Problem 1.11 The kinetic energy of a particle of mass
m is defined to be 12 mv 2 , where v is the magnitude of
the particle’s velocity. If the value of the kinetic energy
of a particle at a given time is 200 when m is in kilograms
and v is in meters per second, what is the value when m
is in slugs and v is in feet per second?
Solution:
200
kg-m2
s2
0.0685 slug
1 kg
= 147.46 = 147
1 ft
0.3048 m
2
slug-ft2
s2
Problem 1.12 The acceleration due to gravity at sea
level in SI units is g = 9.81 m/s2 . By converting units,
use this value to determine the acceleration due to gravity
at sea level in U.S. Customary units.
Solution:
Use Table 1.2. The result is:
m
1 ft
ft
ft
g = 9.81
=
32.2
=
32.185
.
.
.
s2
0.3048 m
s2
s2
Problem 1.13 A furlong per fortnight is a facetious
unit of velocity, perhaps made up by a student as a satirical comment on the bewildering variety of units engineers must deal with. A furlong is 660 ft (1/8 mile). A
fortnight is 2 weeks (14 days). If you walk to class at
2 m/s, what is your speed in furlongs per fortnight to
three significant digits?
Solution:
Convert the units using the given conversions. Record the first three
digits on the left, and add zeros as required by the number of tens in
the exponent. The result is:
ft
1 furlong
3600 s
24 hr
14 day
5
s
660 ft
1 hr
1 day
1 fortnight
furlongs
= 9160
fortnight
Problem 1.14 The cross-sectional area of a beam is
480 in2 . What is its cross-section in m2 ?
Solution:
Convert units using Table 1.2. The result:
1 ft 2 0.3048 m 2
480 in2
= 0.30967 . . . m2 = 0.310 m2
12 in
1 ft
Problem 1.15 At sea level, the weight density (weight
per unit volume) of water is approximately 62.4 lb/ft3 .
1 lb = 4.448 N, 1 ft = 0.3048 m, and g = 9.81 m/s2 .
Using only this information, determine the mass density
for water in kg/m3 .
Solution:
Get wt. density in N/m3 first.
3
lb
N
4.448 N
1 ft
62.4 3
= 9801.77
1 lb
0.3048 m
m3
ft
(carry extra significant figures till end—then round)
weight = mass · g
mass =
weight
g
9801.77
N
m3
s2
9.81 m
= 999
N-s2
m
1
m3
= 999 kg/m3
Problem 1.16 A pressure transducer measures a value
of 300 lb/in2 . Determine the value of the pressure in
pascals. A pascal (Pa) is one newton per meter squared.
Solution:
Convert the units using Table 1.2 and the definition of the Pascal unit.
The result:
2
lb
4.448 N
1 ft
12 in 2
300
2
1 lb
1 ft
0.3048 m
in
N
= 2.07(106 ) Pa
= 2.0683 . . . (106 )
m2
Problem 1.17 A horsepower is 550 ft-lb/s. A watt
is 1 N-m/s. Determine the number of watts generated
by (a) the Wright brothers’ 1903 airplane, which had a
12-horsepower engine; (b) a modern passenger jet with
a power of 100,000 horsepower at cruising speed.
Solution:
Convert units using inside front cover of textbook derive the conversion
between horsepower and watts. The result
746 watt
(a) 12 hp
= 8950 watt
1 hp
746 watt
(b) 105 hp
= 7.46(107 ) watt
1 hp
Problem 2.5 The magnitudes |FA | = 100 lb and
|FB | = 140 lb. If α can have any value, what are the
minimum and maximum possible values of the magnitude of the sum of the forces F = FA + FB , and what
are the corresponding values of α?
Solution: A graphical construction shows that the magnitude is
a minimum when the two force vectors are opposed, and a maximum
when both act in the same direction. The corresponding values are
|F|max = |FA + FB | = |100 + 140| = 240 lb, and α = 0◦ .
|F|min = |FA + FB | = |100 − 140| = 40 lb, and α = 180◦ .
Problem 2.6 The angle θ = 30◦ . What is the magnitude of the vector rAC ?
150 mm
60 mm
B
rAB
rBC
C
A
rAC
Solution:
B
150
mm
60
30°
A
mm
rAC
C
From the law of sines
BC
AB
AC
=
=
sin 30◦
sin α
sin γ
We know BC and AB. Thus
150
60
=
⇒ α = 11.54◦
sin 30◦
sin α
Also 30◦ + α + γ = 180◦ ⇒ γ = 138.46◦
Now, from the law of sines
150
AC
=
sin 30◦
sin 138.46◦
AC = |rAC | = 199 mm
Problem 2.17 Two snowcats tow a housing unit to a
new location at McMurdo Base, Antarctica. (The top
view is shown. The cables are horizontal.) The sum of
the forces FA and FB exerted on the unit is parallel to
the line L, and |FA | = 1000 lb. Determine |FB | and
|FA +FB | (a) graphically and (b) by using trigonometry.
L
50°
FA
30°
FB
TOP VIEW
Solution: The graphical construction is shown. The sum of
the interior angles must be 180◦ . (a) The magnitudes of |FB | and
|FA + FB | are determined from measurements. (b) The trigonometric solution is obtained from the law of sines:
L
50°
|FA + FB |
|FA |
|FB |
=
=
sin 100
sin 30
sin 50
sin 50
from which |FB | = |FA |
= 1000(1.532) = 1532 lb
sin 30
sin 100
|FA + FB | = |FA |
= 1000(1.9696) = 1970 lb
sin 30
30°
FA
TOP VIEW
FB
FA + FB
38°
38°
FB
156°
38°
50°
50°
FA
Problem 2.18 A surveyor determines that the horizontal distance from A to B is 400 m and that the horizontal
distance from A to C is 600 m. Determine the magnitude
of the horizontal vector rBC from B to C and the angle
α (a) graphically and (b) by using trigonometry.
