Ch. 1--Math Review

1

F

Case 1

FCase 2

FIGURE 1.1

Chapter 1

MATHEMATICS REVIEW

A.) Scalars:

1.) A scalar is a variable that has magnitude (i.e., size) but does not havedirection associated with it.

a.) Example 1: Temperature measurements are not recorded as "25o

C upward." Why? Because temperature is a scalar, it has a magnitudebut it does not have direction.

b.) Other examples of scalars: length, time, mass, and speed.

2.) Scalars add and subtract in the same way dollars and cents add andsubtract.

a.) Example 2: If we know the temperatures T1 = 115o and T2 = 98o,

then T2 - T1 = -17o.

B.) Vectors:

1.) A vector is a variable that has bothmagnitude and direction associated with it.

a.) Example: Force is a vector. Push abox up an incline and you will find that thedirection of the force is as important as themagnitude of force. Specifically, the forcerequired to move the box up the incline inCase 1 (see Figure 1.1) will be considerablygreater than the force required in Case 2.

Both magnitude and direction are im-portant with vectors.

2

net force

= 5 nts

4 nts

3 nts

FIGURE 1.2

F

FIGURE 1.3

2.) Because direction makes a difference, vectors do not add or subtractlike scalars.

a.) Example: A threenewton force added to a fournewton force will not necessar-ily sum to an equivalent forceof seven newtons. Figure 1.2depicts a situation in whichthe vector sum of those twoforces generates an equivalentforce of five newtons at an an-gle of 37o with the horizontal.

3.) A vector can be algebraically represented either by a letter with an

arrow over it ( A→

) or by a letter in bold-face type (A). The former is most oftenused in classroom lectures and when doing problems longhand. The latter isused in texts as it is easier to represent on a computer. This book will use bold-face letters to represent vectors.

4.) There will be times when a vector's magnitude is all that isimportant. Notationally, the magnitude of the vector is represented as A . Onsome occasions, this relatively formal approach is inconvenient. An alternateapproach uses an unadorned letter like "A" (without the quotes) or asubscripted letter such as "Ax."

5.) The graphical representation of a vector isdrawn as an arrow. The arrow's orientation depicts thevector's direction and the arrow's length is scaled toreflect the vector's magnitude.

a.) Example: Figure 1.3 depicts a force vectorF whose magnitude is 36 newtons and whose ori-entation is at 45o with the horizontal. The scalingfactor in the sketch is one-half inch per 12 newtons.

Ch. 1--Math Review

3

A

B

FIGURE 1.7

B

B

A

FIGURE 1.6

FIGURE 1.5

FIGURE 1.4

B

A

C = B + A

6.) Manipulating vectors, even for relatively mundane operations likeaddition or subtraction, requires special approaches. The two most commonlyused are graphical manipulation and algebraic manipulation.

C.) Graphical Vector Manipulation:

1.) Consider the two velocity vectors Aand B shown in Figure 1.4. To graphically addthese, proceed as follows:

a.) Reproduce either vector (we'lluse B), drawing it to scale and keepingits orientation exactly as presented inthe original sketch (Figure 1.5).

The scale used will be 50 meters persecond per inch.

b.) Reproduce the second vector(vector A) so that its tail is positioned atthe head of the first vector (Figure 1.6).Again, make the drawing to scale andkeep the vector's orientation intact.

c.) Draw the resultant vectorC=A+B (Figure 1.7). This new vectorwill begin where the sketch began at thetail of B and end where the sketch endedat the head of A.

d.) To determine the magnitude ofthe resultant vector C, use a ruler orcentimeter stick to measure C's length,then multiply by the scaling factor. Inthis case we get approximately 87newtons.

e.) To determine C's directionrelative to the horizontal, a protractoryields approximately 20 degrees.

4

B

A

C

D

E

B+A+C+D+E

FIGURE 1.8

FIGURE 1.9

2A

1/2A

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

A -A

-2A

Note: Adding more than two vectorsfollows a similar pattern with the tail ofeach successive vector being placed at thehead of the previous one until all the vectorsare coupled (see Figure 1.8). Everythingmust be drawn to scale with relativedirections kept intact. The resultant will bea vector that starts where you started andends where you ended.

2.) Vectors can be multiplied byscalars as shown in Figures 1.9a through1.9e below. Figure 1.9a depicts vector A,then presents vectors (1/2)A, 2A, -A, and-2A respectively.

