Physics Solution Manual for 1100 and 2101

500 m 0750 m 20 reasoning the magnitude of the force

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Unformatted text preview: d on the lid by the outside air. SOLUTION According to Equation 11.3, pressure is defined as P = F / A ; therefore, the magnitude of the force on the lid due to the air pressure is F = (0.85 × 105 N/m 2 )(1.3 × 10 –2 m 2 ) = 1.1× 103 N 12. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface divided by the area of the surface, according to Equation 11.3. The force magnitude, therefore, is equal to the pressure times the area. SOLUTION According to Equation 11.3, we have c F = PA = 8.0 × 10 4 lb / in. 2 6 b hb.1 in.g2.6 in.g= 1.3 × 10 6 lb 13. REASONING According to Equation 11.3, the pressure P exerted on the ground by the stack of blocks is equal to the force F exerted by the blocks (their combined weight) divided by the area A of the block’s surface in contact with the ground, or P = F/A. Since the pressure is largest when the area is smallest, the least number of blocks is used when the surface area in contact with the ground is the smallest. This area is 0.200 m × 0.100 m. SOLUTION The pressure exerted by N blocks stacked on top of one another is P= F N Wone block = A A (11.3) where Wone block is the weight of one block. The least number of whole blocks required to 5 produce a pressure of two atmospheres (2.02 × 10 Pa) is N= PA Wone block ( 2.02 ×105 Pa ) ( 0.200 m × 0.100 m ) = 24 = 169 N 570 FLUIDS 14. REASONING Since the weight is distributed uniformly, each tire exerts one-half of the weight of the rider and bike on the ground. According to the definition of pressure, Equation 11.3, the force that each tire exerts on the ground is equal to the pressure P inside the tire times the area A of contact between the tire and the ground. From this relation, the area of contact can be found. SOLUTION The area of contact that each tire makes with the ground is F A= = P 1 2 (Wperson + Wbike ) = 12 ( 625 N + 98 N ) = 4.76 ×10−4 m2 (11.3) 7.60 × 105 Pa P 15. REASONING The cap is in equilibrium, so the sum of all the +y forces acting on it must be zero. There are three forces in the Foutside vertical direction: the force Finside due to the gas pressure inside the bottle, the force Foutside due to atmospheric pressure outside the bottle, and the force Fthread that the screw thread exerts on Finside F the cap. By setting the sum of these forces to zero, and using the thread relation F = PA, where P is the pressure and A is the area of the cap, we can determine the magnitude of the force that the screw threads exert on the cap. SOLUTION The drawing shows the free-body diagram of the cap and the three vertical forces that act on it. Since the cap is in equilibrium, the net force in the vertical direction must be zero. ΣFy = − Fthread + Finside − Foutside = 0 (4.9b) Solving this equation for Fthread, and using the fact that force equals pressure times area, F = PA (Equation 11.3), we have Fthread = Finside − Foutside = Pinside A − Poutside A ( )( ) = ( Pinside − Poutside ) A = 1.80 ×105 Pa − 1.01×105 Pa 4.10 ×10−4 m 2 = 32 N 16. REASONING The power generated by the log splitter pump is the ratio of the work W done on the piston to the elapsed time t: Power = W t (6.10a) Chapter 11 Problems 571 The work done on the piston by the pump is equal to the magnitude F of the force exerted on the piston by the hydraulic oil, multiplied by the distance s through which the piston moves: ( ) W = F cos0o s = Fs (6.1) We have used θ = 0° in Equation 6.1 because the piston moves in the same direction as the force acting on it. The magnitude F of the force applied to the piston is given by F = PA (11.3) where A is the cross-sectional area of the piston and P is the pressure of the hydraulic oil. SOLUTION The head of the piston is circular with a radius r, so its cross-sectional area is given by A = π r 2 . Substituting Equation 11.3 into Equation 6.1, therefore, yields W = PAs = Pπ r 2 s (1) Substituting Equation (1) into Equation 6.10a gives the power required to operate the pump: ( ) 2.0 × 107 Pa π ( 0.050 m ) W Pπ r 2 s Power = = = t t 25 s 2 ( 0.60 m ) = 3.8 × 103 W 17. REASONING The pressure P due to the force FSonF that the suitcase exerts on the elevator F floor is given by P = SonF (Equation 11.3), where A is the area of the elevator floor A beneath the suitcase (equal to the product of the length and width of that region). According to Newton’s 3rd law, the magnitude FSonF of the downward force the suitcase exerts on the floor is equal to the magnitude FFonS of the upward force the floor exerts on the suitcase. We will use Newton’s 2nd law, ΣF = ma (Equation 4.1), to determine the magnitude FFonS of the upward force on the suitcase, which has a mass m and an upward acceleration of magnitude a = 1.5 m/s2, equal to that of the elevator. SOLUTION There are only two forces acting on the suitcase, the upward force FFonS that the floor exerts on the suitcase, and the downward weight W = mg (Equation 4.5) exerted by the earth, where g is the magnitude of the acceleration due to grav...
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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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