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Problem 2.44 The rope ABC exerts forces FBA and
FBC on the block at B. Their magnitudes are equal:
jFBAjDjFBC j. The magnitude of the total force exerted
on the block at Bby the rope is jFBA CFBC jD920 N.
Determine jFBAjby expressing the forces FBA and FBC
in terms of components.
20°
FBC
FBA
B
C
A
B
Solution:
FBA DF⊲j⊳
FBC CFBA DF⊲cos 20°iC[sin 20°1]j⊳
Therefore
⊲920 N⊳2DF2⊲cos220°C[sin 20°1]2⊳)FD802 N
FBC
Problem 2.45 The magnitude of the horizontal force
F1is 5 kN and F1CF2CF3D0. What are the magni-
tudes of F2and F3?
F3
30˚
y
Problem 2.46 Four groups engage in a tug-of-war. The
magnitudes of the forces exerted by groups B,C, and D
are jFBjD800 lb, jFCjD1000 lb, jFDjD900 lb. If the
vector sum of the four forces equals zero, what are the
magnitude of FAand the angle ˛?
FB
y
FC
FD
FAand then take its magnitude. The force vectors are
28
Problem 2.47 In Example 2.5, suppose that the attach-
ment point of cable Ais moved so that the angle between
the cable and the wall increases from 40°to 55°. Draw
a sketch showing the forces exerted on the hook by the
two cables. If you want the total force FACFBto have
a magnitude of 200 lb and be in the direction perpen-
dicular to the wall, what are the necessary magnitudes
of FAand FB?
A
40⬚
Solution: Let FAand FBbe the magnitudes of FAand FB.
FAD195 lb
FBD119 lb
Problem 2.48 The bracket must support the two forces
shown, where jF1jDjF2jD2 kN. An engineer deter-
mines that the bracket will safely support a total force
of magnitude 3.5 kN in any direction. Assume that 0
˛90°. What is the safe range of the angle ˛?
F2
F1
α
30
Problem 2.49 The figure shows three forces acting on
a joint of a structure. The magnitude of Fcis 60 kN, and
FACFBCFCD0. What are the magnitudes of FAand
FB?
y
x
F
B
F
C
F
A
15°
40°
FBDjFBjcos 195°iCjFBjsin 195j⊲kN⊳
FCDjFCjcos 270°iCjFCjsin 270°j⊲kN⊳
Thus FCD60jkN
Since FACFBCFCD0, their components in each direction must also
sum to zero.
FAx CFBx CFCx D0
FAy CFBy CFCy D0
Thus,
jFAjcos 40°CjFBjcos 195°C0D0
jFAjsin 40°CjFBjsin 195°60 ⊲kN⊳D0
Solving for jFAjand jFBj, we get
jFAjD137 kN,jFBjD109 kN
F
B
F
C
270°
Problem 2.50 Four forces act on a beam. The vector
sum of the forces is zero. The magnitudes jFBjD
10 kN and jFCjD5 kN. Determine the magnitudes of
FAand FD.
FD
30°
FBFC
FA
Solution: Use the angles and magnitudes to determine the vectors,
and then solve for the unknowns. The vectors are:
Problem 2.51 Six forces act on a beam that forms part
of a building’s frame. The vector sum of the forces
is zero. The magnitudes jFBjDjFEjD20 kN, jFCjD
16 kN, and jFDjD9 kN. Determine the magnitudes of
FAand FG.
50⬚
70⬚
40⬚40⬚
FE
FB
FG
FCFD
FA
Solution: Write each force in terms of its magnitude and direction
as
y
32
Problem 2.52 The total weight of the man and parasail
is jWjD230 lb. The drag force Dis perpendicular to
the lift force L. If the vector sum of the three forces is
zero, what are the magnitudes of Land D?
x
W
Solution: Let Land Dbe the magnitudes of the lift and drag
forces. We can use similar triangles to express the vectors Land D
in terms of components. Then the sum of the forces is zero. Breaking
into components we have
Problem 2.53 The three forces acting on the car are
shown. The force Tis parallel to the xaxis and the
magnitude of the force Wis 14 kN. If TCWCND0,
what are the magnitudes of the forces Tand N?
