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2-1 CHAPTER 2 The correspondence between the problem set in this fifth edition versus the problem set in the 4'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod). New Ol d New Ol d New Ol d 1 4 mod 21 13 4 1 E 33E mod 2 new 22 14 42 E 34E mod 3 new 23 15 4 3 E 3 5 E 4 7 mo d 2 4 1 7 44E 36E 5 2 mo d 2 5 1 8 45E 37E 6 new 26 new 4 6 E 3 8 E 7 new 27 19 4 7 E 3 9 E 8 new 28 20 4 8 E 4 0 E 9 5 mo d 2 9 2 1 49E 41E 10 6 30 22 11 8 mod 31 23 12 new 32 24 13 9 mod 33 new 14 10 mod 34 25 mod 15 11 35 26 mod 16 new 36 27 mod 17 new 37 28 18 16 mod 38 29 19 n ew 3 9 E 31E mod 2 0 1 2 40E 32E 2-2 2.1 The \u201cstandard\u201d acceleration (at sea level and 45 ° latitude) due to gravity is 9.80665 m/s 2 . What is the force needed to hold a mass of 2 kg at rest in this gravitational field ? How much mass can a force of 1 N support ? Solution: ma = 0 = \u2211 F = F - mg F = mg = 2 × 9.80665 = 19.613 N F = mg => m = F/g = 1 / 9.80665 = 0.102 kg 2.2 A model car rolls down an incline with a slope so the gravitational \u201cpull\u201d in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg. Find the acceleration. Solution: ma = \u2211 F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s 2 2.3 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 1075 kg. Find the necessary force. Solution: Acceleration is the time rate of change of velocity. ma = \u2211 F ; a = dV / dt = (60 × 1000) / (3600 × 5) = 3.33 m/s 2 F net = ma = 1075 × 3.333 = 3583 N 2.4 A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s 2 . What is the force needed to hold the clothes? Solution: F = ma = 2 kg × 24 m/s 2 = 48 N 2.5 A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s 2 up to a speed of 75 km/h. What are the force and total time required? Solution: a = dV / dt => \u2206 t = dV/a = [ ( 75 \u2212 20 ) / 4 ] × ( 1000 / 3600 ) \u2206 t = 3.82 sec ; F = ma = 1200 × 4 = 4800 N 2-3 2.6 A steel plate of 950 kg accelerates from rest with 3 m/s 2 for a period of 10s. What force is needed and what is the final velocity? Solution: Constant acceleration can be integrated to get velocity. a = dV / dt => \u222b dV = \u222b a dt => \u2206 V = a \u2206 t = 3 × 10 = 30 m/s V = 30 m/s ; F = ma = 950 × 3 = 2850 N 2.7 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration? Solution: ma = \u2211 F \u21d2 a = \u2211 F / m m = m steel + m propane = 15 + (1.75 × 44.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s 2 2.8 A rope hangs over a pulley with the two equally long ends down. On one end you attach a mass of 5 kg and on the other end you attach 10 kg. Assuming standard gravitation and no friction in the pulley what is the acceleration of the 10 kg mass when released? Solution: Do the equation of motion for the mass m 2 along the downwards direction, in that case the mass m 1 moves up (i.e. has -a for the acceleration) m 2 a = m 2 g \u2212 m 1 g \u2212 m 1 a (m 1 + m 2 ) a = (m 2 \u2212 m 1 )g This is net force in motion direction a = (10 \u2212 5) g / (10 + 5) = g / 3 = 3.27 m/s 2 g 1 2 2.9 A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s 2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s 2 . Find the required force. Solution: F = ma = F up \u2212 mg F up = ma + mg = 200 ( 2 + 9.5 ) = 2300 N 2-4 2.10 On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is \u201cweighed\u201d with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading? Solution: Moon gravitation is: g = g earth /6 m m m m Beam Balance Reading is 5 kg This is mass comparison Spring Balance Reading is in kg units length \u221d F \u221d g Reading will be 5 6 kg This is force comparison 2.11 One kilogram of diatomic oxygen (O 2 molecular weight 32) is contained in a 500- L tank. Find the specific volume on both a mass and mole basis ( v and v ). Solution: v = V/m = 0.5/1 = 0.5 m 3 /kg v = V/n = V m/M = Mv = 32 × 0.5 = 16 m 3 /kmol 2.12 A 5 m 3 container is filled with 900 kg of granite (density 2400 kg/m 3 ) and the rest of the volume is air with density 1.15 kg/m 3 . Find the mass of air and the overall (average) specific volume. Solution: m air = \u03c1 V = \u03c1 air ( V tot \u2212 m granite / \u03c1 ) = 1.15 [ 5 - (900 / 2400) ] = 1.15 × 4.625 = 5.32 kg v = V / m = 5 / (900 + 5.32) = 0.00552 m 3 /kg 2.13 