North
B
α
rBC
C
60°
20°
East
A
Solution: (a) The graphical solution is obtained by drawing the
figure to scale and measuring the unknowns. (b) The trigonometric
solution is obtained by breaking the figure into three separate right
triangles. The magnitude |rBC | is obtained by the cosine law:
2
2
2
|rBC | = (400) + (600) − 2(400)(600) cos 40
or
B
F
◦
C
|rBC | = 390.25 = 390.3 m
The three right triangles are shown. The distance BD is BD =
(400) sin 60◦ = 346.41 m. The distance CE is CE =
600 sin 20◦ = 205.2 m. The distance F C is F C = (346.4 −
205.2) = 141.2 m.
The angle α is sin α =
141.2
390.3
= 0.36177 . . ., or α = 21.2◦
60°
20°
A
D
E
FB
Problem 2.23 A support is subjected to a force F =
Fx i + 80j (N). If the support will safely support a force
of 100 N, what is the allowable range of values of the
component Fx ?
Solution:
Use the definition of magnitude in Eq. (2.8) and reduce algebraically.
100 ≥ (Fx )2 + (80)2 , from which (100)2 − (80)2 ≥ (Fx )2 .
√
Thus |Fx | ≤ 3600, or −60 ≤ (Fx ) ≤ +60 (N)
Problem 2.24 If FA = 600i − 800j (kip) and FB = Solution: Take the scalar multiple of FB , add the components of the two
200i − 200j (kip), what is the magnitude of the force forces as in Eq. (2.9), and use the definition of the magnitude. F = (600 −
2(200))i + (−800 − 2(−200))j = 200i − 400j
F = FA − 2FB ?
|F| =
(200)2 + (−400)2 = 447.2 kip
Problem 2.25 If FA = i − 4.5j (kN) and FB = Solution: Take the scalar multiples and add the components.
−2i − 2j (kN), what is the magnitude of the force F = F = (6 + 4(−2))i + (6(−4.5) + 4(−2))j = −2i − 35j, and
6FA + 4FB ?
|F| =
(−2)2 + (−35)2 = 35.1 kN
Problem 2.26 Two perpendicular vectors U and V lie
in the x-y plane. The vector U = 6i − 8j and |V| = 20.
What are the components of V?
Solution: The two possible values of V are shown in the sketch.
The strategy is to (a) determine the unit vector associated with U,
(b) express this vector in terms of an angle, (c) add ±90◦ to this
angle, (d) determine the two unit vectors perpendicular to U, and
(e) calculate the components of the two possible values of V. The unit
vector parallel to U is
eU = 6i
62 + (−8)2
−
8j
62 + (−8)2
y
V2
6
V1
Expressed in terms of an angle,
U
◦
◦
eU = i cos α − j sin α = i cos(53.1 ) − j sin(53.1 )
Add ±90◦ to find the two unit vectors that are perpendicular to this
unit vector:
ep1 = i cos(143.1◦ ) − j sin(143.1◦ ) = −0.8i − 0.6j
ep2 = i cos(−36.9◦ ) − j sin(−36.9◦ ) = 0.8i + 0.6j
Take the scalar multiple of these unit vectors to find the two vectors
perpendicular to U.
V1 = |V|(−0.8i − 0.6j) = −16i − 12j.
The components are Vx = −16, Vy = −12
V2 = |V|(0.8i + 0.6j) = 16i + 12j.
The components are Vx = 16, Vy = 12
x
= 0.6i − 0.8j
8
Problem 2.34 (a) Express the position vector from
point A of the front-end loader to point B in terms of
components.
(b) Express the position vector from point B to point C
in terms of components.
(c) Use the results of (a) and (b) to determine the distance
from point A to point C.
y
98 in.
45 in.
C
B
A
55 in.
50 in.
35 in.
x
50 in.
The coordinates are A(50, 35); B(98, 50); C(45, 55).
The vector from point A to B:
Solution:
(a)
y
rAB = (98 − 50)i + (50 − 35)j = 48i + 15j (in.)
(b)
The vector from point B to C is
rBC = (45 − 98)i + (55 − 50)j = −53i + 5j (in.).
(c)
98 in.
45 in.
The distance from A to C is the magnitude of the sum of the
vectors,
C
55 in.
B
A
50 in.
35 in.
x
rAC = rAB + rBC = (48 − 53)i + (15 + 5)j = −5i + 20j.
The distance from A to C is
|rAC | = (−5)2 + (20)2 = 20.62 in.
50 in.
Problem 2.35 Consider the front-end loader in Problem 2.34. To raise the bucket, the operator increases the
length of the hydraulic cylinder AB. The distance between points B and C remains constant. If the length
of the cylinder AB is 65 in., what is the position vector
from point A to point B?
Solution: Assume that the two points A and C are fixed. The
strategy is to determine the unknown angle θ from the geometry.
