The design of this rocket and gantry structure requires a basic knowledgeof both statics and dynamics, which forms the subject matter ofengineering mechanics.

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3

CHAPTER OBJECTIVES

General Principles

• To provide an introduction to the basic quantities and idealizations ofmechanics.

• To give a statement of Newton’s Laws of Motion and Gravitation.

• To review the principles for applying the SI system of units.

• To examine the standard procedures for performing numericalcalculations.

• To present a general guide for solving problems.

C H A P T E R

1

1.1 MechanicsMechanics can be defined as that branch of the physical sciencesconcerned with the state of rest or motion of bodies that are subjectedto the action of forces. In general, this subject is subdivided into threebranches: rigid-body mechanics, deformable-body mechanics, and fluidmechanics. This book treats only rigid-body mechanics since it forms asuitable basis for the design and analysis of many types of structural,mechanical, or electrical devices encountered in engineering. Also, rigid-body mechanics provides part of the necessary background for the studyof the mechanics of deformable bodies and the mechanics of fluids.

Rigid-body mechanics is divided into two areas: statics and dynamics.Statics deals with the equilibrium of bodies, that is, those that are eitherat rest or move with a constant velocity; whereas dynamics is concernedwith the accelerated motion of bodies.Although statics can be consideredas a special case of dynamics, in which the acceleration is zero, staticsdeserves separate treatment in engineering education since many objectsare designed with the intention that they remain in equilibrium.

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4 • CHAPTER 1 General Principles

Historical Development. The subject of statics developed very earlyin history because the principles involved could be formulated simplyfrom measurements of geometry and force. For example, the writings ofArchimedes (287–212 B.C.) deal with the principle of the lever. Studies ofthe pulley, inclined plane, and wrench are also recorded in ancientwritings—at times when the requirements of engineering were limitedprimarily to building construction.

Since the principles of dynamics depend on an accurate measurementof time, this subject developed much later. Galileo Galilei (1564–1642)was one of the first major contributors to this field. His work consistedof experiments using pendulums and falling bodies. The most significantcontributions in dynamics, however, were made by Issac Newton(1642–1727), who is noted for his formulation of the three fundamentallaws of motion and the law of universal gravitational attraction. Shortlyafter these laws were postulated, important techniques for theirapplication were developed by Euler, D’Alembert, Lagrange, and others.

1.2 Fundamental Concepts

Before we begin our study of engineering mechanics, it is important tounderstand the meaning of certain fundamental concepts and principles.

Basic Quantities. The following four quantities are used throughoutmechanics.

Length. Length is needed to locate the position of a point in space andthereby describe the size of a physical system. Once a standard unit oflength is defined, one can then quantitatively define distances andgeometric properties of a body as multiples of the unit length.

Time. Time is conceived as a succession of events.Although the principlesof statics are time independent, this quantity does play an important rolein the study of dynamics.

Mass. Mass is a property of matter by which we can compare the actionof one body with that of another. This property manifests itself as agravitational attraction between two bodies and provides a quantitativemeasure of the resistance of matter to a change in velocity.

Force. In general, force is considered as a “push” or “pull” exerted by onebody on another. This interaction can occur when there is direct contactbetween the bodies, such as a person pushing on a wall, or it can occurthrough a distance when the bodies are physically separated. Examples ofthe latter type include gravitational, electrical, and magnetic forces. In anycase, a force is completely characterized by its magnitude, direction, andpoint of application.

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SECTION 1.2 Fundamental Concepts • 5

Equilibrium

Accelerated motion

Action — reaction

force of A on B

force of B on A

v

a

F2F1

F3

F

F FA B

*Stated another way, the unbalanced force acting on the particle is proportional to thetime rate of change of the particle’s linear momentum.

Idealizations. Models or idealizations are used in mechanics in orderto simplify application of the theory. A few of the more importantidealizations will now be defined. Others that are noteworthy will bediscussed at points where they are needed.

Particle. A particle has a mass, but a size that can be neglected. Forexample, the size of the earth is insignificant compared to the size of itsorbit, and therefore the earth can be modeled as a particle when studyingits orbital motion. When a body is idealized as a particle, the principles ofmechanics reduce to a rather simplified form since the geometry of thebody will not be involved in the analysis of the problem.