Note: A vector multiplied by a positive scalar either increases ordecreases the magnitude of the original vector but does not change itsorientation. Multiplying by -1 effectively flips the vector so that its orientationis opposite that of the original vector, and multiplying by a negative scalarchanges both the magnitude and the direction.

3.) Although it has its place, graphical vector manipulation is generallyawkward. From a mathematical standpoint, it is usually more convenient todeal with vectors algebraically in the context of coordinate axes. We will makeuse of two such grid systems: one that uses Cartesian coordinates and one thatuses polar coordinates.

Ch. 1--Math Review

5

30

40

o

o

4

2

F = 12 nts

= 180o

= 210o

F = 12 nts

C =

D =

y

y

y

x

x

x

FIGURE 1.13

x

y

A

FIGURE 1.11

FIGURE 1.12

FIGURE 1.10

0

0

0

D.) Algebraic Vector Manipulation--Polar Notation:

1.) Polar notation defines a vector by des-ignating the vector's magnitude A and angle θrelative to the + x axis (see Figure 1.10). Usingthat notation, the vector is written A = A ∠θ .

Note: Some math texts use an orderedpaired vector notation like ( A , θ ) to present polarinformation. As physicists do not usually use thisnotation, we will not use it in this book.

a.) Example 1: A force vector F with amagnitude of 12 newtons oriented at 210o

with the +x axis would be characterized as F= 12 ∠210o (Figure 1.11).

b.) Example 2: A force vector F with amagnitude of 12 newtons oriented along the-x axis would be characterized as F =12 ∠180o (Figure 1.12).

2.) Students should be able to characterizea graphically presented vector in polar notation.

a.) Example: From the graph in Figure1.13, characterize vectors C and D in polarnotation (Answer: C = 2 ∠30o, D = 4 ∠-50o).

Note: Negative angles are measured fromthe +x axis clockwise.

3.) Students should be able to graph avector characterized in polar notation.

a.) Example: Graph A = 4 ∠150o and B= 6 ∠-30o (the solution is shown in Figure1.14 on next page).

6

150o

-30o

A = 4

B = 6

y

x

FIGURE 1.14

60 +180 = 240o

o

A

-A

Angle for vector A is 0 = 60

Angle for vector -A is

0 + 180 , oro

oo

FIGURE 1.15

4.) Multiplying a vector by a positive scalar characterized in polarnotation changes the magnitude of the vector but not the direction (i.e., theangle).

a.) Example: If A = 12 ∠ 210o, then 3A = 36 ∠210o.

5.) Multiplying a vector by (-1) does not change the vector's magnitude,but it does reverse its direction (see Figure 1.15). As direction in polar notation

is denoted by the vector's angle, a reversal of direction effectively adds 180o tothe angle.

a.) Example 1: If D = 4 ∠ 110o, then -D = 4 ∠ 290o.

Note: -D is not -4 ∠ 110o.

b.) Example 2: If D = 4 ∠ 110o, then -(1/2)D = 2 ∠ 290o.

Note: Angles should never be greater than 360o. If A = 12 ∠ 220o, -A =12 ∠ (220o+180o) = 12 ∠ 400o = 12 ∠ 40o.

Ch. 1--Math Review

7

A

2 4

1

2

3

x

y

i = vector of magnitude 1 in "x" direction

y

x

4 in "x" direction

3 in "y" dir.

FIGURE 1.18

FIGURE 1.17

FIGURE 1.16

A

E.) Algebraic Manipulation in Cartesian Coordinates:

1.) An effective if not obscuresystem called unit vector notation is usedto denote vectors presented within aCartesian (x-y-z) coordinate system.Within that system, the vector shown inFigure 1.16 would be characterized as A= 4i +3j.

As this notation will be new tomany students, the following willhopefully explain the rationale behind it:

a.) Consider Figure 1.17.Graphical addition suggests thatthe vector A is equal to the sum of"a vector in the +x direction whosemagnitude is 4" and "a vector inthe +y direction whose magnitudeis 3."

b.) We now define a specialvector ̂i whose magnitude isalways one and whose direction isalways in the +x direction (Figure1.18). Such a vector is called aunit vector in the +x direction.

