1. Engineering Mechanics: Statics
Chapter 4:
Force System Resultants
Engineering Mechanics: Statics

2. Chapter Objectives
To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions.
To provide a method for finding the moment of a force about a specified axis.
To define the moment of a couple.
To present methods for determining the resultants of non-concurrent force systems.
To indicate how to reduce a simple distributed loading to a resultant force having a specified location.

3. Chapter Outline Moment of a Force – Scalar Formation Cross Product
Moment of Force – Vector Formulation
Principle of Moments
Moment of a Force about a Specified Axis

4. Chapter Outline Moment of a Couple Equivalent System
Resultants of a Force and Couple System
Further Reduction of a Force and Couple System
Reduction of a Simple Distributed Loading

5. 4.1 Moment of a Force – Scalar Formation
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
Case 1
Consider horizontal force Fx,
which acts perpendicular to
the handle of the wrench and
is located dy from the point O

6. 4.1 Moment of a Force – Scalar Formation
Fx tends to turn the pipe about the z axis
The larger the force or the distance dy, the greater the turning effect
Torque – tendency of
rotation caused by Fx
or simple moment (Mo) z

7. 4.1 Moment of a Force – Scalar Formation
Moment axis (z) is perpendicular to shaded
plane (x-y)
Fx and dy lies on the shaded plane (x-y)
Moment axis (z) intersects
the plane at point O

8. 4.1 Moment of a Force – Scalar Formation
Case 2
Apply force Fz to the wrench
Pipe does not rotate about z axis
Tendency to rotate about x axis
The pipe may not actually
rotate Fz creates tendency
for rotation so moment
(Mo) x is produced

9. 4.1 Moment of a Force – Scalar Formation
Case 2
Moment axis (x) is perpendicular to shaded
plane (y-z)
Fz and dy lies on the shaded plane (y-z)

10. 4.1 Moment of a Force – Scalar Formation
Case 3
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

11. 4.1 Moment of a Force – Scalar Formation
In General
Consider the force F and the point O which lies in the shaded plane
The moment MO about point O,
or about an axis passing
through O and perpendicular
to the plane, is a vector quantity
Moment MO has its specified
magnitude and direction

12. 4.1 Moment of a Force – Scalar Formation
Magnitude
For magnitude of MO,
MO = Fd
where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force
Units for moment is N.m

13. 4.1 Moment of a Force – Scalar Formation
Direction
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

14. 4.1 Moment of a Force – Scalar Formation
Direction
- Thumb points along the moment axis to give the direction and sense of the moment vector
- Moment vector is upwards and perpendicular to the shaded plane

15. 4.1 Moment of a Force – Scalar Formation
Direction
MO is shown by a vector arrow
with a curl to distinguish it from
force vector
Example (Fig b)
MO is represented by the counterclockwise curl, which indicates the action of F

16. 4.1 Moment of a Force – Scalar Formation
Direction
Arrowhead shows the sense of rotation caused by F
Using the right hand rule, the direction and sense of the moment vector points out of the page
In 2D problems, moment of the force is found about a point O

17. 4.1 Moment of a Force – Scalar Formation
Direction
Moment acts about an axis perpendicular to the plane containing F and d
Moment axis intersects
the plane at point O

18. 4.1 Moment of a Force – Scalar Formation
Resultant Moment of a System of
Coplanar Forces
Resultant moment, MRo = addition of the moments of all the forces algebraically since all moment forces are collinear
MRo = ∑Fd
taking clockwise to be positive

19. 4.1 Moment of a Force – Scalar Formation
Resultant Moment of a System of
Coplanar Forces
A clockwise curl is written along the equation to indicate that a positive moment if directed along the
+ z axis and negative
along the – z axis

20. 4.1 Moment of a Force – Scalar Formation
Moment of a force does not always cause rotation
Force F tends to rotate the beam clockwise about A with moment
MA = FdA
Force F tends to rotate the beam counterclockwise about B with moment
MB = FdB
Hence support at A prevents
the rotation

21. 4.1 Moment of a Force – Scalar Formation
Example 4.1
For each case, determine the moment of the
force about point O

22. 4.1 Moment of a Force – Scalar Formation
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

23. 4.1 Moment of a Force – Scalar Formation
Solution

24. 4.1 Moment of a Force – Scalar Formation
Example 4.2
Determine the moments of
the 800N force acting on the
frame about points A, B, C
and D.