From
Problem 2.34 |rAC | = 20.6 and the angle β is tan β = −20
=
5
√
◦
2
2
−4, β = 76 . Similarly, |rCB | = 53 + 5 = 53.2. The angle a
is found from the cosine law:
cos α =
(20.6)2 + (65)2 − (53.2)2
= 0.6776,
2(20.6)(65)
α = 47.3◦ . Thus the angle θ is
θ = 180◦ − 47.34◦ − 75.96◦ = 56.69 . . . = 56.7◦ . The vector
rAB = 65(i cos θ + j sin θ) = 35.69 . . . i + 54.32 . . . j
= 35.7i + 54.3j (in.)
B
53.2
C
20.6
α
β
A
65
θ
Problem 2.59 The cable from B to A on the sailboat
shown in Problem 2.41 exerts a 230-N force at B. The
cable from B to C exerts a 660-N force at B. What is
the magnitude of the total force exerted at B by the two
cables? What is the magnitude of the downward force
(parallel to the y axis) exerted by the two cables on the
boat’s mast?
Solution: Find unit vectors in the directions of the two forcesexpress the forces in terms of magnitudes times unit vectors-add the
forces.
Unit vectors:
eBA =
=
y
B (4, 13) m
rBA
(xA − xB )i + (yA − yB )j
= |rBA |
(xA − xB )2 + (yA − yB )2
(0 − 4)i + (1.2 − 13)j
√
42 + 11.82
eBA = −0.321i − 0.947j
Similarly,
eBC = 0.385i − 0.923j
FBA = |FBA |eBA = −73.8i − 217.8j kN
FBC = |FBC |eBC = 254.1i − 609.2j kN
Adding
F = FBA + FBC = 180.3i − 827j kN
|F| =
C
(9, 1) m
A
(0, 1.2) m
Fx2 + Fy2 = 846 kN (Total force)
Fy = −827 kN (downward force)
Problem 2.60 The structure shown forms part of a
truss designed by an architectural engineer to support
the roof of an orchestra shell. The members AB, AC,
and AD exert forces FAB , FAC , and FAD on the joint
A. The magnitude |FAB | = 4 kN. If the vector sum of
the three forces equals zero, what are the magnitudes of
FAC and FAD ?
Solution:
eAD = √
eAC = √
eAB = √
y
B
(– 4, 1) m
FAC
C
22 + 3 2
−4
42 + 1 2
4
42 + 2 2
i+ √
i+ √
i+ √
−3
22 + 3 2
1
42 + 1 2
2
42 + 2 2
D
j = −0.5547i − 0.8320j
(–2, – 3) m
j = −0.9701i + 0.2425j
j = 0.89443i + 0.4472j
The forces are FAD = |FAD |eAD , FAC = |FAC |eAC ,
FAB = |FAB |eAB = 3.578i + 1.789j. Since the vector sum of
the forces vanishes, the x- and y-components vanish separately:
Fx = (−0.5547|FAD | − 0.9701|FAC | + 3.578)i = 0, and
Fy = (−0.8320|FAD | + 0.2425|FAC | + 1.789)j = 0
These simultaneous equations in two unknowns can be solved by any
standard procedure. An HP-28S hand held calculator was used here:
The results: |FAC | = 2.108 kN , |FAD | = 2.764 kN
B
C
A
D
(4, 2) m
x
A
FAD
Determine the unit vectors parallel to each force:
−2
FAB
x
Problem 2.61 The distance s = 45 in.
(a) Determine the unit vector eBA that points from B
toward A.
(b) Use the unit vector you obtained in (a) to determine
the coordinates of the collar C.
y
The unit vector from B to A is the vector from B to A divided
by its magnitude. The vector from B to A is given by
Solution:
rBA = (xA − xB )i + (yA − yB )j or rBA = (14 − 75)i + (45 − 12)j in.
Hence, vector from B to A is given by rBA = (−61)i + (33)j in. The
magnitude of the vector from B to A is 69.4 in. and the unit vector from B
toward A is eBA = −0.880i + 0.476j.
y
A
A
(14, 45) in
(14, 45) in
s
C
C
B
(75, 12) in
x
s
B
(75, 12) in
x
Problem 2.62 In Problem 2.61, determine the x and y Solution: The coordinates of the point C are given by
coordinates of the collar C as functions of the distance s. xC = xB + s(−0.880) and yC = yB + s(0.476).
Thus, the coordinates of point C are xC = 75 − 0.880s in. and yC =
12+0.476s in. Note from the solution of Problem 2.61 above, 0 ≤ s ≤ 69.4 in.
Problem 2.63 The position vector r goes from point
A to a point on the straight line between B and C. Its
magnitude is |r| = 6 ft. Express r in terms of scalar
components.
y
B
(7, 9) ft
r
A (3, 5) ft
C
(12, 3) ft
Solution: Determine the perpendicular vector to the line BC
from point A, and then use this perpendicular to determine the angular orientation of the vector r. The vectors are
rAB = (7 − 3)i + (9 − 5)j = 4i + 4j,
|rAB | = 5.6568
rAC = (12 − 3)i + (3 − 5)j = 9i − 2j,
|rAC | = 9.2195
rBC = (12 − 7)i + (3 − 9)j = 5i − 6j,
|rBC | = 7.8102
The unit vector parallel to BC is
eBC =
rBC
= 0.6402i − 0.7682j = i cos 50.19◦ − j sin 50.19◦ .
|rBC |
Add ±90◦ to the angle to find the two possible perpendicular vectors:
x
where s is the semiperimeter, s = 12 (|rAC | + |rAB | + |rBC |). Substituting
values, s = 11.343, and area = 22.0 and the magnitude of the perpendicular
2(22)
is |rAP | = 7.8102 = 5.6333. The angle between the vector r and the perpendicular rAP is β = cos−1 5.6333
= 20.1◦ . Thus the angle between the
6
vector r and the x-axis is α = 39.8 ± 20.1 = 59.1◦ or 19.7◦ . The first angle
is ruled out because it causes the vector r to lie above the vector rAB , which is
at a 45◦ angle relative to the x-axis. Thus:
r = 6(i cos 19.7◦ + j sin 19.7◦ ) = 5.65i + 2.02j
eAP 1 = i cos 140.19◦ − j sin 140.19◦ , or
eAP 2 = i cos 39.8◦ + j sin 39.8◦ .