Note: Physics texts that usearrowheads over vector quantitiesreplace those arrowheads with a hatwhen referring to unit vectors. To betechnically complete, therefore, a hat hasbeen included over the "i" term shownabove in Part b. Although you must usea hat whenever writing unit vectors outlonghand (you will notice that I alwaysuse a hat in class), this book will assumethat all boldface i, j, and k vectors areUNIT VECTORS and will notsubsequently include the hat in the text.

8

4 i

y

x

FIGURE 1.22

j = vector of magnitude 1 in "y" direction

y

x

3 j

(ie, vector of magnitude 3 in the "y" direction)

y

x

4 i

3 j

A = 4 i + 3

j

y

x

FIGURE 1.21

FIGURE 1.20

FIGURE 1.19

Put another way, when you see aboldface j in this text, know that it is beingused to denote a unit vector in the y directioneven though it has no hat above it.

c.) With i as defined above,notice that 4i is a vector "whosemagnitude is 4 and whose directionis along the +x axis" (Figure 1.19).This is half of the vector sum neededto carry out the vector additionrequired for the production of vectorA as denoted in Part a above.

d.) Now define a unit vectordirected along the +y axis; call it j(Figure 1.20).

e.) Notice that 3j is a vector"whose magnitude is 3 and whose di-rection is along the +y axis" (Figure1.21). This is the other half of thevector sum needed to carry out theaddition required for the productionof vector A as denoted in Part a.

f.) Having defined the idea of aunit vector, the unit vector character-ization of vector A is A = 4i + 3j(Figure 1.22).

Note 1: For three dimensionalsituations, the +z direction unit vector isdefined as k.

Note 2: Consider the vector A = -4k.Technically, the direction should be associ-ated with the unit vector k. That is, the for-mally correct way of characterizing the vec-tor would be as A = 4 (-k). Unfortunately,embedded signs can be overlooked, so it is

Ch. 1--Math Review

9

G

H-2

-4

-4 -2 2 4x

y

2

4

2 4x

y

2

4J

K

FIGURE 1.24

FIGURE 1.23

acceptable to put the negative sign in front ofthe expression as first depicted.

Note 3: In the vector A = -4i + 3j, the"-4" part of the expression is called "the xcomponent of A," or Ax, and the "+3" term iscalled "the y component of A," or Ay.Components are not really magnitudes--theycan be negative.

2.) Students should be able tocharacterize a graphically presented vector,writing it out in unit vector notation.

a.) Example: From the graph inFigure 1.23, characterize vectors Jand K in unit vector notation (Answer:J = 2i + 4j and K = 2i - 5j).

3.) Students should be able to grapha vector characterized in unit vector notation.

a.) Example: Graph vectors G =-2i + 3j and H = -3i- 2j (solutionshown in Figure 1.24).

F.) Conversion: Unit Vector to PolarNotation:

x

y

A y

xA

A

0

FIGURE 1.25

1.) Consider the known vector A = Axi + Ayjshown in Figure 1.25. To characterize it in the polarnotation, A = A ∠θ .

a.) To determine A : the right triangleshown in Figure 1.25 coupled with thePythagorean relationship yields a vectormagnitude:

10

-2

-4

-4 -2

4x

y

2

-36.9o

FIGURE 1.27

-A = (4i - 3j)

= 5 -36.9

-2

-4

-4 -2 2 4x

y

2

4

A

FIGURE 1.26

o

A = -4i + 3j

= 5 (-36.9 + 180 )o o

A = [Ax2+ Ay

2] 1/2.

b.) To determine θ : The tangent of θ is defined as the ratio of theside opposite θ (i.e., Ay) divided by the side adjacent (Ax), or tan θ =

Ay/Ax. This implies that:

θ = tan-1 (Ay/Ax).

Note: This is expressed as θ , the angle whose tangent is (Ay/Ax).

c.) Example: The conversionof the velocity vector A = (-4i +3j)m/s to polar notation (see Figure1.26):

A = A ∠θ = [Ax

2+ Ay2] 1/2

∠ [tan-1(Ay/Ax)]

= [(-4)2 + (3)2] 1/2

∠ [tan-1 (3/(-4))]

= 5 m/s ∠-36.9o.

Note: There is SOMETHINGWRONG here. A is a second quadrantvector, while the angle given by yourcalculator implies a fourth quadrantangle.

