Page 208 - Schaum's Outline of Theory and Problems of Applied Physics
P. 208

CHAP. 16]                             FLUIDS AT REST                                  193



            3. An external pressure exerted on a fluid is transmitted uniformly throughout the fluid. This does not mean
               that pressures in a fluid are the same everywhere, because the weight of the fluid itself exerts pressures
               that increase with increasing depth. The pressure at a depth h in a fluid of density d due to the weight
               of fluid above is

                                                     p = dgh
               Hence the total pressure at that depth is

                                                 p = p external + dgh
               When a body of fluid is in an open container, the atmosphere exerts an external pressure on it.


        SOLVED PROBLEM 16.8
              The interior of a submarine located at a depth of 50 m in seawater is maintained at sea-level atmospheric
                                                                                                3
                                                                                          3
              pressure. Find the force acting on a window 20 cm square. The density of seawater is 1.03 × 10 kg/m .
                  The pressure outside the submarine is p = p atm + dgh, and the pressure inside is p atm . Hence the net pressure
               p acting on the window is

                                                 3
                                                             2
                                                                             5
                                                     3

                                p = dgh = (1.03 × 10 kg/m )(9.8 m/s )(50 m) = 5.05 × 10 Pa
                                                           2
              Since the area of the window is A = (0.2m)(0.2m) = 0.04 m , the force acting on it is
                                                   5
                                                                           4
                                                                2
                                  F = p A = (5.05 × 10 Pa)(4 × 10 −2  m ) = 2.02 × 10 N

        SOLVED PROBLEM 16.9
              What is the pressure at the bottom of a swimming pool 6 ft deep that is filled with freshwater? Express
              the answer in pounds per square inch.
                                  p = p atm + (dg)h
                                          lb       lb        1ft 2
                                    = 14.7   + 62     (6ft)        = 17.3 lb/in. 2
                                          in. 2   ft 3      144 in. 2
                                                  2
                                            2
              Note the use of the conversion factor 1 ft /144 in. .
        ARCHIMEDES’ PRINCIPLE
        An object immersed in a fluid is acted on by an upward force that arises because pressures in a fluid increase
        with depth. Hence the upward force on the bottom of the object is more than the downward force on its top. The
        difference between the two, called the buoyantforce, is equal to the weight of a body of the fluid whose volume
        is the same as that of the object. This is Archimedes’ principle: The buoyant force on a submerged object is equal
        to the weight of fluid the object displaces.
            If the buoyant force is less than the weight of the object itself, the object sinks; if the buoyant force equals
        the weight of the object, the object floats in equilibrium at any depth in the fluid; if the buoyant force is more
        than the weight of the object, the object floats with part of its volume above the surface.

        SOLVED PROBLEM 16.10

              An aluminum canoe is floating in a swimming pool. After a while it begins to leak and sinks to the bottom.
              What, if anything, happens to the water level in the pool?
                  The water level falls. The floating canoe displaced a volume of water V whose weight equaled its own weight.
              The sunken canoe, however, displaces a volume of water equal to the volume of its aluminum shell, which is smaller
              than V because a given volume of aluminum weighs more than the same volume of water.
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