2–78.

Three forces act on the ring. If the resultant force FR has a z

magnitude and direction as shown, determine the
magnitude and the coordinate direction angles of force F3.
F3
FR 120 N
F2 110 N

SOLUTION
Cartesian Vector Notation: F1 80 N 5 45
y
3 4
FR = 120{cos 45°sin 30°i + cos 45°cos 30°j + sin 45°k} N 30

= {42.43i + 73.48j + 84.85k} N

4 3
F1 = 80 b i + k r N = {64.0i + 48.0k} N
5 5 x

F2 = { - 110k} N

F3 = {F3x i + F3y j + F3z k} N

Resultant Force:

FR = F1 + F2 + F3

{42.43i + 73.48j + 84.85k} = E A 64.0 + F3 x B i + F3 y j + A 48.0 - 110 + F3 z B k F

Equating i, j and k components, we have

64.0 + F3 x = 42.43 F3x = -21.57 N

F3 y = 73.48 N

48.0 - 110 + F3 z = 84.85 F3 z = 146.85 N

The magnitude of force F3 is

F3 = 2F 23 x + F 23 y + F 23 z

= 2(- 21.57)2 + 73.482 + 146.852

= 165.62 N = 166 N Ans.

The coordinate direction angles for F3 are

F3 x - 21.57
cos a = = a = 97.5° Ans.
F3 165.62
F3 y 73.48
cos b = = b = 63.7° Ans.
F3 165.62
F3 z 146.85
cos g = = = g = 27.5° Ans.
F3 165.62
2–79.

Determine the coordinate direction angles of F1 and FR. z

F3
FR 120 N
F2 110 N

SOLUTION
Unit Vector of F1 and FR : F1 80 N 5 45
y
3 4
4 3 30
u F1 = i + k = 0.8i + 0.6k
5 5

uR = cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k

= 0.3536i + 0.6124j + 0.7071k x

Thus, the coordinate direction angles F1 and FR are

cos aF1 = 0.8 aF1 = 36.9° Ans.

cos b F1 = 0 b F1 = 90.0° Ans.

cos gF1 = 0.6 gF1 = 53.1° Ans.

cos aR = 0.3536 aR = 69.3° Ans.

cos b R = 0.6124 b R = 52.2° Ans.

cos gR = 0.7071 gR = 45.0° Ans.

2–85.

The pole is subjected to the force F which has components z

Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, determine the
magnitudes of F and Fy. Fz

g F
SOLUTION
b Fy
y
cos2 a + cos2 b + cos2 g = 1
a
1.5 2 1.25 2
a b + cos2 75° + a b = 1 Fx
F F

F = 2.02 kN Ans. x

Fy = 2.02 cos 75° = 0.523 kN Ans.

*2–116.

Determine the magnitude of the projected component of z

force FAB acting along the z axis.
A
FAC 600 lb
36 ft

FAB 700 lb
SOLUTION
Unit Vector: The unit vector uAB must be determined first. From Fig. a, D
18 ft
rAB (18 - 0)i + (- 12 - 0)j + (0 - 36)k 3 2 6
uAB = = = i - j - k O 12 ft
rAB 2(18 - 0) + (- 12 - 0) + (0 - 36)
2 2 2 7 7 7
B

Thus, the force vector FAB is given by 12 ft

12 ft C
30 y
3 2 6 x
FAB = FAB uAB = 700a i - j - k b = {300i - 200j - 600k} lb
7 7 7

Vector Dot Product: The projected component of FAB along the z axis is

(FAB)z = FAB # k = A 300i - 200j - 600k B # k

= -600 lb

The negative sign indicates that ( FAB)z is directed towards the negative z axis. Thus

(FAB)z = 600 lb Ans.

*2–117.

Determine the magnitude of the projected component of z

force FAC acting along the z axis.
A
FAC 600 lb
36 ft

FAB 700 lb

D
18 ft
O 12 ft
B

SOLUTION 12 ft
12 ft C
30 y
Unit Vector: The unit vector uAC must be determined first. From Fig. a,
x
rAC (12 sin 30° - 0)i + (12 cos 30° - 0)j + (0 - 36)k
uAC = = = 0.1581i + 0.2739j - 0.9487k
rAC 2(12 sin 30° - 0)2 + (12 cos 30° - 0)2 + (0 - 36)2

Thus, the force vector FAC is given by

FAC = FACuAC = 600 A 0.1581i + 0.2739j - 0.9487k B = {94.87i + 164.32j - 569.21k} N

Vector Dot Product: The projected component of FAC along the z axis is

(FAC)z = FAC # k = A 94.87i + 164.32j - 569.21k B # k

= -569 lb

The negative sign indicates that ( FAC)z is directed towards the negative z axis. Thus

(FAC)z = 569 lb Ans.

2–118.

Determine the projection of the force F along the pole. z

F = {2i + 4j + 10k} kN

O y

SOLUTION
2m

Proj F = F # ua = 12 i + 4 j + 10 k2 # a i + j - kb
2 2 1
3 3 3
Proj F = 0.667 kN Ans. 2m 1m
x
2–125.