The problem lies in your calcula-tor's inability to tell the difference be-

tween tan-1[3/(-4)]--a second quadrant

vector--and tan-1[(-3)/4]--a fourthquadrant vector. As a consequence, allcalculators assume they are dealingwith fourth quadrant values wheneverthey are fed tangent arguments that arenegative. In such cases, the correctsecond quadrant angle can be generated

Ch. 1--Math Review

11

x

y

0

Ay

xA

A

FIGURE 1.28

(magnitude of )A

by adding 180o to the calculator's value (see Figure 1.27 on previous page).This means A = 5 m/s ∠ (-36.9o+180o) = 5 m/s ∠ 143.1o.

Note: A similar problem exists between first and third quadrantvectors. That is, third quadrant vectors have a tangent ratio of (-Ay)/(-Ax), whichis positive just as are first quadrant angles. All calculators assume thatpositive tangents are first quadrant vectors. Again, 180o must be added to thecalculator-generated angle if that is not the case.

G.) Conversion: Polar to Unit Vector Notation:

1.) There are two ways to do this. One is bymemorizing a formula that always works; the otheris by using your head (ooh, scary). As you'llundoubtedly forget the memorized equation sooneror later, understanding the seat of your pantsapproach is important. I'll outline both.

2.) Consider the known vector A= A ∠θshown in Figure 1.28. To characterize it formally (i.e.in memorizable form) in unit vector notation:

a.) To determine Ax: Cosine ∠θ1 isdefined as the ratio of the side adjacent to θ

A = 7

0 = 130o

FIGURE 1.29

(i.e., Ax) divided by the hypotenuse ( A ), or cos θ = (Ax)/( A ). This impliesthat Ax = A cos θ .

b.) Similar reasoning produces Ay asAy = A sin θ .

c.) These expressions will work for any angle(hence, if memorized, will never fail you . . .unless your memory fails you). As an example,the conversion of the velocity vector A = 7 m/s∠ 130o into unit vector notation (Figure 1.29)becomes:

12

A = 7 cos 40yo

FIGURE 1.30

0A = 7

A = 7 sin 40xo

= 4.5

= 5.4

12

A

B

y

x

O

00

FIGURE 1.31

A = Axi +Ayj

= ( A cosθ ) i +( A sin θ ) j

= [(7 m/s) cos130o] i + [(7 m/s) sin130o] j = (-4.5 i + 5.4 j) m/s.

Note: The only nice thing about the memorized approach is that italways gives each component's correct sign.

3.) Assuming your memory is as miserable asmine, the more intelligent way to do the conversionoutlined above is by the seat of your pants approach.That is:

a.) Create a convenient right triangle likethe one shown in Figure 1.30.

b.) Use the appropriate trig functions todetermine the vector's components.

c.) Once you have the magnitude of thecomponents, add signs and you're done (i.e., writethe x component as negative). Doing so yields A = (-4.5 i + 5.4 j) m/s.

H.) Dot Product In Polar Notation:

1.) In polar notation, consider the twovectors: A = A ∠θ1 and B = B ∠θ 2 (see Figure1.31). The dot product between A and B producesa scalar quantity. The magnitude of the scalarproduct is defined as:

A . B = A B cos φ ,

where φ is the net angle between the line of the twovectors.

2.) Example: Let A = 5 nt ∠ 30o and B =12 m ∠ 180o (see Figure 1.32 on next page). Whatis A . B?

Ch. 1--Math Review

13

B =12 180o

y

x

A = 5 30o

FIGURE 1.32

a.) Following the definitionof the dot product:

A . B = A B cos θ ,

= (5 nt) (12 m) cos (150o)= - 52 nt.m.

I.) Dot Product In Unit Vector Notation:

1.) In unit vector notation, consider the vectors: A = Axi + Ayj + Azk and B= Bxi + Byj + Bzk. The dot product between A and B produces a scalar quantitythat is mathematically equal to:

A . B = AxBx + AyBy + AzBz.

Note that this is easily derived:

a.) Beginning with A . B = (Axi + Ayj + Azk) . (Bxi + Byj + Bzk).

b.) Calling upon the distributive nature of dot products, we get awhole string of mini dot products:

A . B = [(Axi) . (Bxi )] + [(Axi) . (Byj)] + . . .

c.) As the angle between two vectors in the i direction is zerodegrees, the first mini dot product shown above is equal to (Ax)(Bx)(cos 0o)

= AxBx. The implication is that like-termed products will not be zero(assuming neither Ax nor Bx are zero).

d.) As the angle between the i direction and the j direction is 90o,the second mini dot product is equal to (Ax)(By)(cos 90o) = 0. Theimplication is that all cross-termed products will be zero.

e.) Bottom line: In unit vector notation, A . B = AxBx + AyBy + AzBz.

