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CHAPTER 1
INTRODUCTION AND
MATHEMATICAL CONCEPTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
- (d) The resultant vector R is drawn from the tail of the first vector to the head of the last
vector.
- (c) Note from the drawing that the magnitude R of the resultant vector R is equal to the
shortest distance between the tail of A and the head of B. Thus, R is less than the magnitude
(length) of A plus the magnitude of B.
- (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides are
known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
- (b) The angle is found by using the inverse tangent function,
1 4.0 km
tan 53
3.0 km
θ
− ^ = = °
- (b) In this drawing the vector – C is reversed relative to C , while vectors A and B are not
reversed.
- (c) In this drawing the vectors – B and – C are reversed relative to B and C , while vector A
is not reversed.
- (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
- (c) When the two vector components A x
and A y
are added by the tail-to-head method, the
sum equals the vector A. Therefore, these vector components are the correct ones.
- (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as
2 2
x y
A = A + A. If A x
and A y
double in size, then the magnitude of
A doubles: ( ) ( )
2 2 2 2 2 2 2 2 4 4 2 2. x y x y x y
A + A = A + A = A + A = A
- (a) The angle θ is determined by the inverse tangent function,
1 tan
y
x
A
A
θ
−
. If A x
and
A
y
both become twice as large, the ratio does not change, and θ remains the same.
- (b) The displacement vector A points in the – y direction. Therefore, it has no scalar
component along the x axis ( A x
= 0 m) and its scalar component along the y axis is negative.
2 INTRODUCTION AND MATHEMATICAL CONCEPTS
- (e) The scalar components are given by A x ′
= −(450 m) sin 35.0° = −258 m and A y ′
= −(450 m) cos 35.0° = −369 m.
- (d) The distance (magnitude) traveled by each runner is the same, but the directions are
different. Therefore, the two displacement vectors are not equal.
- (c) A x
and B x
point in opposite directions, and A y
and B y
point in the same direction.
- (d)
16. A
y
= 3.4 m, B y
= 3.4 m
17. R
x
= 0 m, R y
= 6.8 m
- R = 7.9 m, θ = 21 degrees
4 INTRODUCTION AND MATHEMATICAL CONCEPTS
4. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need
to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km =
1 mi (see the page facing the inside of the front cover of the text).
b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the conversions
1 mi = 1609 m and 1 h = 3600 s.
SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi)
= 1, we find the speed of the bicyclists is
( )
mi mi
Speed = 34.0 1 34.
h
1.609 km
h 1 mi
km
h
b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is
( )( )
mi mi
Speed = 34.0 1 1 34.
h
h
1609 m
1 mi
1 h
m
3600s s
______________________________________________________________________________
- REASONING In order to calculate d , the units of a and b must be, respectively, cubed and
squared along with their numerical values, then combined algebraically with each other and
the units of c. Ignoring the values and working first with the units alone, we have
( )
( ) ( )
3 3 3
2 2
m (^) m
=
m/s s
a
d
cb
2
m / s ( )
2 ⋅ s
2
1
m
=
s
Therefore, the units of d are m
2 /s.
SOLUTION With the units known, the numerical value may be calculated:
( )
( ) ( )
3
2 2
2
m /s 0.75 m /s
d = =
- REASONING AND SOLUTION x has the dimensions of [L], v has the dimensions of
[L]/[T], and a has the dimensions of [L]/[T]
2
. The equation under consideration is v
n = 2 ax.
Chapter 1 Problems 5
The dimensions of the right hand side are
L
T
L
L
T
2
2
2
= , while the dimensions of the left
hand side are
[ ]
[ ]
[ ]
[ ]
L L
T
T
n (^) n
n
. The right side will equal the left side only when n = 2.
- SSM REASONING This problem involves using unit conversions to determine the
number of magnums in one jeroboam. The necessary relationships are
1.0 magnum = 1.5 liters
1.0 jeroboam = 0.792 U. S. gallons
1.00 U. S. gallon = 3.785 × 10
3 = 3.785 liters
These relationships may be used to construct the appropriate conversion factors.
