Unformatted text preview: W = mg (Equation 4.5). 97. REASONING AND SOLUTION The figure at the right shows the two
forces that initially act on the box. Since the box is not accelerated, the
two forces must have zero resultant: FB0 − Wbox = 0 . Therefore, FB0 = Wbox B0 (1)
W From Archimedes' principle, the buoyant force on the box is equal to
the weight of the water that is displaced by the box: FB0 = Wdisp F box (2) Combining (1) and (2) we have Wbox = Wdisp, or mbox g = ρwatergVdisp. Therefore, mbox = ρwaterVdisp
Since the box floats with onethird of its height beneath the water, Vdisp = (1/3)Vbox, or
Vdisp = (1/3)L3. Therefore, Chapter 11 Problems mbox = ρ water L3 (3) 3 The figure at the right shows the three forces that act on the box after
water is poured into the box. The box begins to sink when
Wbox + Wwater > FB 621 F B (4) The box just begins to sink when the equality is satisfied. From
Archimedes' principle, the buoyant force on the system is equal to the
weight of the water that is displaced by the system: FB = Wdisplaced. W box W water The equality in (4) can be written as
mboxg + mwaterg = mdisplacedg (5) When the box begins to sink, the volume of the water displaced is equal to the volume of the
box; Equation (5) then becomes
mbox + ρwaterVwater = ρwaterVbox. The volume of water in the box at this instant is Vwater = L2h, where h is the depth of the
water in the box. Thus, the equation above becomes
mbox + ρwaterL2h = ρwaterL3 Using Equation (3) for the mass of the box, we obtain ρ water L3
3 + ρ water L2 h = ρ water L3 Solving for h gives
2
3 2
3 h = L = (0.30 m) = 0.20 m 98. REASONING AND SOLUTION The mercury, being more dense, will flow from the right
container into the left container until the pressure is equalized. Then the pressure at the
bottom of the left container will be P = ρwghw + ρmghmL and the pressure at the bottom of
the right container will be P = ρmghmR. Equating gives ρwghw + ρmg(hmL − hmR) = 0 (1) 622 FLUIDS Both liquids are incompressible and immiscible so
hw = 1.00 m and hmL + hmR = 1.00 m
Using these in (1) and solving for hmL gives, hmL = (1/2)(1.00 − ρw/ρm) = 0.46 m.
So the fluid level in the left container is 1.00 m + 0.46 m = 1.46 m from the bottom. 99. SSM WWW
container is REASONING AND SOLUTION The pressure at the bottom of the
P = Patm + ρwghw + ρmghm We want P = 2Patm, and we know h = hw + hm = 1.00 m. Using the above and rearranging
gives
hm = Patm − ρ w gh ( ρm − ρw ) g = ( )( ) 1.01 × 105 Pa − 1.00 × 103 kg/m3 9.80 m/s 2 (1.00 m ) (13.6 × 10 3 kg/m − 1.00 × 10 kg/m
3 3 3 )(9.80 m/s2 ) = 0.74 m 100. REASONING AND SOLUTION
We refer to Table 11.1 for values of
3
3
ρsoda = ρwater = 1.000 × 10 kg/m and ρAl = 2700 kg/m3 for the density of aluminum. The
mass of aluminum required to make the can is mAl = mtotal − msoda
where the mass of the soda is
3 3 msoda = ρsodaVsoda = (1.000 × 10 kg/m )(3.54 × 10 –4 Therefore, the mass of the aluminum used to make the can is mAl = 0.416 kg − 0.354 kg = 0.062 kg
Using the definition of density, ρ = m/V, we have VAl = mAl ρ Al = 0.062 kg
= 2.3 ×10 –5 m3
2700 kg/m3 3 m ) = 0.354 kg Chapter 11 Problems 623 101. REASONING The number of gallons per minute of water that the fountain uses is the
volume flow rate Q of the water. According to Equation 11.10, the volume flow rate is
Q = Av , where A is the crosssectional area of the pipe and v is the speed at which the
water leaves the pipe. The area is given. To find a value for the speed, we will use
Bernoulli’s equation. This equation states that the pressure P, the fluid speed v, and the
elevation y at any two points (1 and 2) in an ideal fluid of density ρ are related according to
2
2
P + ρ v1 + ρ gy1 = P2 + ρ v2 + ρ gy2 (Equation 11.11). Point 1 is where the water leaves
1
1
2 1
2 the pipe, so we are seeking v1. We must now select point 2. Note that the problem specifies
the maximum height to which the water rises. The maximum height occurs when the water
comes to a momentary halt at the top of its path in the air. Therefore, we choose this point
as point 2, so that we can take advantage of the fact that v2 = 0 m/s.
We know that P = P2 = Patmospheric and v2 = 0 m/s.
1 SOLUTION Substituting this information into Bernoulli’s equation, we obtain
2
Patmospheric + ρ v1 + ρ gy1 = Patmospheric + ρ ( 0 m/s ) + ρ gy2
1
2 2 1
2 Solving for v1 gives v1 = 2 g ( y2 − y1 ) , which we now substitute for the speed v in the
expression Q = Av for the volume flow rate:
Q = Av1 = A 2 g ( y2 − y1 ) ( = 5.00 ×10−4 m 2 ) 2 (9.80 m/s2 ) (5.00 m − 0 m ) = 4.95 ×10−3 m3 / s To convert this result to gal / min, we use the facts that 1 gal = 3.79 × 10−3 m3 and
1 min = 60 s: m3 1 gal Q = 4.95 ×10−3 3.79 × 10−3 m3 s 60 s = 78.4 gal / min 1 min 102. REASONING AND SOLUTION The upward buoyant force must equal the weight of the
shell if it is floating, ρwgV = W. The submerged volume of the shell is V = (4/3)π R23,
where R2 is its outer radius. Now R23 = (3/4)m/(ρwπ ) gives R2 = 3 3m
4π ρ w = 3 ( 3 (1.00 kg ) 4π 1.00 × 10 kg/m...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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