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Vectors and Coordinate Systems

Wind has both a speed and a

direction hence the motion of the wind is described

by a vector

looking Ahead The goals of Chapter 3 are to learn how vectors are represented and used In this chapter you will learn to

bull Understand and use the basic properties of vectors

bull Decompose a vector into its components and reassemble vector components into a magnitude and direction

bull Add and subtract vectors both graphically and using components

Looking Back This chapter continues the development of vectors that was begun in Chapter 1 Please review

bull Section 13 Vector addition and subtraction

Many of the quantities that we use to describe the physical world are simply numbers For example the mass of an object is 2 kg its temperature is 21 degC and it occupies a volume of 250 cm3

A quantity that is fully described by a single number (with units) is called a scalar quantity Mass temperature and volume are all scalars Other scalar quantities include pressure density energy charge and voltage We will often use an algebraic symbol to represent a scalar quantity Thus m will represent mass T temperature V volume E energy and so on Notice that scalars in printed text are shown in italics

Our universe has three dimensions so some quantities also need a direction for a full descriptionlfyou ask someone for directions to the post office the reply Go three blocks will not be very helpful A full description might be Go three blocks south A quantity having both a size and a direction is called a vector quantity

You met examples of vector quantities in Chapter I position displacement velocity and acceleration You will soon make the acquaintance of others such as force momentum and the electric field Now before we begin a study of forces its worth spending a little time to look more closely at vectors

31 Scalars and Vectors Suppose you are assigned the task of measuring the temperature at various points throughout a building and then showing the infonnation on a building floor plan To do this you could put little dots on the floor plan to show the points at which you made measurements then write the temperature at that point beside the dot

78

Exercises and Problems 77

74 64 m = 0 m + (32 mfs)(4 s - 0 s) + ~a(4 s - 0 S)2 ~

75 (IOms)2 = VOl - 2(98 ms2)(IOm - Om)II 76 (0 mlS)2 = (5 ms)2 - 2(98 ms2)(sin IOO)(XI - 0 m) ~

77 VI = 0 mfs + (20 ms2)(S s - 0 s) Ii XI = 0 m + (0 mfs)(S s - 0 s) + ~(20 ms2)(S s - 0 S)2

X2 = XI + v l(10 s - S s)

Challenge Problems

78 Jill has just gotten out of her car in the grocery store parking Ii lot The parking lot is on a hill and is tilted 3deg Fifty meters

downhill from Jill a little old lady lets go of a fully loaded shopping cart The cart with frictionless wheels starts to roll straight downhill Jill immediately starts to sprint after the cart with her top acceleration of 20 mfs2 How far has the cart

rolled before Jill catches it 79 As a science project you drop a watennelon off the top of theiI Empire State Building 320 m above the sidewalk It so hapshy

pens that Superman flies by at the instant you release the watennelon Superman is headed straight down with a speed of 3S ms How fast is the watennelon going when it passes Supennan

80 Your school science club has devised a special event for ~ homecomjng Youve attached a rocket to the rear of a small

car that has been decorated in the blue-and-gold school colors The rocket provides a constant acceleration for 90 s As the rocket shuts off a parachute opens and slows the car at a rate of SO mls2

The car passes the judges box in the center of the

grandstand 990 m from the starting line exactly 12 s after you fUe the rocket What is the cars speed as it passes the judges

81 Careful measurements have been made of Olympic sprinters

II in the I QO-meter dash A simple but reasonably accurate model

is that a sprinter accelerates at 36 mfs2 for 3t s then runs at

constant velocity to the finish line a What is the race time for a sprinter who follows this

model b A sprinter could run a faster race by accelerating faster at

the beginning thus reaching top speed sooner If a sprinters top speed is the same as in part a what accelerashytion would he need to run the 1 QO-meter dash in 99 s

c By what percent did the sprinter need to increase his accelshyeration in order to decrease his time by I

82 A sprinter can accelerate with constant acceleration for 40 s II before reaching top speed He can run the lOO-meter dash in

IO s What is his speed as he crosses the finish line 83 The Starship Enterprise returns from warp drive to ordinary II space with a forward speed of SO kmls To the crews great

surprise a Kljngon ship is 100 km directly ahead traveling in the same direction at a mere 20 kmfs Without evasive action the Enterprise will overtake and collide with the Klingons in just Slightly over 30 s The Enterprises computers react instantly to brake the ship What acceleration does the Entershyprise need to just barely avoid a collision with the KJingon ship Assume the acceleration is constant Hint Draw a position-versus-tirne graph showing the motions of both the Enterprise and the Klingon ship Let Xo = 0 kIn be the location of the Enterprise as it returns from warp drive How do you show graphically the situation in which the collishysion is barely avoided Once you decide what it looks like graphically express that situation mathematically

STOP TO THINK ANSWERS

Stop to Think 21 d The particle starts with positive X and moves to negative x

Stop to Think 22 c The velocity is the slope of the position graph The slope is positive and constant until the position graph crosses the axis then positive but decreasing and finally zero when the position graph is horizontal

Stop to Think 23 b A constant positive V corresponds to a linshyearly increasing x starting from Xi = -10 m The constant negashytive V then corresponds to a linearly decreasing x

Stop to Think 24 a or b The velocity is constant while a = 0 it decreases linearly while a is negative Graphs a b and c all have

the same acceleration but only graphs a and b have a positive inishytial velocity that represents a particle moving to the right

Stop to Think 25 d The acceleration vector Gil points downhill (negative s-direction) and has the constant value - g sin ethroughshyout the motion

Stop to Think 26 c Acceleration is the slope of the graph The slope is zero at B Although the graph is steepest at A the slope at that point is negative and so a A lt aBo Only C has a positive slope soae gt aBo

(a) (b)

220

200

190

200 bull

bull 18 0 14

Temperature in DC Velocities in rns

FIGURE 31 Measurements of scalar and vector quantities

In other words as Figure 3la shows you can represent the temperature at each point with a simple number (with units) Temperature is a scalar quantity

Having done such a good job on your fust assignment you are next assigned the task of measuring the velocities of several employees as they move about in their work Recall from Chapter I that velocity is a vector it has both a size and a direction Simply writing each employees speed is not sufficient because speed doesnt take into account the direction in which the person moved After some thought you conclude that a good way to represent the velocity is by drawing an arrow whose length is proportional to the speed and that points in the direction of motion Further as Figure 31 b shows you decide to place the taiL of an arrow at the point where you measured the velocity

As this example illustrates the geometric representation of a vector is an arrow with the tail of the arrow (not its tip) placed at the point where the meashysurement is made The vector then seems to radiate outward from the point to which it is attached An arrow makes a natural representation of a vector because it inherently has both a length and a direction

The mathematical term for the length or size of a vector is magnitude so we can say that a vector is a quantity having a magnitude and a diIection As an example Figure 32 shows the geometric representation of a particles velocity vector Ii The particles speed at this point is 5 mIs and it is moving in the direcshytion indicated by the arrow The arrow is drawn with its tail at the point where the velocity was measured

NOTE ~ Although the vector arrow is drawn across the page from its tail to its tip this does not indicate that the vector stretches across this distance Instead the vector arrow tells us the value of the vector quantity only at the one point where the tail of the vector is placed

Arrows are good for pictures but we also need an aLgebraic representation of vectors to use in labels and in equations We do this by drawing a small arrow over the letter that represents the vector r for position Ii for velocity afor accelshyeration and so on

The magnitude of a vector is indicated by the letter without the arrow For example the magnitude of the velocity vector in Figure 32 is v = 5 ms This is the objects speed The magnitude of the acceleration vector ais written a The magnitude of a vector is a scalar quantity

NOTE ~ The magnitude of a vector cannot be a negative number it must be positive or zero with appropriate units