Solution:
Fx:TNsin 20°D0
Problem 2.54 The cables A,B, and Chelp support a
pillar that forms part of the supports of a structure. The
magnitudes of the forces exerted by the cables are equal:
jFAjDjFBjDjFCj. The magnitude of the vector sum of
the three forces is 200 kN. What is jFAj?
Solution: Use the angles and magnitudes to determine the vector
components, take the sum, and solve for the unknown. The angles
between each cable and the pillar are:
ADtan14m
D33.7°,
Problem 2.55 The total force exerted on the top of the
mast Bby the sailboat’s forestay AB and backstay BC is
180i820j(N). What are the magnitudes of the forces
exerted at Bby the cables AB and BC ?
Solution: We first identify the forces:
FAB DTAB
⊲4.0mi11.8mj⊳
⊲4.0m⊳2C⊲11.8m⊳2
⊲5.0mi12.0mj⊳
34
Problem 2.56 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 jFABjD4 kN. If the vector sum of the
three forces equals zero, what are the magnitudes of FAC
and FAD?
FAB
A
(– 4, 1) m B
C
D
x
y
FAC
FAD
(–2, –3) m
(4, 2) m
Solution: Determine the unit vectors parallel to each force:
eAB D4
p42C22iC2
p42C22jD0.89443iC0.4472j
The forces are FAD DjFADjeAD ,FAC DjFACjeAC,
FAB DjFABjeAB D3.578iC1.789j. Since the vector sum of the forces
vanishes, the x- and y-components vanish separately:
FxD⊲0.5547jFADj0.9701jFACjC3.578⊳iD0,and
FyD⊲0.8320jFADjC0.2425jFACjC1.789⊳jD0
These simultaneous equations in two unknowns can be solved by any
standard procedure. An HP-28S hand held calculator was used here:
The results: jFACjD2.108 kN ,jFAD jD2.764 kN
B
Problem 2.57 The distance sD45 in.
(a) Determine the unit vector eBA that points from B
x
A
(14, 45) in
Solution:
(a) The unit vector is the position vector from Bto Adivided by its
69.35 in ⊲61iC33j⊳in D⊲0.880iC0.476j⊳
C0.476j⊳
Problem 2.58 In Problem 2.57, determine the xand y
coordinates of the collar Cas functions of the distance s.
Solution: The coordinates of the point Care given by
Problem 2.59 The position vector rgoes from point
Ato a point on the straight line between Band C. Its
magnitude is jrjD6 ft. Express rin terms of scalar
components.
x
y
A
(7, 9) ft
(12, 3) ft
(3, 5) ft
B
C
r
Solution: Determine the perpendicular vector to the line BC from
point A, and then use this perpendicular to determine the angular orien-
tation of the vector r. The vectors are
y
B[7,9]
P
36
Problem 2.60 Let rbe the position vector from point
Cto the point that is a distance smeters along the
straight line between Aand B. Express rin terms of
components. (Your answer will be in terms of s).
y
(10, 9) m
B
r
s
Problem 2.61 A vector UD3i4j12k. What is its
magnitude?
Solution: Use definition given in Eq. (14). The vector magni-
tude is
Problem 2.62 The vector eD1
3iC2
3jCezkis a unit
vector. Determine the component ez. (Notice that there
Solution:
Problem 2.63 An engineer determines that an attach-
ment point will be subjected to a force FD20iCFyj
45k⊲kN⊳. If the attachment point will safely support a
force of 80-kN magnitude in any direction, what is the
Solution:
802½F2
xCF2
yCF2
z
Problem 2.64 A vector UDUxiCUyjCUzk. Its
magnitude is jUjD30. Its components are related by
the equations UyD2Uxand UzD4Uy. Determine the
components. (Notice that there are two answers.)
