Physics Solution Manual for 1100 and 2101

This is because no water is added to or taken out of

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Unformatted text preview: agnitude of buoyant force (2) Magnitude of buoyant force These equations contain two unknowns, the volume V of the object and its mass m. By subtracting Equation (2) from Equation (1), we can eliminate the mass algebraically. The result is 15.2 N − 13.7 N = gV ( ρ water − ρ alcohol ) Solving this equation for the volume, and using the densities from Table 11.1, we have V= 48. 15.2 N − 13.7 N 1.5 N = = 7.9 × 10−4 m3 2 3 3 3 g ( ρ water − ρalcohol ) 9.80 m/s 1.00 × 10 kg/m − 806 kg/m ( )( ) REASONING The free-body diagram shows the two forces acting on the balloon, its weight W and the buoyant force FB. Newton’s second law, Equation 4.2b, states that the net force ΣFy in the y direction is equal to the mass m of the balloon times its acceleration ay in that direction: FB − W = ma y 1 24 43 ΣFy Our solution is based on this statement. (4.2b) FB W Free-body diagram for the balloon +y Chapter 11 Problems 591 SOLUTION Solving Equation 4.2b for the acceleration ay gives ay = FB − W m (1) The weight W of an object is equal to its mass m times the magnitude g of the acceleration due to gravity, or W = mg (Equation 4.5). The mass, in turn, is equal to the product of an object’s density ρ and its volume V, so m = ρ V (Equation 11.1). Combining these two relations, the weight can be expressed as W = ρ V g. According to Archimedes’ principle, the magnitude FB of the buoyant force is equal to the weight of the cool air that the balloon displaces, so FB = mcool airg = (ρcool airV)g. Since we are neglecting the weight of the balloon fabric and the basket, the weight of the balloon is just that of the hot air inside the balloon. Thus, m = mhot air = ρhot airV and W = mhot airg = (ρhot airV)g . Substituting the expressions FB = (ρcool airV)g, m = ρhot airV, and W = (ρhot airV)g into Equation (1) gives ay = Vg − ρ hot airVg ( ρcool air − ρ hot air ) g ρ FB − W = cool air = m ρhot airV ρhot air (1.29 kg/m3 − 0.93 kg/m3 )(9.80 m/s2 ) = 3.8 m/s2 = 0.93 kg/m3 49. SSM REASONING AND SOLUTION The weight of the coin in air is equal to the sum of the weights of the silver and copper that make up the coin: e je j Wair = mcoin g = msilver + mcopper g = ρ silver Vsilver + ρ copper Vcopper g (1) The weight of the coin in water is equal to its weight in air minus the buoyant force exerted on the coin by the water. Therefore, Wwater = Wair − ρ water (Vsilver + Vcopper ) g Solving Equation (2) for the sum of the volumes gives V silver + V copper = Wair − Wwater ρ water g = mcoin g − Wwater ρ water g (2) 592 FLUIDS or (1.150 × 10 –2 kg) (9.80 m / s 2 ) − 0.1011 N = 1.184 × 10 –6 m 3 3 3 2 (1.000 × 10 kg / m ) (9.80 m / s ) Vsilver + Vcopper = From Equation (1) we have V silver = mcoin − ρ copper V copper ρ silver = 1.150 × 10 –2 kg − (8890 kg / m 3 )Vcopper 10 500 kg / m 3 or Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )Vcopper (3) Substitution of Equation (3) into Equation (2) gives Wwater = Wair − ρ water 1.095 × 10 –6 m 3 − ( 0.8467 )Vcopper + Vcopper g Solving for Vcopper gives Vcopper = 5.806 × 10 –7 m 3 . Substituting into Equation (3) gives Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )( 5.806 × 10 –7 m 3 ) = 6.034 × 10 –7 m 3 From Equation 11.1, the mass of the silver is msilver = ρsilverVsilver = (10 500 kg/m3 )(6.034 × 10−7 m 3 ) = 6.3 × 10 −3 kg 50. REASONING AND SOLUTION The buoyant force exerted by the water must at least equal the weight of the logs plus the weight of the people, FB = WL + WP ρwgV = ρLgV + WP Now the volume of logs needed is V= MP ρW − ρL = 4 ( 80.0 kg ) 1.00 × 10 kg/m − 725 kg/m 3 3 3 = 1.16 m3 The volume of one log is VL = π(8.00 × 10 –2 2 m) (3.00 m) = 6.03 × 10 –2 m 3 Chapter 11 Problems 593 The number of logs needed is –2 N = V/VL = (1.16)/(6.03 × 10 ) = 19.2 Therefore, at least 20 logs are needed . 51. SSM REASONING The height of the cylinder that is in the oil is given by hoil = Voil / ( π r 2 ) , where V oil is the volume of oil displaced by the cylinder and r is the radius of the cylinder. We must, therefore, find the volume of oil displaced by the cylinder. After the oil is poured in, the buoyant force that acts on the cylinder is equal to the sum of the weight of the water displaced by the cylinder and the weight of the oil displaced by the cylinder. Therefore, the magnitude of the buoyant force is given by F = ρ water gV water + ρ oil gV oil . Since the cylinder floats in the fluid, the net force that acts on the cylinder must be zero. Therefore, the buoyant force that supports the cylinder must be equal to the weight of the cylinder, or ρ water gV water + ρ oil gV oil = mg where m is the mass of the cylinder. Substituting values into the expression above leads to V water + ( 0.725)Voil = 7 .00 × 10 –3 m 3 From the figure in the text, Vcylinder = Vwater + (1) Voil . Substituting values into the expression for Vcylinder gives V water + Voil = 8.48 × 10 –3 m 3 (2) Subtracting Equation (1) from E...
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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.

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