It is important to get in the habit of using the arrow symbol for vectors If you omit the vector arrow from the velocity vector Ii and write only v then youre referring only to the objects speed not its velocity The symbols rand r or Ii and v do not represent the same thing so if you omit the vector arrow from vector symbols you will soon have confusion and mistakes

31 Scalars and Vectors 79

Magnitude Direction of vector of vector

b~Of~ru v r The cC(Ir i ~ dlt11 Lh t)~~

Ih~ rJg~ hut il repr~st n l rhe pan ide velolt il I t

11m one ~xim

FIGU RE 32 The velocity vector vhas both a magnitude and a direction

- -

80 C HAP T E R 3 Vectors and Coordinate Systems

The boats displacement is the straightshyline connection from its initial to its final position

Net displacement S N

~ End ~

8 3 mi

e II StaI1 ~~

p 4mi ~

Individual displacements

FIGUR E 3 4 The net displacement C resulting from two displacements A and ii

32 Properties of Vectors Recall from Chapter 1 that the displacement is a vector drawn from an objects initial position to its position at some later time Because displacement is an easy concept to think about we can use it to introduce some of the properties of vecshytors However these properties apply to all vectors not just to displacement

Suppose that Sam starts from his front door walks across the street and ends up 202 ft to the northeast of where he started Sams displacement which we will label 5 is shown in Figure 33a The displacement vector is a straight-line conshynection from his initial to his final position not necessarily his actual path The dotted line indicates a possible route Sam might have taken but his displacement is the vector S

To describe a vector we must specify both its magnitude and its direction We can write Sams displacement as

S = (200 ft northeast)

where the first number specifies the magnitude and the second number is the direction The magnitude of Sams displacement is 5 = 200 ft the distance between hi s initial and final points

Sams next-door neighbor Bill also walks 200 ft to the northeast starting from his own front door Bills displacement E= (200 ft northeast) has the same magshynitude and direction as Sams displacement S Because vectors are defined by their magnitude and direction two vectors are equal if they have the same magnitude and direction This is true regardless of the starting points of the vectors hus ~e two displacements in Figure 33b are equal to each other and we can write B = 5

(a) (b)

I

I

BillsSams actual palh~ Ii dIUJ llIlt Ihe

Sams 5 gt (ll11e TnJln lluJt ltshy lI1d dire lion di splacementy

8=

nirl tctm~11 1 i Ihe -traigh l-li ne CC1 Ilne-c nll Jro1llllc Ill il ia l ill Ille tinalpOllliuli

FIGURE 33 Displacement vectors

NOTE A vector is unchanged if you move it to a different point on the page as long you dont change its length or the direction it points We used this idea in Chapter 1 when we moved velocity vectors around in order to find the avershyage acceleration vector a ~

Vector Addition

Figure 34 shows the displacement of a hiker who starts at point P and ends at point S She ftrst hikes 4 miles to the east then 3 miles to the north The first leg of the hike is described by the displacement Ii = (4 mi east) The second leg of the hike has displacement E = (3 mi north) Now by definition a vector from the initial position P to the final position S is also a displacement This is vector C on the figure Cis the net displacement because it describes the net result of the hikers ftrst having displacement Ii then displacement E

If you earn $50 on Saturday and $60 on Sunday your net income for the weekshyend is the sum of $50 and $60 With scalars the word net implies addition The

32 Properties of Vectors 81

same is true with vectors ~he net displacement Cis an initial displacement A plus a second displacement B or

- raquo - - )

C=A+B (31 )

The sum of two vectors is called the resultant vector It s not hard to show that vector addition is conunutative A + 8 = 8 + A That is you can add vectors in any order you wish

Look back at Tactics Box LIon page 10 to see the three-step procedure for adding two vectors This tip-to-tail method for adding vectors which is used to find C= A + 8 in Figure 34 is called graphical addition Any two vectors of the same type-two velocity vectors or two force vectors--can be added in exactly the same way

The graphical method for adding vectors is straightforward but we need to ~ a little geometry to come up with a complete description of the resultant vector C Vector Cof Figure 34 is defined by its magnitude C and by its direction Because the three vectors A ii and Cform a right triangle the magnitude or length of C is given by the Pythagorean theorem

C = VA 2 + 8 2 = V(4 mi)2 + (3 mi)2 = S mi (32)

Notice that Equation 32 uses the magnitudes A and B of the vectors Aand 8 The angle e which is used in Figure 34 to describe the direction of C is easily found for a right triangle

(33)

A together the hikers net displacement is ~ ~ ~

C = A + B = (S mi 37deg north of east) (34)

NOTE [gt Vector mathematics makes extensive use of geometry and trigonometry Appendix A at the end of this book contains a brief review of these topics ~

EXAMPU 3 1 Using graphical addition to find SOLVE The two displacements are A = (100 m east) and a displacement B = (200 m northwest) The net displacement C= II + Bis A bird flies 100 m due east from a tree then 200 m northwest found by drawing a vector from the initial to the final position (that is 45deg north of west) What is the birds net di splacement But describing C is a bit trickier than the example of the hiker

VISUA li ZE Figure 35 shows the two individual displacements because Aand ii are not at right angles First we can find the

which we ve called II and ii The net displacement is the vector magnitude of Cby using the law of cosines from trigonometry sum C= A + B which is found graphically C2 = A2 + B2 - 2ABcos(45deg )

End = (lOOm)2 + (200m)2 - 2(100m)(200m)cos(45deg)

= 21720 m2

T Ill Mirel IIcr Thus C = Y2l720 m2 = 147 m Then a second use of the law

C = i -- B di~p l acemcllll

of cosines can determine angle ltgt (the Greek letter phi)

8 2 = A2 + C2 - 2ACcosltgt

Angle (1 d cnb~ s Ille dilctlO[) o f vector E

It is easier to describe C with the angle (J = 180deg shyStart 100 m ltgt = 74deg The bird s net displacement is

FIGURE 3 5 The birds net displacement is C= A+ ii C= (147 m 74deg north of west)

- -

82 C H A PT E R 3 Vectors and Coordinate Systems

3

FIGURE 1 7 The net displacement aher four individual displacements

Ti le kngtlt 11II1 lretlhd by l lt~ raclor ( I bOIl h fl = -

N poin in Ihe am dill liI) A

FIGURE 3 8 Multiplication of a vector by a scalar

When two vectors are to be added it is often convenient to draw them with

their tails together as shown in Figure 36a To ~valuate 0 + E you could move vector E over to where its tail is on the tip of D then use the tip-to-tail rule of

graphical addition The gives vector t = ~ + E in Figure 36b Alternatively Figure 36c shows that the ector_~um D + E can be found as the diagonal of the parallelogram defined by D and E This method for vector addition which some of you may have learned is called the parallelogram rule of vector addition

(a) (b)

h2JELshyDD

What is 8 + pound Tip-to-tail rule Parallelogram rule Slide the tail of E Find the diagonal of to the tip of 8 the parallelogram

formed by 8 and E

FIGURE 36 Two vectors can be added using the tip-to-tail rule or the parallelogram rule

Vector addition is easily extended to more than two vectors Figure 37 shows

a hiker moving from initial position a to position I then position 2 then position 3 and finally arriving at pos ition 4 These four segments are described by disshy

placement vectors 0 O2 Oh and 04 The hikers net displacement an arrow

from position ato position 4 is the vector Oe( In this case

---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)

The vector sum is found by using the tip-to-tail method three times in succession

STOP TO THINK 31 Which figure shows A + A2 + A3

(a) (b) (e) (d) (e)

Multiplication by a Scalar

Suppose a second bird flies twice as far to the east as the bird in Example 31 The first birds di splacement was A = (100 m east) where a subscript has been added to denote the first bird The second birds displacement will then cershytainly be 12 = (200 m east) The words twice as indicate a multiplication so we can say

