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Unformatted text preview: be expressed as F = P − Patm A . If we assume that the plunger remains
perpendicular to the floor of the load bed, the torque that the plunger creates about the axis
shown in the figure in the text is c hc h τ = Fl = P − Patm Al = P − Patm ( π r 2 ) l
SOLUTION Direct substitution of the numerical data into the expression above gives c h τ = 3.54 × 10 6 Pa − 1.01 × 10 5 Pa π ( 0.150 m ) 2 ( 3.50 m ) = 8.50 × 10 5 N ⋅ m 40. REASONING The ice with the bear on it is floating, so that the upwardacting buoyant
force balances the downwardacting weight Wice of the ice and weight Wbear of the bear. The
magnitude FB of the buoyant force is the weight WH 2 O of the displaced water, according to Chapter 11 Problems Archimedes’ principle. Thus, we have FB = WH 2 O 585 = Wice + Wbear , the expression with which we will obtain Wbear. We can express each of the weights WH 2 and Wice as mass times the O magnitude of the acceleration due to gravity (Equation 4.5) and then relate the mass to the
density and the displaced volume by using Equation 11.1.
SOLUTION Since the ice with the bear on it is floating, the upwardacting buoyant force
FB balances the downwardacting weight Wice of the ice and the weight Wbear of the bear.
The buoyant force has a magnitude that equals the weight WH 2 O of the displaced water, as stated by Archimedes’ principle. Thus, we have FB = WH 2 O = Wice + Wbear or Wbear = WH 2 O − Wice In Equation (1), we can use Equation 4.5 to express the weights WH 2 O (1)
and Wice as mass m times the magnitude g of the acceleration due to gravity. Then, the each mass can be
expressed as m = ρ V (Equation 11.1). With these substitutions, Equation (1) becomes
Wbear = mH O g − mice g = ( ρ H OVH
2 2 2 O )g − ( ρiceVice ) g (2) When the heaviest possible bear is on the ice, the ice is just below the water surface and
displaces a volume of water that is VH O = Vice . Substituting this result into Equation (2), we
2 find that Wbear = ( ρ H OVice ) g − ( ρiceVice ) g = ( ρH
2 ( )( 2 O − ρice )Vice g )( ) = 1025 kg/m3 − 917 kg/m3 5.2 m3 9.80 m/s2 = 5500 N 41. SSM REASONING The buoyant force exerted on the balloon by the air must be equal in
magnitude to the weight of the balloon and its contents (load and hydrogen). The magnitude
of the buoyant force is given by ρairV g . Therefore, ρairV g = Wload + ρhydrogenV g
where, since the balloon is spherical, V = (4 / 3)π r 3 . Making this substitution for V and
solving for r, we obtain
1/ 3 3Wload
r= 4π g(ρair − ρ hydrogen ) 586 FLUIDS SOLUTION Direct substitution of the data given in the problem yields
1/ 3 3(5750 N)
r=
2
3
3 4π (9.80 m/s ) (1.29 kg/m − 0.0899 kg/m ) = 4.89 m 42. REASONING When the cylindrical tube is floating, it is in equilibrium, and there is no net
force acting on it. Therefore, the upwarddirected buoyant force must have a magnitude that
equals the magnitude of the tube’s weight, which acts downward. Since the magnitude FB of
the buoyant force equals the weight W of the tube, we have FB = W. This fact, together with
Archimedes’ principle, will guide our solution.
SOLUTION According to Archimedes’ principle, the magnitude of the buoyant force
equals the weight of the displaced fluid, which is the mass m of the displaced fluid times the
magnitude g of the acceleration due to gravity, or W = mg (Equation 4.5). But the mass is
equal to the density ρ of the fluid times the displaced volume V, or m = ρV (Equation 11.1).
The result is that the weight of the displaced fluid is ρVg. Therefore, FB = W becomes ρVg = W
The volume V of the displaced fluid equals the crosssectional area A of the cylindrical tube
times the height h beneath the fluid surface, or V = Ah. With this substitution, our previous
result becomes ρAhg = W. Thus, the height h to which the fluid rises is h= W
ρ Ag a. The height hacid that the acid rises is
hacid = W ρacid Ag = 5.88 ×10−2 N (1280 kg/m )( 7.85 ×10
3 −5 m 2 )(9.80 m/s )
2 = 5.97 × 10−2 m b. The antifreeze rises to a height of
hantifreeze =
= W ρantifreeze Ag
5.88 ×10−2 N (1073 kg/m )( 7.85 ×10
3 −5 m 2 )(9.80 m/s )
2 = 7.12 ×10−2 m Chapter 11 Problems 587 43. REASONING Since the duck is in equilibrium, its downwardacting weight is balanced by
the upwardacting buoyant force. According to Archimedes’ principle, the magnitude of the
buoyant force is equal to the weight of the water displaced by the duck. Setting the weight
of the duck equal to the magnitude of the buoyant force will allow us to find the average
density of the duck.
SOLUTION Since the weight Wduck of the duck is balanced by the magnitude FB of the buoyant force, we have that Wduck = FB. The duck’s weight is Wduck = mg = (ρduckVduck)g, where ρduck is the average density of the duck and Vduck is its volume. The magnitude of the
buoyant force, on the other hand, equals the weight of the water displaced by the duck, or
FB = mwaterg, where mwater is the mass of the displaced water. But mwater = ρ water ( 1 Vduck ) ,
4 since onequarter of the duck...
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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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