In fact, we can have the arrow answer both the amount and direction questions by having the length of the arrow indicate “how far” or “how fast” and the direction the arrow is pointing i
Trang 120.7 A BRIEF INTRODUCTION TO VECTORS
Questions of displacement and velocity have figured prominently in our studies When we talk about displacement, we ask
“how far?” and “in what direction?”
When we talk about velocity, we ask
“how fast?” and “in what direction?”
We’ve been answering the question “in what direction?” by using a plus or a minus sign This has restricted the type of motion we describe Our objects are going up or down; our boats are headed east or west; our cars travel on long straight highways; our trains are either inbound or outbound If we wander along a meandering path, we can report our velocity only with reference to some benchmark point on the path; we’re either heading toward it or away from it
We can free ourselves from this world of dichotomy by using an arrow to indicate
direction In fact, we can have the arrow answer both the amount and direction questions by
having the length of the arrow indicate “how far” or “how fast” and the direction the arrow
is pointing indicate the direction of the displacement or motion An arrow used in this way
is called a vector
Defining Vectors
D e f i n i t i o n
A vector has two defining characteristics:
length (or magnitude) and direction
A vector is often denoted byv, a boldface letter with an arrow above it, and represented
by an arrow The length ofv is denoted by |v|.
Vectors can be used in a large variety of situations; for example, vectors can be used
to model velocity, force, and displacement A velocity vector is a vector whose length represents speed and whose direction indicates the direction of motion A force vector is
a vector whose magnitude corresponds to the magnitude of the force and whose direction indicates the direction of application of the force We can give a similar interpretation for
a displacement vector We refer to the head of the vector (the tip of the arrow) and the tail
of the vector (the other end) A vector represented by an arrow whose tail is at A= (a1, a2) and whose head is at B= (b1, b2)can be denoted by−−−→AB
Trang 2head
v
→
tail
v
→
A = (a1, a2)
B = (b1, b2)
|b1– a1|
|b2– a2|
Figure 20.46
Using the Pythagorean Theorem we calculate the length of−−−→ABas
(b2− a2)2+ (b1− a1)2 Two vectors,u and v are equivalent if they have the same magnitude and the same direction.
For example, all the vectors drawn in Figure 20.47 are equivalent
2
1
1
2
y
v
→
v
→
v
→
Figure 20.47
The simplest way of determining that these vectors are all equivalent is to notice that each is constructed (from tail to head) by a displacement of−2 units in the horizontal direction followed by a displacement of+1 unit in the vertical direction We say that each
of these vectors has a horizontal component of −2 and a vertical component of +1 The
horizontal and vertical components of a vector together completely determine the vector They determine its direction, and its magnitude is given by
(horizontal component)2+ (vertical component)2
Horizontal and Vertical Components of Vectors
Consider the vector−−−→OP, where O= (0, 0) and P is a point on the unit circle The length
of−−−→OP is 1 Let’s suppose that −−−→OP makes an angle of θ with the positive x-axis, where
θ∈ [0, 2π] and θ is swept out counterclockwise from the positive x-axis The coordinates
of P are (cos θ , sin θ ), where cos θ is the horizontal component of−−−→OP and sin θ is the vertical component of−−−→OP
Trang 3x
P
1
1
O
θ
Figure 20.48
Given any vectorv of length L we can position v so that its tail is at the origin Its head
lies somewhere on a circle of radius L centered at the origin, as shown in Figure 20.49 Let
θbe the anglev makes with the positive x-axis, where θ ∈ [0, 2π].
The horizontal component ofv = L cos θ, and
the vertical component ofv = L sin θ.
θ
y
x L
v
Lsin θ
Lcos θ
→
Figure 20.49
If the tail of a vector is at the origin, the horizontal component is the x-coordinate of the head and the vertical component is the y-coordinate of the head
EXERCISE 20.10 Dis the displacement vector for a turtle Its horizontal component is+3 and its vertical
component is−4
(a) How far has the turtle moved?
