1. An-Najah National University College of Engineering
Statics-61110
Chapter [2]
Force Systems
Dr. Sameer Shadeed
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

2. Chapter Objectives Students will be able to:
Resolve a 2D and 3D vector into components
Work with 2D and 3D vectors using Cartesian vector notations
Estimate the resultant of forces and couples in 2D and 3D
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

3. Applications
There are four concurrent cable forces acting on the bracket
How do you determine the resultant force acting on the bracket ?
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

4. Applications
Given the forces in the cables, how will you determine the resultant force acting at D, the top of the tower?
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

5. Scalars and Vectors
Scalars: A mathematical quantity possessing magnitude only (e.g. area, volume, mass, energy)
Vectors: A mathematical quantity possessing magnitude and direction (e.g. forces, velocity, displacement)
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

6. Types of Vectors
A Free vector: is a vector whose action is not confined to or associated with a unique line in space.
A Sliding vector: is a vector which has a unique line of action in space but not a unique point of application.
A Fixed vector: is a vector for which a unique point of application is specified and thus cannot be moved without modifying the conditions of the problem.
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

7. Working with Vectors
For two vectors to be equal they must have the same:
1. Magnitude
2. Direction V
But they do not need to have the same point of application
A negative vector of a given vector has same magnitude but opposite direction V -V
V and -V are equal and opposite V + (-V) = 0
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

8. Vector Operations Product of a scalar and a vector
V + V + V = 3V (the number 3 is a scalar)
This is a vector in the same direction as V but 3 times as long
(+n)V = vector same direction as V, n times as long
(-n)V = vector opposite direction as V, n times as long
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

9. Vector Addition
The sum of two vectors can be obtained by attaching the two vectors to the same point and constructing a parallelogram (Parallelogram law)
As vector: V = V1+ V2
As scalar: V ≠ V1+ V2
Addition of vectors is commutative: V1 + V2 = V2 + V1
The sum of three vector (Parallelogram law) V1 V2 V3 R R1
R1 = V1 + V2
R = R1 + V3 = V1 + V2 + V3
Vector addition is associative: V1 + V2 + V3 = (V1 + V2 ) + V3
= V1 + (V2 + V3 )
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

10. Vector Subtraction
Vector Subtraction: is the addition of the corresponding negative vector
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

11. Two-Dimensional Force System
Rectangular Components: is the most common two-dimensional resolution of a force vector
where Fx and Fy are the vector components of F in the x- and y-directions
In terms of the unit vectors i and j, Fx = Fxi and Fy = Fyj
where the scalers Fx and Fy are the x and y scaler components of the vector F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

12. Two-Dimensional Force System
The scalar components can be positive or negative, depending on the quadrant into which F points
The magnitude and direction of F is expressed by:
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

13. Determining the Components of a Force
Dimensions are not always given in horizontal and vertical directions
Angles need not be measured counterclockwise from the x-axis
The origin of coordinates need not be on the line of action of a force
Therefore, it is essential that we be able to determine the correct components of a force no matter how the axes are originated or how angles are measured
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

14. Resultant of Two Concurrent Forces
From which we can conclude that:
The term means “the algebric sum of x-scalar components”
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

15. Example 1
Statics Notes 2013 [Force Systems]
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16. Example 1 (Solution) Statics Notes 2013 [Force Systems]
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17. Example 1 (Solution) Statics Notes 2013 [Force Systems]
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18. Example 2
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19. Example 2 (Solution) Statics Notes 2013 [Force Systems]
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20. Example 2 (Solution) Statics Notes 2013 [Force Systems]
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21. Example 3
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22. Example 3 (Solution) Statics Notes 2013 [Force Systems]
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23. Example 4
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24. Example 4 (Solution) Statics Notes 2013 [Force Systems]
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25. Example 4 (Solution) Statics Notes 2013 [Force Systems]
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26. Example 4 (Solution) Statics Notes 2013 [Force Systems]
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27. Moment (Moment about a Point)
Moment: is the measure of the tendency of the force to make a rigid body rotate about a point or fixed axis
The magnitude of the moment M is proportional both to the magnitude of the force F and the moment arm d
The basic units of the moment in SI units are newton-meters (N.m), and in the U.S. Custamary system are pound-feet (Ib.ft)
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

28. Moment (Moment about a Point)
The moment is a vector M perpendicular to the plane of the body
The sense of M depends on the direction in which F tends to rotate the body
The right hand rule is used to identify this sense
Moment direction: a plus sign (+) for counterclockwise moments and a minus sign (−) for clockwise moments, or vice versa
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