A2 = 2A

Multiplying a vector by a positive scalar gives another vector of different magnishyrude but pointing ~n the same direcTion

Let the vector A be

(3 6)

~here we ve specified the vectors magnitude A and direction eA Now let B = cA where c is a positive scalar constant We define the multiplica tion of a

vector by a scalar such that

B = cA means that (B ell) = (cA (JA) (37)

- -

- -


In other words the vector is s~retched or compressed by the fa~tor c (ie vector B has magnitude B = cA) but B points in the same direction as A This is illustrated

in Figure 38 on the previous page

We used this property of vectors in Chapter I when we asserted that vector a points in the same direction as ~v From the definition

_ ~v (1) _a= - = - ~v (38)

~I ~I

where (l~t) is a scalar constant we see that apoints in the same direction as ~v

but differs in length by the factor ( 1M) Tail of-i r - (41111 ill ma)lIitlldc

Suppose we multiply Aby zero Using Equation 37 at tip of it hut nrrlre in iJirectlon tn Thu ~ -- (-4) = (1 I o A= 0 = (0 m direction undefined) (39) Til )1 - r~turn ~ rn Ih~ ral1l1tg

The product is a vector having zero length or magnitude This vector is known as poill The reu Itart cclnr O the zero vector denoted O The direction of the zero vector is irrelevant you canshy

not descri be the direction of an arrow of zero length I

What happens if we multipl y a vector by a negative number Equation 37 does not apply if c_lt 0 because vector jj cannot have a ~egative magnitude Conshy

sider the vector - A which is equivalent to multiplying A by - Because FIGURE 39 Vector -A -- -- -A+(-A)=O (3 10)

the vector -A must be s uch that when it is added to A the resultant is the zero vector O In other words the lip of -A must return to the tail of A as shown in

Figure 39 This will be true only if -A is equal in magnitude to A but opposite in direction Thus we can conclude that

-A = (A direction opposite A) (311) That is multiplying a vector by -1 reverses its direction without changing its

length

As a n example Figure 310 shows vectors A 24 and - 3A Multiplication by

2 doubles the length of the vector but does not change its direction Multiplicatio n

by - 3 stretches the length by a factor of 3 and reverses the direction FIGURE 31 0 Vectors A 24 and - 34

EXAMPLE 32 Velocity and displacement This addition of the three vectors is shown in Figure 311 using Carolyn drives her car north at 30 kmlhr for I hour east at the tip-to-tail method trnltl stretches from Carolyns initial 60 kmlhr for 2 hours then north at 50 kInlhr for I hour What is position to her final position The magnitude of her net disshyCarolyns net displacement placement is found using the Pythagorean theorem

SOLVE Chapter I defined velocity as (net = V(120 km)2 + (80 kIn)2 = 144 km _ tr v=- The direction of tlrnet is described by angle e which is

t

so the displacement tl r during the time interval tlt is tlr = (tl1) V 80 kIn )e = tan- --- = 337deg(This is multiplication of the vector v by the scalar tl Carolyns 120 km

velocity during the first hour is V = (30 kmhr north) so her Thus Carolyns net displacement is tlrne1 = (144 kIn 337deg north displacement during this interval is of east)

tlr = (1 hour)(30 kInhr north) = (30 kIn north)

Similarly

tlr2 = (2 hours)(60 kInhr east) = (120 kIn east)

tr] = (J hour)(50 kInlhr north) = (50 kIn north)

In this case multiplication by a sca lar changes not only the length of the vector but also its units from kmlhr to km The direction however is unchanged Carolyn s net di splacement is

FIGURE 3 11 The net displacement is the tlrnet = tr + tlr2 + tlr3 vector sum tlrnet = tlr + tl r2 + tlr3


Vector Subtraction (a) (b) Figure 312a shows two vectors Pand Q What is R= P- Q Look back at

trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij

FIGURE 312 Vector subtraction

y

II

--------~--L---x

III IV

FIGURE 313 A conventional Cartesian coordinate system and the quadrants of the xy-plane

Tactics Box 12 on page 11 which showed how to perform vector subtraction graphically Figure 312b finds P - Qby writing R= P+ (-Q) then using the rules of vector addition

STOP TO THINK 32 Which figure shows 2A - Ii

(a) (b) (c) (d) (e)

33 Coordinate Systems and Vector Components

Thus far our discussion of vectors and their properties has not used a coordinate system at all Vectors do not require a coordinate system We can add and subtract vectors graphically and we will do so frequently to clarify our understanding of a situation But the graphical addition of vectors is not an especially good way to find quantitative results In this section we will introduce a coordinate description of vectors that will be the basis of an easier method for doing vector calculations

Coordinate Systems

As we noted in the first chapter the world does not come with a coordinate sysshytem attached to it A coordinate system is an artificially imposed grid that you place on a problem in order to make quantitative measurements It may be helpful to think of drawing a grid on a piece of transparent plastic that you can then overshylay on top of the problem This conveys the idea that you choose

Where to place the origin and bull How to orient the axes

Different problem solvers may choose to use different coordinate systems that is perfectly acceptable However some coordinate systems will make a problem easier to solve Part of our goal is to learn how to choose an appropriate coordishynate system for each problem

We will generally use Cartesian coordinates This is a coordinate sys tem with the axes perpendicular to each other forming a rectangular grid The standard xy-coordinate system with which you are fami Iiar is a Cartesian coordinate sysshytem An xyz-coordinate system would be a Cartesian coordinate system in three dimensions There are other possible coordinate systems such as polar coordishynates but we will not be concerned with those for now

The placement of the axes is not entirely arbitrary By convention the positive y-axis is located 90deg counterclockwise (ccw) from the positive x-axis as illusshytrated in Figure 313 Figure 313 also identifies the four quadrants of the coordishynate system I through IV Notice that the quadrants are counted ccw from the positive x-axis

33 Coordinate Systems and Vector Components 85

Coordinate axes have a positive end and a negative end separated by zero at the origin where the two axes cross When you draw a coordinate system it is important to label the axes This is done by placing x and y labels at the positive ends of the axes as in Figure 3] 3 The purpose of the labels is twofold

To identify which axis is which and bull To identify the positive ends of the axes

This will be important when you need to determine whether the quantities in a problem should be assigned positive or negative values

Component Vectors

Lets see how ~e can use a coordinate system to describe a vector Figure 314 shows a vector A and an xy-coordinate system that weve chosen Once the direcshytions of the axes are known we can d~fine two Eew vectors parallel to the axes that we call the comp~nent vectors of A Vect~ AX called the x-component vector is the projection of A along the x-axis Vector AI the y-component vector is the proshyjection of A along the y-axis Notice that the component vectors are perpendicular to each other

You can see using the parallelogram rule that A is the vector sum of the two component vectors

(312)

In essence we have broken vector A into two perpendicular vectors that are pa allel to the coordinate axes This process is called the decomposition of vector A

into its component vectors

NOTE ~ It is not necessary for the tail of Ato be at the origin All we need to know is the oriellfation of the coordinate system so that we can draw Ax and Ay parallel to the axes

Components

You learned in Chapter 2 to give the one-dimensional kinematic variable v a posshyitive sign if the velocity vector Ii points toward the positive end of the x-axis a negative sign if Ii points in the negative x-direction The basis of that rule is that v is what we call the x-component of the velocity vector We need to extend this idea to vectors in general

Suppose vector Ahas been decomposed into component vectors Ax and AI parshyallel to the coordinate axes We can describe each component vector with a single number (a scalar) called the component The x-component and y-component of vector A denoted Ax and A are determined as follows