SOLUTION By multiplying one jeroboam by the appropriate conversion factors we can
determine the number of magnums in a jeroboam as shown below:
1.0 jeroboam ( )
0.792 gallons
1.0 jeroboam
3.785 liters
1.0 gallon
1.0 magnum
1.5 liters
2.0 magnums
- REASONING In the expression for the volume flow rate, the dimensions on the left side of
the equals sign are [L]
3 /[T]. If the expression is to be valid, the dimensions on the right side
of the equals sign must also be [L]
3 /[T]. Thus, the dimensions for the various symbols on
the right must combine algebraically to yield [L]
3 /[T]. We will substitute the dimensions for
each symbol in the expression and treat the dimensions of [M], [L], and [T] as algebraic
variables, solving the resulting equation for the value of the exponent n.
SOLUTION We begin by noting that the symbol π and the number 8 have no dimensions.
It follows, then, that
( ) (^) [ ]
[ ]
[ ]
[ ]
3
2 1
M
L
L
or
8 T
n
n R P P
Q
L
π
η
[ ][ ]
[ ]
2
L T
M
[ L]^ [ ]
[ L]
T
[ ] [ ]
[ ][ ]
[ ]
[ ][ ]
[ ]
[ ]
2
3
L T L
L T
L T
L
T
n n
[ ]
[ ] [ ]
L
L T
n
= (^) [ ]
[ ]
[ ]
[ ] [ ] [ ] [ ]
3 L 3 4
or L or L L L L
L
n
n
= = =
Chapter 1 Problems 7
- SSM REASONING The dimension of the spring constant k can be determined by first
solving the equation T = 2 π m / k for k in terms of the time T and the mass m. Then, the
dimensions of T and m can be substituted into this expression to yield the dimension of k.
SOLUTION Algebraically solving the expression above for k gives
2 2 k = 4 π m T /. The
term
2 4 π is a numerical factor that does not have a dimension, so it can be ignored in this
analysis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is
[ ]
[ ]
2
M
Dimension of
T
k =
- REASONING The shortest distance between the tree and the termite mound is equal to the
magnitude of the chimpanzee's displacement r.
SOLUTION
a. From the Pythagorean theorem, we have
r = (51 m) + (39 m) = 64 m
2 2
b. The angle θ is given by
θ =
F
H
G
I
K
J=^ °
− tan
39 m
51 m
37 south of east
1
- SSM WWW REASONING The shortest distance between the two towns is along the
line that joins them. This distance, h , is the hypotenuse of a right triangle whose other sides
are h o
= 35.0 km and h a
= 72.0 km, as shown in the figure below.
SOLUTION The angle θ is given by tan θ = h o
/ h a
so that
θ = tan
− 1
- 0 km
72.0 km
= 25.9° S of W
We can then use the Pythagorean theorem to find h.
h h h o a
2 2 (35.0 km) 72 0 km) 80 1km
2 2 (..
51 m
r
θ
39 m
W
S
θ
o
h
a
h
h
θ
8 INTRODUCTION AND MATHEMATICAL CONCEPTS
- REASONING When the monkey has climbed as far up the
pole as it can, its leash is taut, making a straight line from the
stake to the monkey, that is, L = 3.40 m long. The leash is
the hypotenuse of a right triangle, and the other sides are a
line drawn from the stake to the base of the pole
( d = 3.00 m), and a line from the base of the pole to the
monkey (height = h ).
SOLUTION These three lengths are related by the Pythagorean theorem (Equation 1.7):
( ) ( )
2 2 2 2 2 2
2 2 2 2
or
3.40 m 3.00 m 1.6 m
h d L h L d
h L d
- REASONING Using the Pythagorean theorem (Equation 1.7), we find that the relation
between the length D of the diagonal of the square (which is also the diameter of the circle)
and the length L of one side of the square is
2 2 D = L + L = 2 L.
SOLUTION Using the above relation, we have
0.35 m
2 or 0.25 m
D
D = L L = = =
______________________________________________________________________________
- REASONING In both parts of the drawing the line of sight, the horizontal dashed line, and
the vertical form a right triangle. The angles θ a
= 35.0° and θ b
= 38.0° at which the person’s
line of sight rises above the horizontal are known, as is the horizontal distance d = 85.0 m
from the building. The unknown vertical sides of the right triangles correspond,
respectively, to the heights H a
and H b
of the bottom and top of the antenna relative to the
person’s eyes. The antenna’s height H is the difference between H b
and H a
b a
H = H − H.