Choose the latter, since it points from A to the line.
Given the triangle defined by vertices A, B, C, then the magnitude of
the perpendicular corresponds to the altitude when the base is the line
2(area)
BC. The altitude is given by h =
. From geometry, the area
base
of a triangle with known sides is given by
area = s(s − |rBC |)(s − |rAC |)(s − |rAB |),
y
B[7,9]
P
r
A[3,5]
C[12,3]
x
Problem 2.76 The position vector from a point A to
a point B is 3i + 4j − 4k (ft). The position vector from
point A to point C is −3i + 13j − 2k. ft
(a) What is the distance from point B to point C?
(b) What are the direction cosines of the position vector
from point B to point C?
Solution:
The vector from point B to point C is rBC = rAC − rAB .
Thus
rBC = (−3 − 3)i + (13 − 4)j + (−2 − (−4))k = −6i + 9j + 2k.
The distance between
points B and C is
√
(a) |rBC | = 62 + 92 + 22 = 11 (ft). The direction cosines are
cos θx = −6
= −0.5454,
11
2
cos θz = 11
= 0.1818
(b)
cos θy =
9
11
= 0.8182,
Problem 2.77 A vector U = 3i − 2j + 6k. Deter- Solution: By definition, the unit vector is the vector whose components
mine the components of the unit vector that has the same are the direction cosines
√ of U. (See discussion following Eq. (2.15)). The
magnitude is |U| = 32 + 22 + 62 = 7. Thus the unit vector is
direction as U.
u=
U
|U|
=
3
i
7
− 72 j + 67 k
Problem 2.78 A force vector F = 3i − 4j − 2k (N). Solution: By definition, the unit vector is the vector whose components are
the direction cosines of F. The magnitude is
(a) What is the magnitude of F?
(b) Determine the components of the unit vector that (a) |F| = 32 + 42 + 22 = 5.385 (N) The unit vector is
has the same direction as F.
(b) e =
4
2
3
i−
j−
k
5.385
5.385
5.385
= 0.5571i − 0.7428j − 0.3714k
Problem 2.79 A force vector F points in the same Solution: By definition, F = |F|e, where e is a unit vector in the direction
direction as the unit vector e = 27 i − 67 j − 37 k. The of F. (See discussion following Eq. (2.16).) Thus
magnitude of F is 700 lb. Express F in terms of com- F = 700 2 i − 6 j − 3 k = 200i − 600j − 300k
7
7
7
ponents.
Problem 2.80 A force vector F points in the same
direction as the position vector r = 4i + 4j − 7k (m).
The magnitude of F is 90 kN. Express F in terms of
components.
Solution: By definition, F = |F|e, where e is a unit vector in the direction
of
The magnitude is |r| =
√ F. Find the unit vector from the position4vector.
42 + 42 + 72 = 9; the unit vector is e = 9 i + 49 j − 79 k. The components
are
4
4
7
F = 90
i + j − k = 40i + 40j − 70k (kN)
9
9
9
Problem 2.93 A cable extends from point C to point
E. It exerts a 50-lb force T on plate C that is directed
along the line from C to E. Express T in terms of scalar
components.
y
6 ft
A
E
D
z
20°
B
C
4 ft
Find the unit vector eCE and multiply it times the magnitude of the force to get the vector in component form,
Solution:
eCE =
rCE
(xE − xC )i + (yE − yC )j + (zE − zC )k
= |rCE |
(xE − xC )2 + (yE − yC )2 + (zE − zC )2
The coordinates of point C are (4, −4 sin 20◦ , 4 cos 20◦ ) or
(4, −1, 37, 3.76) (ft) The coordinates of point E are (0, 2, 6) (ft)
eCE =
(0 − 4)i + (2 − (−1.37))j + (6 − 3.76)k
√
42 + 3.372 + 2.242
eCE = −0.703i + 0.592j + 0.394k
T = 50eCE (lb)
T = −35.2i + 29.6j + 19.7k (lb)
y
6 ft
A
E
D
T
2 ft
x
4 ft
T
z
B
C
20°
4 ft
Problem 2.94 What are the direction cosines of the
force T in Problem 2.93?
Solution:
From the solution to Problem 2.93,
eCE = −0.703i + 0.592j + 0.394k
However
eCE = cos θx i + cos θy j + cos θz k
Hence,
cos θx = −0.703
cos θy = 0.592
cos θz = 0.394
4 ft
T
2 ft
x
Problem 2.97 The 70-m-tall tower is supported by
three cables that exert forces FAB , FAC , and FAD on it.
The magnitude of each force is 2 kN. Express the total
force exerted on the tower by the three cables in terms
of scalar components.
A
y
FAD
A
FAB
FAC
D
60 m
60 m
B
x
40 m
C
40 m
Solution: The coordinates of the points are A (0, 70, 0), B (40,
0, 0), C (−40, 0, 40) D (−60, 0, −60).
