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4.2 Moments in Three-Dimensional Force Systems
4.2 Moments in Three-Dimensional Force Systems Example 1, page 1 of 3
200 mm
200 mm
A
F = 50 N
1
1. Use the cross-product definition of the moment of a force to determine the
moment of the force about point A. Also, compare the sign of the result with
that obtained from the scalar definition of positive moment, M = Fd.
B
HEAVY DUTY 250MM CHINA
Introduce a position vector rAB
with tail at A and head at B.
B
A
rAB = { 200i} mm
x
y
+
2 Express the force in rectangular components.
z
y
x
F = 50 N
30°
(50 N) sin 30° = 25 N
(50 N) cos 30° = 43.30 NF = {43.30i 25j} N (1)
HEAVY DUTY 250MM CHINA
y
x
z
z
30°
4.2 Moments in Three-Dimensional Force Systems Example 1, page 2 of 3
6 MA = ( 200 mm)( 25 N)k
= {5000k} N·mm
= {5k} N·m Ans.
3 Use the vector cross product to compute the moment
about point A.
MA = rAB F
= { 200i} mm {43.30i 25j} N
= [ 200(43.30) i i + ( 200)( 25) i j ] N·mm
= 0, because cross
product of parallel
vectors is zero
= k, by the right-hand rule
4
j
k
i
A memory aid for cross products:5
Assign a plus sign if the product is in
the order indicated by the arrowheads;
minus sign otherwise
k
i
j
i j = +k
i k = j
from i to j
thumb points out
of i-j plane
fingers of right hand
curl from i to j
ik
j
Apply the
right-hand rule.
4.2 Moments in Three-Dimensional Force Systems Example 1, page 3 of 3
k
50 N
8 The 50-N force tends to rotate the
wrench counterclockwise about
the axis (through A) defined by
the unit vector k. By the
right-hand rule, this definition of
positive moment reduces to the
usual sign convention for positive
moment in coplanar problems.
7 Display the MA vector (a double-headed arrow)
z
y
x
MA = {+5k} N·m
y
50 N
MA = {+5k} N·m
x
+
direction of fingersthumb points
out-of-plane
Ans.
HEAVY DUTY 250MM CHINA
HEAVY DUTY 250MM CHINA
A
A
z
30°
30°
4.2 Moments in Three-Dimensional Force Systems Example 2, page 1 of 4
1
2. A force F = 20 N is applied to the end of a string of length L. The other end of
the string is tied to the handle of a wrench as shown. Use the cross-product
definition of the moment to determine the moment of F about point A. Discuss the
effect of distance L on your answer.
Introduce a position vector rAC
with head at C and tail at A.
F = 20 N
B
A
C
rAC
z
y
x
HEAVY DUTY 250MM CHINA
30°x
z
y
C
L
HEAVY DUTY 250MM CHINA
B
F = 20 N
A
200 mm
30°
4.2 Moments in Three-Dimensional Force Systems Example 2, page 2 of 4
200 mm
A
2
B
Determine the rectangular components of rAC.
C
y
z
x
rAC
L cos 30°
rAC = (L cos 30° + 200 mm)i (L sin 30°)j (1)53
L sin 30°4
L
x
y
z
C
F = 20 N
30°
Determine the rectangular
components of the force F
6
F = (20 N) cos 30° i (20 N) sin 30° j (2)
HEAVY DUTY 250MM CHINA
30°
4.2 Moments in Three-Dimensional Force Systems Example 2, page 3 of 4
7
MA = rAC F
= {( L cos 30° 200)i L sin 30° j} mm { 20 cos 30° i 20 sin 30° j} N
= [( L cos 30° 00)( 20 cos 30°)(i i)
= 0
+ ( L cos 30° 00)( 20 sin 30°)(i j)
= k
+ ( L sin 30° )( 20 cos 30°)(j i)
= k
+ ( L sin 30°)( 20 sin 30°)(j j) ] N·mm
= 0
j
k
i
= k[( L cos 30° 200)( 20 sin 30°) ( L sin 30°)( 20 cos 30°)] N·mm
= k[( L cos 30°)( 20 sin 30°) 200( 20 sin 30°)
( L sin 30°)( 20 cos 30°)] N·mm
The terms involving L drop out
= {2000k} N·mm = {2k} N·m Ans.
Use the cross product to compute the moment.
4.2 Moments in Three-Dimensional Force Systems Example 2, page 4 of 4
Ans.