Rigid Body. A rigid body can be considered as a combination of a largenumber of particles in which all the particles remain at a fixed distance fromone another both before and after applying a load.As a result, the materialproperties of any body that is assumed to be rigid will not have to beconsidered when analyzing the forces acting on the body. In most cases theactual deformations occurring in structures, machines, mechanisms, and thelike are relatively small,and the rigid-body assumption is suitable for analysis.

Concentrated Force. A concentrated force represents the effect of aloading which is assumed to act at a point on a body. We can represent aload by a concentrated force, provided the area over which the load isapplied is very small compared to the overall size of the body.An examplewould be the contact force between a wheel and the ground.

Newton’s Three Laws of Motion. The entire subject of rigid-bodymechanics is formulated on the basis of Newton’s three laws of motion, thevalidity of which is based on experimental observation.They apply to themotion of a particle as measured from a nonaccelerating reference frame.With reference to Fig. 1–1, they may be briefly stated as follows.

First Law. A particle originally at rest, or moving in a straight line withconstant velocity, will remain in this state provided the particle is notsubjected to an unbalanced force.

Second Law. A particle acted upon by an unbalanced force F experiencesan acceleration a that has the same direction as the force and a magnitudethat is directly proportional to the force.* If F is applied to a particle ofmass m, this law may be expressed mathematically as

(1–1)

Third Law. The mutual forces of action and reaction between twoparticles are equal, opposite, and collinear.

F = ma

Fig. 1–1

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6 • CHAPTER 1 General Principles

Newton’s Law of Gravitational Attraction. Shortly after formulatinghis three laws of motion, Newton postulated a law governing thegravitational attraction between any two particles. Stated mathematically,

(1-2)

where

Weight. According to Eq. 1–2, any two particles or bodies have amutual attractive (gravitational) force acting between them. In the case ofa particle located at or near the surface of the earth, however, the onlygravitational force having any sizable magnitude is that between the earthand the particle. Consequently, this force, termed the weight, will be theonly gravitational force considered in our study of mechanics.

From Eq. 1–2, we can develop an approximate expression for findingthe weight W of a particle having a mass If we assume the earthto be a nonrotating sphere of constant density and having a mass

then if r is the distance between the earth’s center and theparticle, we have

Letting yields

(1–3)

By comparison with we term g the acceleration due to gravity.Since it depends on r, it can be seen that the weight of a body is not anabsolute quantity. Instead, its magnitude is determined from where themeasurement was made. For most engineering calculations, however, gis determined at sea level and at a latitude of 45°, which is consideredthe “standard location.”

F = ma,

W = mg

g = GMe>r2

W = G

mMe

r2

m2 = Me,

m1 = m.

r = distance between the two particles

m1, m2 = mass of each of the two particles

G = 66.73110-122 m3>1kg # s22experimental evidence,

G = universal constant of gravitation; according to

F = force of gravitation between the two particles

F = G

m1m2

r2

1.3 Units of MeasurementThe four basic quantities—force, mass, length and time—are not allindependent from one another; in fact, they are related by Newton’ssecond law of motion, Because of this, the units used to measurethese quantities cannot all be selected arbitrarily. The equality is maintained only if three of the four units, called base units, arearbitrarily defined and the fourth unit is then derived from the equation.

F = maF = ma.

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SECTION 1.3 Units of Measurement • 7

TA B L E 1 – 1 • Systems of Units

Name Length Time Mass Force

International meter second kilogram newton*System of Units (m) (s) (kg) (N)

(SI)

U.S. Customary foot second slug* pound(FPS) (ft) (s) (lb)

unit.*Derived

a lb # s2

ftb

akg # m

s2 b

SI Units. The International System of units, abbreviated SI after theFrench “Système International d’Unités,” is a modern version of the metricsystem which has received worldwide recognition.As shown in Table 1–1,the SI system specifies length in meters (m), time in seconds (s), and massin kilograms (kg). The unit of force, called a newton (N), is derived from

Thus, 1 newton is equal to a force required to give 1 kilogram ofmass an acceleration of

If the weight of a body located at the “standard location” is to bedetermined in newtons, then Eq. 1–3 must be applied. Here

however, for calculations, the value willbe used. Thus,

(1–4)

Therefore, a body of mass 1 kg has a weight of 9.81 N, a 2-kg body weighs19.62 N, and so on, Fig. 1–2a.