14

A

B

y

x

B sin O (component of B perpendicular to line-of-A)

B cos O (component of B along line-of-A)

O

FIGURE 1.33

2.) Example: Let A = (3i - 4j - 5k) newtons and B = (2i + 7j +3k) meters.What is A . B ? Following the derived expression for the dot product expressedin unit vector notation:

A . B = AxBx + AyBy + AzBz= (3 nt) (2 m) + (-4 nt) (7 m) + (-5 nt) (3 m)= -37 nt.m.

J.) Dot Product In General:

1.) To understand the physicalsignificance of the dot product, consider Figure1.33. In it, the vector B has been split into twocomponents--one parallel to the line of A and oneperpendicular to the line of A. Notice that thecomponent parallel to the line of A has amagnitude of A B cos φ .

a.) Conclusion: When the dot productis taken between two vectors, it generatesa quantity equal to the product of:

i.) The magnitude of one vector( A in this case) and,

B

A

y

xline-of-B

0

A sin 0 (component of A perpendicular to line-of-B)

FIGURE 1.34

A cos 0 (component of A along line-of-B)

ii.) The magnitude of the second vector's component that runsparallel to the first vector (i.e., B cos φ ).

Note 1: It doesn't matterwhether you take ( A )( B cos φ )or ( A cos φ )( B ). Both willwork.

Note 2: For the sake ofvisualization, if A = 5 nt ∠ 30o

and B = 12 m ∠ 180o, thecomponent of A along the line of Bis shown in Figure 1.34 and the

Ch. 1--Math Review

15

BA

y

x

B cos 0 (component of B along line-of-A)

line of A

line of A

FIGURE 1.35

component of B along theline of A is shown in Figure1.35.

2.) Example of asituation in which the dotproduct comes in handy:Consider Fig-ure 1.36.

a.) A force F isapplied to a movingobject as it traversesover a frictionless

FIGURE 1.36

F

0

d

F sin 0

F cos 0

surface for a distance d. Notice thatthe force will increase the object'sspeed. Notice also that the forcecomponent that makes the objectincrease its speed is the componentof F along the line of d, or F cosθ .

b.) As the amount of speed theobject picks up is dependent onlyupon the distance over which theforce acts (i.e., the magnitude of thedisplace-ment vector d) and thecomponent of F along the line of d (i.e., F cos θ ), the product of thosetwo quantities is deemed important enough to be given a special name--WORK.

c.) In short, the work done by the force F acting on an object whose dis-placement is defined by a distance d is mathematically defined as WF = F . d.

K.) Cross Product In Polar Notation:

1.) Consider two vectors A = A ∠θ1 and B = B ∠θ 2. The cross productbetween A and B produces a vector whose magnitude is mathematically defined as:

A . B = A B sin φ ,

where φ is the angle between the two vectors.

16

2.) Example: Let A = 5 newtons ∠ 30o and B = 12 meters ∠ 180o (lookback at Figure 1.32). What is BxA ?

a.) Following the definition of the magnitude of the cross product:

BxA = B A sin φ ,

= (12 m)(5 nt) sin (150o) = 30 nt.m.

b.) When dealing with vectors in polar notation, the direction of across product vector can be found using the right-hand rule. This rule isoutlined below:

i.) The line of both vectors should be extended until theyintersect (they already intersect in the problem we are examining);

ii.) With the wrist of the right hand placed at the intersection ofthe two vectors, the straightened fingers of the open right handshould be positioned so that they are parallel to the direction of thefirst vector (vector B in this case);

iii.) The fingers of the right hand should then be curled (waved) inthe direction of the second vector (vector A in this case). Note that youmay have to flip your hand over to do this.

iv.) If the thumb of the right hand is held out at a right angle tothe fingers, the direction of the thumb will point in the direction ofthe cross product vector.

Note: Both vectors A and B are in the plane of the paper, whereas thedirection of the right thumb alluded to above will be out of or into that plane.As peculiar as this may seem now, the direction of a cross product is alwaysperpendicular to the plane defined by the vectors being crossed.

c.) In the problem above, your thumb should end up pointingdownward into the page. Assuming the +i direction is to the right andthe +j direction is upward toward the top of the page, the cross product'sdirection will be in the negative z direction characterized by a -k in unitvector notation.