25. 4.1 Moment of a Force – Scalar Formation
Solution
Scalar Analysis
Line of action of F passes through C

26. 4.1 Moment of a Force – Scalar Formation
Example 4.3
Determine the resultant moment of the four
forces acting on the rod about point O

27. 4.1 Moment of a Force – Scalar Formation
Solution
Assume positive moments acts in the +k
direction, CCW
View Free Body Diagram

28. 4.2 Cross Product
Cross product of two vectors A and B yields C, which is written as
C = A X B
Read as “C equals A cross B”

29. 4.2 Cross Product Magnitude
Magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle θ between their tails
For angle θ, 0° ≤ θ ≤ 180°
Therefore,
C = AB sinθ

30. 4.2 Cross Product Direction
Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right hand rule
- Curling the fingers of the right
hand form vector A (cross) to
vector B
- Thumb points in the direction of
vector C

31. 4.2 Cross Product
Expressing vector C when magnitude and direction are known
C = A X B = (AB sinθ)uC
where scalar AB sinθ defines the magnitude of vector C unit vector uC defines the direction of vector C

32. 4.2 Cross Product Laws of Operations 1. Commutative law is not valid
A X B ≠ B X A
Rather,
A X B = - B X A
Shown by the right hand rule
Cross product A X B yields a vector opposite in direction to C
B X A = -C

33. 4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar
a( A X B ) = (aA) X B = A X (aB) = ( A X B )a
3. Distributive Law
A X ( B + D ) = ( A X B ) + ( A X D )
Proper order of the cross product must be maintained since they are not commutative

34. 4.2 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

35. 4.2 Cross Product Laws of Operations In a similar manner,
i X j = k i X k = -j i X i = 0
j X k = i j X i = -k j X j = 0
k X i = j k X j = -i k X k = 0
Use the circle for the results.
Crossing CCW yield positive
and CW yields negative results

36. 4.2 Cross Product 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)
= (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k

37. 4.2 Cross Product
Laws of Operations
In determinant form,

38. 4.3 Moment of Force - Vector Formulation
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

39. 4.3 Moment of Force - Vector Formulation
Magnitude
For magnitude of cross product,
MO = rF 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

40. 4.3 Moment of Force - Vector Formulation
Direction
Direction and sense of MO are determined by right-hand rule
- Extend r to the dashed position
- Curl fingers from r towards F
- Direction of MO is the same
as the direction of the thumb

41. 4.3 Moment of Force - Vector Formulation
Direction
*Note:
- “curl” of the fingers indicates the sense of rotation
- Maintain proper order of r
and F since cross product
is not commutative

42. 4.3 Moment of Force - Vector Formulation
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

43. 4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
For force expressed in Cartesian
form,
where rx, ry, rz represent the x, y, z
components of the position vector
and Fx, Fy, Fz represent that of the
force vector

44. 4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
With the determinant expended,
MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k
MO is always perpendicular to
the plane containing r and F
Computation of moment by cross
product is better than scalar for
3D problems

45. 4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
Resultant moment of forces about point O can be determined by vector addition
MRo = ∑(r x F)

46. 4.3 Moment of Force - Vector Formulation
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,
MA = rBC x F

47. 4.3 Moment of Force - Vector Formulation
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.

48. 4.3 Moment of Force - Vector Formulation
Solution
Either one of the two position vectors can be used for the solution, since MA = rB x F or MA = rC x F
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

49. 4.3 Moment of Force - Vector Formulation
Solution
Substitute into determinant formulation

50. 4.3 Moment of Force - Vector Formulation
Solution Or
Substitute into determinant formulation
For magnitude,

51. 4.3 Moment of Force - Vector Formulation
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.

52. 4.3 Moment of Force - Vector Formulation
View Free Body
Diagram
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,

53. 4.3 Moment of Force - Vector Formulation
Solution
For magnitude
For unit vector defining the direction of moment axis,

54. 4.3 Moment of Force - Vector Formulation
Solution
For the coordinate angles of the moment axis,