U.S. Customary. In the U.S. Customary system of units (FPS) length ismeasured in feet (ft), force in pounds (lb), and time in seconds (s),Table 1–1.The unit of mass, called a slug, is derived from Hence,1 slug is equal to the amount of matter accelerated at when actedupon by a force of

In order to determine the mass of a body having a weight measuredin pounds, we must apply Eq. 1–3. If the measurements are made at the“standard location,” then will be used for calculations.Therefore,

(1–5)

And so a body weighing 32.2 lb has a mass of 1 slug, a 64.4-lb body hasa mass of 2 slugs, and so on, Fig. 1–2b.

m =

Wg 1g = 32.2 ft>s22

g = 32.2 ft>s2

1 lb 1slug = lb # s2>ft2. 1 ft>s2F = ma.

W = mg 1g = 9.81 m>s22

g = 9.81 m>s2g = 9.806 65 m>s2;

1 m>s2 1N = kg # m>s22.F = ma.

9.81 N

32.2 lb

1 kg

1 slug

Fig. 1–2

(a)

(b)

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8 • CHAPTER 1 General Principles8 • CHAPTER 1 General Principles

TA B L E 1 – 2 • Conversion Factors

Unit of Unit of Quantity Measurement (FPS) Equals Measurement (SI)

Force lb 4.448 2 NMass slug 14.593 8 kgLength ft 0.304 8 m

Conversion of Units. Table 1–2 provides a set of direct conversionfactors between FPS and SI units for the basic quantities.Also, in the FPSsystem, recall that (inches), (mile),

(kilo-pound), and 2000 lb = 1 ton.1000 lb = 1 kip5280 ft = 1 mi1 ft = 12 in.

1.4 The International System of Units

*The kilogram is the only base unit that is defined with a prefix.

The SI system of units is used extensively in this book since it is intendedto become the worldwide standard for measurement. Consequently, therules for its use and some of its terminology relevant to mechanics willnow be presented.

Prefixes. When a numerical quantity is either very large or very small,the units used to define its size may be modified by using a prefix. Someof the prefixes used in the SI system are shown in Table 1–3. Eachrepresents a multiple or submultiple of a unit which, if appliedsuccessively, moves the decimal point of a numerical quantity to everythird place. * For example,

or Notice that the SIsystem does not include the multiple deca (10) or the submultiple centi(0.01), which form part of the metric system. Except for some volume andarea measurements, the use of these prefixes is to be avoided in scienceand engineering.

5 mm (milli-meter).0.005m =(mega-newton),4 000 000 N = 4 000 kN(kilo-newton) = 4MN

TA B L E 1 – 3 • Prefixes

Exponential Form Prefix SI Symbol

Multiple1 000 000 000 giga G1 000 000 mega M1 000 kilo k

Submultiple0.001 milli m0.000 001 micro0.000 000 001 nano n10-9

m10-610-3

103106109

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SECTION 1.4 The International System of Units • 9

Rules for Use. The following rules are given for the proper use of thevarious SI symbols:

1. A symbol is never written with a plural “s,” since it may be confusedwith the unit for second (s).

2. Symbols are always written in lowercase letters, with the followingexceptions: symbols for the two largest prefixes shown in Table 1–3,giga and mega, are capitalized as G and M, respectively; and symbolsnamed after an individual are also capitalized, e.g., N.

3. Quantities defined by several units which are multiples of oneanother are separated by a dot to avoid confusion with prefixnotation, as indicated by Also,

whereas ms (milli-second).

4. The exponential power represented for a unit having a prefix refersto both the unit and its prefix.For example,Likewise, represents

5. Physical constants or numbers having several digits on either sideof the decimal point should be reported with a space between everythree digits rather than with a comma; e.g., 73 569.213 427. In thecase of four digits on either side of the decimal, the spacing isoptional; e.g., 8537 or 8 537. Furthermore, always try to use decimalsand avoid fractions; that is, write 15.25 not

6. When performing calculations, represent the numbers in terms of theirbase or derived units by converting all prefixes to powers of 10. Thefinal result should then be expressed using a single prefix. Also, aftercalculation, it is best to keep numerical values between 0.1 and 1000;otherwise, a suitable prefix should be chosen. For example,

7. Compound prefixes should not be used; e.g., (kilo-micro-second) should be expressed as ms (milli-second) since

8. With the exception of the base unit the kilogram, in general avoidthe use of a prefix in the denominator of composite units. Forexample, do not write N/mm, but rather kN/m; also, m/mg shouldbe written as Mm/kg.

