1. Engineering Mechanics : STATICS
BDA10203
Lecture #09
Group of Lecturers
Faculty of Mechanical and Manufacturing Engineering
Universiti Tun Hussein Onn Malaysia

3. MOMENT OF A FORCE (Section 4.1)
Today’s Objectives :
Students will be able to:
a) understand and define moment, and,
b) determine moments of a force in 2-D and 3-D cases.
Learning Topics :
Applications
Moment in 2-D
Moment in 3-D

4. APPLICATIONS (continued)
What is the effect of the 30 N force on the lug nut?

5. MOMENT IN 2-D The moment of a force about a point or axis
A measure of the tendency of the force to cause a body to rotate about the point or axis
Sometimes called a torque

6. When apply force Fy to the wrench No moment is produced about point O
MOMENT IN 2-D
Note that
When apply force Fy to the wrench
No moment is produced about point O
Lack of tendency to rotate
as line of action passes
through O

7. MOMENT IN 2-D (continued)
Moment MO is vector quantity and has its specified magnitude and direction
In the 2-D case, the magnitude of the moment is
Mo = F d
As shown, d is the perpendicular distance from point O to the line of action of the force, F.
Units for moment is N.m

8. MOMENT IN 2-D (continued)
In 2-D, the direction of MO is either clockwise or counter-clockwise depending on the tendency for rotation.
Direction of MO is specified by using “right hand rule”
- fingers of the right hand are curled to follow the sense of rotation when force rotates about point O
-Using the right hand rule, the direction and sense of the moment vector points out of the page
The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive

9. Example 4.1
For each case, determine the moment of the
force about point O

10. Solution
Line of action is extended as a dashed line to establish moment arm d
Tendency to rotate is indicated and the orbit is shown as a colored curl
The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. Mo +

11. Solution

12. Example 4.2
Determine the moments of
the 800N force acting on the
frame about points A, B, C
and D.

13. Solution Line of action of F passes through C
The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. Mx +

14. RESULTANT MOMENT IN 2-D
Resultant moment of a system of coplanar forces can be determined
by simply adding the moments of all forces algebraically since all
the moment vectors are collinear.
+ MRo = ∑Fd

15. Determine the resultant moment of the four
Example 4.3
Determine the resultant moment of the four
forces acting on the rod about point O
Solution
Assume positive moments acts in the +k
direction, CCW

16. MOMENT IN 3-D
Moments in 3-D can be calculated using scalar (2-D) approach but it can be difficult and time consuming. Thus, it is often easier to use a mathematical approach called the vector cross product.
Read as “Momen equals vector r cross vector F”

17. CROSS PRODUCT
In general, the cross product of two vectors A and B results in another vector C , i.e., C = A  B. The magnitude and direction of the resulting vector can be written as
C = A  B = A B sin  UC
Here UC is the unit vector perpendicular to both A and B vectors as shown (or to the plane containing the A and B vectors).

18. Cartesian Vector Formulation
CROSS PRODUCT
Cartesian Vector Formulation
Use C = AB sinθ on pair of Cartesian unit vectors
Example
For i X j, (i)(j)(sin90°)
= (1)(1)(1) = 1
For i X i, (i)(i)(sin0°)
= (1)(1)(0) = 0

19. CROSS PRODUCT
The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product.
For example: i  j = k , j x i = -k
Note that a vector crossed into itself is zero, e.g., i  i = 0

20. Consider cross product of vector A and B
Laws of Operations
Consider cross product of vector A and B
A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk)
= AxBx (i X i) + AxBy (i X j) + AxBz (i X k)
+ AyBx (j X i) + AyBy (j X j) + AyBz (j X k)
+ AzBx (k X i) +AzBy (k X j) +AzBz (k X k)
A X B = (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k

21. CROSS PRODUCT (continued)
Of even more utility, the cross product can be written as
Each component can be determined using 2  2 determinants.