The horizontal side d of the triangle is adjacent to the angles θ a
and θ b
, while the vertical
sides H a
and H b
are opposite these angles. Thus, in either triangle, the angle θ is related to
the horizontal and vertical sides by Equation 1.
o
a
tan
h
h
θ
a
a
tan
H
d
θ = (1)
b
b
tan
H
d
θ = (2)
d
h
L
Stake
10 INTRODUCTION AND MATHEMATICAL CONCEPTS
The required distance c is also found using the Pythagorean theorem.
c
2 = L
2
2 = (0.397 nm)
2
2 = 0.237 nm
2
Then,
c = 0.487 nm
- REASONING The drawing shows the heights of the two
balloonists and the horizontal distance x between them.
Also shown in dashed lines is a right triangle, one angle of
which is 13.3°. Note that the side adjacent to the 13.3°
angle is the horizontal distance x , while the side opposite
the angle is the distance between the two heights, 61.0 m −
48.2 m. Since we know the angle and the length of one side
of the right triangle, we can use trigonometry to find the
length of the other side.
SOLUTION The definition of the tangent function, Equation 1.3, can be used to find the
horizontal distance x , since the angle and the length of the opposite side are known:
length of opposite side
tan13.
length of adjacent side (= x )
Solving for x gives
length of opposite side 61.0 m 48.2 m
54.1 m
tan13.3 tan13.
x
- REASONING Note from the drawing that the shaded
right triangle contains the angle θ , the side opposite the
angle (length = 0.281 nm), and the side adjacent to the
angle (length = L ). If the length L can be determined, we
can use trigonometry to find θ. The bottom face of the
cube is a square whose diagonal has a length L. This
length can be found from the Pythagorean theorem,
since the lengths of the two sides of the square are
known.
SOLUTION The angle can be obtained from the inverse tangent function, Equation 1.6, as
( )
1 θ tan 0.281 nm / L
− =
. Since L is the length of the hypotenuse of a right triangle whose
sides have lengths of 0.281 nm, its value can be determined from the Pythagorean theorem:
( ) ( )
2 2 L = 0.281 nm + 0.281 nm =0.397 nm
48.2 m
61.0 m
13.3°
x
0.281 nm
0.281 nm
0.281 nm
θθ θθ
L
Chapter 1 Problems 11
Thus, the angle is
1 1 0.281 nm 0.281 nm
tan = tan 35.
L 0.397 nm
θ
− ^ ^ − ^
______________________________________________________________________________
- REASONING There are two right
triangles in the drawing. Each
contains the common side that is
shown as a dashed line and is
labeled D , which is the distance
between the buildings. The
hypotenuse of each triangle is one
of the lines of sight to the top and
base of the taller building. The
remaining (vertical) sides of the
triangles are labeled H 1
and H 2
Since the height of the taller building is H 1
+ H
2
and the height of the shorter building is H 1
the ratio that we seek is ( H 1
+ H
2
)/ H
1
. We will use the tangent function to express H 1
in
terms of the 52° angle and to express H 2
in terms of the 21° angle. The unknown distance D
will be eliminated algebraically when the ratio ( H 1
+ H
2
)/ H
1
is calculated.
SOLUTION The ratio of the building heights is
1 2
1
Height of taller building
Height of shorter building
H H
H
Using the tangent function, we have that
1
1
2
2
tan 52 or tan 52
tan 21 or tan 21
H
H D
D
H
H D
D
Substituting these results into the expression for the ratio of the heights gives
1 2
1
Height of taller building tan 52 tan 21
Height of shorter building tan 52
tan 21
1 1.
tan 52
H H D D
H D
21 °
52 ° D
H
1
H
2
Chapter 1 Problems 13
θ 41°
L
41°
θ
L
0.75 m
h 1
h 2
- REASONING The trapeze cord is L =
8.0 m long, so the trapeze is initially h 1
L cos 41° meters below the support. At
the instant he releases the trapeze, it is h 2
= L cos θ meters below the support. The
difference in heights is d = h 2
0.75 m. Given that the trapeze is released
at a lower elevation than the platform,
we expect to find θ < 41°.