Ch. 1--Math Review

17

B

A

y

x

thumb in direction of first vector (vector B)

straight fingers in direction of second vector (vector A)

palm facing direction of cross product (into the page)

FIGURE 1.37

Note: Assuming we are not dealing with an angle of 180o, there are two

angles between any given vectors--one less than 180o and one more than 180o.The "wave" should always be through the angle less than 180o.

d.) Putting everything together, we get BxA = 30 nt.m (-k). Or, if youdo not like having the negative sign embedded in the middle of anexpression, the cross product could be written BxA = -30 nt.m (k).

Note 1: There will be times when the two vectors being crossed will nothave a common starting point. When that occurs, it is very important that youextend the line of the two vectors until they intersect before trying to executethe right-hand rule.

Note 2: An alternate approach todetermining cross product directions:Position your open right hand so that yourthumb is in the direction of the first vector(B in this case) and your straightenedfingers are in the direction of the secondvector (you may have to flip your handover to accomplish this). The direction thepalm of your right hand faces (i.e., along avector coming out of the palm) will be thedirection of the cross product. Try it forthe above situation (Figure 1.37) and youwill find that your palm faces downward.

Note 3: If you are wondering why anyone would want an operation thattakes two x-y plane vectors and produces a third vector whose direction isperpendicular to the x-y plane, an example is coming in the section CROSSPRODUCT IN GENERAL.

L.) Cross Product In Unit Vector Notation:

1.) In unit vector notation, consider the vectors A = Axi + Ayj + Azk and B= Bxi + Byj + Bzk. The cross product between B and A produces a vector themagnitude and direction of which can be determined by evaluating the followingmatrix:

18

B x A = yB Bz

xA

Bx

A y

i j k

=

A z

i [(B )(A ) - (B )(A )] + j [(B )(A ) - (B )(A )] + k [(B )(A ) - (B )(A )]

z

x

y y

y

z

z x

x x

z

y .

Note: Because there are unit vectors directly placed within the matrix,its evaluation will automatically give you both the magnitude and direction ofthe cross product. You do not have to mess with the right-hand rule whenevaluating cross products in u.v.n.

2.) To begin with, memorizing the end-result of the cross product matrixis a bit of a waste of time. It is much better to simply learn how to evaluatesuch a matrix (you will need to know how to do this later when we get to circuitanalysis). The obvious question is, how do you do that?

3.) Consider the following example. Let A = i - 2j + 3k and B = -4i - 5j -6k. What is B x A?

a.) The matrix requires that you array the unit vectors across the toprow as shown. The first vector (B in this case) in the cross product has itscomponents (signs included) placed in the second row of the matrix, andthe second vector has its components placed in the third row. So:

B x A = -4 -5 -6

3-21

i j k

b.) There are two ways to proceed from here. I prefer the approachthat is, in my opinion, the simplest. To follow that technique, the firsttwo columns must be reproduced to the right of the matrix as shownbelow (the reason for doing this will become evident shortly).

-4 -5 -6

3-21

i j k

-4 -5

-21

i j

Ch. 1--Math Review

19

c.) Once set up, circle the first i unit vector, then cross out the rowand column in which it resides.

-4 -5 -6

3-21

i j k

-4 -5

-21

i j

d.) The x component of the cross product will be i (i.e., the unit vectorthat is circled) times the evaluation of the 2x2 matrix shown in thesketch. Note that the upper left-hand piece of that 2x2 matrix is onecolumn over, one row down from the circled i).

e.) To evaluate a 2x2 matrix, take the upper left-hand piece timesthe bottom right-hand piece MINUS the upper right-hand piece timesthe lower left-hand piece. In our example, that will be (-5)(3) - (-6)(-2) =-27. Once you multiply that by i to get -27i, you have the x component ofthe cross product.

f.) The exact same thing is done with the j column and row (this isusually overlaid on top of the previous set-up--it isn't likely that you aregoing to want to recreate the entire matrix to do the next step). That isshown below.