The position vectors corresponding to the cables are:
40 m
z
The resultant force exerted on the tower by the cables is:
FR = FAB + FAC + FAD = −0.9872i − 4.5654j − 0.2022k kN
rAD = (−60 − 0)i + (0 − 70)j + (−60 − 0)k
A
rAD = −60i − 70k − 60k
rAC = (−40 − 0)i + (0 − 70)j + (40 − 0)k
A
rAC = −40i − 70j + 40k
rAB = (40 − 0)i + (0 − 70)j + (0 − 0)k
D
rAB = 40i − 70j + 0k
60 m
60 m
The unit vectors corresponding to these position vectors are:
uAD =
rAD
−60
70
60
=
i−
j−
k
|rAD |
110
110
110
B
40 m
= −0.5455i − 0.6364j − 0.5455k
uAC =
rAC
40
70
40
=− i−
j+
k
|rAC |
90
90
90
E
A
rAB
40
70
=
i−
j + 0k = 0.4963i − 0.8685j + 0k
|rAB |
80.6
80.6
The forces are:
40 m
40 m
= −0.4444i − 0.7778j + 0.4444k
uAB =
C
FAD
FAB = |FAB |uAB = 0.9926i − 1.737j + 0k
FAC = |FAC |uAC = −0.8888i − 1.5556j + 0.8888
FAD = |FAD |uAD = −1.0910i − 1.2728j − 1.0910k
FAC
FAB
x
When the vectors are perpendicular, U · V ≡ 0.
Problem 2.107 Two perpendicular vectors are given Solution:
Thus
in terms of their components by
U · V = Ux Vx + Uy Vy + Uz Vz = 0
U = Ux i − 4j + 6k
= 3Ux + (−4)(2) + (6)(−3) = 0
and V = 3i + 2j − 3k.
3Ux = 26
Use the dot product to determine the component Ux .
Problem 2.108
Three vectors
Ux = 8.67
Solution:
For mutually perpendicular vectors, we have three equations,
i.e.,
U = Ux i + 3j + 2k
U·V =0
V = −3i + Vy j + 3k
U·W =0
W = −2i + 4j + Wz k
V·W =0
are mutually perpendicular. Use the dot product to determine the components Ux , Vy , and Wz
Thus

= 0
3 Eqns
=0
 3 Unknowns
=0
−3Ux + 3Vy + 6
−2Ux + 12 + 2Wz
+6 + 4Vy + 3Wz
Solving, we get
Ux
Vy
Wz
Problem 2.109 The magnitudes |U| = 10 and |V| =
20.
(a) Use the definition of the dot product to determine
U · V.
(b) Use Eq. (2.23) to obtain U · V.
= 2.857
= 0.857
= −3.143
y
V
U
45°
30°
x
Solution:
(a)
The definition of the dot product (Eq. (2.18)) is
U · V = |U||V| cos θ. Thus
◦
y
◦
U · V = (10)(20) cos(45 − 30 ) = 193.2
(b)
U
V
The components of U and V are
U = 10(i cos 45◦ + j sin 45◦ ) = 7.07i + 7.07j
45°
V = 20(i cos 30◦ + j sin 30◦ ) = 17.32i + 10j
From Eq. (2.23)
U · V = (7.07)(17.32) + (7.07)(10) = 193.2
30°
x
Problem 2.116 The force F = 21i + 14j (kN). Resolve it into vector components parallel and normal to the
line OA.
y
F
O
x
z
Solution:
A (6, – 2, 3) m
The position vector of point A is
y
rA = 6i − 2j + 3k
√
The magnitude is |rA | = 62 + 22 + 32 = 7. The unit vector
rA
parallel to OA is eOA = |r | = 67 i − 27 j + 37 k
A
(a) The component of F parallel to OA is
1
(6i − 2j + 3k)
(F · eOA ) eOA = ((3)(6) + (−2)(2))
7
F
x
FP = 12i − 4j + 6k (kN)
(b)
The component of F normal to OA is
FN
z
= F − Fp = (21 − 12)i + (14 − (−4))j + (0 − 6)k
A
(6, – 2, 3) m
= 9i + 18j − 6k (kN)
Problem 2.117 At the instant shown, the Harrier’s
thrust vector is T = 3800i + 15,300j − 1800k (lb), and
its velocity vector is v = 24i + 6j − 2k (ft/s). Resolve T
into vector components parallel and normal to v. (These
are the components of the airplane’s thrust parallel and
normal to the direction of its motion.)
y
v
T
x
Solution:
|v| =
The magnitude of the velocity vector is given by
vx2 + vy2 + vz2 = 242 + 62 + (−2)2 .
y
Thus, |v| = 24.8 ft/s. The components of the unit vector in the
direction of the velocity vector are given by
v
vx
vy
vz
, ey =
, and ez =
.
ex =
|v|
|v|
|v|
Substituting numerical values, we get ex = 0.967, ey = 0.242, and
ez = −0.0806. The dot product of T and this unit vector gives the
component of T parallel to the velocity. The resulting equation is
Tparallel = Tx ex + Ty ey + Tz ez . Substituting numerical values,
we get Tparallel = 7232.12 lb. The magnitude of the vector T is
15870 lb. Using the Pythagorean Theorem, we get
Tnormal =
|T |2 − (Tparallel )2 = 14130 lb.