SOLUTION Putting the above relationships together, we have
2 1
1 1
cos cos 41 or cos 41 cos
cos cos 41
0.75 m
cos cos 41 cos cos 41 32
8.0 m
d h h L L d L L
d
L
d
L
θ θ
θ
θ
− −
- SSM REASONING AND SOLUTION A single rope must supply the resultant of the
two forces. Since the forces are perpendicular, the magnitude of the resultant can be found
from the Pythagorean theorem.
a. Applying the Pythagorean theorem,
F = ( 475 ) = 5 70. × 10
2 N + (315 N)
2 2 N
b. The angle θ that the resultant makes with the
westward direction is
θ =
F
H
G
I
K
J=^ °
− tan.
1
N
475 N
Thus, the rope must make an angle of 33.6° south of west.
θ
F
475 N
315 N
W
S
14 INTRODUCTION AND MATHEMATICAL CONCEPTS
- REASONING The Pythagorean theorem (Equation 1.7) can be used to find the magnitude
of the resultant vector, and trigonometry can be employed to determine its direction.
a. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the
resultant a westerly direction and vector B gives the resultant a southerly direction.
Therefore, the resultant A + B points south of west.
b. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the
resultant a westerly direction and vector – B gives the resultant a northerly direction.
Therefore, the resultant A + (– B) points north of west.
SOLUTION Using the Pythagorean theorem and trigonometry, we obtain the following
results:
( ) ( )
( ) ( )
2 2
1
2
a. Magnitude of 63 units 63 units 89 units
63 units
tan 45 south of west
63 units
b. Magnitude of 63 units 63 units
θ
−
A B
A B
2
1
89 units
63 units
tan 45 north of west
63 units
θ
−
______________________________________________________________________________
25. SSM WWW REASONING
a. Since the two force vectors A and B have directions due west and due north, they are
perpendicular. Therefore, the resultant vector F = A + B has a magnitude given by the
Pythagorean theorem: F
2 = A
2
2
. Knowing the magnitudes of A and B , we can calculate
the magnitude of F. The direction of the resultant can be obtained using trigonometry.
b. For the vector F ′′′′ = A – B we note that the subtraction can be regarded as an addition in
the following sense: F ′′′′ = A + (–B). The vector –B points due south, opposite the vector B ,
so the two vectors are once again perpendicular and the magnitude of F ′′′′ again is given by
the Pythagorean theorem. The direction again can be obtained using trigonometry.
16 INTRODUCTION AND MATHEMATICAL CONCEPTS
2 2 2 d = s + s = 2 s = s 2 (2)
SOLUTION First, we apply the Pythagorean theorem to the right triangle formed by the
three displacement vectors, using Equations (1) for J and K :
( ) ( )
2 2 2 2 2 2 2 D = J + K = 4 d + 2 d = 16 d + 4 d = 20 d = d 20 (3)
Substituting Equation (2) into Equation (3) gives
( ) (^ ) D = d 20 = s 2 20 = s 40 = 4.0 cm 40 = 25 cm
- REASONING For convenience, we can assign due east to be the positive direction and due
west to be the negative direction. Since all the vectors point along the same east-west line,
the vectors can be added just like the usual algebraic addition of positive and negative
scalars. We will carry out the addition for all of the possible choices for the two vectors and
identify the resultants with the smallest and largest magnitudes.
SOLUTION There are six possible choices for the two vectors, leading to the following
resultant vectors:
50.0 newtons 10.0 newtons 60.0 newtons 60.0 newtons, due east
50.0 newtons 40.0 newtons 10.0 newtons 10.0 newtons, due east
50.0 newtons 30.0 newtons 20.0 newtons 20.0 newtons, due
1 2
1 3
1 4
F F
F F
F F east
10.0 newtons 40.0 newtons 30.0 newtons 30.0 newtons, due west
10.0 newtons 30.0 newtons 20.0 newtons 20.0 newtons, due west
40.0 newtons 30.0 newtons 70.0 newtons 70.0 newtons,
2 3
2 4
3 4
F F
F F
F F due west
The resultant vector with the smallest magnitude is + =10.0 newtons, due east 1 3
F F.
The resultant vector with the largest magnitude is + =70.0 newtons, due west 3 4
F F.