Determine the projected component of force FAC along the z

axis of strut AO. Express the result as a Cartesian vector.
3 ft B

C 4 ft

5 ft
FAB ⫽ 70 lb
60⬚ FAC ⫽ 60 lb
SOLUTION O
A

Unit Vectors: The unit vectors uAC and uAO must be determined first. From Fig. a, x 6 ft 2 ft
(5 cos 60° - 0)i + (0 - 6)j + (5 sin 60° - 2)k
uAC = = 0.3621i - 0.8689j + 0.3375k y
3(5 cos 60° - 0)2 + (0 - 6)2 + (0 - 2)2

(0 - 0)i + (0 - 6)j + (0 - 2)k

uAO = = -0.9487j - 0.3162 k
3(0 - 0)2 + (0 - 6)2 + (0 - 2)2

Thus, the force vectors FAC is given by

FAC = FAC uAC = 60(0.3621i - 0.8689j + 0.3375k) = {21.72i - 52.14j + 20.25k} lb

Vector Dot Product: The magnitude of the projected component of FAC along strut
AO is

(FAC)AO = FAC # uAO = (21.72i - 52.14j + 20.25k) # (-0.9487j - 0.3162k)

= (21.72)(0) + (-52.14)(-0.9487) + (20.25)(-0.3162)

= 43.057 lb

Thus, (FAC)AO expressed in Cartesian vector form can be written as

(FAC)AO = (FAC)AOuAO = 43.057(-0.9487j - 0.3162k)

= {-40.8j - 13.6k} lb Ans.

2–130.

Determine the angle u between the pipe segments BA and BC. z

3 ft A
4 ft
4 ft 2 ft
B
x y

SOLUTION F ⫽ {30i ⫺ 45j ⫹ 50k} lb

Position Vectors: The position vectors rBA and rBC must be determined first. From 4 ft
Fig. a,
C
rBA = (0 - 3)i + (0 - 4)j + (0 - 0)k = { - 3i - 4j} ft
rBC = (7 - 3)i + (6 - 4)j + (- 4 - 0)k = {4i + 2j - 4k} ft

The magnitude of rBA and rBC are

rBA = 3(- 3)2 + (- 4)2 = 5 ft

rBC = 342 + 22 + (- 4)2 = 6 ft

Vector Dot Product:

rBA # rBC = (- 3i - 4j) # (4i + 2j - 4k)

= (- 3)(4) + (-4)(2) + 0(-4)

= - 20 ft2
Thus,
rBA # rBC
u = cos-1 a b = cos-1 c d = 132°
-20
Ans.
rBA rBC 5(6)
2–131.

Determine the angles u and f made between the axes OA z

of the flag pole and AB and AC, respectively, of each cable.
1.5 m

B 2m C

4m
SOLUTION FC 40 N
FB 55 N
rA C = { - 2i - 4j + 1k} m ; rA C = 4.58 m 6m u A
f
rAB = {1.5i - 4j + 3k} m; rAB = 5.22 m

rA O = { - 4j - 3k} m; rA O = 5.00 m O

rA B # rA O = (1.5)(0) + ( -4)( -4) + (3)(- 3) = 7

3m

rAB # rAO 4m
u = cos - 1 ¢ ≤ x
rAB rAO
y
7
= cos - 1 ¢ ≤ = 74.4° Ans.
5.22(5.00)

rAC # rAO = (-2)(0) + (- 4)(- 4) + (1)( -3) = 13

rAC # rAO
f = cos - 1 a b
rAC rAO

13
= cos - 1 a b = 55.4° Ans.
4.58(5.00)
*2–136.

Determine the components of F that act along rod AC and z

perpendicular to it. Point B is located at the midpoint of A
the rod.

4m B

4m
O F 600 N
SOLUTION x
C

3m
rAC = (- 3i + 4j - 4k), rAC = 2(- 3) + 4 + ( - 4) = 241 m
2 2 2
6m

rAC - 3i + 4j + 4k 4m
rAB = = = -1.5i + 2j - 2k D y
2 2

rAD = rAB + rBD

rBD = rAD - rAB

= (4i + 6j - 4k) - ( - 1.5i + 2j - 2k)

= {5.5i + 4j - 2k} m

rBD = 2(5.5)2 + (4)2 + (- 2)2 = 7.0887 m

rBD
F = 600a b = 465.528i + 338.5659j - 169.2829k
rBD

Component of F along rAC is F| |

F # rAC (465.528i + 338.5659j - 169.2829k) # ( -3i + 4j - 4k)

F| | = =
rAC 241
F| | = 99.1408 = 99.1 N Ans.

Component of F perpendicular to rAC is F

F2 + F2|| = F2 = 6002

F2 = 6002 - 99.14082

F = 591.75 = 592 N Ans.

2–137.

Determine the components of F that act along rod AC and z

perpendicular to it. Point B is located 3 m along the rod A
from end C.

4m B

4m
O F 600 N
SOLUTION x
C

3m
rCA = 3i - 4j + 4k 6m

rCA = 6.403124 4m y
D
3
rCB = (r ) = 1.40556i - 1.874085j + 1.874085k
6.403124 CA

rOB = rOC + rCB

= -3i + 4j + r CB

= -1.59444i + 2.1259j + 1.874085k

rOD = rOB + rBD

rBD = rOD - rOB = (4i + 6j) - rOB

= 5.5944i + 3.8741j - 1.874085k

rBD = 2(5.5944)2 + (3.8741)2 + ( -1.874085)2 = 7.0582

rBD
F = 600( ) = 475.568i + 329.326j - 159.311k
rBD

rAC = (- 3i + 4j - 4k), rAC = 241

Component of F along rAC is F| |

F # rAC (475.568i + 329.326j - 159.311k) # (- 3i + 4j - 4k)

F| | = =
rAC 241
F| | = 82.4351 = 82.4 N Ans.

Component of F perpendicular to rAC is F

F2 + F2|| = F2 = 6002

F2 = 6002 - 82.43512

F = 594 N Ans.