the fact that F = P A, will now enable us to compute the total force that
water exerts on a dam, as we see in the following example.
Example 6.5. Consider a trapezoid-shaped dam that is 60 feet wide at its base and 90
feet wide at its top, and assume the dam is 25 feet tall with water that rises to within 5
feet of the top of its face. Water weighs 62.5 pounds per cubic foot. How much force does
the water exert against the dam?
Solution. First, we sketch a picture of the dam, as shown in Figure 6.18. Note that, as in
problems involving the work to pump out a tank, we let the positive x-axis point down.
It is essential to use the fact that pressure is constant at a fixed depth. Hence, we
consider a slice of water at constant depth on the face, such as the one shown in the
figure. First, the approximate area of this slice is the area of the pictured rectangle. Since
the width of that rectangle depends on the variable x (which represents the how far the
slice lies from the top of the dam), we find a formula for the function y = f (x) that
determines one side of the face of the dam. Since f is linear, it is straightforward to find
that y = f (x) = 45 −
3
5 x. Hence, the approximate area of a representative slice is
A slice = 2 f (x)△x = 2(45 −
3
5
x)△x.
At any point on this slice, the depth is approximately constant, and thus the pressure can
be considered constant. In particular, we note that since x measures the distance to the
top of the dam, and because the water rises to within 5 feet of the top of the dam, the
depth of any point on the representative slice is approximately (x − 5). Now, since pressure
373
where P represents pressure, F represents force, and A the area of the region being
considered. Of course, in the equation F = P A, we assume that the pressure is constant
over the entire region A.
Most people know from experience that the deeper one dives underwater while
swimming, the greater the pressure that is exerted by the water. This is due to the fact
that the deeper one dives, the more water there is right on top of the swimmer: it is the
force that “column” of water exerts that determines the pressure the swimmer experiences.
To get water pressure measured in its standard units (pounds per square foot), we say that
the total water pressure is found by computing the total weight of the column of water that
lies above a region of area 1 square foot at a fixed depth. Such a rectangular column with
a 1 × 1 base and a depth of d feet has volume V = 1 · 1 · d ft 3 , and thus the corresponding
weight of the water overhead is 62.4d. Since this is also the amount of force being exerted
on a 1 square foot region at a depth d feet underwater, we see that P = 62.4d (lbs/ft 2 ) is
the pressure exerted by water at depth d.
The understanding that P = 62.4d will tell us the pressure exerted by water at a depth
of d, along with the fact that F = P A, will now enable us to compute the total force that
water exerts on a dam, as we see in the following example.
Example 6.5. Consider a trapezoid-shaped dam that is 60 feet wide at its base and 90
feet wide at its top, and assume the dam is 25 feet tall with water that rises to within 5
feet of the top of its face. Water weighs 62.5 pounds per cubic foot. How much force does
the water exert against the dam?
Solution. First, we sketch a picture of the dam, as shown in Figure 6.18. Note that, as in
problems involving the work to pump out a tank, we let the positive x-axis point down.
It is essential to use the fact that pressure is constant at a fixed depth. Hence, we
consider a slice of water at constant depth on the face, such as the one shown in the
figure. First, the approximate area of this slice is the area of the pictured rectangle. Since
the width of that rectangle depends on the variable x (which represents the how far the
slice lies from the top of the dam), we find a formula for the function y = f (x) that
determines one side of the face of the dam. Since f is linear, it is straightforward to find
that y = f (x) = 45 −
3
5 x. Hence, the approximate area of a representative slice is
A slice = 2 f (x)△x = 2(45 −
3
5
x)△x.
At any point on this slice, the depth is approximately constant, and thus the pressure can
be considered constant. In particular, we note that since x measures the distance to the
top of the dam, and because the water rises to within 5 feet of the top of the dam, the
depth of any point on the representative slice is approximately (x − 5). Now, since pressure