(b) If the turtle started at (1, 3), where did it end up?
ANSWERS
Trang 4EXERCISE 20.11 v is the velocity vector at a certain instant for an object in motion At this instant the object
is traveling southwest at 50 miles per hour Let due east correspond to the positive x-axis What are the horizontal and vertical components of the velocity?
v
→
horizontal component
vertical component
Figure 20.50
EXAMPLE 20.15 A wheel 60 inches in diameter is oriented vertically and spinning steadily at a rate of 1
revolution every π/2 minutes There’s a piece of gum stuck on the rim Let−−−−→G(t )be the position vector pointing from the center of the wheel to the wad of gum At time t= 0, the vector−−−−→G(t )is pointing in the same direction as the positive x-axis Find a formula for h(t), the vertical component of−−−−→G(t )
SOLUTION This is simply the Ferris wheel question formulated using different language
30
t
G→(t )
30 sin t
Figure 20.51
Since h(0)= 0, a formula of the form 30 sin(Bt) can be used One revolution is completed
in π/2 minutes, so the period is π/2.2πB = π/2 => B = 4
h(t )= 30 sin(4t) The purpose of Example 20.15 is to show that we already know how to find the vertical and horizontal components of a vector Given a vector (such as a velocity vector, force vector, or displacement vector), physicists, engineers, and architects often find it useful to resolve15the vector into its vertical and horizontal components The next example asks for the resolution of a force vector into its horizontal and vertical components
EXAMPLE 20.16 A suitcase is being pulled along a horizontal airport floor A force of 50 pounds is being
applied at an angle of 35◦with the horizontal
(a) What is the component of the force in the direction of motion?
(b) What is the component of the force perpendicular to the direction of motion?
15
Trang 5SOLUTION (a) Let x= the component of the force in the direction of motion.
cos 35◦= x
50
x= 50 cos 35◦
x= 50 cos
35◦ π
180◦
≈ 40.96 Approximately 40.96 pounds of force are being applied in the direction of motion
x ≈40.96 lbs 35°
50 lbs
y ≈28.68 lbs
Figure 20.52
(b) Let y= the component of the force perpendicular to the direction of motion
sin 35◦= y
50
y= 50 sin 35◦
y≈ 28.68 Approximately 28.68 pounds of force are being applied perpendicular to the direction
of motion What is the significance of what we have calculated? We’ve resolved the force into two perpendicular components If a force of (50 sin 35◦) pounds is applied upwards and simultaneously a force of (50 cos 35◦) pounds is applied in the direction of motion, the sum
of these two forces is equivalent to the original force
EXAMPLE 20.17 Consider the trajectory made by some thrown or launched object The object could be a
tennis ball, a javelin, a basketball, or a stone The exact nature of the trajectory is of critical importance to athletes and marksmen alike If we neglect air resistance and spin and consider only the force of gravity, then the trajectory of the object is determined by the initial velocity
of the object, given by the velocity vectorv0 Let’s denote the length ofv0by v0(with no arrow above it) v0is the initial speed We’ll denote the angle of launch by θ , where θ is in the interval (0, π/2) For simplicity’s sake, we’ll assume the object is launched from ground level The only force acting on the object is the force of gravity, resulting in a downward acceleration of g meters per second squared
(a) Find the vertical position of the object at time t
(b) When will the object hit the ground?
(c) How far away will the object be from its launching spot when it hits the ground?