Chapter 1 Problems 17
- REASONING Both P and Q and the
vector sums K and M can be drawn with
correct magnitudes and directions by
counting grid squares. To add vectors,
place them tail-to-head and draw the
resultant vector from the tail of the first
vector to the head of the last. The vector
2 P is equivalent to P + P , and − Q is a
vector that has the same magnitude as Q ,
except it is directed in the opposite
direction.
The vector M runs 11 squares
horizontally and 3 squares vertically, and
the vector K runs 4 squares horizontally
and 9 squares vertically. These distances can be converted from grid squares to centimeters
with the grid scale: 1 square = 4.00 cm. Once the distances are calculated in centimeters, the
Pythagorean theorem (Equation 1.7) will give the magnitudes of the vectors.
SOLUTION
a. The vector M = P + Q runs 11 squares horizontally and 3 squares vertically, and these
distances are equivalent to, respectively, (^) ( )
cm
4.00 11 squares 44.0 cm
square
and
( )
cm
4.00 3 squares 12.0 cm
square
. Thus, the magnitude of M is
( ) ( )
2 2 M = 44.0 cm + 12.0 cm = 45.6 cm
b. Similarly, the lengths of the horizontal and vertical distances of K = 2 P − Q are
4 horizontal squares and 9 vertical squares, or 16.0 cm and 36.0 cm, respectively. The
magnitude of K is then
( ) ( )
2 2 K = 16.0 cm + 36.0 cm =39.4 cm
8.00 cm
P
P
− Q
Q
M
K
P
Chapter 1 Problems 19
Using trigonometry, we can see that the direction of the resultant is
tan tan
θ= θ
F
H
G
I
K
J=^ °
−
B
A
or =
2.8 km
km
east of south
1
b. Referring to the drawing and following the same procedure as in part a, we find
F
H
G
I
K
J=^ °
−
R A B B R A
B
A
2 2 2 2 2 2 2
1
b g or^ b km^ g b km^ g km
or =
2.8 km
km
west of south
tan tan
θ θ
- SSM WWW REASONING AND SOLUTION The single force needed to produce
the same effect is equal to the resultant of the forces provided by the two ropes. The figure
below shows the force vectors drawn to scale and arranged tail to head. The magnitude and
direction of the resultant can be found by direct measurement using the scale factor shown
in the figure.
a. From the figure, the magnitude of the resultant is 5600 N.
b. The single rope should be directed along the dashed line in the text drawing.
32. REASONING
a. and b. The drawing shows the two vectors A and B , as well as the resultant vector A + B.
The three vectors form a right triangle, of which two of the sides are known. We can employ
the Pythagorean theorem, Equation 1.7, to find the length of the third side. The angle θ in
the drawing can be determined by using the inverse cosine function, Equation 1.5, since the
side adjacent to θ and the length of the hypotenuse are known.
2900 N 2900 N
Resultant
Scale:
1000 N
N
E
S
W
A + B
B
A
units
12.3 units
θ
20 INTRODUCTION AND MATHEMATICAL CONCEPTS
c. and d. The drawing illustrates the two vectors A and − B , as well as the resultant vector
A − B. The three vectors form a right triangle, which is identical to the one above, except
for the orientation. Therefore, the lengths of the hypotenuses and the angles are equal.
SOLUTION
a. Let R = A + B. The Pythagorean theorem (Equation 1.7) states that the square of the
hypotenuse is equal to the sum of the squares of the sides, so that
2 2 2 R = A + B. Solving for
B yields
( ) ( )
2 2 2 2 B = R − A = 15.0 units − 12.3 units =8.6 units
b. The angle θ can be found from the inverse cosine function, Equation 1.5:
1 12.3 units cos 34.9 north of west
15.0 units
θ
− ^
c. Except for orientation, the triangles in the two drawings are the same. Thus, the value for
B is the same as that determined in part (a) above: B =8.6 units
d. The angle θ is the same as that found in part (a), except the resultant vector points south
of west, rather than north of west: (^) θ = 34.9 °south of west
______________________________________________________________________________
- REASONING AND SOLUTION The figures below are scale diagrams of the forces drawn
tail-to-head. The scale factor is shown in the figure.
a. From the figure on the left, we see that F A
− F
B
= 142 N, θ = 67° south of east .
b. Similarly, from the figure on the right, F B
− F
A
= 142 N, θ = 67° north of west .
N
E
S
W
A −−−− B −−−− B
A
units
12.3 units
θ
