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Unformatted text preview: ual to the net outward force divided by the area of the
roof, (P1 − P2) = ΣF /A, the speed of the wind is
v2 = 2 ( ΣF ) ρA = 2(22 000 N)
=
(1.29 kg/m3 )(5.0 m × 6.3 m) 33 m/s 74. REASONING AND SOLUTION As seen in the figure, the lower pipe is at the level of zero
potential energy.
v2
A2 h
v1 A1 2 2 If the flow rate is uniform in both pipes, we have (1/2)ρv1 = (1/2)ρv2 + ρgh (since
P1 = P2) and A1v1 = A2v2. We can solve for v1, i.e., v1 = v2(A2/A1), and plug into the
previous expression to find v2, so that v2 = 2gh
2 A2 −1 A1 = ( ) 2 9.80 m/s 2 (10.0 m )
2 π ( 0.0400 m )2 −1 π ( 0.0200 m )2 = 3.61 m/s The volume flow rate is then given by
Q = A2v2 = π r2 v2 = π (0.0400 m) (3.61 m/s) = 1.81× 10 –2 m3 /s
2 2 608 FLUIDS 75. SSM REASONING AND SOLUTION Bernoulli's equation (Equation 11.12) is
2
2
1
1
P1 + 2 ρv 1 = P2 + 2 ρv 2 If we let the right hand side refer to the air above the plate, and the left hand side refer to the
air below the plate, then v 1 = 0 m/s, since the air below the plate is stationary. We wish to
find v 2 for the situation illustrated in part b of the figure shown in the text. Solving the
equation above for v 2 ( with v 2 = v 2 b and v 1 = 0 ) gives
v2 b = 2 ( P1 − P2 ) (1) ρ In Equation (1), P1 is atmospheric pressure and P2 must be determined. We must first
consider the situation in part a of the text figure.
The figure at the right shows the forces that act on the rectangular plate
in part a of the text drawing. F1 is the force exerted on the plate from
the air below the plate, and F2 is the force exerted on the plate from the
air above the plate. Applying Newton's second law, we have (taking
"up" to be the positive direction), F 1 mg
F1 − F2 − mg = 0 F 2 F1 − F2 = mg Thus, the difference in pressures exerted by the air on the plate in part a of the drawing is P1 − P2 = F1 − F2
A = mg
A (2) where A is the area of the plate. From Bernoulli's equation ( Equation 11.12) we have, with
v2 = v2a and v1a = 0 m/s,
2
1
P1 − P2 = 2 ρ v 2 a (3) where v 2 a is the speed of the air along the top of the plate in part a of the text drawing.
Combining Equations (2) and (3) we have mg 1 2
= ρv
A 2 2a (4) Chapter 11 Problems 609 The figure below, on the left, shows the forces that act on the plate in part b of the text
drawing. The notation is the same as that used when the plate was horizontal (part a of the
text figure). The figure at the right below shows the same forces resolved into components
along the plate and perpendicular to the plate. F 1 F 1 mg cosθ
F F 2 2 θ mg sinθ mg Applying Newton's second law we have F1 − F2 − mg sin θ = 0 , or
F1 − F2 = mg sin θ Thus, the difference in pressures exerted by the air on the plate in part b of the text figure is P1 − P2 = F1 − F2
A = mg sin θ
A Using Equation (4) above,
2
1
P1 − P2 = 2 ρv 2 a sin θ Thus, Equation (1) becomes v2 b = 2 c ρv
1
2 2
2a sin θ h ρ Therefore,
2
v 2 b = v 2 a sin θ = (11.0 m / s) 2 sin 30.0 ° = 7.78 m / s 76. REASONING AND SOLUTION
a. At the surface of the water (position 1) and at the exit of the hose (position 2) the
pressures are equal (P1 = P2) to the atmospheric pressure. If the hose exit defines y = 0 m,
2 2 we have from Bernoulli's equation (1/2)ρv1 + ρgy = (1/2)ρv2 . If we take the speed at the
surface of the water to be zero (i.e., v1 = 0 m/s) we find that
v2 = 2 gy 610 FLUIDS b. The siphon will stop working when v2 = 0 m/s, or y = 0 m , i.e., when the end of the
hose is at the water level in the tank.
c. At point A we have
2 PA + ρg(h + y) + (1/2)ρvA = P0 + (1/2)ρv2 2 But vA = v2 so that PA = P0 – ρ g ( y + h) 77. SSM WWW REASONING AND SOLUTION
a. If the water behaves as an ideal fluid, and since the pipe is horizontal and has the same
radius throughout, the speed and pressure of the water are the same at all points in the pipe.
Since the right end of the pipe is open to the atmosphere, the pressure at the right end is
atmospheric pressure; therefore, the pressure at the left end is also atmospheric pressure, or
1.01 × 10 5 Pa . b. If the water is treated as a viscous fluid, the volume flow rate Q is described by
Poiseuille's law (Equation 11.14):
Q= π R 4 ( P2 − P1 )
8η L Let P1 represent the pressure at the right end of the pipe, and let P2 represent the pressure
at the left end of the pipe. Solving for P2 (with P1 equal to atmospheric pressure), we
obtain
8η LQ
P2 =
+ P1
π R4
Therefore,
P2 = 8(1.00 × 10 −3 Pa ⋅ s)(1.3 m)(9.0 × 10 −3 m 3 / s )
+ 1.013 × 10 5 Pa = 1.19 × 10 5 Pa
π ( 6.4 × 10 −3 m) 4 78. REASONING AND SOLUTION
The Reynold's number, Re, can be written as
Re = 2v ρ R / η . To find the average speed v , ( ) ( 2000 ) 4.0 × 10−3 Pa ⋅ s
(Re)η
v=
=
= 0.5 m/s
2 ρ R 2 1060 kg/m3 8.0 × 10−3 m ( )( ) Chapter 11 Problems 611 π R 4 ( P2 − P )
1
79. REASONING The volume flow rate Q is governed by Poiseuille’s law: Q =
8η L
(Equation 11.14). We will obtain the factor Rdilated/Rnormal by applying this law to the dilated
and to the normal blood vessel.
SOLUTION Solving Poiseuille’s law for the radius R gives R = 4 8η LQ
. When the
π ( P2 − P )
1 vessel dilates, the viscosity η, the le...
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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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