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Unformatted text preview: end, the liquid collects at the other end and experiences a
centripetal force given by (see Equation 8.11, and use F = ma ) F = mrω 2 = mLω 2 .
Since there is no air in the tube, this is the only radial force experienced by the liquid, and it
results in a pressure of
F mLω 2
P= =
A
A
where A is the crosssectional area of the tube. The mass of the liquid can be expressed in
terms of its density ρ and volume V: m = ρV = ρAl . The pressure may then be written as P= ρ AlLω 2
A = ρ lLω 2 (1) If the tube were completely filled with liquid and allowed to hang vertically, the pressure at
the bottom of the tube (that is, where h = L ) would be given by P = ρgL (2) SOLUTION According to the statement of the problem, the quantities calculated by
Equations (1) and (2) are equal, so that ρlLω 2 = ρgL . Solving for ω gives ω= g
=
l 9.80 m / s 2
= 31.3 rad / s
0.0100 m Chapter 11 Problems 579 30. REASONING According to Equation 11.4, the pressure Pmercury at a point 7.10 cm below
the ethyl alcoholmercury interface is
Pmercury = Pinterface + ρ mercury gh mercury where Pinterface is the pressure at the alcoholmercury interface, and (1)
h mercury = 0.0710 m . The pressure at the interface is
Pinterface = Patm + ρ ethyl ghethyl (2) Equation (2) can be used to find the pressure at the interface. This value can then be used in
Equation (1) to determine the pressure 7.10 cm below the interface. SOLUTION Direct substitution of the numerical data into Equation (2) yields
Pinterface = 1.01 × 10 5 Pa + (806 kg / m 3 )( 9.80 m / s 2 )( 1.10 m) = 1.10 × 10 5 Pa
Therefore, the pressure 7.10 cm below the ethyl alcoholmercury interface is Pmercury = 1.10 × 10 5 Pa + (13 600 kg / m 3 )( 9.80 m / s 2 )( 0.0710 m) = 1.19 × 10 5 Pa 31. REASONING Pressure is defined as the magnitude of the force acting perpendicular to a
surface divided by the area of the surface. Thus, the magnitude of the total force acting on
the vertical dam surface is the pressure times the area A of the surface. But exactly what
pressure should we use? According to Equation 11.4, the pressure at any depth h under the
water is P = Patm + ρ gh , where Patm is the pressure of the atmosphere acting on the water at
the top of the reservoir. Clearly, P has different values at different depths. As a result, we
need to use an average pressure. In Equation 11.4, it is only the term ρgh that depends on
depth, and the dependence is linear. In other words, the value of ρgh is proportional to h.
Therefore, the average value of this term is ρ g
water in the full reservoir and 1
2 ( H ) , where H is the total depth of the
1
2 H is the average depth. The average pressure acting on the vertical surface of the dam in contact with the water is, then,
P = Patm + ρ g ( H)
1
2 (1) SOLUTION According to the definition of pressure given in Equation 11.3, the magnitude
Ftotal of the total force acting on the vertical surface of the dam is Ftotal = PA , where P is
given by Equation (1). Using Equation (1) to substitute for P gives 580 FLUIDS Ftotal = PA = Patm + ρ g ( H ) A
1
2 The area is A = (120 m)(12 m), and Patm = 1.01×105 Pa . Table 11.1 gives the density of
water as ρ = 1.00 ×103 kg/m3 . With these values, we find that
Ftotal = Patm + ρ g ( H ) A
1
2 ( )( = 1.01×105 Pa + 1.00 × 103 kg/m3 9.80 m/s 2 ) 1 (12 m ) (120 m )(12 m ) = 2.3 ×108 N
2 32. REASONING AND SOLUTION The force exerted on the top surface is
F2 = P2A2 = Patmπ R2 2 The force exerted on the bottom surface is F1 = P1A1 = (Patm + ρgh)π R12. Equating and
rearranging yields
R22 = R12(1 + ρgh/Patm)
or
2 2 R2 = 1.485 R1 (1) Consider a right triangle formed by drawing a vertical line from a point on the
circumference of the bottom circle to the plane of the top circle so that two sides are equal to
R2 – R1 and h. Then tan 30.0° = (R2 – R1)/h.
a. Now, R1 = R2 −h tan 30.0° = R2 – 2.887 m (2) Substituting (2) into (1) and rearranging yields
0.485 R22 – 8.57 R2 + 12.38 = 0
which has two roots, namely, R2 = 16.1 m and 1.59 m. The value R2 = 1.59 m leads to a
negative value for R1. Clearly, a radius cannot be negative, so we can eliminate the root
R2 = 1.59 m, and we conclude that R2 = 16.1 m .
b. Now that R2 is known, Equation (2) gives R1 = 13.2 m . Chapter 11 Problems 581 33. SSM REASONING According to Equation 11.4, the initial pressure at the bottom of the
P
c h+ ρgh , while the final pressure is P = c h+ ρgh . pool is P0 = Patm f 0 atm Therefore, the f change in pressure at the bottom of the pool is ∆P = Pf − P0 = P
P
P
P
c h+ ρgh − c h+ ρgh = c h− c h
atm atm f atm 0 f atm 0 According to Equation 11.3, F = PA , the change in the force at the bottom of the pool is ∆F = ( ∆P ) A = P
P
c h− c h A
atm f atm 0 SOLUTION Direct substitution of the data given in the problem into the expression above
yields
133 Pa
∆F = 765 mm Hg − 755 mm Hg (12 m)(24 m)
= 3.8 × 10 5 N
1.0 mm Hg b F
G
H g I
J
K Note that the conversion factor 133 Pa = 1.0 mm Hg is used to convert mm Hg to Pa. 34. REASONING...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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