TACTICS BOX 3 1 Determining the components of a vector

o The absolute valueJAxl of the x-component Ax is the magnitude of the component vector A _

f The sign of Ax is positive if Ax points in the positive x-direction negative if Ax points in the negative x-direction

~ The y-component AI is determined similarly

In other words the component Ax tell s us two things how big Ax is and with its sign which end of the axis Ax points toward Figure 315 on the next page shows three examples of determining the components of a vector

y A - -

A

--~----------~-------x

The rQmlfllWnf nl~ (PlIlrtgtIlCI1I

enor i raralkl ecror j parlllI 10 the ax i 10 til r-axilt

tGURE 3 14 Component vectors Ax and AI are drawn parallel to the coordinate axes such that A= Ax + A

86 CHAPTER 3 Vectors and Coordinate Systems

y (m) B POIJ1 t) n Ih~ n~ A Magnitude = ~ m _~jj piNt drecllllll +- mY (m)

-i 101laquo ill A B 3 -0 B = +2 (i C I

rhe phit e 3 I dir~ltmiddotlioll

MagnItude = 2 m 2 = ~~ III 2

~ MagnItude = 3 m (lt (

~ - ~ - A B Maonitude = 2 m Magnitude =

------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C

8 POillb mlhe llegIlI e )

plill in the ptie -2- 2 Ji r~(l(l n 0 = L 111 -2 I-dir middottioll 0 8 = 2 II I

(l11Pt)IICnt

= ~ ~1lI

The -component lit I= i~ ~ =

~ (

3 m

C shy I

4

(a) Y Magnitude A = VA + A

A ~A =ASino

A = AcosO -+-----j-----------x

Direction of A () = tan - (Al A)

(b) Y --1----------x

Magnitude Direction of C C = VC7- -~ +-C cJgt = lan- (CICI)

c = - Ccosltgt

Cex = Csinltgt

FtGURE 316 Moving between the graphical representation and the component representation

FIGUR E 315 Determining the components of a vector

NOTE ~ Beware of the somewhat confusing terminology e and Ay are called component vectors whereas Ax and Ay are simply called components The components Ax and A are scalars-just numbers (with units)-so make sure you do not put arrow symbols over the components

Much of physics is expressed in the language of vectors We will frequently need to decompose a vector into its components We will also need to reassemshyble a vector from its components In other words we need to move back and forth between the graphical and the component representations of a vector To do so we apply geometry and trigonometry

Consider first the problem of decomposing a vector into its x- and y-components Figure 316a shows a vector Aat angle 0 from the x-axis It is essential to use a picture or diagram such as this to define the angle you are using to describe the vectors direction

Apoints to the right and up so Tactics Box 31 tells us that the components Ax and A are both positive We can use trigonometry to find

Ax = AcosO (3 13)

Ay = AsinO

where A is the magnitude or length of A These equations convert the length and angle d~scription of vector Ainto the vectors components but they are correct only if A is in the first quadran~

Figure 316~ shows vector C in the fourth quadrant In this case where the comshyponent vector AI is pointing down in the negative y-direction the y-component Cy

is a negative number The angle cent is measured from the y-axis so the components of Care

C = Csincent (314)

C = -Ccoscent

The role of sine and cosine is reversed from that in Equations 313 because we are using a different angle

NOTE Each decomposition requires that you pay close attention to the direction in which the vector points and the angles that are defined The minus sign when needed must be inserted manuaJiy ~

We can also go in the opposite direction and determine the length and angle of a vector from its x- and y-components Because A in Figure 316a is the hypotenuse of a right triangle its length is given by the Pythagorean theorem

A = VA + A (315)

33 Coordinate Sys tems and Vector Components 87

Similarly the tangent of angle 8 is the ratio of the far side to the adjacent side so

Al)8 = tan-I Ax (316)(

where tan -I is the inverse tangent function Equations 315 and 316 can be thought of as the reverse of Equations 313

Equation 3 J5 always works for finding the length or magnitude of a vector because the squares eliminate any concerns over the signs of the components But finding the angle just like finding the components requires close attention to how the angle is def~ned and to the signs of the components For example finding the angle of vector C in Figure 3 J6b requires the length of C) without the minus sign Thus vector Chas magnitude and direction

C = YC + CJ l

(317)

cP = tan -I ( I~ I )

Notice that the roles of x and y differ from those in Equation 3 J6

EXAMP LE 3 3 Finding the components (a) y

of an acceleration vector -------------------~---x

Find the x- and y-components of the accele ration vector a shown in Fig ure 317a

VISUALIZE It s important to draw vectors Figure 317b shows the or iginal vector adecomposed into components paraliel to

the axes

SOLVE The acceleration vector a= (6 ms2 30deg below the (b)

negative x-a xis) points to the left (negative x-d irection) and a is negative a (ms2

)down (negative y-directi on) so the components ax and a are

both negalive -- a (ms2)

-4 or = -acos30deg = -(6 ms2)cos30deg = -52 ms2

a y = - asin 30deg = -(6 ms2)si n30deg = -30 ms2 2 -2shya = 6 ms

negative ASS ESS The units of or and a r are the same as the units of vecshy ~ ---tor a N otice that we had to insert the minus signs manually by -- -------~~l observing that the vector is in the third quadrant

FIG URE 3 1 7 The acceleration vector aof Example 33

EXAMPLE 3 4 Finding the direction of motion VIS UALIZE Figure 3 18b show s the components v and Vy and

Figure 3 18a shows a particles velocity vector V Determine the defines an angle e with which we can specify the direction of

partic le s speed and directio n of motion motion

(a) v (nus) (b) v (ms) L4

4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2

Vx = -6 ms Direction e= lan-(vl lvi)

FIGURE 31 8 The velocity vector v of Example 34


SOLVE We can read the components of Ii directly from the axes The absolute value signs are necessary because Vx is a negashy11 = -6 mls and vl = 4 ms Notice that v is negative This is tive number The velocity vector Ii can be written in terms of the enough information to find the panicle s speed v which is the speed and the direction of motion as magnitude of Ii

Ii = (72 mis 337deg above the negative x-axis) v = Vv2 + v = V( -6 mJS)2 + (4 ms)2 = 72 mJs

From trigonometry angle (J is

---------IIc---+----X (e m)

y

2 The unir ~ClO I hJ e kn~IJI 1 10 ullit- mid roillt 1[1 111lt r middotd ircClinl1 and -ry-diredlon

~ J 0middotmiddotmiddotmiddotmiddot

--~~~------X

2

FIGURE 319 The unit vectors land j

STOP TO THINK 33 What are the x- and y-components C and C of vector Cx y

y (em)

2

34 Vector Algebra Vector components are a powerful tool for doing mathematics with vectors In this section you II learn how to use components to add and subtract vectors First we I introduce an efficient way to write a vector in terms of its components

Unit Vectors

The vectors (l + x-direction) and (l + y-direction) shown in Figure 319 have some interesting and useful properties Each has a magnitude of I no units and is parallel to a coordinate axis A vector with these properties is called a unit vector These unit vectors have the special symbols

l == (I + x-direction)

== (I +y-direction)

The notation l (read i hat) and (read j hat) indicates a unit vector with a magnitude of J

Unit vectors establish the directions of the pos itive axes of the coordinate sysshyte m Our choice of a coordinate system may be arbitrary but once we decide to place a coordinate system on a problem we need something to tell us That direcshytion is the positive x-direction This is what the unit vectors do

The unit vector~ provide a useful way~ to write compo nent vec tors The component vector Ax is the piece of vector A that is para llel to the x-axis Simishylarly Ar is paralle l to the y-axi s Because by definition the vector l points along the x-axis and points along the y-ax is we can write

Ax = Ax1 (318)

34 Vector Algebra 89

Equations 318 separate each component vector into a scalar piece of length Ax (or Ay) and a directional piece I (or ) The full decomposition of vector Ii can then be written