9. Although not expressed in multiples of 10, the minute, hour, etc.,are retained for practical purposes as multiples of the second.Furthermore, plane angular measurement is made using radians(rad). In this book, however, degrees will often be used, where180° = p rad.

1 kms = 111032110-62 s = 1110-32 s = 1 ms.

kms

= 3000110-62 N # m = 3110-32 N # m = 3 mN # m 150 kN2160 nm2 = [5011032 N][60110-92 m]

15 14.

1mm22 = mm # mm.mm2mN2

= 1mN22 = mN # mN.

m # s 1meter-second2, N = kg # m>s2= kg # m # s-2.

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10 • CHAPTER 1 General Principles

Computers are often used in engineering foradvanced design and analysis.

10 • CHAPTER 1 General Principles10 • CHAPTER 1 General Principles

1.5 Numerical Calculations

Numerical work in engineering practice is most often performed by usinghandheld calculators and computers. It is important, however, that theanswers to any problem be reported with both justifiable accuracy andappropriate significant figures. In this section we will discuss these topicstogether with some other important aspects involved in all engineeringcalculations.

Dimensional Homogeneity. The terms of any equation used todescribe a physical process must be dimensionally homogeneous; that is,each term must be expressed in the same units. Provided this is the case,all the terms of an equation can then be combined if numerical values aresubstituted for the variables. Consider, for example, the equation

where, in SI units, s is the position in meters, m, t is time inseconds, s, is velocity in m/s, and a is acceleration in Regardless ofhow this equation is evaluated, it maintains its dimensional homogeneity.In the form stated, each of the three terms is expressed in meters

] or solving for a, the terms areeach expressed in units of [ (m/s)/s].

Since problems in mechanics involve the solution of dimensionallyhomogeneous equations, the fact that all terms of an equation arerepresented by a consistent set of units can be used as a partial checkfor algebraic manipulations of an equation.

Significant Figures. The accuracy of a number is specified by thenumber of significant figures it contains. A significant figure is any digit,including a zero, provided it is not used to specify the location of thedecimal point for the number. For example, the numbers 5604 and 34.52each have four significant figures.When numbers begin or end with zeros,

m>s2,m>s2,m>s2a = 2s>t2

- 2v>t,1m>s22s2,[m, 1m>s2s,

m>s2.vs = vt +

12 at2,

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SECTION 1.5 Numerical Calculations • 11

*Of course, some numbers, such as or numbers used in derived formulas are exactand are therefore accurate to an infinite number of significant figures.

e,p,

however, it is difficult to tell how many significant figures are in thenumber. Consider the number 400. Does it have one (4), or perhaps two(40), or three (400) significant figures? In order to clarify this situation, thenumber should be reported using powers of 10. Using engineering notation,the exponent is displayed in multiples of three in order to facilitateconversion of SI units to those having an appropriate prefix. Thus, 400expressed to one significant figure would be Likewise, 2500 and0.00546 expressed to three significant figures would be and

Rounding Off Numbers. For numerical calculations, the accuracyobtained from the solution of a problem generally can never be betterthan the accuracy of the problem data. This is what is to be expected, butoften handheld calculators or computers involve more figures in theanswer than the number of significant figures used for the data. For thisreason, a calculated result should always be “rounded off” to anappropriate number of significant figures.

To convey appropriate accuracy, the following rules for rounding offa number to n significant figures apply:

• If the digit is less than 5, the digit and others followingit are dropped. For example, 2.326 and 0.451 rounded off to significant figures would be 2.3 and 0.45.

• If the digit is equal to 5 with zeros following it, then roundoff the nth digit to an even number. For example, and0.8655 rounded off to significant figures become and 0.866.

• If the digit is greater than 5 or equal to 5 with any nonzerodigits following it, then increase the nth digit by 1 and drop the digit and others following it. For example, 0.723 87 and 565.500 3rounded off to significant figures become 0.724 and 566.

Calculations. As a general rule, to ensure accuracy of a final resultwhen performing calculations on a pocket calculator, always retain agreater number of digits than the problem data. If possible, try to work outthe computations so that numbers which are approximately equal are notsubtracted since accuracy is often lost from this calculation.