Trang 60
v
0 sin θ θ
→
Figure 20.53
SOLUTION (a) To analyze the vertical position of the object at time t, we can restrict our attention to
the vertical component of velocity We can express the vertical component of the initial velocity,v0, as v0sin θ
Let v(t) be the vertical velocity of the object (in meters per second) at time t and let s(t )be the vertical position (in meters) of the object at time t
Let t= 0 be the moment of launch
The vertical acceleration of the object is −g, where g is the gravitational constant; therefore v(t )= −g
Our strategy is as follows:
Using information about acceleration, (the derivative of v(t)), find v(t)
Using information about velocity, (the derivative of s(t)), find s(t)
Let’s guess a function whose derivative is−g; we know that any other such function differs from it by a constant.16The derivative of−gt is −g so
v(t )= −gt + C We know that v(0)= v0sin θ ; this information is the initial
condition that allows us to solve for the constant C
v0sin θ= −g · 0 + C
v0sin θ= C (Keep in mind that v0sin θ is a constant! Both v0and θ are
fixed.) v(t )= −gt + v0sin θ
The velocity is the derivative of the displacement function; v(t)= s(t )
s(t )= −gt + v0sin θ Let’s guess a function whose derivative is−gt + v0sin θ ; we know that any other such function differs from it by a constant
s(t )= −gt 2
2 + (v0sin θ )t+ C1 We know that s(0)= 0; this information is the
initial condition that allows us to solve for the constant C1
0= 0 + 0 + C1
0= C Therefore s(t)= −gt2
2 + (v0sin θ )t
(b) The object hits the ground when s(t)= 0, so we must set s(t) equal to zero
−gt
2
2 + (v0sin θ )t= 0 Solve for t
16
Trang 7t (−gt
2+ v0sin θ )= 0
t= 0 or gt
2 = v0sin θ
t= 0 or t=2v0sin θ
g The object hits the ground2v0 sin θ
g seconds after it is launched
(c) To find the horizontal distance the object will travel we direct our attention to the horizontal component of velocity Unlike the vertical component of velocity (which
is affected by the force of gravity), the horizontal component of velocity is constant Therefore, the horizontal distance traveled is given by
(the horizontal component of velocity)· (time) The horizontal component of v 0 is
v0cos θ
v0
v0 cos θ θ
→
Figure 20.54
The time the object travels is the time between launch and when it hits the ground:
2v 0 sin θ
g seconds
horizontal distance traveled= v0cos θ·2v0sin θ
2
0cos θ sin θ g
So, for instance, if an object is launched at ground level at an angle of 30◦and with an initial speed of 28 meters per second it will hit the ground after2(28 m/sec) sin(π/6)
9.8m/sec 2 ≈ 2.86 sec It travels a horizontal distance of (28m/sec) cos(π/6)·2(28 m/sec) sin(π/6)9.8m/sec2 ≈ 69.28
EXERCISE 20.12 Suppose an object is launched from the ground with an initial velocity of 96 feet per second
at an angle of 40◦ The gravitational constant is 32 ft/sec2 Answer questions (a), (b), and (c) from the last example by working through all the steps using these concrete numbers
ANSWERS (a) v(t)= −32t + 96 sin 40◦, so s(t)= −16t2+ (96 sin 40◦)t+ 5
s(t )≈ −16t2+ 61.7t + 5 in ft/sec (b) The object hits the ground after about 3.936 seconds
(c)≈ 289.463 feet
Finding the Component of a Vector in an Arbitrary Direction
In many situations it is useful to find the component of a vector not in the horizontal and vertical directions but in some other direction For instance, a sailor might be interested
in the component of the wind’s velocity in the direction of his boat’s motion If an object
Trang 8is being pulled up an incline we are interested in the component of the force applied in the direction of the object’s motion A swimmer swimming in a current is interested in the component of the force of the water in her direction of motion
The component of v in the direction of u gives a measure of how much of v is in
the direction ofu Below we give some examples (Recall that |v| denotes the length or
magnitude ofv.)
The component ofv
in the direction ofu
is
|v| if u and v point in the same direction
−|v| if u and v point in the opposite directions.
0 ifu and v are perpendicular.
