**ch02**

(Parte **1** de 4)

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).

2.1The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. 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 = å F = F - mg

F = mg=> | m = F/g = 1 / 9.80665 = 0.102 kg |

F = mg = 2 · 9.80665 = 19.613 N

2.2A model car rolls down an incline with a slope so the gravitational “pull” 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 = å F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2

2.3A 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 = å F ;a = dV / dt = (60 · 1000) / (3600 · 5) = 3.3 m/s2

Fnet = ma = 1075 · 3.3 = 3583 N

2.4A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s2. What is the force needed to hold the clothes?

Solution:

F = ma = 2 kg · 24 m/s2 = 48 N

2.5A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required?

Solution:

a = dV / dt => Dt = dV/a = [ ( 75 - 20 ) / 4 ] · ( 1000 / 3600 ) Dt = 3.82 sec ; F = ma = 1200 · 4 = 4800 N

2.6A steel plate of 950 kg accelerates from rest with 3 m/s2 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 dt | => DV = a Dt = 3 · 10 = 30 m/s |

V = 30 m/s ; | F = ma = 950 · 3 = 2850 N |

a = dV / dt => ò dV = ò

2.7A 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 = å F Þ a = å F / m m = msteel + mpropane = 15 + (1.75 · 4.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s2

2.8A 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 m2 along the downwards direction, in that case the mass m1 moves up (i.e. has -a for the acceleration)

m2 a = m2 g - m1 g - m1a | |

(m1 + m2 ) a = (m2 - m1 )g |

This is net force in motion direction

a = (10 - 5) g / (10 + 5) = g / 3 = 3.27 m/s2 |

2.9A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force.

Solution:

F = ma = Fup - mg Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N

2.10On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” 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?

Moon gravitation is: g = gearth/6 |

Solution: m m m m

Beam Balance Reading is 5 kg This is mass comparison Spring Balance Reading is in kg units length µ F µ g

Reading will be |

6 kg

This is force comparison

Solution: v = V/m = 0.5/1 = 0.5 m3/kg v = V/n = V m/M = Mv = 32 · 0.5 = 16 m3/kmol

2.12A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume.

Solution:

= 1.15 [ 5 - (900 / 2400) ] = 1.15 · 4.625 = 5.32 kg |

mair = r V = rair ( Vtot - mgranite / r ) v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg

2.13A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 6 m/s2 what is the needed force?

Solution:

m = mtank + mgasoline = 15 + 0.3 · 800 = 255 kg F = ma = 255 · 6 = 1530 N

2.14A vertical hydraulic cylinder has a 125-m 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› = PA = Ffl = P0A + mpg ; | P0 = 1 bar = 100 kPa |

A = (p/4) D2 = (p/4) · 0.1252 = 0.01227 m2 mp = (P-P0)A/g = ( 1500 - 100 ) · 1000 · 0.01227 / 9.80665 = 1752 kg

2.15A barometer to measure absolute pressure shows a mercury column height of 725 m. The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient pressure.

Solution:

Hg : Dl = 725 m = 0.725 m; | r = 13550 kg/m3 |

P = r gDl = 13550 · 9.80665 · 0.725 · 10-3 = 96.34 kPa

2.16A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gunpowder 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.

2.17Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction.

Solution:

= 121.9 - 31.52 = 90.4 N |

a = 90.4 / 5 = 18.08 m/s2

2.18A piston/cylinder with cross sectional area of 0.01 m2 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› = Ffl = PA = mpg + P0A

= 100 kPa + 98.07 = 198 kPa |

P = P0 + mpg/A = 100 kPa + (100 · 9.80665) / (0.01 · 1000)

2.19The 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:

Ffl = ma = mg = 740 · 9.80665 = 7256.9 N

Force balance:F› = ( P - P0 ) A = Ffl=> P = P0 + Ffl / A A = p D2 (1 / 4) = 0.031416 m2

P = 101 + 7256.9 / (0.031416 · 1000) = 332 kPa

2.20A 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:

Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa

2.21The 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/m3, is attached to the tank to measure the vacuum, what column height difference would it show?

Solution:

DP = P0 - Ptank = rgDl

Dl = ( P0 - Ptank ) / rg = [(97 - 85 ) · 1000 ] / (13550 · 9.80665) = 0.090 m = 90 m

2.22A 5-kg piston in a cylinder with diameter of 100 m 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 = PoA + mpg and the initial state. These two points determine the straight line shown in the P-V diagram.