-4 -5 -6

3-21

i j k

-4 -5

-21

i j

g.) For this part, multiply the circled j times the evaluation of the2x2 matrix shown in the sketch above (the upper left-hand piece is -6).The evaluation for this part will be (-6)(1) - (-4)(3) = +6. Multiplying bythe j yields +6j, and you have the y component of the cross product.

h.) The z component is taken care of similarly (try it) yielding a finalcross product, unit vectors and all, of:

B xA = i [(-5)(3) - (-6)(-2)] + j [(-6)(1) - (-4)(3)] + k [(-4)(-2)-(-5)(1)] = -27i + 6j + 13k.

20

A

B

y

x

O

FIGURE 1.38

(component of A perpendicular to line-of-B)

A sin O

line of B

M.) Cross Product In General:

1.) To understand the physical significance of the cross product, re-consider Figure 1.33.

a.) Vector B has been split into two components--one parallel to theline of A and one perpendicular to the line of A. Notice that thecomponent perpendicular to the line of A has a magnitude of B sin φ .

b.) Evidently, cross products generate a vector whose magnitude isequal to the product of:

i.) The magnitude of one of the vectors ( A in Fig 1.33), and

ii.) The magnitude of the second-vector's-component that runsperpendicular to the first vector ( B sin φ ).

c.) An example of such a situation follows in Part 3 of this section.

Note: Again, it makes nodifference whether the magnitude isobtained by determining ( A )( Bsin φ ) or ( A sin φ )( B ). A sketchof the information required todetermine the former is shown inFigure 1.33, whereas the informa-tion required to determine thelatter is shown in Figure 1.38.

2.) The significance of across product's direction dependsupon the situation in which thecross product is used. For instance,a charged particle moving withvelocity v in a magnetic field B will feel a magnetic force F that is proportionalto vxB. In this case, the direction of the cross product is the direction of themagnetic force as it is applied to the particle (the apparently odd fact that thedirection of a magnetic force on a moving charged particle is always

Ch. 1--Math Review

21

FIGURE 1.39

r

F

F sin 0

F cos 0

wrench0

perpendicular to the plane defined by the velocity and magnetic field vectors wasdeduced experimentally).

Note: Concerning this "perpendicular-to-the-plane-of-the-two-vectors"characteristic of cross-product directions, notice that if the vectors being crossedare in the x-y plane--a situation that will always be the case when working inpolar notation--the cross product direction will be either in the +k or -k direction.

3.) Example of a situation inwhich the cross product comes in handy:Consider Figure 1.39.

a.) A force F is applied to awrench at a distance r units fromthe axis of rotation. Note that:

i.) The greater r , theless difficult it will be torotate the bolt;

ii.) The greater F , theless difficult it will be to rotate the bolt; and

iii.) The force component that will make the bolt rotate will be thecomponent perpendicular to the line of r (i.e., F sin φ ).

b.) As ease of rotation is related to r and ( F sin φ ), the product ofthose two variables is deemed important enough to be given a specialname--torque ( ΓΓ ). In short, the magnitude of the torque applied by Fabout the axis of rotation will be ΓΓ = rxF .

c.) Assuming r and F are in the x-y plane, the direction of the crossproduct will either be in the +k or -k direction (using the right-hand ruleoutlined above, it turns out to be the +k direction).

Note 1: For straight-line motion, the use of the i, j, and k unit vectors todenote, say, a velocity's direction, is easy to decipher. In such a case, they quiteliterally tell you the direction in which the body is moving. Unfortunately, whendealing with planar rotational motion and the cross products that define them,the idea of direction is a little more complicated.

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Note 2: If Part d (below) does not make perfect sense now, don't worry aboutit. All you really need to be able to do at this point is to calculate a cross product(both magnitude and direction). The significance of the calculation will becomemuch more evident when we begin using the math in specific situations.

d.) Direction of a cross product:

i.) To begin with, notice that the motion of a spinning object(we'll assume it is spinning in the x-y plane) does not really goanywhere--it just sits there spinning. That means that the directionof, say, the rotational velocity of the object can have nothing to do withlinear displacement.

ii.) In fact, direction as related to spinning objects addresses onequestion only: Is the rotation clockwise or counterclockwise?

iii.) As there are no unit vectors defined in the clockwise direction,a little fancy footwork is required to convey the sense of rotation, usingthe unit vector notation we already have at our disposal.

iv.) The vectors used to identify the torque that started the objectspinning in the first place (i.e., F and r) are in the x-y plane, but no-tice that the object rotates about a line parallel to the z-axis.

v.) Physicists and mathematicians traditionally define thesense-of-rotation by denoting the axis about which the object rotates.

vi.) If, therefore, a rotation in the x-y plane is clockwise (this kindof rotation would screw a bolt INTO the x-y plane), the axis ofrotation is along the z axis (inward) and the unit vector assigned todenote that fact would be -k.