(319)

Figure 20 shows how the unit vectors and the components fit together to form vector A

~OTE In three dimens~ns the unit vector along the +z-direction is called k and todescribe vector A we would include an additional component vector A z = Azk

You may have learned in a math class to think of vectors as pajrs or triplets of numbers such as (4 -25) This is another and completely equivalent way to write the components of a vector Thus we could write for a vector in three dimensions

jj = 41- 2 + 5k = (4-25)

You will find the notation using unit vectors to be more convenient for the equashytions we will use in physics but rest assured that you already know a lot about vectors if you learned about them as pairs or triplets of numbers

y

4 =A)

~~__~----~~------x

~lulllp1i(lllnJI r I jOf

by ~ ala JO~Il t 11Jrl~e

I llIil dll 111 Jirllillfl V(tllf i Ilkfllol~ lh~ ha Icnllitl I and POtllt llld dirn 111m tlllh ~ til rl1 lt1 II r i

FIGURE 320 The decomposition of vector Ii isAl + AJ

EXAMPlU5 Run rabbit run A rabbit escaping a fox runs 40deg north of west at 10 mls A coordinate system is establi shed with the positive x-axis to the

east and the positive y-axi s to the north Write the rabbits

velocity in terms of components and unit vectors

VISUALIZE Figure 321 shows the rabbit s velocity vector and the coordinate axes Were showing a velocity vector so the

axes are labeled v x and Vy rather than x and y

V N

I v v=IOmls~ ~

SOL ve 10 mls is the rabbit s speed not its velocity The velocshy

ity which includes directional information is

v= (10 mis 40deg north of west)

Vector vpoints to the left and up so the components Vx and Vv

are negative and positive respectively The components are

Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms

With Vr and Vy now known the rabbit s velocity vector is

v= Vxl + vyj = (-7661 + 643j) mls

Notice that weve pulled the units to the end rather than writingv = vsin40deg I them with each component40deg L _ _ _ _ - -e

ASSESS Notice that the minus sign for Vx was inserted manuallyv = -vcos40deg -----+--~-----------vx Signs dont occur automatically you have to set them after

checking the vectors direction FIGURE 32 1 The velocity vector vis decomposed into components Vx and vy-

Working with Vectors

You learned in Section 32 how to add vectors graphically but it is a tedious probshylem jn geometry and trigonometry to find precise values for the magnitude and direction of the resultant The addition and subtraction of vectors becomes much easier if we use components and unit vectors

To see this lets evaluate the vector sum jj = A + B + C To begin write this

sum in terms of the components of each vector

jj = DJ + Dy = Ii + 13 + C= (A) + AJ) + (Brl + By) + (Cl + Cy)

(320)


We can group together all the x-components and all the y-components on the right

side in which case Equation 320 is

(Dx)i + (D)] = (Ar + B + cJi + (Ay + By + Cy)] (321)

Comparing the x- and y-components on the left and right sides of Equation 321 we find

(322) Dy = Ay + By + Cy

Stated in words Equation 322 says that we can perform vector addition by

adding the x-components of the individual vectors to give the x-component of

the resultant and by adding the y-components of the individual vectors to give

the y-component of the resultant This method of vector addition is called

algebraic addition

EXAMPLE 36 Using algebraic addition to find a displacement Example 31 was about a bird that flew 100 m to the east then 200 m to the northwest Use the algebraic addition of vectors to find the birds net displacement Compare the result to Example 31

VISUA LI ZE Figure 322 shows displacement vectors A = (100 m east) and jj = (200 m northwest) We draw vectors tip-to-tail if we are going to add them graphically but its usually easier to draw them all from the origin if we are going to use algebraic addition

y N

displacement C==A+B________-L__

~

Net

____ A__ x-L~~ ~

1()() m

FIGURE 322 The net displacement is C= A + B

SOLVE To add the vectors algebraically we mu st know their components From the figure these are seen to be

A= 100 1m

jj = (-200 cos 45deg I + 200 sin 45deg j) m

= (-141 i + 141 j) m

Notice that vector quantities must include u~its Also notice as you would expect from the figure that B has a negative x-component Adding Aand jj by components gives

C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m

This would be a perfectly acceptable answer for many purshyposes l-0wever we need to calculate the magnitude and direcshytion of C if we want to compare this result to our earlier answer The magnitude of Cis

C = C + c = Y(-41 m)2 + (141 m)2 = 147 m

The angle e as defined in Figure 322 is

e = tan-I(I~I) = lan-(44 ) = 74deg

Thus C= (147 m 74deg north of west) in perfect agreement with Example 31

Vector subtraction and the multiplication of a vector by a scalar using composhy

nents are very much like vector addition To find R= P- Qwe would compute

(323)

Similarly T = cS would be

(324)


The next few chapters will make frequent use of vector equations For example you will learn that the equation to calculate the force on a car skidding to a stop is

if = Ii + W+ Jtl (325)

The following general rule is used to evaluate such an equation

The x-component of the left-hand side of a vector equation is found by doing scalar calculations (addition subtraction multiplication) with just the x-components of all the vectors on the right-hand side A separate set of calculations uses just the y-components and if needed the z-components

Thus Equation 325 is really just a shorthand way of writing three simultaneous equations

Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth

In other words a vector equation is interpreted as meaning Equate the x-components on both sides of the equals sign then equate the y-components and then the z-components Vector notation allows us to write these three equations in a much more compact form

Tilted Axes and Arbitrary Directions

As weve noted the coordinate system is entirely your choice It is a grid that you impose on the problem in a manner that will make the problem easiest to solve We will soon meet problems where it will be convenient to tilt the axes of the coordinate system such as those shown in Figure 323 Although you may not have seen such a coordinate system before it is perfectly legitimate The axes are perpendicular and the y-axis is oriented correctly with respect to the x-axis While we are used to having the x-axis horizontal there is no requirement that it has to be that way

Finding components with tilted axes is no harder than what we have done so far Vector C in Figure 323 can be decomposed C= C) + Cy where Cx = C cos 8 and Cy = C sin 8 Note that the unit vectors i and correspond to the axes not to horizontal and vertical so they are also tilted

Tilted axes are useful if you need to determine component vectors parallel to and perpendicular to an arbitrary line or surface For example we will soon need to decompose a force vector into component vectors parallel to and perpenshydicular to a surface

Figure 324a shows a vctor Aand a tilted line Suppose we would like to find the component vectors of A parallel and perpendicular to the line To do so estabshylish a tilted coordinate system with the x-axis parallel to the Jine and the y-axis perpendicular to the line as shown in Figure 324b Then A is equivalent to vector All the component of Aparallel to the line and Ay is equivalent to the perpendicular component vec~r Ai Notice that A= All + Ai

If ltJ is the angle between A an~ the line we can easily calculate the parallel and perpendicular components of A

A = Ax = AcosltJ (327)

A i = Ay = AsinltJ

It was not necessary to have the tail of A on the line in order to find a component of Aparallel to the line The line simply indicates a direction and the component vector All points in that direction

rh~ componenl~ 0 C ~rc found wilh repec t ttl the tilted IC

middotmiddotUnit e(lor~ i ~1I1ltl j Ieti nl lfh - lilt Y a I

FIGURE 323 A coordinate system with tilted axes

(b) y I

FIGUR E 324 Finding the components of A parallel to and perpendicular to the line


Y Component of F perpendicular to the suJiace H aI onzont

10 ~ force vector F

ltgt - 20deg Surface

20deg x

FIGURE 325 Finding the component of a force vector perpendicular to a surface

EXAMPLE 37 Finding the force perpendicular to a surface A horizontal force Fwith a strength of ION is applied to a surface (You II learn in Chapter 4 that force is a vector quantity measured in units of newtons abbreviated N) The surface is tilted at a 20deg angle Find the component of the force vector pershypendicular to the surface