T
x
Problem 2.144 The bar AB is 6 m long and is perpendicular to the bars AC and AD. Use the cross product
to determine the coordinates xB , yB , zB of point B.
y
B
(0, 3, 0) m
A
C
D
(0, 0, 3) m
(4, 0, 0) m
z
Solution: The strategy is to determine the unit vector perpendicular to both AC and AD, and then determine the coordinates that will
agree with the magnitude of AB. The position vectors are:
y
rOA = 0i + 3j + 0k, rOD = 0i + 0j + 3k, and
rAD = (0 − 0)i + (0 − 3)j + (3 − 0)k = 0i − 3j + 3k, rAC
[4,0,0]
= (4 − 0)i + (0 − 3)j + (0 − 0)k = 4i − 3j + 0k.
The magnitude |R| = 19.21 (m). The unit vector is
eAB =
R
= 0.4685i + 0.6247j + 0.6247k.
|R|
Thus the vector collinear with AB is
rAB = 6eAB = +2.811i + 3.75j + 3.75k.
Using the coordinates of point A:
xB = 2.81 + 0 = 2.81 (m)
yB = 3.75 + 3 = 6.75 (m)
zB = 3.75 + 0 = 3.75 (m)
B
A
[0,3,0]
rOC = 4i + 0j + 0k. The vectors collinear with the bars are:
The vector collinear with rAB is
i
j k
R = rAD × rAC = 0 −3 3 = 9i + 12j + 12k
4 −3 0 (xB, yB, zB)
C
D
z
[0,0,3]
x
x
Problem 3.7 The two springs are identical, with unstretched lengths 250 mm and spring constants k =
1200 N/m.
(a) Draw the free-body diagram of block A.
(b) Draw the free-body diagram of block B.
(c) What are the masses of the two blocks?
300 mm
A
280 mm
B
Solution: The tension in the upper spring acts on block A in the
positive Y direction, Solve the spring force-deflection equation for
the tension in the upper spring. Apply the equilibrium conditions
to block A. Repeat the steps for block B.
TU A = 0i +
N
1200
m
300 mm
(0.3 m − 0.25 m)j = 0i + 60j N
A
Similarly, the tension in the lower spring acts on block A in the negative
Y direction
TLA = 0i −
N
1200
m
280 mm
(0.28 m − 0.25 m)j = 0i − 36j N
B
The weight is WA = 0i − |WA |j
The equilibrium conditions are
F=
Fx +
Fy = 0,
F = WA + TU A + TLA = 0
Tension,
upper spring
Collect and combine like terms in i, j
Solve
Fy = (−|WA | + 60 − 36)j = 0
A
|WA | = (60 − 36) = 24 N
Tension,
lower
spring
The mass of A is
mA =
Weight,
mass A
|WL |
24 N
=
= 2.45 kg
|g|
9.81 m/s2
The free body diagram for block B is shown.
The tension in the lower spring TLB = 0i + 36j
y
The weight: WB = 0i − |WB |j
Apply the equilibrium conditions to block B.
B
x
F = WB + TLB = 0
Collect and combine like terms in i, j:
Solve:
Fy = (−|WB | + 36)j = 0
|WB | = 36 N
The mass of B is given by mB =
|WB |
|g|
=
Tension,
lower spring
36 N
9.81 m/s2
= 3.67 kg
Weight,
mass B
Problem 3.18 A 10-kg painting is suspended by a
wire. If α = 25◦ , what is the tension in the wire?
α
α
Solution: Isolate support pin fixed to the wall or other support.
The angle of the right hand wire with the positive x axis is −α, hence
the tension is
α
α
F2 = |F2 |(i cos α − j sin α)
The angle of the left hand wire is (180◦ + α) hence
F1 = |F1 |(−i cos α − j sin α).
The weight is W = 0i + |W|j
The equilibrium conditions are
y
W
F = W + F1 + F2 = 0
x
Substitute the vector forces, and collect like terms,
Fx = (|F2 | cos α − |F1 | cos α)i = 0,
Fy = (|W| − |F2 | sin α − |F1 | sin α)j = 0.
Thus |F1 | = |F2 |, and
|F1 | = |F2 | =
1
2
|W|
sin α
.
F1
α
α
F2
With α = 25◦ and |W| = (10 kg) 9.81 sm2 = 98.1 N
1
98.1
|F1 | = |F2 | =
= 116.06 N
2
0.423
Problem 3.19 If the wire supporting the suspended
painting in Problem 3.18 breaks when the tension exceeds 150 N and you want a 100 percent safety factor
(that is, you want the wire to be able to support twice the
actual weight of the painting), what is the smallest value
of α you can use?
Solution:
From Problem 3.18
1
W
|F1 | = |F2 | =
2 sin α
m
and |W| = (10 kg) 9.81 2 = 98.1 N.
s
y
W
x
Thus
1
98.1
150
For a tension |F| =
= 75,
2
|F|
2
1
98.1
sin α =
= 0.654 or α = 40.8◦
2
75
sin α =
F1
α
α
F2
Problem 3.27 The mass of the suspended crate is 5 kg.
What are the tensions in the cables AB and AC?
10 m
B
C
5m
7m
Solution:
Find the interior angles in the figure, then apply the
equilibrium conditions to the isolated crate. Given the triangle shown,
with known sides A, B, and C, find the unknown interior angles α, β,
and γ using the. law of cosines
A
B 2 = A2 + C 2 − 2AC cos β
Solve: cos β =
cos γ =
A2 + C 2 − B 2
. Similarly,
2AC
A2 + B 2 − C 2
.
2AB
10 m
27.66◦
For A = 10, B = 7, C = 5, γ =
third angle is
and β =
40.54◦ .