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m 3 . If the system is decelerated with 6 m/s 2 what is the needed force? Solution: m = m tank + m gasoline = 15 + 0.3 × 800 = 255 kg F = ma = 255 × 6 = 1530 N 2-5 2.14 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 1500 kPa. Solution: Force balance: F \u2191 = PA = F \u2193 = P 0 A + m p g ; P 0 = 1 bar = 100 kPa A = ( \u03c0 /4) D 2 = ( \u03c0 /4) × 0.125 2 = 0.01227 m 2 m p = (P-P 0 )A/g = ( 1500 \u2212 100 ) × 1000 × 0.01227 / 9.80665 = 1752 kg 2.15 A barometer to measure absolute pressure shows a mercury column height of 725 mm. The temperature is such that the density of the mercury is 13550 kg/m 3 . Find the ambient pressure. Solution: Hg : \u2206 l = 725 mm = 0.725 m; \u03c1 = 13550 kg/m 3 P = \u03c1 g \u2206 l = 13550 × 9.80665 × 0.725 × 10 -3 = 96.34 kPa 2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gun- powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally? Solution: The cannon ball has 101 kPa on the side facing the atmosphere. ma = F = P 1 × A - P 0 × A a = (P 1 - P 0 ) × A / m = ( 7000 - 101 ) \u03c0 [ ( 0.15 2 /4 )/5 ] = 24.38 m/s 2 2.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction. Solution: ma = F = ( P 1 - P 0 ) A - mg sin 40 0 ma = ( 7000 - 101 ) × \u03c0 × ( 0.15 2 / 4 ) - 5 × 9.80665 × 0.6428 = 121.9 - 31.52 = 90.4 N a = 90.4 / 5 = 18.08 m/s 2 2-6 2.18 A piston/cylinder with cross sectional area of 0.01 m 2 has a piston mass of 100 kg resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Solution: Force balance: F \u2191 = F \u2193 = PA = m p g + P 0 A P = P 0 + m p g/A = 100 kPa + (100 × 9.80665) / (0.01 × 1000) = 100 kPa + 98.07 = 198 kPa 2.19 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700 kg of a car? Solution: F \u2193 = ma = mg = 740 × 9.80665 = 7256.9 N Force balance: F \u2191 = ( P - P0 ) A = F \u2193 => P = P 0 + F \u2193 / A A = \u03c0 D 2 (1 / 4) = 0.031416 m 2 P = 101 + 7256.9 / (0.031416 × 1000) = 332 kPa 2.20 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel. Solution: P gauge = 1.25 MPa = 1250 kPa; P 0 = 0.96 bar = 96 kPa P = P gauge + P 0 = 1250 + 96 = 1346 kPa 2.21 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is 97 kPa. If a U-tube with mercury, density 13550 kg/m 3 , is attached to the tank to measure the vacuum, what column height difference would it show? Solution: \u2206 P = P 0 - P tank = \u03c1 g \u2206 l \u2206 l = ( P 0 - P tank ) / \u03c1 g = [(97 - 85 ) × 1000 ] / (13550 × 9.80665) = 0.090 m = 90 mm 2-7 2.22 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa. The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown, the pressure is 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing the piston to rise 2 cm. Find the new pressure. Solution: A linear spring has a force linear proportional to displacement. F = k x, so the equilibrium pressure then varies linearly with volume: P = a + bV, with an intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V -> 0) when there is no spring force F = PA = P o A + m p g and the initial state. These two points determine the straight line shown in the P-V diagram. Piston area = A P = ( \u03c0 /4) × 0.1 2 = 0.00785 m 2 400 106.2 2 1 0 0.4 P V 0.557 2 P a = P 0 + m p g A p = 100 + 5 × 9.80665 0.00785 = 106.2 kPa intersect for zero volume. V 2 = 0.4 + 0.00785 × 20 = 0.557 L P 2 = P 1 + dP dV \u2206 V = 400 + (400-106.2) 0.4 - 0 (0.557 - 0.4) = 515.3 kPa 2.23 A U-tube manometer filled with water, density 1000 kg/m 3 , shows a height difference of 25 cm. What is the gauge pressure? If the right branch is tilted to make an angle of 30 ° with the horizontal, as shown in Fig. P2.23, what should the length of the column in the tilted tube be relative to the U-tube? Solution: h H 30 ° \u2206 P = F/A = mg/A = V \u03c1 g/A = h \u03c1 g = 0.25 × 1000 × 9.807 = 2452.5 Pa = 2.45 kPa h = H × sin 30 ° \u21d2 H = h/sin 30 ° = 2h = 50 cm 2-8 2.24 The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kg/m 3 . What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m 3 , as manometer fluid? Solution: \u2206 P = \u03c1 1 gh 1 = 900 × 9.807 × 0.2 = 1765.26 Pa = 1.77 kPa h hg = \u2206 P/ ( \u03c1 hg g) = ( \u03c1 1 gh 1 ) / ( \u03c1 hg g) = 900 13600 × 0.2 = 0.0132 m= 13.2 mm 2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercury