PROBLEM 2.62
Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k.
SOLUTION
222
22 2
(690 lb) (300 lb) (580 lb)
(690 lb) (300 lb) ( 580 lb)
950 lb
xyz
F FFF
=+−
= ++
= + +−
=
F i jk
950 lbF=
690 lb
cos 950 lb
x
x
F
F
θ
= =
43.4
x
θ
= °
300 lb
cos 950 lb
y
y
F
F
θ
= =
71.6
y
θ
= °
580 lb
cos 950 lb
z
zF
F
θ
−
= =
127.6
z
θ
= °
PROBLEM 2.63
Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k.
SOLUTION
222
2 22
(650 N) (320 N) (760 N)
(650 N) ( 320 N) (760 N)
xyz
F FFF
=−+
= ++
= +− +
Fijk
1050 NF=
650 N
cos 1050 N
x
x
F
F
θ
= =
51.8
x
θ
= °
320 N
cos 1050 N
y
y
F
F
θ
−
= =
107.7
y
θ
= °
760 N
cos 1050 N
z
z
F
F
θ
= =
43.6
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.64
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
x = 69.3° and
θ
z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle
θ
y, (b) the
other components and the magnitude of the force.
SOLUTION
222
2 22
cos cos cos 1
cos (69.3 ) cos cos (57.9 ) 1
cos 0.7699
xyz
y
y
θθθ
θ
θ
++=
°+ + °=
= ±
(a) Since
0,
y
F<
we choose
cos 0.7699
y
θ
= −
140.3
y
θ
∴= °
(b)
cos
174.0 lb ( 0.7699)
yy
FF
F
θ
=
−=−
226.0 lbF=
226 lbF=
cos (226.0 lb)cos69.3
xx
FF
θ
= = °
79.9 lb
x
F=
cos (226.0 lb)cos57.9
zz
FF
θ
= = °
120.1lb
z
F=
consent of McGraw–Hill Education.
PROBLEM 2.65
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
x = 70.9° and
θ
y = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle
θ
z, (b) the
other components and the magnitude of the force.
SOLUTION
222
22 2
cos cos cos 1
cos 70.9 cos 144.9 cos 1
cos 0.47282
xyz
z
z
θθθ
θ
θ
++=
+ °+ °=
= ±
(a) Since
0,
z
F<
we choose
cos 0.47282
z
θ
= −
118.2
z
θ
∴= °
(b)
cos
52.0 ( 0.47282)
zz
FF
lb F
θ
=
−=−
110.0 lbF=
110.0 lbF=
cos (110.0 lb)cos70.9
xx
FF
θ
= = °
36.0 lb
x
F=
cos (110.0 lb)cos144.9
yy
FF
θ
= = °
90.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.66
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
y = 55° and
θ
z = 45°.
Knowing that the x component of the force is − 500 lb, determine (a) the angle
θ
x, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
222 2 22
(cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )
xyz y yz
θθθ θ θθ
++=⇒=−−
PROBLEM 2.67
A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that
θ
x = 65°,
θ
y = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle
θ
z.
SOLUTION
222
222
cos cos cos 1
cos 65 cos 40 cos 1
cos 0.48432
xyz
z
z
θθθ
θ
θ
++=
+ °+ °=
= ±
z
zz
z
(a)
1200 NF=
cos (1200 N) cos65
xx
FF
θ
= =
507 N
x
F=
cos (1200 N)cos40
yy
FF
θ
= = °
919 N
y
F=
cos (1200 N)cos61.032
zz
FF
θ
= = °
582 N
z
F=
consent of McGraw–Hill Education.
PROBLEM 2.68
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AB is 408 N, determine the
components of the force exerted on the plate at B.
SOLUTION
We have:
(320 mm) (480 mm) (360 mm) 680 mmBA BA=+=i+ j- k
Thus:
8 12 9
F17 17 17
B BA BA BA BA
BA
TTT
BA
= = =
λ i+ j- k
8 12 9 0
17 17 17
BA BA BA
TTT
+− =
ijk
Setting
408 N
BA
T=
yields,
192.0 N, 288 N, 216 N
x yz
F FF=+ =+=−
consent of McGraw–Hill Education.
PROBLEM 2.69
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 429 N, determine the
components of the force exerted on the plate at D.
SOLUTION
We have:
(250 mm) (480 mm) (360 mm) 650 mmDA DA=−=i+ j+ k
Thus:
5 48 36
F13 65 65
D DA DA DA DA
DA
TTT
DA
= = = −
λ i+ j+ k
5 48 36 0
13 65 65
DA DA DA
TTT
−+ + =
ijk
Setting
429 N
DA
T=
yields,
165.0 N, 317 N, 238 N
x yz
F FF=− =+=+
consent of McGraw–Hill Education.
PROBLEM 2.70
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AB is 2 kips, determine the
components of the force exerted at A by the cable.
SOLUTION
Cable AB:
( 46.765 ft) (45 ft) (36 ft)
74.216 ft
46.765 45 36
74.216
AB
AB AB AB
AB
AB
T
− ++
= =
− ++
= =
ijk
λ
ijk
Tλ
( ) 1.260 kips
AB x
T= −
( ) 1.213 kips
AB y
T= +
( ) 0.970 kips
AB z
T= +
consent of McGraw–Hill Education.
PROBLEM 2.62
Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k.