22. MOMENT IN 3-D
Moment of force F about point O can be expressed using cross product
MO = r X F
where r represents position
vector from O to any point
lying on the line of action
of F

23. Magnitude Direction MOMENT IN 3-D For magnitude of cross product,
MO = r F sinθ
where θ is the angle measured between tails of r and F.Treat r as a sliding vector.
Since d = r sinθ,
MO = rF sinθ = F (rsinθ) = Fd
Direction
- “curl” of the fingers indicates the sense of rotation

24. MOMENT IN 3-D (continued)
So, using the cross product, a moment can be expressed as
By expanding the above equation using 2  2 determinants we get (sample units are N - m)
MO = (r y FZ - rZ Fy) i - (r x Fz - rz Fx ) j (rx Fy - ry Fx ) k
The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D formulation.

25. PRINCIPLE OF TRANSMISSIBILITY
For force F applied at any point A, moment created about O is MO = rA x F
F has the properties of a sliding vector and
therefore act at any point along its line of action and still create the same moment about O
MO = rA x F = rB x F = rC x F

26. PRINCIPLE OF TRANSMISSIBILITY
Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant
Given the perpendicular distance from A to cable is rd
MA = rdF
In 3D problems, using cross product,
MA = rAB x F = rAC x F

27. RESULTANT MOMENT IN 3-D MRo = ∑(r x F)
Resultant moment of a System of Forces :
If a body is acted upon by a system of forces as shown in the
figure, the resultant moment of the forces about point O can be
determined by vector addition as
MRo = ∑(r x F)

28. EXAMPLE
Example 4.4
The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.
Solution
Either one of the two position vectors can be used for the solution,
MA = rB x F or MA = rC x F

29. Position vectors are represented as rB = {1i + 3j + 2k} m and
Solution
Position vectors are represented as
rB = {1i + 3j + 2k} m and
rC = {3i + 4j} m
Force F has magnitude 60N and is
directed from C to B
Substitute into determinant formulation

30. Solution Or
For magnitude,

31. EXAMPLE Given: The pole support a 100N (≈ 10kg) traffic light.
Find: Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A.
Plan: 1) Find a position vector of traffic light.
2) Using Cartesian vector notation, determine the force that the traffic light acted on the pole.
3) MA = r X F
5.4m=

32. EXAMPLE (continued) F = F (UCA)
Vector = Magnitude x direction
F = F (UCA)
UCA = rCA / | rCA | = -5.4 k / 5.4 = -1 k
F = 100{-1k} = -100k B
5.4m=
A(0,0,0) F
rAD = {( XD – XA)i +(YD – YA)j +(ZD – ZA) k}m
= {(3.6sin30 – 0)i + (3.6cos30 – 0)j
+ (5.4 – 0)k}m
= {1.8i j + 5.4k}m
D (3.6 sin 30, 3.6 kos 30, 0)
MA = rAD X F = {(1.8i j + 5.4k) x (-100k)} Nm
= (3.12 x -100)i - (1.8 x -100)j + (0) k
= -312 i j

33. IN CLASS TUTORIAL (GROUP PROBLEM SOLVING)
Given: The man pulls on the rope with a force of F = 20N.
Find: Determine the moment that this force exerts about the base of the pole at O.
Plan: 1) Find a position vector of rOA and rAB.
2) Using Cartesian vector notation, determine the force that acted on the pole.
3) MA = r X F

34. GROUP PROBLEM SOLVING (Continue)
rAB = {( XB – XA)i +(YB – YA)j +(ZB – ZA) k}m
= {(4 – 0)i + (-3 – 0)j + (1.5 – 10.5)k}m
= {4i - 3j - 9k}m
rOA = {( XA – XO)i +(YA – YO)j +(ZA – ZO) k}m
= {(0 – 0)i + (0 – 0)j + (10.5 – 0)k}m
= {0i + 0j k}m
F = FuAB = F(rAB /rAB )
= 20{0.4i - 0.3j - 0.9k}
={8i – 6j - 18k} N
MO = rOA X F = {63i + 84j} Nm

35. HOMEWORK TUTORIAL Q1 (4-39):
The curved rod lies in the x-y plane and has a radius r. If a force F acts at its end as shown, determine the moment of this force about point B.

36. HOMEWORK TUTORIAL (continued)
Q2 (4-40):
The force F acts at the end of the beam. Determine the moment of the force about point A.