30°
30°
HEAVY DUTY 250MM CHINA
Note: All position vectors
and force vectors are in
the xy plane.
z
x
y
3
2
1
F
rA3
rA1rA2
rAC
C
Discussion: Distance L does not appear in the answer
for MA. Thus the result is valid for all values of L,
and so the head of the position vector can be located
anywhere (for example, points 1, 2, 3, or C in the
figure) on the line of action of the force.
B
8
A
4.2 Moments in Three-Dimensional Force Systems Example 3, page 1 of 6
3. A shower/bathtub grab bar is being pulled by a force F = 30 lb as
shown. Determine the moment of F about the support A. Also determine
the coordinate direction angles of the moment vector and interpret the
result.
16 in.
8 in.
5 in.
A
B
z
x
y
F = 30 lb
40°
60°
4.2 Moments in Three-Dimensional Force Systems Example 3, page 2 of 6
16 in.
8 in.
5 in.
A
B
z
x
y
40°
60°
Introduce a position vector with tail at A and head at B.
rAB = { 16i + 8j + 5k} in. (1)
1
rAB
F = 30 lb
4.2 Moments in Three-Dimensional Force Systems Example 3, page 3 of 6
Determine the rectangular
components of the force F.
2
F y = (30 lb) cos 60°
= 15 lb (2)
3
(30 lb) sin 60°4
F z = (30 lb)(sin 60°)(sin 40°)
= 16.70 lb (3)
5
F x = (30 lb)(sin 60°)(cos 40°)
= 19.90 lb (4)
6
Component form of F,
from Eqs. 2-4:
F = { 19.90i 15j + 16.70k} lb (5)
7
40°
60°
x
y
z
F = 30 lb
4.2 Moments in Three-Dimensional Force Systems Example 3, page 4 of 6
8
MA = rAB F
= { 16i + 8j + 5k } { 19.90i 15j + 16.70k}
Because both vectors each have three non-zero
components, evaluating the cross products is
easier if we use the determinant form.
9
i j k
MA = 16 8 5
19.90 15 16.70
Expand the determinant in terms of the first row,
remembering to insert the minus sign for the j term.
10
8 5 16 5 16 8MA = i j + k
15 16.70 19.90 16.70 19.90 15
= i[8(16.70) 5( 15)] j[ 16(16.70) 5( 19.90)] + k[ 16( 15) 8( 19.90)]
= {208.6i +167.7j + 399.2k} lb·in. Ans.
Calculate the moment.
4.2 Moments in Three-Dimensional Force Systems Example 3, page 5 of 6
Observation: The above procedure is tedious and error prone. A much
better approach is to use a calculator with a built-in function for evaluating
the cross product of two vectors. If such a calculator is available, all you
need to do is enter the components of the vectors and then let the calculator
perform the arithmetic. As a result, typical errors such as missing a minus
sign or mis-copying a number from one line to the next are avoided.
11
To determine the coordinate direction angles, first determine the
magnitude of the moment.
M A = (208.6)2 + (167.7)2 + (399.2)2
= 480.6 lb·in.
480.6
399.2
480.6
167.7
480.6
208.6
Coordinate direction angles
= cos-1 = cos-1 = 64.3° Ans.
= cos-1 = cos-1 = 69.6° Ans.
= cos-1 = cos-1 = 33.8° Ans.zM
A
M A
M A
M A
M Ay
M Ax
4.2 Moments in Three-Dimensional Force Systems Example 3, page 6 of 6
16 in.
5 in.
A
B
z
x
y
40°
60°
64.3°
69.6°
33.8°
M A = 480.6 lb·in.
Interpretation: The force F = 30 lb
produces a moment of 480.6 lb·in. about
point A. This moment tends to rotate the
grab bar about an axis defined by the moment vector.
12
Axis of rotation
Ans.
F = 30 lb
4.2 Moments in Three-Dimensional Force Systems Example 4, page 1 of 3
4. A force F = 15 N acting parallel to the z axis is applied to the
handle of a socket wrench to turn a bolt at A. Determine the
moment of the force about the point A. Also, state which
component of the moment tends to turn the bolt.
100 mm
80 mm
A
B
x
y
z
F = 15 N
C
4.2 Moments in Three-Dimensional Force Systems Example 4, page 2 of 3
Introduce a position vector
with head at B and tail at A.
rAB = {80i 100j} mm
1
F = 15 N
80 mm
z
A
y
100 mm
B
x
Calculate the moment.