In engineering we generally round off final answers to three significantfigures since the data for geometry, loads, and other measurements areoften reported with this accuracy.* Consequently, in this book theintermediate calculations for the examples are often worked out to foursignificant figures and the answers are generally reported to threesignificant figures.

n = 3

n + 1n + 1

1.2411032n = 31.24511032n + 1

n = 2n + 1n + 1

5.46110-32. 2.50110320.411032.

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12 • CHAPTER 1 General Principles

E X A M P L E 1.1

Convert 2 km/h to m/s. How many ft/s is this?

SolutionSince and the factors of conversion arearranged in the following order, so that a cancellation of the units canbe applied:

Ans.

From Table 1–2, Thus

Ans. = 1.82 ft>s 0.556 m>s =

0.556 ms

1 ft

0.3048 m

1 ft = 0.3048 m.

=

2000 m3600 s

= 0.556 m>s 2 km>h =

2 kmh

a1000 mkm

b a 1 h3600 s

b

1 h = 3600 s,1 km = 1000 m

E X A M P L E 1.2

Convert the quantities and to appropriate SI units.

SolutionUsing Table 1–2,

Ans.

Also, and

Ans. = 26.8 Mg>m3

= 26.811032kg>m3

52 slug>ft3=

52 slug

ft3 a14.593 8 kg

1 slugb a 1 ft

0.304 8 mb3

1 ft = 0.304 8 m.1 slug = 14.593 8 kg

= 1334.5 N # s = 1.33 kN # s

300 lb # s = 300 lb # sa4.448 2 Nlb

b1 lb = 4.448 2 N.

52 slug>ft3300 lb # s

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SECTION 1.5 Numerical Calculations • 13

E X A M P L E 1.3

Evaluate each of the following and express with SI units having anappropriate prefix: (a) (50 mN)(6 GN), (b) (c)

SolutionFirst convert each number to base units, perform the indicated opera-tions, then choose an appropriate prefix (see Rule 6 on p. 9).

Part (a)

Ans.

Note carefully the convention (Rule 4 on p. 9).

Part (b)

Ans.

We can also write

Part (c)

Ans.

Here we have used Rules 4 and 8 on p. 9.

= 50 kN3>kg

= 0.0511032 kN3>kg

= 0.05110122 N3a 1 kN

103 Nb3 1

kg

= 0.05110122 N3>kg

45 MN3>900 Gg =

451106 N2390011062 kg

= 0.144 m # MN2

14411092m # N2= 14411092m # N2a1 MN

106 Nb a1 MN

106 Nb

= 144 Gm # N2

= 14411092 m # N2

= [400110-32 m][0.36110122 N2]

1400 mm210.6 MN22 = [400110-32 m][0.611062 N]2

kN2= 1kN22 = 106 N2

= 300 kN2

= 30011062 N2a 1 kN

103 Nb a 1 kN

103 Nb

= 30011062 N2

150 mN216 GN2 = [50110-32 N][611092 N]

45 MN3>900 Gg.1400 mm210.6 MN22,

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14 • CHAPTER 1 General Principles

1.6 General Procedure for Analysis

IMPORTANT POINTS• Statics is the study of bodies that are at rest or move with constant

velocity.

• A particle has a mass but a size that can be neglected.

• A rigid body does not deform under load.

• Concentrated forces are assumed to act at a point on a body.

• Newton’s three laws of motion should be memorized.

• Mass is a property of matter that does not change from onelocation to another.

• Weight refers to the gravitational attraction of the earth on a bodyor quantity of mass. Its magnitude depends upon the elevation atwhich the mass is located.

• In the SI system the unit of force, the newton, is a derived unit.The meter, second, and kilogram are base units.

• Prefixes G, M, k, m, n are used to represent large and smallnumerical quantities. Their exponential size should be known,along with the rules for using the SI units.

• Perform numerical calculations to several significant figures andthen report the final answer to three significant figures.

• Algebraic manipulations of an equation can be checked in part byverifying that the equation remains dimensionally homogeneous.

• Know the rules for rounding off numbers.

m,

When solving problems, do the work asneatly as possible. Being neat generallystimulates clear and orderly thinking,and vice versa.