VISUALIZE Figure 325 shows a horizontal force Fapplied to the surface A tilted coordinate system has its y-axis perpendicular to the surface so the perpendicular component is F1 = Fybull

SOLVE From geometry the force vector Fmakes an angle ltgt = 20deg with the tilted x-axis The perpendicular component of F is thus

F1 = Fsin20deg = (ION)sin20deg = 342N

STOP TO THINK 34 Angle 4gt that specifies the direction of Cis given by

- --f---x

Y a tan-ICCCy) b tan-ICCICyl)

c tan - ICICIICyl) d tan-ICCCx)

e tan-ICCICI) f tan - ICICyIICI)

Summary 93

SUMMARY

The goal of Chapter 3 has been to learn how vectors are represented and used

GENERAL PRINCIPLES

A vector is a quantity described by both a magnitude and a direction Unit Vectors y

U nit vectors have magnitude 1 ~eclion and no units Unit vectors

The ~lO r i and J define the directions dc-cl it~ Iht of the x- and y-axes ~1t1jtll)lI at ~ lellglh Qr n1gnilUdc i l-Ihi PO lilt ~ ~~ot(d A JltlgrlHudl d lor

USING VECTORS

Components The components Ax and Ay are the magnitudes ~ the

The component vectors are parallel to the x- and y-axes

A = Ax + Ay = Ax + AyJ component vectors Ax and A( = Ao Ay ami a plus or minusIn the figure at the right for example -r--~--~-------X

sign to show whether the Ax = Acos() y component vector points

AltO

Ay = AsiDO () = tan - I (AAx) A gt 0

Alt O gt Minus signs need to be included if the vector points

AltOdown or left

Working Graphically

A gt 0 toward the positive end or AgtO the negative end of the axis x A gt0

A lt 0

Negative Subtraction

Addition -t Al shyBlLt-B A - 8

Working Algebraically Vector calculations are done component by component

Cx 2Ax + BxC= 2A + B means Cy - 2Ay + By

The magnitude of Cis then C = VCx 2 + Cy

2 and its direction is found usi ng tan - I

TERMS AND NOTATION

scalar quantity zero vector 6 decomposition vector quantity componentCartesian coordinates magnitude unit vector lor]quadrants resultant vector algebraic addition component vector graphical addition

94 C HAP T E R 3 Vectors and Coordi nate Systems

EXERCISES AND PROBLEMS

Exercises

Section 32 Properties of Vectors

I a Can a vector have nonzero magnitude if a component is

zero If no why not If yes give an example

b Can a vector have zero magnitude and a nonzero composhy

nent ~f no_~hy ~~ot If yes give an example 2 Suppose C = A + B

a Under what circumstances does C = A + B b Could C = A - B If so how If not why not

3 Suppose C = A- 8 a Under what circumstances does C = A - B b Cou Id C = A + B If so how If not why not

4 Tra~e the_vectors i~ Fig~re Ex34 onto your paper Then find (a) A + B and (b) A - B

FIGURE EX3 4

5 Tra~e the_vectors i~ Fig~re Ex35 onto your paper Then find (a) A + B and (b) A-B

FIGURE EX3 S

Section 33 Coordinate Systems and Vector Components

6 A position vector in the first quadrant has an x-component of 6 m and a magnitude of 10m What is the value of its y-component

7 A velocity vector 40deg below the positive x-axis ha~ ay-component of 10 mls What is the value of its x-component

8 a What are the x- and y-components

of vector E in terms of the angle f) and the magnitude E shown in

Figure Ex38 b For the same vector what are the

x- and y-components in terms of

the angle 4gt and the magnitude E FIGURE EX38

9 Draw each of the following vectors then find its x- and

y-components a r= (100m 45deg below +x-axis ) b Ii = (300 mis 20deg above + x-axis)

c a= (50 ms2 -y-direction)

d F = (50 N 369deg above -x-axis)


y-components a r = (2 km 30deg left of +y-axis) b Ii = (5 cmls -x-direction)

c a = (10 mls2 40deg left of -y-axis)

d F ~ (50 N 369deg rightof +y-axis)

11 let C = (315 m 15deg above the negative x-axis) and

D = (256730deg to the right of the negative y-axis) Find the magnitude the x-component and the y-component of each

vector

12 The quantity called the electricfieUi is a vector The electric field

inside a scientific instrument is E == (125l - 250 j) V1m where V1m stands for volts per meter What are the magnitude and direction of the electric field

Section 34 Vector Algebra

13 Draw each of the following vectors label an angle that specishy

fies the vectors direction then find the vectors magnitude

and direction a Ii == 4l - 6j b r= (SOL + 80j) m

c Ii = (-20l + 40j) mls d a= (2l - 6j) mls2

14 Draw each of the following vectors label an angle that specifies

the -ecrors direction then find its magnitude and direction a B = -4l + 4j b r= ( - 21 - j) cm

c Ii == (-lOl- looj) mph d a= (20l + loj) ms2

15 LetA == 21 + 3jandB = 41- 2j a Draw a coordinate system and on it show vectors Aand 8 b Use graphical vector subtraction to find C = A - 8

16 LetA = 51 + 2j8 == -31- 5jandC = A + 8 a Write vector C in component form

b Draw a coordinate system and on it show vectors A 8 and C

c What are the magnitude and direction of vector C 17 Let Ii = 51 + 2j8 = - 31 - 5jandD = A - 8

a Write vector D in component form

b Draw a coordinate system and on it show vectors Ii ii and D

c What are the magnitude and direction of vector D

18 LetA = 51 + 2j 8 == -31 - 5jandE = 2A + 38 a Write vector Ein component form

b Draw a coordinate system and on it show vectors A B and E

c What are the magnitude and direction of vector E 19 LetA = 51 + 2j B = -31 - 5jand F = A - 4B

a Write vector Fin component form b Draw a coordinate system and on it show vectors A 8

and F c What are the magnitude and direction of vector F

20 Are the following statements true or false Explain your

answer

a The magnitude of a vector can be different in different coordinate systems

b The direction of a vector can be different in different coorshydinate systems

c The components of a vector can be different in different

coordinate systems 21 Let A = (40 ffivertically downward) and 8 == (so m 120deg

clockwise from A) Find the x- and y-components of Aand B in each of the two coordinate systems shown in Figure Ex321

__Y_ ~y x30deg x - - -- -- - shy

Coordinate Coordinate fiGURE EX3 21 s y~l em I syste m 2

- -

22 What are the x- and y-components of the velocity vector shown in Figure Ex322

N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems

23 Figure P323 shows vectors Aand ii Let C= A+ ii a Reproduce the figure on your page as accurateIY_as possishy

ble using a ruler and protractor Draw vector C on your figure using the graphical addition of A and ii Then determine the magnitude and direction of Cby measuring it with a ruler and protractor

b Based on your figure of part a use geometry and trigonometry to calculale the magnitude and direction of C

c Decompose vectors Aand ii into components then use these to calculate algebraically the magni- FIGURE Pl23

tude and direction of C 24 a What is the angle ltP between vectors E and F in Figshy

ure P324 b Use geometry and tri~onolletryo determine the magnishy

tude and direction of C = pound + F c US~compone~ts to determine the magnitude and direction

ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I

FIGURE P324 FIGURE P3 25

25 For the three vectors shown above in Figure P325 A + ii + C= -2i What is vector ii a Write ii in component form b Write ii as a magnitude and a direction

26 Figure P326 shows vectors Aand ii Find vector Csuch that A+ ii + C= O Write your answer in component form

y

100

FIGURE P3 26 FIGURE Pl 27

27 Figure P327 shows vectors A and B Find b = 2A + ii Write your answer in component form

Exercises and Problems 9S

28 Let A= (30 m 20deg south of east) ii = (20 m north) and C= (50 m 70deg south of west) a Draw and label A ii and Cwith their tails at the origin