The
B
C
α = (180 − 27.66 − 40.64) = 111.8◦
5m
7m
A
Isolate the cable juncture at A. The angle between the positive x axis
and the tension TAC is γ. The tension is
TAC = |TAC |(i cos γ + j sin γ).
The angle between the positive x axis and the tension TAB is
(180◦ − β),
A
TAB = |TAB |(i cos(180 − β) + j sin(180 − β))
β
TAB = |TAB |(−i cos β + j sin β).
C
◦
◦
γ
α
B
The weight is W = 0i − |W|j.
The equilibrium conditions are
y
F = W + TAB + TAC = 0.
Fx = (|TAC | cos γ − |TAB | cos β)i = 0
Fy = (|TAC | sin γ + |TAB | sin β − |W|) = 0
Solve:
|TAC | =
|TAB | =
cos β
cos γ
|TAB |,
|W| cos γ
sin(β + γ)
,
|W| cos β
.
sin(β + γ)
m
|W| = (5 kg) 9.81 2 = 49.05 N,
s
|TAC | =
For
and
β = 40.54◦ ,
|TAB | = 46.79 N,
|TAC | = 40.15 N
C
γ
β
Substitute and collect like terms,
x
B
γ = 27.66◦
A
W
Problem 3.39 While working on another exhibit, a curator at the Smithsonian Institution pulls the suspended
Voyager aircraft to one side by attaching three horizontal cables as shown. The mass of the aircraft is 1250 kg.
Determine the tensions in the cable segments AB, BC,
and CD.
D
30°
C
50°
B
Isolate each cable juncture, beginning with A and solve
the equilibrium equations at each juncture. The angle between the
cable AB and the positive x axis is α = 70◦ ; the tension in cable
AB is TAB = |TAB |(i cos α + j sin α). The weight is W =
0i − |W|j. The tension in cable AT is T = −|T|i + 0j. The
equilibrium conditions are
Solution:
A
70°
F = W + T + TAB = 0.
Substitute and collect like terms
Fx (|TAB | cos α − |T|)i = 0,
D
Fy = (|TAB | sin α − |W|)j = 0.
C
Solve: the tension in cable AB is |TAB | =
|W|
.
sin α
m
For |W| = (1250 kg) 9.81 2 = 12262.5 N and α = 70◦
s
12262.5
|TAB | =
= 13049.5 N
0.94
70°
A
T
Isolate juncture B. The angles are α = 50◦ , β = 70◦ , and the tension
cable BC is TBC = |TBC |(i cos α + j sin α). The angle between
the cable BA and the positive x axis is (180 + β); the tension is
TBA = |TBA |(i cos(180 + β) + j sin(180 + β))
y
= |TBA |(−i cos β − j sin β)
B
x
The tension in the left horizontal cable is T = −|T|i + 0j. The
equilibrium conditions are
α
A
T
F = TBA + TBC + T = 0.
Substitute and collect like terms
W
Fx = (|TBC | cos α − |TBA | cos β − |T|)i = 0
y
Fy = (|TBC | sin α − |TBA | sin β)j = 0.
Solve: |TBC | =
sin β
sin α
|TBA |.
For |TBA | = 13049.5 N, and α =
|TBC | = (13049.5)
0.9397
0.7660
β=
70◦ ,
|TCD | =
sin β
sin α
β
= 16007.6 N
A
y
D
x
|TCB |.
Substitute: |TCD | = (16007.6)
α
B
T
Isolate the cable juncture C. The angles are α = 30◦ , β = 50◦ . By
symmetry with the cable juncture B above, the tension in cable CD is
C
x
50◦ ,
0.7660 0.5
= 24525.0 N.
50°
B
T
α
C
β
This completes the problem solution.
B
30°
Problem 3.50 The 50-kg sphere is at rest on the
smooth horizontal surface. The horizontal force F =
500 N. What is the normal force exerted on the sphere
by the surface?
30°
F
Solution: Isolate the sphere and solve the equilibrium equations.
The angle between the cable and the positive x is (180 − α). The
tension:
30°
T = |T|(−i cos α + j sin α).
The other forces are
F = |F|i + 0j,
F
N = 0i + |N|j,
W = 0i − |W |j.
The equilibrium conditions are
F = T + F + N + W = 0.
T
α
Substitute and collect like terms,
Fx = (−|T| cos α + |F|)i = 0,
F
Fy = (|N| − |W| + |T| sin α)j = 0.
Solve: |T| =
|F|
cos α
, and |N| = |W| − |F| tan α.
W
N
For |W| = (50)(9.81) = 490.5 N, |F| = 500 N, and α = 30◦
|N| = 490.5 − (500)(0.577) = 201.8 N
Problem 3.51
Consider the stationary sphere in
Problem 3.50.
(a) Draw a graph of the normal force exerted on the
sphere by the surface as a function of the force F
from F = 0 to F = 1 kN.
(b) In the result of (a), notice that the normal force decreases to zero and becomes negative as F increases.
What does that mean?
Solution:
From the solution of Problem 3.50,
|N| = |W| − |F| tan α.
(a)
(b)
The commercial package TK Solver Plus was used to produce
the graph of the normal force vs. the applied force, for |W| =
(50)(9.81) = 490.5 N and α = 30◦ , as shown.
The normal force becomes negative when the cylinder is lifted
from the surface (it would take a negative force to keep it in
contact with the surface).