The most effective way of learning the principles of engineeringmechanics is to solve problems. To be successful at this, it is importantto always present the work in a logical and orderly manner, assuggested by the following sequence of steps:

1. Read the problem carefully and try to correlate the actual physicalsituation with the theory studied.

2. Draw any necessary diagrams and tabulate the problem data.3. Apply the relevant principles, generally in mathematical form.4. Solve the necessary equations algebraically as far as practical, then,

making sure they are dimensionally homogeneous, use a consistentset of units and complete the solution numerically.Report the answerwith no more significant figures than the accuracy of the given data.

5. Study the answer with technical judgment and common sense todetermine whether or not it seems reasonable.

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PROBLEMS • 15

P R O B L E M S1-1. Round off the following numbers to threesignificant figures: (a) 4.65735 m, (b) 55.578 s, (c) 4555 N,(d) 2768 kg.

1-2. Wood has a density of What is itsdensity expressed in SI units?

1-3. Represent each of the following quantities in thecorrect SI form using an appropriate prefix: (a) 0.000431kg, (b) (c) 0.00532 km.

*1-4. Represent each of the following combinations ofunits in the correct SI form using an appropriate prefix:(a) m/ms, (b) (c) ks/mg, and (d)

1-5. If a car is traveling at determine its speed inkilometers per hour and meters per second.

1-6. Evaluate each of the following and express with anappropriate prefix: (a) (b) and(c)

1-7. A rocket has a mass of slugs on earth.Specify (a) its mass in SI units, and (b) its weight in SI units.If the rocket is on the moon, where the acceleration dueto gravity is determine to three significantfigures (c) its weight in SI units, and (d) its mass in SI units.

*1-8. Represent each of the following combinations ofunits in the correct SI form: (a) (b) Mg/mN, and(c)

1-9. The pascal (Pa) is actually a very small unit ofpressure. To show this, convert to Atmospheric pressure at sea level is Howmany pascals is this?

1-10. What is the weight in newtons of an object that hasa mass of: (a) (b) (c) 4.50 Mg? Express theresult to three significant figures. Use an appropriate prefix.

1-11. Evaluate each of the following to three significantfigures and express each answer in SI units using anappropriate prefix: (a) 354 mg(45 km)/(0.035 6 kN),(b) (.004 53 Mg)(201 ms), (c) 435 MN/23.2 mm.

0.5 g,10 kg,

14.7 lb>in2.lb>ft2.1 Pa = 1 N>m2

MN>1kg # ms2.kN>ms,

gm = 5.30 ft>s2,

250110321230 m23.

10.002 mg22,1430 kg22,

55 mi/h,

km # mN.mkm,

35.311032 N,

4.70 slug>ft3.

*1-12. Convert each of the following and express theanswer using an appropriate prefix: (a) to

(b) 6 ft/h to mm/s, and (c) to

1-13. Convert each of the following to three significantfigures. (a) to (b) to and(c) 15 ft/h to mm/s.

1-14. If an object has a mass of 40 slugs, determine itsmass in kilograms.

1-15. Water has a density of What is thedensity expressed in SI units? Express the answer to threesignificant figures.

*1-16. Two particles have a mass of 8 kg and 12 kg,respectively. If they are 800 mm apart, determine the forceof gravity acting between them. Compare this result withthe weight of each particle.

1-17. Determine the mass of an object that has a weightof (a) 20 mN, (b) 150 kN, (c) 60 MN. Express the answerto three significant figures.

1-18. If a man weighs 155 lb on earth, specify (a) hismass in slugs, (b) his mass in kilograms, and (c) hisweight in newtons. If the man is on the moon, wherethe acceleration due to gravity is determine (d) his weight in pounds, and (e) his mass inkilograms.

1-19. Using the base units of the SI system, show thatEq. 1–2 is a dimensionally homogeneous equation whichgives F in newtons. Determine to three significant figuresthe gravitational force acting between two spheres thatare touching each other.The mass of each sphere is 200 kgand the radius is 300 mm.

*1-20. Evaluate each of the following to threesignificant figures and express each answer in SI unitsusing an appropriate prefix: (a) (b) (35 mm)2 (48 kg)3.

(0.631 Mm) / (8.60 kg)2,

gm = 5.30 ft>s2,

1.94 slug>ft3.

kN>m3,450 lb>ft3N # m,20 lb # ft

kN # m.835 lb # ftkN>m3,175 lb>ft3

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