MA = rAB F
= (80i 100j) { 15k}
= 80( 15)(i k) + ( 100)( 15)(j k)
= j = i
= ( 1200)( j) + (1500)i
= {1500i + 1200j} N·mm
= {1.5i + 1.2j} N·m Ans.
2
i
k
j
(F points in negative z-direction.)
rAB
C
4.2 Moments in Three-Dimensional Force Systems Example 4, page 3 of 3
Display the moment vector.3
The component that tends to
rotate the shaft AC of the
wrench about the x axis
(and thus turn the bolt) is
M A = 1.5 N·m Ans.
4
100 mm
15 N
B
x
y
z
A
80 mm
M A = 1.5 N·m
x
M A = 1.2 N·m
y
M A
x
C
4.2 Moments in Three-Dimensional Force Systems Example 5, page 1 of 6
5. Pulley B is used to drive pulley C. Determine the resultant
moment about bearing A produced by the belt forces acting on
pulley B. Also, interpret your result.
z
x
y
30°
Belt forces
A
BC
D
E
F
Q = 55 N
P = 30 N
40 mm
Radius = 70 mm
4.2 Moments in Three-Dimensional Force Systems Example 5, page 2 of 6
z
x
y
30°
Belt forces
A
BC
D
E
F
Q = 55 N
P = 30 N
40 mm
Radius = 70 mm
1 Introduce position vectors with
tails at A and heads at E and F.
rAF
rAE
2 From the figure,
rAE = {40i 70j} mm (1)
4.2 Moments in Three-Dimensional Force Systems Example 5, page 3 of 6
30°
z
x
y
30°
Belt forces
ABC D
E
F
Q = 55 N
P = 30 N
40 mm
Radius = 70 mm
3 To determine the components of r AF, consider a
view of the pulley from the positive x axis:
rAF
rAE
7 In component form, from Eqs. 2 and 3,
rAF
= {40i + 60.62j 35k} mm (4)
Q = 55 N
P = 30 N
z
y
B
F
6
4
5
rAFz = (70 mm) cos 60° = 35 mm (3)
Angle = 90° 30° = 60°
rAFy
= (70 mm) sin 60°
= 60.62 mm (2)
Radius = 70 mm
4.2 Moments in Three-Dimensional Force Systems Example 5, page 4 of 6
z
x
y
30°
Belt forces
AB
CD
E
F
Q = 55 N
P = 30 N
40 mm
Radius = 70 mmrAF
rAE
8 Express the forces in rectangular components.
Q = (55 N) sin 30°j + (55 N) cos 30°k
= {27.5j + 47.63k} N (5)
P = {30k} N (6)
4.2 Moments in Three-Dimensional Force Systems Example 5, page 5 of 6
MA= rAF × Q + rAE × P
9
i j k i j k
MA = 40 60.62 35 + 40 70 0
0 27.5 47.63 0 0 30
60.62 35 40 35 40 60.62 = i j + k
27.5 47.63 0 47.63 0 27.5
= i[60.62(47.63) ( 35)(27.5)] j[40(47.63) ( 35)(0)] + k[40(27.5) 60.62(0)]
+ i[ 70(30) 0(0)] j[40(30) 0(0)] + k[40(0) 70)(0)]
= {1750i 3105j + 1100k} N·mm
= {1.750i 3.105j + 1.100k} N·m Ans.
70 0 40 0 40 70 + i j + k
0 30 0 30 0 0
Calculate the resultant moment.
4.2 Moments in Three-Dimensional Force Systems Example 5, page 6 of 6
40 mm
z
x
y
30°
Belt forces
AB
C
55 N
30 N
M x = 1.750 N·m
Interpretation: The belt forces acting on pulley B tend to
rotate the entire structure with a 3.73 N·m moment about an
axis defined by the direction of the moment vector MA.
The 1.750 N·m x component of MA is the component of
moment that rotates the shaft and drives pulley C.
11
Ans.