Piston area = AP = (p/4) · 0.12 = 0.00785 m2

V 0.557 a = P0 + mpg

Ap = 100 + 5 · 9.80665

= 106.2 kPa | intersect for zero volume. |

P2 = P1 + dP dV D V

= 400 + |

0.4 - 0 (0.557 - 0.4)

= 515.3 kPa |

2.23A U-tube manometer filled with water, density 1000 kg/m3, 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

= 0.25 · 1000 · 9.807 = 2452.5 Pa | |

= 2.45 kPa |

DP = F/A = mg/A = Vrg/A = hrg h = H · sin 30° Þ H = h/sin 30° = 2h = 50 cm

2.24The difference in height between the columns of a manometer is 200 m with a fluid of density 900 kg/m3. What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m3, as manometer fluid?

Solution:

DP = r1gh1 = 900 · 9.807 · 0.2 = 1765.26 Pa = 1.7 kPa hhg = DP/ (rhg g) = (r1 gh1) / (rhg g) = 900

13600 ·0.2 = 0.0132 m= 13.2 m

as shown in Fig. P2.25. Assuming that you know rA, rHg and measure the heights h1, h2 , and h3, find the density rB.

Solution:

Balance forces on each side: P0 + rAg(h3 - h2) + rHggh2 = P0 + rBg(h3 - h1) + rHggh1

Þ rB = rAŁ ç æ h3 - h1 + rHgŁ

2.26Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3, 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:

F = mg = r V g = 1000 · 31.416 · 9.80665 = 308086 N Tanks have same net force up (holds same m in gravitation field)

Pbot = P0 + r H g Pbot,A = 101 + (1000 · 10 · 9.80665 / 1000) = 199 kPa Pbot,B = 101 + (1000 · 2.5 · 9.80665 / 1000) = 125.5 kPa

rHg = 13595 - 2.5 T kg/ m3 | T in Celsius |

2.27The density of mercury changes approximately linearly with temperature as 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:

DP = rgh Þ h = DP/rg ;rsu = 13507.5 ;rw = 13632.5 hsu = 100·103/(13507.5 · 9.807) = 0.7549 m hw = 100·103/(13632.5 · 9.807) = 0.7480 m Dh = hsu - hw = 0.0069 m = 6.9 m

2.28Liquid water with density r 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› = Ffl

PA = P0A + mH2Og P = P0 + mH2Og/A

P = P0 + (H-h)rgh, Vair

P P 0

2.29A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 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› = Ffl = P0A + mpg = PA |

P = P0 + mpg/A = 101.325 + 5 · 25 / (1000 · 0.0015) = 184.7 kPa

2.30A piece of experimental apparatus is located where g = 9.5 m/s2 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 m. What is the pressure drop in kPa?

DP = rgh ; | rHg = 13600 |

Solution: DP = 13600 · 9.5 · 0.2 = 25840 Pa = 25.84 kPa

1000 kg/m3, 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 - h2) - (H - h1) = DhHg = h1 - h2

= P2A + rH2Oh2gA + rHg(H - h2)gA |

P1A + rH2Oh1gA + rHg(H - h1)gA

P1 - P2 = rHgDhHgg - rH2ODhHgg = 13600 · 0.2 · 9.5 - 1000 · 0.2 · 9.5 | |

= 25840 - 1900 = 23940 Pa = 23.94 kPa |

2.32Two piston/cylinder arrangements, A and B, have their gas chambers connected by a pipe. Cross-sectional areas are A = 75 cm2 and AB = 25 cm2 with the piston mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation. Find the mass mB so that none of the pistons have to rest on the bottom.

Solution:

Force balance for both pistons: | F› = Ffl |

A: mPAg + P0AA = PAA | |

B: mPBg + P0AB = PAB |

P P 0 0 Same P in A and B gives no flow between them.

mPAg

A + P0 = mPBg

AB + P0

=> mPB = mPA A/ AB = 25 · 25/75 = 8.3 kg

2.33Two hydraulic piston/cylinders are of same size and setup as in Problem 2.32, but with neglible piston masses. A single point force of 250 N presses down on piston A. Find the needed extra force on piston B so that none of the pistons have to move.

Solution: No motion in connecting pipe: PA = PB & Forces on pistons balance

A = 75 cm2 ; | AB = 25 cm2 |

PA = P0 + FA / A = PB = P0 + FB / AB

FB = FA AB / A = 250 · 25 / 75 = 83.3 N

2.34 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation. Assume the density of water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do you feel at each place?