UDUxiC⊲2Ux⊳jC⊲4⊲2Ux⊳⊳kDUx⊲1i2j8k⊳
where Uxcan be factored out since it is a scalar. Take the magnitude,
noting that the absolute value of jUxjmust be taken:
30 DjUxjp12C22C82DjUxj⊲8.31⊳.
Solving, we get jUxjD3.612, or UxDš3.61. The two possible
vectors are
UD3.61iC⊲2⊲3.61⊳⊳j
Problem 2.65 An object is acted upon by two
forces F1D20iC30j24k(kN) and F2D60iC
20jC40k(kN). What is the magnitude of the total force
acting on the object?
Solution:
F1D⊲20iC30j24k⊳kN
Problem 2.66 Two vectors UD3i2jC6kand
VD4iC12j3k.
Solution: The magnitudes:
(a) jUjDp32C22C62D7and jVjDp42C122C32D13
38
Problem 2.67 In Active Example 2.6, suppose that
you want to redesign the truss, changing the position
of point Dso that the magnitude of the vector rCD from
point Cto point Dis 3 m. To accomplish this, let the
coordinates of point Dbe ⊲2,y
D,1⊳m, and determine
the value of yDso that jrCDjD3 m. Draw a sketch of
the truss with point Din its new position. What are the
new directions cosines of rCD?
rCD
(2, 3, 1) m
(4, 0, 0) m
Cx
y
D
Solution: The vector rCD and the magnitude jrCDjare
Problem 2.68 A force vector is given in terms of its
components by FD10i20j20k(N).
(a) What are the direction cosines of F?
Solution:
FD⊲10i20j20k⊳N
Problem 2.69 The cable exerts a force Fon the hook
at Owhose magnitude is 200 N. The angle between the
vector Fand the xaxis is 40°, and the angle between
the vector Fand the yaxis is 70°.
(a) What is the angle between the vector Fand the
zaxis?
(b) Express Fin terms of components.
Strategy: (a) Because you know the angles between
the vector Fand the xand yaxes, you can use Eq. (2.16)
to determine the angle between Fand the zaxis.
(Observe from the figure that the angle between Fand
the zaxis is clearly within the range 0 <
z<180°.) (b)
The components of Fcan be obtained with Eqs. (2.15).
y
x
z
O
70°
40°
F
Solution:
Problem 2.70 A unit vector has direction cosines
cos xD0.5 and cos yD0.2. Its zcomponent is posi-
Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
1⊲0.5⊳2⊲0.2⊳2DC0.8426.
Problem 2.71 The airplane’s engines exert a total thrust
force Tof 200-kN magnitude. The angle between Tand
the xaxis is 120°, and the angle between Tand the yaxis
is 130°. The zcomponent of Tis positive.
Solution: The x- and y-direction cosines are
lDcos 120°D0.5,mDcos 130°D0.6428
from which the z-direction cosine is
40
Problem 2.72 Determine the components of the posi-
tion vector rBD from point Bto point D. Use your result
B (5, 0, 3) m
C (6, 0, 0) m
z
x
Solution: We have the following coordinates: A⊲0,0,0⊳,
B⊲5,0,3⊳m, C⊲6,0,0⊳m, D⊲4,3,1⊳m
Problem 2.73 What are the direction cosines of the
position vector rBD from point Bto point D?
Solution:
Problem 2.74 Determine the components of the unit
vector eCD that points from point Ctoward point D.
Solution: We have the following coordinates: A⊲0,0,0⊳,
B⊲5,0,3⊳m, C⊲6,0,0⊳m, D⊲4,3,1⊳m
3.74 m ⊲2iC3jCk⊳mD⊲0.535iC0.802jC0.267k⊳
Problem 2.75 What are the direction cosines of the
unit vector eCD that points from point Ctoward point D?
Problem 2.76 In Example 2.7, suppose that the
caisson shifts on the ground to a new position. The
magnitude of the force Fremains 600 lb. In the new
position, the angle between the force Fand the xaxis
is 60°and the angle between Fand the zaxis is 70°.
Express Fin terms of components.