The magnitude of the moment is
M A = (1.750)2 + ( 3.105)2 + (1.100)2
= 3.73 N·m
10
4.2 Moments in Three-Dimensional Force Systems Example 6, page 1 of 3
6. A child on a bicycle collides with a mailbox and
exerts the force F shown. If the base of the pole at O
will fail if the magnitude of the moment there exceeds
60 N·m, determine if the mailbox will fall over.
250 mm75 mm
A
O
x
y
z
F = {80i + 12j 10k} N
900 mm
4.2 Moments in Three-Dimensional Force Systems Example 6, page 2 of 3
Introduce a position vector with tail at O and
head at A.
rOA = {250i + 900j + 75k} mm
1
i j k
= 250 900 75
80 12 10
MO = rOA × F
2
rOA
= i[900( 10) (12)(75)] j[250( 10) 80(75)] + k[250(12) 80(900)]
= { 9900i + 8500j 69000k} N·mm
= { 9.9i + 8.5j 69.0k} N·m
900 75 250 75 250 900 = i j + k
12 10 80 10 80 12
Calculate the moment.
250 mm75 mm
A
O
x
y
z
F = {80i + 12j 10k} N
900 mm
4.2 Moments in Three-Dimensional Force Systems Example 6, page 3 of 3
Magnitude of moment
M O = ( 9.9)2 + (8.5)2 + ( 69.0)2
= 70.2 N·m
3
Because 70.2 N·m exceeds the 60 N·m
maximum allowable moment, the
mailbox will fall over.
4
Ans.
4.2 Moments in Three-Dimensional Force Systems Example 7, page 1 of 6
7. Copper tubing emerges from the wall at A and is subjected to a
force F at its free end B. The tubing will fail if the magnitude of
the moment at A exceeds 3 N·m. Determine the largest value of
the force F that can be applied to the free end of the tubing.
30°
40°250 mm
300 mm
A
Bx
z
y
F
35°
200 mm
4.2 Moments in Three-Dimensional Force Systems Example 7, page 2 of 6
250 mm
200 mm
300 mm35°A
B
x
z
y
F
1 Introduce a position vector rAB
with tail at A and head at B.
r AB
30°
40°
35°
4.2 Moments in Three-Dimensional Force Systems Example 7, page 3 of 6
250 mm
200 mm
300 mm
A
B
x
z
y
F
2 Determine the
components of r AB.
r AB
3 r ABx = 200 mm + 250 mm = 450 mm (1)
250 mm
D
C
30°
40°
35°
4.2 Moments in Three-Dimensional Force Systems Example 7, page 4 of 6
7 In component form, from Eqs. 1-3,
r AB = {450i 172.1j + 245.7k} mm (4)
r ABz = (300 mm) cos 35° = 245.7 mm (3)6
4
r ABy
= (300 mm) sin 35°
= 172.1 mm (2)
5
To determine the y and z components of rAB,
consider a view from the positive x axis.
r AB
35°
300 mm
y
z
x, B, D
C, A
4.2 Moments in Three-Dimensional Force Systems Example 7, page 5 of 6
8 Determine the components of F.
F sin 40°10
F z = F sin 40° sin 30°
= 0.3214F (6)
11B x
z
y
F
F y = F cos 40°
= 0.7660F (5)
9
F x = F sin 40° cos 30°
= 0.5567F (7)
12
13 In component form, from Eqs. 5-7,
F = F{0.5567i 0.7660j + 0.3214k} (8)
30°
40°
4.2 Moments in Three-Dimensional Force Systems Example 7, page 6 of 6
MA = rAB × F
14
i j k
= 450 172.1 245.7
0.5567F 0.7660F 0.3214F
172.1 245.7 450 245.7 450 172.1 = i j + k
0.7660F 0.3214F 0.5567F 0.3214F 0.5567F 0.7660F
= (132.9F)i (7.8F)j (248.9F)k
M A = (132.9F)2 + ( 7.8F)2 + ( 248.9F)2
= F (132.9)2 + ( 7.8)2 + ( 248.9)2
= 282.3F N·mm
= 0.2823F N·m
15
Equate M A to the largest allowable moment, 3 N·m:
M A = 3 N·m
0.2823F N·m
Solving gives
F = 10.6 N Ans.
16
Calculate the resultant moment. Compute the magnitude.
4.2 Moments in Three-Dimensional Force Systems Example 8, page 1 of 6
8. Two forces, P = 60 N and Q = 80 N act on the vertices of a cube as
shown. Determine the moment of each force about point O, if the length
of each edge of the cube is 2 m. Also, determine the shortest distance
from O to the line BF.
x
z
y
A B
CD
E F
G
O
P = 60 N Q = 80 N
Each edge is
2 m long.
4.2 Moments in Three-Dimensional Force Systems Example 8, page 2 of 6
x
z
y
A B
CD
E F
G
O
P = 60 N Q = 80 N
Each edge is
2 m long.