manometer. Reservoir A is moved up/down so the two top surfaces are level at h 3 as shown in Fig. P2.25. Assuming that you know \u03c1 A , \u03c1 Hg and measure the heights h 1 , h 2 , and h 3 , find the density \u03c1 B . Solution: Balance forces on each side: P 0 + \u03c1 A g(h 3 - h 2 ) + \u03c1 Hg gh 2 = P 0 + \u03c1 B g(h 3 - h 1 ) + \u03c1 Hg gh 1 \u21d2 \u03c1 B = \u03c1 A \uf8ed \uf8ec \uf8eb \uf8f8 \uf8f7 \uf8f6 h 3 - h 2 h 3 - h 1 + \u03c1 Hg \uf8ed \uf8ec \uf8eb \uf8f8 \uf8f7 \uf8f6 h 2 - h 1 h 3 - h 1 2.26 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m 3 , the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall with diameter 4m. What is the total force from the bottom of each tank to the water and what is the pressure at the bottom of each tank? Solution: V A = H × \u03c0 D 2 × (1 / 4) = 10 × \u03c0 × 2 2 × ( 1 / 4) = 31.416 m 3 V B = H × \u03c0 D 2 × (1 / 4) = 2.5 × \u03c0 × 4 2 × ( 1 / 4) = 31.416 m 3 Tanks have the same volume, so same mass of water F = mg = \u03c1 V g = 1000 × 31.416 × 9.80665 = 308086 N Tanks have same net force up (holds same m in gravitation field) P bot = P 0 + \u03c1 H g P bot,A = 101 + (1000 × 10 × 9.80665 / 1000) = 199 kPa P bot,B = 101 + (1000 × 2.5 × 9.80665 / 1000) = 125.5 kPa 2-9 2.27 The density of mercury changes approximately linearly with temperature as \u03c1 Hg = 13595 - 2.5 T kg/ m 3 T in Celsius so the same pressure difference will result in a manometer reading that is influenced by temperature. If a pressure difference of 100 kPa is measured in the summer at 35 ° C and in the winter at -15 ° C, what is the difference in column height between the two measurements? Solution: \u2206 P = \u03c1 gh \u21d2 h = \u2206 P/ \u03c1 g ; \u03c1 su = 13507.5 ; \u03c1 w = 13632.5 h su = 100 × 10 3 /(13507.5 × 9.807) = 0.7549 m h w = 100 × 10 3 /(13632.5 × 9.807) = 0.7480 m \u2206 h = h su - h w = 0.0069 m = 6.9 mm 2.28 Liquid water with density \u03c1 is filled on top of a thin piston in a cylinder with cross- sectional area A and total height H . Air is let in under the piston so it pushes up, spilling the water over the edge. Deduce the formula for the air pressure as a function of the piston elevation from the bottom, h. Solution: Force balance H h P 0 Piston: F \u2191 = F \u2193 PA = P 0 A + m H 2 O g P = P 0 + m H 2 O g/A P = P 0 + (H-h) \u03c1 g h, V air P P 0 2.29 A piston, m p = 5 kg, is fitted in a cylinder, A = 15 cm 2 , that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s 2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: P 0 g Force balance: F \u2191 = F \u2193 = P 0 A + m p g = PA P = P 0 + m p g/A = 101.325 + 5 × 25 / (1000 × 0.0015) = 184.7 kPa 2-10 2.30 A piece of experimental apparatus is located where g = 9.5 m/s 2 and the temperature is 5 ° C. An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer (see Problem 2.27 for density) showing a height difference of 200 mm. What is the pressure drop in kPa? Solution: \u2206 P = \u03c1 gh ; \u03c1 Hg = 13600 \u2206 P = 13600 × 9.5 × 0.2 = 25840 Pa = 25.84 kPa 2.31 Repeat the previous problem if the flow inside the apparatus is liquid water, \u03c1 \u2245 1000 kg/m 3 , instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns. Solution: Balance forces in the manometer: P P 1 2 · · h h 1 2 H (H - h 2 ) - (H - h 1 ) = \u2206 h Hg = h 1 - h 2 P 1 A + \u03c1 H 2 O h 1 gA + \u03c1 Hg (H - h 1 )gA = P 2 A + \u03c1 H 2 O h 2 gA + \u03c1 Hg (H - h 2 )gA \u21d2 P 1 - P 2 = \u03c1 H 2 O (h 2 - h 1 )g + \u03c1 Hg (h 1 - h 2 )g P 1 - P 2 = \u03c1 Hg \u2206 h Hg g - \u03c1 H 2 O \u2206 h Hg g = 13600 × 0.2 × 9.5 - 1000 × 0.2 × 9.5 = 25840 - 1900 = 23940 Pa = 23.94 kPa 2.32 Two piston/cylinder arrangements, A and B, have their gas chambers connected by a pipe. Cross-sectional areas are A A = 75 cm 2 and A B = 25 cm 2 with the piston mass in A being m A = 25 kg. Outside pressure is 100 kPa and standard gravitation. Find the mass m B so that none of the pistons have to rest on the bottom. Solution: A B P P 0 0 Force balance for both pistons: F \u2191 = F \u2193 A: m PA g + P 0 A A = PA A B: m PB g + P 0 A B = PA B Same P in A and B gives no flow between them. m PA g A A + P 0 = m PB g A B + P 0 => m PB = m PA A A / A B = 25 × 25/75 = 8.33 kg

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