40⬚F
y
x
54⬚
z
Solution: We need to find the angle ybetween the force Fand
the yaxis. We know that
Problem 2.77 Astronauts on the space shuttle use radar
to determine the magnitudes and direction cosines of the
position vectors of two satellites Aand B. The vector rA
from the shuttle to satellite Ahas magnitude 2 km, and
direction cosines cos xD0.768, cos yD0.384, cos zD
0.512. The vector rBfrom the shuttle to satellite Bhas
magnitude 4 km and direction cosines cos xD0.743,
cos yD0.557, cos zD0.371. What is the distance
between the satellites?
x
rB
B
y
Solution: The two position vectors are:
42
Problem 2.78 Archaeologists measure a pre-Colum-
bian ceremonial structure and obtain the dimensions
shown. Determine (a) the magnitude and (b) the
direction cosines of the position vector from point Ato
point B.
4 m
y
10 m
A
10 m
4 m
8 m
Problem 2.79 Consider the structure described in
Problem 2.78. After returning to the United States,
an archaeologist discovers that a graduate student has
erased the only data file containing the dimension b.
But from recorded GPS data he is able to calculate that
the distance from point Bto point Cis 16.61 m.
(a) What is the distance b?
(b) Determine the direction cosines of the position
vector from Bto C.
4 m
y
10 m
z
b
x
A
C
10 m
B
4 m
8 m
8 m
Problem 2.80 Observers at Aand Buse theodolites to
measure the direction from their positions to a rocket
in flight. If the coordinates of the rocket’s position at a
given instant are (4, 4, 2) km, determine the direction
cosines of the vectors rAR and rBR that the observers
would measure at that instant.
A
rAR
rBR
y
Solution: The vector rAR is given by
44
Problem 2.81 In Problem 2.80, suppose that the coor-
dinates of the rocket’s position are unknown. At a given
instant, the person at Adetermines that the direction
cosines of rAR are cos xD0.535, cos yD0.802, and
cos zD0.267, and the person at Bdetermines that the
direction cosines of rBR are cos xD0.576, cos yD
0.798, and cos zD0.177. What are the coordinates of
the rocket’s position at that instant.
The magnitude of rAB is given by jrABjD⊲5⊳2C⊲2⊳2D5.39 km.
The unit vector along the line AR,
uAR Dcos xiCcos yjCcos zkD0.535iC0.802jC0.267k.
rAR D0.535rARiC0.802rARjC0.267rARk,and
4.489 km. Calculating the components, we get
Problem 2.82* The height of Mount Everest was orig-
inally measured by a surveyor in the following way.
He first measured the altitudes of two points and the
horizontal distance between them. For example, suppose
that the points Aand Bare 3000 m above sea level
and are 10,000 m apart. He then used a theodolite to
measure the direction cosines of the vector rAP from
point Ato the top of the mountain Pand the vector rBP
from point Bto P. Suppose that the direction cosines of
rAP are cos xD0.5179,cos yD0.6906, and cos zD
0.5048, and the direction cosines of rBP are cos xD
0.3743,cos yD0.7486, and cos zD0.5472. Using
this data, determine the height of Mount Everest above
sea level.
P
y
A
z
Bx
Solution: We have the following coordinates A⊲0,0,3000⊳m,
Equating components gives us five equations (one redundant) which
we can solve for the five unknowns.
Problem 2.83 The distance from point Oto point Ais
20 ft. The straight line AB is parallel to the yaxis, and
point Bis in the x-zplane. Express the vector rOA in
terms of scalar components.
Strategy: You can resolve rOA into a vector from Oto
Band a vector from Bto A. You can then resolve the
vector form Oto Binto vector components parallel to
the xand zaxes. See Example 2.8.
y
x
z
30°
B
O
A
rOA
60°
Solution: See Example 2.8. The length BA is, from the right triangle
The vector rOA is given by rOA DrOB CrBA, from which
Problem 2.84 The magnitudes of the two force vectors
are jFAjD140 lb and jFBjD100 lb. Determine the mag-
nitude of the sum of the forces FACFB.
y
FB
FA
46