37. Resultant Moment of Forces
Example 4.5
Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.

38. Solution Position vectors are directed from point O to each force
rA = {5j} m and
rB = {4i + 5j - 2k} m
For resultant moment about O,

39. Solution For magnitude
For unit vector defining the direction of moment axis,

40. Solution
For the coordinate angles of the moment axis,

41. PRINCIPLE OF MOMENTS F a b d O
For example, MO = F d and the direction is counter-clockwise.
Often it is easier to determine MO by using the components of F as shown. a b O F
F x
F y
Using this approach, MO = (FY a) – (FX b).
Note the different signs on the terms!
The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive.

42. PRINCIPLE OF MOMENTS
Principle of moments is also known as Varignon’s Theorem
“Moment of a force about a point is equal to the
sum of the moments of the forces’ components
about the point”
For F = F1 + F2,
MO = r X F1 + r X F2
= r X (F1 + F2)
= r X F

43. Principles of Moments Example 1.0
A 200N force acts on the bracket. Determine
the moment of the force about point A.

44. Principles of Moments Solution Method 1:
From trigonometry using triangle BCD,
CB = d = 100cos45° = 70.71mm
= m
Thus,
MA = Fd = 200N( m)
= 14.1N.m (CCW)
As a Cartesian vector,
MA = {14.1k}N.m

45. Principles of Moments Solution Method 2:
Resolve 200N force into x and y components
Principle of Moments
MA = ∑Fd
MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m)
= 14.1 N.m (CCW)
Thus,
MA = {14.1k}N.m

46. Principles of Moments Example 2.0
The force F acts at the end of the angle
bracket. Determine the moment of the force
about point O.

47. Principles of Moments Solution Method 1
MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.6N.m
= 98.6N.m (CCW)
As a Cartesian vector,
MO = {-98.6k}N.m

48. Principles of Moments Solution Method 2: Express as Cartesian vector
r = {0.4i – 0.2j}N
F = {400sin30°i – 400cos30°j}N
= {200.0i – 346.4j}N
For moment,

49. EXAMPLE 3.0 Given: A 400 N force is applied to the frame and  = 20°.
Find: The moment of the force at A.
Plan:
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.

50. EXAMPLE (continued) Solution +  Fy = 400 sin 20° N
+  Fx = 400 cos 20° N
+ MA = {(400 sin 20°)(3) + (400 cos 20°)(2)} N·m
= N·m

51. IN CLASS TUTORIAL (GROUP PROBLEM SOLVING)
Given: A 40 N force is applied to the wrench.
Find: The moment of the force at O.
Plan: 1) Resolve the force along x and y axes.
2) Determine MO using scalar analysis.
Solution: Fy = 40 cos 20° N
Fx = 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= N·mm = N·m

52. ATTENTION QUIZ
10 N
5 N
3 m P m
1. Using the CCW direction as positive, the net moment of the two forces about point P is
A) 10 N ·m B) 20 N ·m C) N ·m
D) 40 N ·m E) N ·m
2. If r = { 5 j } m and F = { 10 k } N, the moment
r x F equals { _______ } N·m.
A) 50 i B) 50 j C) –50 i
D) – 50 j E) 0
Answers:
1. B
2. A

53. CONCEPT QUIZ
1. If a force of magnitude F can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut. (Max, Min).
A) (Q, P) B) (R, S)
C) (P, R) D) (Q, S)
Answers:
1. D
2. A
2. If M = r  F, then what will be the value of M • r ?
A) 0 B) 1
C) r 2 F D) None of the above.

54. HOMEWORK TUTORIAL Q1 (4-4):
Determine the magnitude and directional sense of the resultant moment
of the forces A and B about point O.
=45°
=3m
=5m
=30°
=40kN
3m=
=13m
=6m
=60kN
(Ans: 858 kN.m CW)

55. HOMEWORK TUTORIAL (continued)
Q2 (4-5):
Determine the magnitude and directional sense of the resultant moment
of the forces A and B about point P.
=45°
=3m
=5m
=30°
=40kN
3m=
=13m
=6m
=60kN
(Ans: 1165 kN.m CW)

56. HOMEWORK TUTORIAL (continued)
Q3 (4-31):
The worker is using the bar to pull two pipes together in order to
complete the connection. If he applies a horizontal force F to the
handle of the lever, determine the moment of this force about the
end A. What would be the tension T in the cable needed to cause
the opposite moment about point A.
Ans: MA=460N.m CCW
T = 3.26kN