To determine the
moment of the force
P about point O, we
have several choices
of position vector
(Recall that the head
of the vector can lie
anywhere on the line
of action of P).
1
Select the position vector with
the simpler form.
rOD = {2i + 2j +2k} m
rOE = {2j} m (1)
2
Simpler (only one component)
rOD
rOE
4.2 Moments in Three-Dimensional Force Systems Example 8, page 3 of 6
x
z
y
A B
CD
E F
G
O
P = 60 NQ = 80 N
Each edge is
2 m long.
Determine the components of
the force P by introducing a
position vector rDE:
rDE = { 2i 2k} m
3
P = (60 N)
= (60 N)
= { 42.43i 42.43k} N (2)
4
rDE
2i 2k
( 2)2 + ( 2)2
rDE
r DE
The force P has magnitude 60 N and
points in the direction from D to E, so
4.2 Moments in Three-Dimensional Force Systems Example 8, page 4 of 6
5 Calculate the moment.
MOP = rOE P
= {2j} { 42.43i 42.43k}
= (2)( 42.43)(j i) (2)(42.43)(j k)
= { 84.86i + 84.86k} N·m Ans.
j
k
i= k = i
Since rOE has only one component, it is easier to
calculate cross products of base vectors individually
rather than use the determinant approach to calculating
the cross product.
4.2 Moments in Three-Dimensional Force Systems Example 8, page 5 of 6
Calculate the moment.
M OQ
= r OB × Q
8
i j k
= 2 0 2
0 56.57 56.57
0 2 2 2 2 0 = i j + k
56.57 56.57 0 56.57 0 56.57
= {113.14i 113.14j 113.14k} N·m (3) Ans.
To determine the moment of the force Q
about point O, we could introduce either
position vector rOB or rOF. Let's
arbitrarily choose rOB.
rOB = {2i + 2k} m
6
Determine the components of
force Q by noting it has
magnitude 80 N and points from F to B so
Q = (80 N)
= (80 N)
= { 56.57j + 56.57k} N
7
2j + 2k
( 2)2 + (2)2
rFB
r FB
x
z
y
A B
CD
E F
GO
P = 60 N Q = 80 N
rFB
rOB
rOF
Each edge is
2 m long.
4.2 Moments in Three-Dimensional Force Systems Example 8, page 6 of 6
x
z
y
A B
CD
E F
G
O
P = 60 N Q = 80 N
Each edge is
2 m long.
Finally, to determine the
shortest distance from point O
to line FB, consider the plane
defined by O and FB:
9
From the figure it is clear that of the various distances
d 1, d
2, d
3, ..., from O to line FB, the perpendicular
distance d is the shortest. But we also know that, by
the scalar definition of moment, the moment of Q
about point O equals the perpendicular distance from O
to the line of action of Q times Q. That is,
M QO
= Qd
Thus,
d =
= = 2.45 m Ans.
10
Q = 80 N
B
F
O
d 1
d 2
d 3
d
(113.14)2 + ( 113.14)2 + ( 113.14)2
80
M QO
Q
d
4.2 Moments in Three-Dimensional Force Systems Example 9, page 1 of 6
9. Determine the moment about the screw at A of the force F = 2 N
applied to the sheet-metal bracket shown. Also, determine the shortest
distance from A to the line connecting B and C.
x
y
z
60 mm
100 mm
70 mm
80 mm
30 mm
60 mm50 mm
A
B
C
F = 2 N
25°
4.2 Moments in Three-Dimensional Force Systems Example 9, page 2 of 6
To calculate the moment
we could use either the
position vector rAB or rAC
.
1
rAC has a single component,
rAC = (50 mm + 60 mm)i
= { 110i} mm (1)
while rAB has two components, so let's
choose the simpler, rAC.
2
x
y
z
AC
F = 2 N
rAC
rAB
B 25°
50 mm 60 mm
4.2 Moments in Three-Dimensional Force Systems Example 9, page 3 of 6
25°
B 25°
x
30 mm
B
A
Next, we need to calculate the components
of the force, F. To do this, we first need to
calculate the coordinates of point B. By
inspection, the x coordinate of B is 60 mm.
3
To find the y and z
coordinates of point B,
consider a view from
the positive x-axis.
