Ingeniero

1-34

Review Problems
1-85 A hydraulic lift is used to lift a weight. The
diameter of the piston on which the weight to be
placed is to be determined.
Assumptions 1 The cylinders of the lift are
vertical. 2 There are no leaks. 3 Atmospheric
pressure act on both sides, and thus it can be
disregarded.
Analysis Noting that pressure is force per unit area,
the pressure on the smallerpiston is determined
from
F1
m1g
P
1
A1
D12 / 4
(25 kg)(9.81 m/s 2 )
2

(0.10 m) / 4

Weight
2500 kg

F1
25 kg

F2

10 cm

D2

1 kN
1000 kg m/s 2

31.23 kN/m 2 31.23 kPa
From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the
needed diameter is determined from
P1

P2

F2
A2

m2 g
2

D2 / 4

31.23 kN/m 2(2500 kg)(9.81 m/s 2 )

1 kN

2

1000 kg m/s 2

D2 / 4

D2

1.0 m

Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal’s
principle.

1-86 A vertical piston-cylinder device contains a gas. Some weights are to
be placed on the piston to increase the gas pressure. The local atmospheric
pressure and the mass of the weights thatwill double the pressure of the
gas are to be determined.
Assumptions Friction between the piston and the cylinder is negligible.
Analysis The gas pressure in the piston-cylinder device initially depends
on the local atmospheric pressure and the weight of the piston. Balancing
the vertical forces yield

Patm

P

m piston g
A

100 kPa

(5 kg)(9.81 m/s 2 )
2

(0.12 m )/4

1 kN1000 kg m/s 2

WEIGTHS

GAS

95.66 kN/m 2

The force balance when the weights are placed is used to determine the mass of the weights
(m piston m weights ) g
P Patm
A
(5 kg m weights )(9.81 m/s 2 )
1 kN
200 kPa 95.66 kPa
m weights
2
(0.12 m )/4
1000 kg m/s 2
A large mass is needed to double the pressure.

95.7 kPa

115.3 kg

1-35

1-87 An airplane is flying over acity. The local atmospheric pressure in that city is to be determined.
Assumptions The gravitational acceleration does not change with altitude.
Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3.
Analysis The local atmospheric pressure is determined from
Patm Pplane
gh
58 kPa (1.15 kg/m 3 )(9.81 m/s 2 )(3000 m)

1 kN
1000 kg m/s 2

91.84 kN/m 2

91.8kPa

The atmospheric pressure may be expressed in mmHg as
Patm
g

hHg

91.8 kPa
(13,600 kg/m 3 )(9.81 m/s 2 )

1000 Pa
1 kPa

1000 mm
1m

688 mmHg

1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a
body at different locations are to be determined.
Analysis The weight of an 80-kg man at various locations is obtained bysubstituting the altitude z (values
in m) into the relation

W

mg

(80kg)(9.807 3.32 10

Sea level:
Denver:
Mt. Ev.:

6

z m/s 2 )

1N
1kg m/s 2

(z = 0 m): W = 80 (9.807-3.32x10-6 0) = 80 9.807 = 784.6 N
(z = 1610 m): W = 80 (9.807-3.32x10-6 1610) = 80 9.802 = 784.2 N
(z = 8848 m): W = 80 (9.807-3.32x10-6 8848) = 80 9.778 = 782.2 N

1-89 A man is considering buying a 12-ozsteak for $3.15, or a 320-g steak for $2.80. The steak that is a
better buy is to be determined.
Assumptions The steaks are of identical quality.
Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let
us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is
determined to be

12 ounce steak:Unit Cost =

$3.15
12 oz

16 oz
1 lbm

1 lbm
0.45359 kg

320 gram steak:
Unit Cost =

$2.80
320 g

1000 g
1 kg

$8.75/kg

Therefore, the steak at the international market is a better buy.

$9.26/kg

1-36

1-90 The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to
be expressed in N and kgf.
Analysis Noting that 1...