Use a coordinate system with the x-axis to the east b Write 1 ii and Cin component form using unit vectors c Find the magnitude and the direction of b = A + ii + C

29 Trace the vectors in Figure P329 onto your paper Use the graphical method of vector addition

and subtraction to find the following D- 3- Fshya jj + pound + F 0 F b b + 2pound FIGURE P3 29 c D - 2pound + F

30 LetE = 2 + 3] and F = 2i - 2] Find the magnitude of a pound and F b pound + F c - pound - 2F

31 Find a vector that points in the same direction as the vector (i + ]) and whose magnitude is I

32 The position of a particle as a function of time is given by r = (5i + 4]) m where I is in seconds a What is the particles distance from the origin at I = 02

and 5 s b Find an expression for the particles velocity II as a funcshy

tion of time c What is the particles speed at t = 0 2 and 5 s

33 While vacationing in the mountains you do some hiking In

the morning your displacement is SmOming = (2000 m east)

+(3000 m north) + (200 m vertical) After lunch your displacement is Sanemoon = (1500 m west) + (2000 m north) -(300 m vertical) a At the end of the hike how much higher or lower are you

compared 0 your starting point b What is your total displacement

34 The minute hand on a watch is 20 cm in length What is the displacement vector of the tip of the minute hand a From 800 to 820 AM b From 800 to 900 AM

35 Bob walks 200 m south then jogs 400 m southwest then walks 200 m in a direction 30deg east of north a Draw an accurate graphical representation of Bobs motion

Use a ruler and a protractor b Use either trigonometry or components to find the displaceshy

mentthat will return Bob to his starting point by the most direct route Give your answer as a distance and a direction

c Does your answer to part b agree with what you can meashy

sure on your diagram of part a 36 Jims dog Sparky runs 50 m northeast to a tree then 70 m

west to a second tree and finally 20 m south to a third tree a Draw a picture and establish a coordinate system b Calculate Sparkys net displacement in component form c Calculate Sparkys net displacement as a magnitude and

an angle 37 A field mouse trying to escape a hawk runs east for 50 m

darts southeast for 30 m then drops 10 m down a hole into its burrow What is the magnitude of the net displacement of the mouse

38 Carlos runs with velocity II = (5 mis 25deg north of east) for 10 minutes How far to the north of his starting position does Carlos end up

39 A cannon tilted upward at 30deg fires a cannonball with a speed of 100 mls What is the component of the cannonballs velocshyity parallel to the ground

96 C H APT E R 3 Vectors and Coordinate Systems

40 Jack and Jill ran up the hill at 30 mis The horizontal composhynent of Jills velocity vector was 25 mis a What was the angle of the hill b What was the vertical component of Jills velocity

41 The treasure map in Figshyure P341 gives the followshying directions to the buried -x Treasure treasure Start at the old oak tree walk due north for 500

paces then due east for 100 paces Dig But when you

Yellow brick road arrive you find an angry dragon just north of the tree To avoid the dragon you set off along the yellow brick road at an angle 600 east of FIGURE P341

north After walking 300 paces you see an opening through the woods Which direction should you go and how far to reach the treasure

42 Mary needs to row her boat across a I OO-m-wide river that is flowing to the east at a speed of 10 ms Mary can row the boat with a speed of20 mls relative to the water a If Mary rows straight north how far downstream will she

land b Draw a picture showing Marys displacement due to rowshy

ing her displacement due to the rivers motion and her net displacement

43 A jet plane is flying horizontally with a speed of 500 mis over a hill that slopes upward with a 3 grade (ie the rise is 3 of the run) What is the component of the planes velocshyity perpendicular to the ground

44 A flock of ducks is trying to migrate south for the winter but they keep being blown off course by a wind blowing from the west at 60 ms A wise elder duck finally realizes that the solution is to fly at an angle to the wind If the ducks can fly at 80 mls relative to the air what direction should they head in order to move directly south

45 A pine cone falls straight down from a pine tree growing on a 200 slope The pine cone hits the ground with a speed of 10 ms What is the component of the pine cones impact velocity (a) parallel to the ground and (b) perpendicular to the ground

46 The car in Figure P346 speeds up as it turns a quarter-circle curve from north to east When exactly halfway around the curve the cars acceleration is a= (2 ms2 150 south of east) At this point what is the component of Q (a) tangent to the circle and (b) perpendicular to the circle

47 Figure P347 shows three ropes tied together in a knot One of your friends pulls on a rope with 3 units of force and another pulls on a secshyond rope with 5 units of force How hard and in what direction must you pull on the third rope to keep the knot from moving FIGURE P347

48 Three forces are exerted on an object placed on a tilted floor in Figure P348 The forces are meashysured in newtons (N) Assuming that forces are vectors

a What i ~ the c0fipon~nt of_he net force Fnel = FI + F2 + F) parshyallel to the floor 5N

b What is the component of Fnci perpendicular to the floor

c What are the magnitude and FIGURE P3 48

direction of Fnel 49 Figure P349 shows four electrical charges located at the

corners of a rectangle Like charges you will recall repel each other while opposite charges attract Charge B exerts a repul sive force (directly away from B) on charge A of 3 N Charge C exerts an attractive force (directly toward C) on A

141 em B

--- - ----- --- + charge A of 6 N Finally charge D exerb an attractive

lOOcmforce of 2 N on charge A Assuming that forces are vecshytors what is the magnitude

C Dand direction of the net force Fnci exerted on charge A FIGURE P349

FIGURE P346


Stop to Think 33 Cx = -4 cm C = 2 cmStop to Think 31 c The graphical construction of 4 I + 42 + 43 y

is shown below Stop to Think 34 c Vector Cpoints to the left and down so both

Stop to Think 32 a The graphical construction of 2A - B is C and C are negative C is in the numerator because it is the side opposite cPshown below

ParaJlelogram of (4 I + A) and 1

STOP TO THINK 31 STOP TO THINK 32

Exercises and Problems 77

74 64 m = 0 m + (32 mfs)(4 s - 0 s) + ~a(4 s - 0 S)2 ~

75 (IOms)2 = VOl - 2(98 ms2)(IOm - Om)II 76 (0 mlS)2 = (5 ms)2 - 2(98 ms2)(sin IOO)(XI - 0 m) ~

77 VI = 0 mfs + (20 ms2)(S s - 0 s) Ii XI = 0 m + (0 mfs)(S s - 0 s) + ~(20 ms2)(S s - 0 S)2

X2 = XI + v l(10 s - S s)

Challenge Problems

78 Jill has just gotten out of her car in the grocery store parking Ii lot The parking lot is on a hill and is tilted 3deg Fifty meters

downhill from Jill a little old lady lets go of a fully loaded shopping cart The cart with frictionless wheels starts to roll straight downhill Jill immediately starts to sprint after the cart with her top acceleration of 20 mfs2 How far has the cart

rolled before Jill catches it 79 As a science project you drop a watennelon off the top of theiI Empire State Building 320 m above the sidewalk It so hapshy

pens that Superman flies by at the instant you release the watennelon Superman is headed straight down with a speed of 3S ms How fast is the watennelon going when it passes Supennan

80 Your school science club has devised a special event for ~ homecomjng Youve attached a rocket to the rear of a small

car that has been decorated in the blue-and-gold school colors The rocket provides a constant acceleration for 90 s As the rocket shuts off a parachute opens and slows the car at a rate of SO mls2

The car passes the judges box in the center of the

grandstand 990 m from the starting line exactly 12 s after you fUe the rocket What is the cars speed as it passes the judges