4
y coordinate of B
= 30 mm 70 mm 42.26 mm
= 142.26 mm
5
z coordinate of B
= 90.63 mm + 80 mm
= 170.63 mm
6
y
z
60 mm
100 mm
70 mm
80 mm
60 mm50 mm
AC
F = 2 N
100 mm
80 mm
30 mm
70 mm
(100 mm) sin 25° = 42.26 mm
(100 mm) cos 25°
= 90.63 mm
z
y
4.2 Moments in Three-Dimensional Force Systems Example 9, page 4 of 6
To determine the components of the force, introduce the
position vector rBC.
rBC = ( 50 mm 60 mm)i
+ [0 ( 142.26 mm)]j
+ (0 170.63 mm)k
= { 110i + 142.26j 170.63k} mm (2)
7
F = (2 N)
= (2 N)
= { 0.887i + 1.148j 1.377k} N
8
110i + 142.26j 170.63k
( 110)2 + (142.26)2 + ( 170.63)2
rBC
r BC
y
z
60 mm
100 mm
70 mm
80 mm
60 mm50 mm
AC
F = 2 N
C coordinates
( 50 mm, 0, 0)
B coordinates (60, 142.26 mm, 170.63 mm)
rBC
B 25°The force F has magnitude 2 N and
points from B to C so
x
30 mm
4.2 Moments in Three-Dimensional Force Systems Example 9, page 5 of 6
9
MA = rAC F
= { 110i} { 0.887i + 1.148j 1.377k}
= ( 110)( 0.887)(i i) + ( 110)(1.148)(i j) + ( 110)( 1.377)(i k)
= { 151.4j 126.3k} N·mm (3) Ans.
= j= k= 0
k
j
i
Calculate the moment.
4.2 Moments in Three-Dimensional Force Systems Example 9, page 6 of 6
To determine the shortest distance
between point A and line BC, consider the
plane formed by A and BC.
10
Magnitude of moment about A = Fd
or,
( 151.4)2 + ( 126.3)2 N mm = (2 N)d
Solving gives,
d = 98.6 mm Ans.
F = 2 N
d
A
B
C
11
by Eq. 3
y
z
60 mm
100 mm
70 mm
80 mm
60 mm50 mm
AC
F = 2 N
B 25°
x
30 mm
4.2 Moments in Three-Dimensional Force Systems Example 10, page 1 of 5
10. If the tension in the cable BC is T = 80 lb,
determine the moment about point A of the cable force
acting on the frame at point B. Also, determine the
shortest distance from A to the line through B and C.
32 in.
18 in. 10 in.
AB
x
y
z
T = 80 lb
C
12 in.
4.2 Moments in Three-Dimensional Force Systems Example 10, page 2 of 5
To calculate the
moment, we could
use either position
vector rAB or rAC.
1
Since rAC has only one
component while r AB has three,
let's use rAC.
rAC = {12j} in. (1)
2
32 in.
18 in.10 in.
AB
x
y
z
T = 80 lb
C
rAC
rAB
12 in.
4.2 Moments in Three-Dimensional Force Systems Example 10, page 3 of 5
To determine the components of the force, T,
introduce the position vector rBC.
3
T = (80 lb)
= (80 lb)
= { 69.62i + 4.35j 39.16k} lb
4
32i + 2j 18k
( 32)2 + 22 + ( 18)2
rBC
r BC
rBC = (0 32 in.)i + (12 in. 10 in.)j + (0 18 in.)k
= { 32i + 2j 18k} in.
32 in.
18 in. 10 in.
AB
x
y
z
T = 80 lb
C
rBC
The force T has magnitude 80 lb and
points from B to C so
12 in.
4.2 Moments in Three-Dimensional Force Systems Example 10, page 4 of 5
5 Calculate the moment.
MA = rAC T
= 12j { 69.62i + 4.35j 39.16k}
= (12)( 69.62)(j i) + (12)(4.35)(j j) + (12)( 39.16)(j k)
= { 469.92i + 835.44k} lb·in. Ans. (2)
= k = i= 0
k
j
i
by Eq. 1
4.2 Moments in Three-Dimensional Force Systems Example 10, page 5 of 5
Finally, to determine the shortest distance
between point A and line BC, consider the
plane formed by A and BC.
6
Magnitude of moment about A = Td
or,
( 469.92)2 + (835.44)2 = (80 lb)d
Solving gives,
d = 11.98 in. Ans.
7
32 in.
18 in. 10 in.
AB
x
y
z
T = 80 lb
C
T = 80 lb
d
Extension of BC
A
B
C
by Eq. 2
12 in.