By the same token, if a rotation in the x-y plane is counterclockwise(i.e., screwing the bolt OUT), the axis of rotation is along the z axis(outward) and the unit vector assigned to denote that fact would be +k.

vii.) Using that designation, torques that produce counterclockwisemotion unscrew bolts out of the page and are designated as having a+k direction. Torques that produce clockwise motion screw a bolt intothe page and are defined as having a -k direction.

Ch. 1--Math Review

23

river moving 2 miles/hr

100 meters

30 m

direction everyone wants to goboat

hotel

FIGURE I

viii.) In light of all this, it shouldn't be surprising to find that atorque that makes a bolt rotate, say, clockwise, will have a crossproduct whose direction is along the -k axis. Put another way, when atorque calculation readings rxF = (3 nt.m)(-k), the -k part of the crossproduct tells you that the torque in question has a clockwise sense.

QUESTIONSNote: If it is now at least three days into the course, I would suggest you

begin with Question #1.3. Do Questions #1.1 and #1.2 only if you have time--they are not terribly important concepts.

1.1) With some urgency, a newlywed couple on their honeymoon hires anot-so-bright boatman to row them across a 100 meter wide river to theirhoneymoon hotel. In calm water, the boatman (we'll call him Jack the idiot, forreasons that will become obvious shortly) can row his dinghy 5 miles per hour.The river moves at 2 miles per hour.

When sitting directly across from the hotel, the boatman points his bowat the hotel and proceeds to row like mad without a second glance. In blissfulignorance, he, the newlyweds, and the boat do not move directly across the riverbut instead move across and down the river--Jack has not compensated for thecurrent. (Landing nowhere close to the hotel, the couple is less than delightedwith Jack's performance, hence tagging him "the idiot.")

a.) Using graphical manipulation, determine the boat's actual ve-locity (as a vector) relative to dry land.

b.) (This one is a stinker--don't spend a lotof time on it). Changing the problem slightly,let's assume the boat starts 30 meters downriver from the original starting position (thatis, 30 meters below a line drawn directlyacross the river to the hotel--see Figure I). IfJack is clever, he can point the boat upstreamto compensate for the moving water and makethe boat's net movement travel directly towardthe hotel. How would he have to orient theboat to do so and, as a consequence, how fastwould the boat move relative to dry land?Again, use graphical means to determine this!

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A

-1x

y

D = 2.6C

B = 1.3

0 = 70

1

-1

o

o

1

FIGURE II

0 = 35

1.2) A plane flies 80 miles north, then makes a north-westerly turn(versus south-westerly turn) at an unknown angle and flies an additional 60miles. Upon landing, the pilot realizes she is exactly 30o west of north, relativeto her starting point.

a.) Use graphical manipulation to determine the "unknown angle" ofher westerly turn.

b.) What was her net distance traveled (i.e., her net displacement)?

Note: Graphical manipulation is only rarely used in physics, but it issomething that is usually included in physics classes for the sake ofcompleteness.

1.3) Graph the vectors T = 8i -12j and P = 7∠ (-60o).

1.4) From the graph in Figure II, characterize vectors A and C in unitvector notation and vectors B and D in polar notation.

1.5) On the graph shown in Figure II, vectors B andD are almost the same length even though their magnitudes are different. Why, and what does the goof do to the angles?

1.6) Assume: A = -8i +12jB = -4i - 3jC = 5i + 6j -7kD = 7 ∠ (-60o)E = 12 ∠ 225o

F = 2 ∠ 105o.

In the problems below, use the above vectors as they are denoted above(that is, if asked to do AxB, do so in unit vector notation if A and B are given inu.v.n.). With that in mind, determine:

a.) -(1/3)A; b.) -6E;c.) A + B - C; d.) E converted to unit vector notation;e.) F converted to u.v.n. f.) A converted to polarg.) B converted to polar h.) A.C;i.) D.E; j.) AxB;k.) CxB; l.) DxE.

1.7) What does A.B really tell you?1.8) What does AxB really tell you?