81 Careful measurements have been made of Olympic sprinters

II in the I QO-meter dash A simple but reasonably accurate model

is that a sprinter accelerates at 36 mfs2 for 3t s then runs at

constant velocity to the finish line a What is the race time for a sprinter who follows this

model b A sprinter could run a faster race by accelerating faster at

the beginning thus reaching top speed sooner If a sprinters top speed is the same as in part a what accelerashytion would he need to run the 1 QO-meter dash in 99 s

c By what percent did the sprinter need to increase his accelshyeration in order to decrease his time by I

82 A sprinter can accelerate with constant acceleration for 40 s II before reaching top speed He can run the lOO-meter dash in

IO s What is his speed as he crosses the finish line 83 The Starship Enterprise returns from warp drive to ordinary II space with a forward speed of SO kmls To the crews great

surprise a Kljngon ship is 100 km directly ahead traveling in the same direction at a mere 20 kmfs Without evasive action the Enterprise will overtake and collide with the Klingons in just Slightly over 30 s The Enterprises computers react instantly to brake the ship What acceleration does the Entershyprise need to just barely avoid a collision with the KJingon ship Assume the acceleration is constant Hint Draw a position-versus-tirne graph showing the motions of both the Enterprise and the Klingon ship Let Xo = 0 kIn be the location of the Enterprise as it returns from warp drive How do you show graphically the situation in which the collishysion is barely avoided Once you decide what it looks like graphically express that situation mathematically


Stop to Think 21 d The particle starts with positive X and moves to negative x

Stop to Think 22 c The velocity is the slope of the position graph The slope is positive and constant until the position graph crosses the axis then positive but decreasing and finally zero when the position graph is horizontal

Stop to Think 23 b A constant positive V corresponds to a linshyearly increasing x starting from Xi = -10 m The constant negashytive V then corresponds to a linearly decreasing x

Stop to Think 24 a or b The velocity is constant while a = 0 it decreases linearly while a is negative Graphs a b and c all have

the same acceleration but only graphs a and b have a positive inishytial velocity that represents a particle moving to the right

Stop to Think 25 d The acceleration vector Gil points downhill (negative s-direction) and has the constant value - g sin ethroughshyout the motion

Stop to Think 26 c Acceleration is the slope of the graph The slope is zero at B Although the graph is steepest at A the slope at that point is negative and so a A lt aBo Only C has a positive slope soae gt aBo

(a) (b)

220

200

190

200 bull

bull 18 0 14
















11m one ~xim


- -




~ End ~

8 3 mi

e II StaI1 ~~

p 4mi ~









(a) (b)

I

I



8=




Vector Addition





- raquo - - )

C=A+B (31 )




C = VA 2 + 8 2 = V(4 mi)2 + (3 mi)2 = S mi (32)


(33)








= 21720 m2








- -


3








(a) (b)

h2JELshyDD


formed by 8 and E






---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)



(a) (b) (e) (d) (e)



A2 = 2A


Let the vector A be

(3 6)




- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







(a) (b)

220

200

190

200 bull

bull 18 0 14
















11m one ~xim


- -




~ End ~

8 3 mi

e II StaI1 ~~

p 4mi ~









(a) (b)

I

I



8=




Vector Addition





- raquo - - )

C=A+B (31 )




C = VA 2 + 8 2 = V(4 mi)2 + (3 mi)2 = S mi (32)


(33)








= 21720 m2








- -


3








(a) (b)

h2JELshyDD


formed by 8 and E






---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)



(a) (b) (e) (d) (e)



A2 = 2A


Let the vector A be

(3 6)




- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







- -




~ End ~

8 3 mi

e II StaI1 ~~

p 4mi ~









(a) (b)

I

I



8=




Vector Addition





- raquo - - )

C=A+B (31 )




C = VA 2 + 8 2 = V(4 mi)2 + (3 mi)2 = S mi (32)


(33)








= 21720 m2








- -


3








(a) (b)

h2JELshyDD


formed by 8 and E






---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)



(a) (b) (e) (d) (e)



A2 = 2A


Let the vector A be

(3 6)




- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









- raquo - - )

C=A+B (31 )




C = VA 2 + 8 2 = V(4 mi)2 + (3 mi)2 = S mi (32)


(33)








= 21720 m2








- -


3








(a) (b)

h2JELshyDD


formed by 8 and E






---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)



(a) (b) (e) (d) (e)



A2 = 2A


Let the vector A be

(3 6)




- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







- -


3








(a) (b)

h2JELshyDD


formed by 8 and E






---- - - -+ --

Dne( = D + D2 + D3 + D4 (3 5)



(a) (b) (e) (d) (e)



A2 = 2A


Let the vector A be

(3 6)




- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







- -

- -





_ ~v (1) _a= - = - ~v (38)

~I ~I











length






t




Similarly







trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









trdQ

- - _1_ _ WhatisP-Q R=P+(-Q)

=1gt-ij


y

II

--------~--L---x

III IV




(a) (b) (c) (d) (e)



Coordinate Systems










Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346











Component Vectors



(312)




Components








y A - -

A

--~----------~-------x











------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346














------+--r----------r--x (m) -----+-o-------------x (m) x (m)

-2 -I ~ 2 3 4 -2 -I 2 3 4 -2 -I - I C



(l11Pt)IICnt

= ~ ~1lI


~ (

3 m

C shy I

4


A ~A =ASino

A = AcosO -+-----j-----------x


(b) Y --1----------x


c = - Ccosltgt

Cex = Csinltgt







Ax = AcosO (3 13)

Ay = AsinO




C = Csincent (314)

C = -Ccoscent




A = VA + A (315)



Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









Al)8 = tan-I Ax (316)(



C = YC + CJ l

(317)







the axes













4

--- v= 4 ms

v - 22

v e- r v (ms) - 6 -4 - 2 - 6 ~4 ~2







---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346











---------IIc---+----X (e m)

y



--~~~------X

2



y (em)

2


Unit Vectors



== (I +y-direction)




Ax = Ax1 (318)



(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









(319)




jj = 41- 2 + 5k = (4-25)


y

4 =A)

~~__~----~~------x










V N

I v v=IOmls~ ~






Vx = - (10 mls) cos 40deg = -766 mls

Vy = + (10 mls) sin40deg = 643 ms


v= Vxl + vyj = (-7661 + 643j) mls









(320)






(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346












(322) Dy = Ay + By + Cy





algebraic addition



y N


~

Net

____ A__ x-L~~ ~

1()() m



A= 100 1m


= (-141 i + 141 j) m


C= A + jj = JOoi m + (-141 i + 141j) m

= (100m - 141 m)i + (141 m)j

= (-411 + 141j) m


C = C + c = Y(-41 m)2 + (141 m)2 = 147 m


e = tan-I(I~I) = lan-(44 ) = 74deg




(323)


(324)



if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









if = Ii + W+ Jtl (325)




Fy = n l + WI + Jt~ (326)

Fe = nz + W z + Jth









A i = Ay = AsinltJ





(b) y I




10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









10 ~ force vector F


20deg x







- --f---x




Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







Summary 93

SUMMARY


GENERAL PRINCIPLES




USING VECTORS





AltO



AltOdown or left

Working Graphically


A lt 0







TERMS AND NOTATION




Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346









Exercises









FIGURE EX3 4


FIGURE EX3 S



















vector
























answer






__Y_ ~y x30deg x - - -- -- - shy


- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346







- -


N Y ~

v = (100 mis w~vt ~_~oo

FIGURE EX3 22

Problems








ofC = pound + F

y y

f

ii

4 A --~f-- x

2

-I




y

100
















































141 em B




FIGURE P346



























141 em B




FIGURE P346







Vectors and Coordinate Systems - Montgomery County · PDF fileVectors and Coordinate Systems ....

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