SOLUTION
222
22 2
(690 lb) (300 lb) (580 lb)
(690 lb) (300 lb) ( 580 lb)
950 lb
xyz
F FFF
=+−
= ++
= + +−
=
F i jk
950 lbF=
690 lb
cos 950 lb
x
x
F
F
θ
= =
43.4
x
θ
= °
300 lb
cos 950 lb
y
y
F
F
θ
= =
71.6
y
θ
= °
580 lb
cos 950 lb
z
zF
F
θ
−
= =
127.6
z
θ
= °
PROBLEM 2.63
Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k.
SOLUTION
222
2 22
(650 N) (320 N) (760 N)
(650 N) ( 320 N) (760 N)
xyz
F FFF
=−+
= ++
= +− +
Fijk
1050 NF=
650 N
cos 1050 N
x
x
F
F
θ
= =
51.8
x
θ
= °
320 N
cos 1050 N
y
y
F
F
θ
−
= =
107.7
y
θ
= °
760 N
cos 1050 N
z
z
F
F
θ
= =
43.6
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.64
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
x = 69.3° and
θ
z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle
θ
y, (b) the
other components and the magnitude of the force.
SOLUTION
222
2 22
cos cos cos 1
cos (69.3 ) cos cos (57.9 ) 1
cos 0.7699
xyz
y
y
θθθ
θ
θ
++=
°+ + °=
= ±
(a) Since
0,
y
F<
we choose
cos 0.7699
y
θ
= −
140.3
y
θ
∴= °
(b)
cos
174.0 lb ( 0.7699)
yy
FF
F
θ
=
−=−
226.0 lbF=
226 lbF=
cos (226.0 lb)cos69.3
xx
FF
θ
= = °
79.9 lb
x
F=
cos (226.0 lb)cos57.9
zz
FF
θ
= = °
120.1lb
z
F=
consent of McGraw–Hill Education.
PROBLEM 2.65
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
x = 70.9° and
θ
y = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle
θ
z, (b) the
other components and the magnitude of the force.
SOLUTION
222
22 2
cos cos cos 1
cos 70.9 cos 144.9 cos 1
cos 0.47282
xyz
z
z
θθθ
θ
θ
++=
+ °+ °=
= ±
(a) Since
0,
z
F<
we choose
cos 0.47282
z
θ
= −
118.2
z
θ
∴= °
(b)
cos
52.0 ( 0.47282)
zz
FF
lb F
θ
=
−=−
110.0 lbF=
110.0 lbF=
cos (110.0 lb)cos70.9
xx
FF
θ
= = °
36.0 lb
x
F=
cos (110.0 lb)cos144.9
yy
FF
θ
= = °
90.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.66
A force acts at the origin of a coordinate system in a direction defined by the angles
θ
y = 55° and
θ
z = 45°.
Knowing that the x component of the force is − 500 lb, determine (a) the angle
θ
x, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
222 2 22
(cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )
xyz y yz
θθθ θ θθ
++=⇒=−−
PROBLEM 2.67
A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that
θ
x = 65°,
θ
y = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle
θ
z.
SOLUTION
222
222
cos cos cos 1
cos 65 cos 40 cos 1
cos 0.48432
xyz
z
z
θθθ
θ
θ
++=
+ °+ °=
= ±
z
zz
z
(a)
1200 NF=
cos (1200 N) cos65
xx
FF
θ
= =
507 N
x
F=
cos (1200 N)cos40
yy
FF
θ
= = °
919 N
y
F=
cos (1200 N)cos61.032
zz
FF
θ
= = °
582 N
z
F=
consent of McGraw–Hill Education.
PROBLEM 2.68
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AB is 408 N, determine the
components of the force exerted on the plate at B.
SOLUTION
We have:
(320 mm) (480 mm) (360 mm) 680 mmBA BA=+=i+ j- k
Thus:
8 12 9
F17 17 17
B BA BA BA BA
BA
TTT
BA
= = =
λ i+ j- k
8 12 9 0
17 17 17
BA BA BA
TTT
+− =
ijk
Setting
408 N
BA
T=
yields,
192.0 N, 288 N, 216 N
x yz
F FF=+ =+=−
consent of McGraw–Hill Education.
PROBLEM 2.69
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 429 N, determine the
components of the force exerted on the plate at D.
SOLUTION
We have:
(250 mm) (480 mm) (360 mm) 650 mmDA DA=−=i+ j+ k
Thus:
5 48 36
F13 65 65
D DA DA DA DA
DA
TTT
DA
= = = −
λ i+ j+ k
5 48 36 0
13 65 65
DA DA DA
TTT
−+ + =
ijk
Setting
429 N
DA
T=
yields,
165.0 N, 317 N, 238 N
x yz
F FF=− =+=+
consent of McGraw–Hill Education.
PROBLEM 2.70
In order to move a wrecked truck, two cables are attached
at A and pulled by winches B and C as shown. Knowing
that the tension in cable AB is 2 kips, determine the
components of the force exerted at A by the cable.
SOLUTION
Cable AB:
( 46.765 ft) (45 ft) (36 ft)
74.216 ft
46.765 45 36
74.216
AB
AB AB AB
AB
AB
T
− ++
= =
− ++
= =
ijk
λ
ijk
Tλ
( ) 1.260 kips
AB x
T= −
( ) 1.213 kips
AB y
T= +
( ) 0.970 kips
AB z
T= +
consent of McGraw–Hill Education.