Solution: DP = rgh

Pocean= P0 + DP = 1025 · 100 + 1000 · 9.81 · 15 | |

= 2.4965 · 105 Pa = 250 kPa | |

Phill = P0 - DP = 1025 · 100 - 1.18 · 9.81 · 250 | |

= 0.9606 · 105 Pa = 9.61 kPa |

2.35In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface. This is illustrated in Fig. P2.35. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level.

Solution:

Pbottom = Ptop + rgl = 125 + 1000 · 9.807 · 25 · 10-3 = 370 kPa

2.36Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B?

Force balance for the piston: | PBAB + mpg + P0(A - AB) = PAA |

A = (p/4)0.12 = 0.00785 m2; | AB = (p/4)0.0252 = 0.000491 m2 |

Solution:

- 100 (0.00785 - 0.000491) = 2.944 kN |

PBAB = PAA - mpg - P0(A - AB) = 500· 0.00785 - (25 · 9.807/1000) PB = 2.944/0.000491 = 5996 kPa = 6.0 MPa

2.37Two cylinders are filled with liquid water, r = 1000 kg/m3, and connected by a line with a closed valve. A has 100 kg and B has 500 kg of water, their cross-sectional areas are A = 0.1 m2 and AB = 0.25 m2 and the height h is 1 m. Find the pressure on each side of the valve. The valve is opened and water flows to an equilibrium.

Find the final pressure at the valve location.

VA = vH2OmA = mA/r = 0.1 = AAhA | => hA = 1 m |

VB = vH2OmB = mB/r = 0.5 = ABhB | => hB = 2 m |

Solution:

PVB = P0 + rg(hB+H) = 101325 + 1000 · 9.81 · 3 = 130 755 Pa

PVA = P0 + rghA = 101325 + 1000 · 9.81 · 1 = 1 135 Pa Equilibrium: same height over valve in both

Vtot = VA + VB = h2AA + (h2 - H)AB Þ h2 = hAAA + (hB+H)AB

A + AB = 2.43 m

PV2 = P0 + rgh2 = 101.325 + (1000 · 9.81 · 2.43)/1000 = 125.2 kPa

2.38Using the freezing and boiling point temperatures for water in both Celsius and

Fahrenheit scales, develop a conversion formula between the scales. Find the conversion formula between Kelvin and Rankine temperature scales.

TFreezing = 0 oC = 32 F; | TBoiling = 100 oC = 212 F |

DT = 100 oC = 180 F Þ ToC = (TF - 32)/1.8 or | TF = 1.8 ToC + 32 |

Solution: For the absolute K & R scales both are zero at absolute zero.

TR = 1.8 · TK

English Unit Problems

2.39EA 2500-lbm car moving at 15 mi/h is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/h. What are the force and total time required?

Solution:

a = dV

= DV / D t | Þ D t = DV / a |

D t = (50 -15) · 1609.34 · 3.28084/(3600 · 15) = 3.42 sec F = ma = 2500 · 15 / 32.174 lbf= 1165 lbf

2.40ETwo pound moles of diatomic oxygen gas are enclosed in a 20-lbm steel container. A force of 2000 lbf now accelerates this system. What is the acceleration?

Solution:

mtot = mO2 + msteel = 64 + 20 = 84 lbm a = Fgc mtot = (2000 · 32.174) / 84 = 766 ft/s2

2.41E A bucket of concrete of total mass 400 lbm is raised by a crane with an acceleration of 6 ft/s2 relative to the ground at a location where the local gravitational acceleration is 31 ft/s2. Find the required force.

Solution:

F = ma = Fup - mg Fup = ma + mg = 400 · ( 6 + 31 ) / 32.174 = 460 lbf

2.42EOne pound-mass of diatomic oxygen (O2 molecular weight 32) is contained in a 100-gal tank. Find the specific volume on both a mass and mole basis (v and v).

Solution:

v = V/m = 15/1 = 15 ft3/lbm v = V/n = V m/M = Mv = 32 ·15 = 480 ft3/lbmol

2.43E A 30-lbm steel gas tank holds 10 ft3 of liquid gasoline, having a density of 50 lbm/ft3. What force is needed to accelerate this combined system at a rate of 15 ft/s2?

Solution:

m = mtank + mgasoline = 30 + 10 · 50 = 530 lbm

F = ma gC = (530 · 15) / 32.174 = 247.1 lbf

2.44EA differential pressure gauge mounted on a vessel shows 185 lbf/in.2 and a local barometer gives atmospheric pressure as 0.96 atm. Find the absolute pressure inside the vessel.

(Parte **1** de 4)