We denote the component ofv in the direction of u by vu (It will be a number, not a vector, so we don’t put an arrow over the v.) To visualize vu, begin by placing the tails of the vectorsv and u together Let θ be the angle between u and v We’ll measure θ as an
angle between 0 and π ; there is no need to distinguish between its terminal and initial sides
The direction ofu matters but its length does not If the angle between u and v is acute,
then vuis positive; if the angle is obtuse, thenv is not in the direction of u, and vuis negative You can visualize what is meant by the component ofv in the direction of u in the examples
shown in Figure 20.55 by tilting your head so thatu looks horizontal and visualizing the
horizontal component ofv.
v
v u
→
v
→
v
→
v
→
u
→
u
→
u
→
u
→
→
v u→
v u→
v u→
(the component of
v in the direction of u)
is negative
Figure 20.55
In the examples shown in Figure 20.56 it is easier to tilt your head so that you’re letting
u be vertical and visualize the vertical component of v.
v
→
v
→
v
→
→
→
u
→
u
→
u
→
v u→
here the component
of v in the direction
of u is negative
Figure 20.56
Trang 9Defining and computing the component of v in the direction of u
The component ofv in the direction of u, vuis defined as vu= |v| cos θ,
where θ is the angle betweenu and v.
v
→
v
→
u
→
u
→
→
v u→
v u→
v u→
cos θ =
length of v
→
v u→
cos θ =
| v |
Figure 20.57
Observations
The definition of vuinvolves the length ofv but not the length of u This is in line with
our earlier remark that the length ofu is irrelevant; only the direction of u matters.
If θ∈ [0, π/2), then cos θ is positive; if θ ∈ (π/2, π], then cos θ is negative This is exactly what we want
|v| cos θ
has a maximum value of|v| when θ = 0. uandvhave the
same direction. has a minimum value of − |v| when θ = π. uandvhave
opposite directions
This is in agreement with our intuition
EXAMPLE 20.18 A suitcase is being pulled up a ramp that makes a 10◦angle with the horizontal A force of
50 pounds is applied at an angle of 35◦with the horizontal What is the component of the force in the direction of motion?
SOLUTION Letu be a vector in the direction of motion and F be the force vector |F| = 50 The angle
betweenu and F is 25◦ Let x= the component of the force in the direction of motion
x= 50 cos 25◦
≈ 45.3
v
→
50 25°
10°
ramp
Figure 20.58
Approximately 45.3 pounds of force are being applied in the direction of motion
Trang 10EXERCISE 20.13 The angle between vectorsu and v is 2π/3 |u| = 3 and |v| = 8.
(a) Find the component ofv in the direction of u Give an exact answer Accompany your
answer by a sketch
(b) Find the component ofu in the direction of v Accompany your answer by a sketch.
P R O B L E M S F O R S E C T I O N 2 0 7
1.v is a vector of length 3 When its tail is at the origin, it makes an angle of 60◦with the positive x-axis What is the horizontal component ofv? The vertical component of v?
For Problems 2 through 5, vectors u and v are described Find
(a) the component of v in the direction of u.
(b) the component of u in the direction of v.
2.|u| = 5, |v| = 7, and the angle between u and v is π
6
3.|u| = 5, |v| = 7, and the angle between u and v is π
3
4.|u| = 5, |v| = 7, and the angle between u and v is 2π3
5.u is a vector of length 2 directed due south v is a vector of length 3 directed northeast.
6 Supposev is a vector of length 8 and the component of v in the direction of u is 4.
(a) Can the angle betweenu and v be determined? If so, what is it?
(b) Can the direction ofu be determined? If so, what is it?
(c) Can the length ofu be determined? If so, what is it?
7 You’re pushing an object along a table, exerting a force of 10 pounds in the direction indicated below What is the component of force in the direction of motion?
direction
of motion
20°
8 A plane is traveling 300 miles per hour There is a 50-mph wind The angle between the velocity vector of the wind and the velocity vector of the plane is 110◦
(a) What is the component of the direction of the plane’s motion?
(b) In the absence of the wind but all else remaining the same, how fast would the plane be traveling?
9 Suppose an object is launched from a height of 64 feet with an initial velocity of 96 feet per second at an angle of π/6 radians Assume that the only force acting on the object is the force of gravity, which results in a downward acceleration of 32 ft/sec (a) Find the vertical position of the object at time t
(b) When will the object hit the ground?