Physics Solution Manual for 1100 and 2101

00 103 kgm3 31102 m3 62 102 m3 81 kg 390 kgm3 31102

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 upward-acting buoyant force balances the downward-acting 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 upward-acting buoyant force FB balances the downward-acting 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 upward-directed 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 cross-sectional 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 downward-acting weight is balanced by the upward-acting 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 one-quarter of the duck...
View Full Document

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.

Ask a homework question - tutors are online