N
o
r
m
a
l
Normal Force vs Force
500
400
300
200
F 100
o
r
0
c
e −100
0
200
400
600
Force
800
1000
Problem 3.64 Prior to its launch, a balloon carrying
a set of experiments to high altitude is held in place by
groups of student volunteers holding the tethers at B, C,
and D. The mass of the balloon, experiments package,
and the gas it contains is 90 kg, and the buoyancy force
on the balloon is 1000 N. The supervising professor conservatively estimates that each student can exert at least
a 40-N tension on the tether for the necessary length of
time. Based on this estimate, what minimum numbers
of students are needed at B, C, and D?
y
A (0, 8, 0) m
C (10,0, –12) m
Solution:
Fy = 1000 − (90)(9.81) − T = 0
D
(–16, 0, 4) m
x
B (16, 0, 16) m
z
T = 117.1 N
y
A(0, 8, 0)
B(16, 0, 16)
C(10, 0, −12)
D(−16, 0, 4)
We need to write unit vectors eAB , eAC , and eAD .
eAB = 0.667i − 0.333j + 0.667k
A (0, 8, 0) m
eAC = 0.570i − 0.456j − 0.684k
eAD = −0.873i − 0.436j + 0.218k
We now write the forces in terms of magnitudes and unit vectors

F

 AB
FAC

 FAD
T
C (10, 0, Ð12) m
D
(Ð16, 0, 4) m
z
x
B (16, 0, 16) m
= 0.667FAB i − 0.333FAB j + 0.667FAB k
= 0.570FAC i − 0.456FAC j − 0.684FAC k
= −0.873FAD i − 0.436FAC j + 0.218FAC k
= 117.1j (N)
1000 N
(90) g
The equations of equilibrium are
Fx = 0.667FAB + 0.570FAC − 0.873FAD = 0
Fy = −0.333FAB − 0.456FAC − 0.436FAC + 117.1 = 0
T
Fz = 0.667FAB − 0.684FAC + 0.218FAC = 0
y
Solving, we get
T
(0, 8, 0)
FAB = 64.8 N ∼ 2 students
A
FAC = 99.8 N ∼ 3 students
FAC
FAD = 114.6 N ∼ 3 students
FAD
C (10, 0, −12) m
D
x
(−16, 0, 4)
z
B (16, 0, 16) m
Problem 3.74 The 200-kg slider at A is held in place
on the smooth vertical bar by the cable AB.
(a) Determine the tension in the cable.
(b) Determine the force exerted on the slider by the bar.
y
2m
B
A
5m
2m
x
2m
z
The coordinates of the points A, B are A(2, 2, 0),
B(0, 5, 2). The vector positions
Solution:
rA = 2i + 2j + 0k,
y
2m
rB = 0i + 5j + 2k
The equilibrium conditions are:
B
F = T + N + W = 0.
Eliminate the slider bar normal force as follows: The bar is parallel to
the y axis, hence the unit vector parallel to the bar is eB = 0i+1j+0k.
The dot product of the unit vector and the normal force vanishes: eB ·
N = 0. Take the dot product of eB with the equilibrium conditions:
eB · N = 0.
eB · F = eB · T + eB · W = 0.
A
2m
5m
x
2m
z
The weight is
eB · W = 1j · (−j|W|) = −|W| = −(200)(9.81) = −1962 N.
T
The unit vector parallel to the cable is by definition,
eAB =
rB − rA
.
|rB − rA |
N
Substitute the vectors and carry out the operation:
W
eAB = −0.4851i + 0.7278j + 0.4851k.
(a)
The tension in the cable is T = |T|eAB . Substitute into the
modified equilibrium condition
eB F = (0.7276|T| − 1962) = 0.
Solve: |T| = 2696.5 N from which the tension vector is
T = |T|eAB = −1308i + 1962j + 1308k.
(b)
The equilibrium conditions are
F = 0 = T + N + W = −1308i + 1308k + N = 0.
Solve for the normal force: N = 1308i − 1308k. The magnitude is |N| = 1850 N.
Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in the y-component
of the equilibrium equation (the slider bar is parallel to the y-axis). However, in
the general case, the slider bar will not be parallel to an axis, and the unknown
normal force will be projected onto all components of the equilibrium equations (see Problem 3.75 below). In this general situation, it will be necessary to
eliminate the slider bar normal force by some procedure equivalent to that used
above. End Note.
Problem 3.84 The mass of the suspended object A is
mA and the masses of the pulleys are negligible. Determine the force T necessary for the system to be in
equilibrium.
T
A
Solution: Break the system into four free body diagrams as
shown. Carefully label the forces to ensure that the tension in any
single cord is uniform. The equations of equilibrium for the four objects, starting with the leftmost pulley and moving clockwise, are:
S − 3T = 0,
R − 3S = 0,
F
F − 3R = 0,
R
and 2T + 2S + 2R − mA g = 0.
R
We want to eliminate S, R, and F from our result and find T in terms
of mA and g. From the first two equations, we get S = 3T , and
R = 3S = 9T . Substituting these into the last equilibrium equation
results in 2T + 2(3T ) + 2(9T ) = mA g.
R
R
S
S
S
Solving, we get T = mA g/26 .
S
T
T
S
S
R
R
T
T
T
A
mAg
T
A
Note: We did not have to solve for F to find the appropriate value of T . The
final equation would give us the value of F in terms of mA and g. We would get
F = 27mA g/26. If we then drew a free body diagram of the entire assembly,
the equation of equilibrium would be F − T − mA g = 0. Substituting in the
known values for T and F , we see that this equation is also satisfied. Checking
the equilibrium solution by using the “extra” free body diagram is often a good
procedure.