29. Moment (The Cross Product)
The moment of F about point A may be represented by the cross-product expression
where r is a posision vector which runs from the moment reference point A to any point on the line of action of F
The order of r × F of the vectors must be maintained because F × r would produce a vector with a sense opposite to that of M
F × r = -M
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

30. Moment (The Cross Product)
The magnitude of the moment M is gevin by:
M = F (r sin α)
From the shown diagram:
r sin α = d
M = Fd
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

31. Moment (Varignon’s Theorem)
The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point
Mo = r × R
R = P + Q
Mo = r × P + r × Q
Mo = Rd = Pp − Qq
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

32. Example 5
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33. Example 5 (Solution) Statics Notes 2013 [Force Systems]
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34. Example 5 (Solution) Statics Notes 2013 [Force Systems]
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35. Example 5 (Solution) Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

36. Example 6 Given: Find: Moment of the 100 N force about O
Magnitude of a horizontal force applied at A which create the same moment about O
The smallest force applied at A which creates the same moment about O
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

37. Example 6 (Solution) 1. Moment of the 100 N force about O
Mo = Fd = 100(5 cos60o) = 250 N.m
2. Magnitude of a horizontal force applied at A which create the same moment about O
Mo = 250 = PL
L = 5 sin 60o = 4.33 m
P = 250/4.33 = 57.7 N
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

38. Example 6 (Solution)
3. The smallest force applied at A which creates the same moment about O
P is smallest when d in M = Fd is a maximum
This occurs when P is perpendicular to the lever
Mo = Pd = P × 5 = 250
P = 250/5 = 50 N
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

39. Couple
Couple: is the moment produced by two equal, opposite, and non-collinear forces
The combined moment of the two forces about axis normal to their plane and passing through any point such as O in their plane is the couple M
M = F(a + d) - Fa
M = Fd
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

40. Couple (Vector Algebra Method)
With the cross-product notation, the combined moment about point O of the forces forming the couple is
M = rA × F + rB × (–F) = (rA – rB) × F
where rA and rB are position vectors which run from point O to arbitrary points A and B on the lines of action of F and –F
Because rA – rB = r, M can be express as:
M = r × F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

41. The Sense of a Couple Vector
The sense of a couple vector M can represent as clockwise or counterclockwise by one of the shown conventions
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

42. Equivalent Couple
Changing the values of F and d does not change a given couple as long the product Fd remains the same
A couple is not affected if the forces act in a different but parallel plane
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

43. Force – Couple System
The combination of the force and couple in the shown figure is refered to as a force – couple system
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

44. Example 7
Statics Notes 2013 [Force Systems]
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45. Example 7 (Solution) Statics Notes 2013 [Force Systems]
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46. Example 8
Statics Notes 2013 [Force Systems]
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47. Example 8 (Solution) Statics Notes 2013 [Force Systems]
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48. Resultants
The Resultant of a system of forces is the simplest force combination which can replace the original forces without altering the external effect on the rigid body to which the forces are applied
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

49. Resultants Algebraic Method Statics Notes 2013 [Force Systems]
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50. Resultants Algebraic Method Statics Notes 2013 [Force Systems]
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51. Resultants
Principle of Moments: This process is summarized in equation form by
If the resultant force R for a given force system is zero, the resultant of the systam need not be zero because the resultant may be a couple
The resultant force in the shown figure have a zero resultant force but have a resultant clockwise couple M = F3d
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

52. Example 9
Statics Notes 2013 [Force Systems]
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53. Example 9 (Solution) Statics Notes 2013 [Force Systems]
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54. Example 9 (Solution) Statics Notes 2013 [Force Systems]
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55. Example 9 (Solution) Statics Notes 2013 [Force Systems]
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56. Example 9 (Solution) Statics Notes 2013 [Force Systems]
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57. Three Dimensional Force System
Rectangular Components
The Force F acting at point O in the shown figure has the rectangular components Fx, Fy and Fz where
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

58. Rectangular Components
The above equation may write as
nF is a unit vector which characterizes the direction of F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

59. Rectangular Components
(a) Specification by two points on the line of action of the force
If the coordinates of points A and B of the shown figure are known, the force F may be written as
Thus the x, y, and z scalar components of F are the scalar coefficients of the unit vectors i, j, and k, respectively
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

60. Rectangular Components
(b) Specification by two angles which orient the line of action of the force
Consider the geometry of the shown figure and assuming that the angles θ and Φ are known. First Resolve F into horizantal and vertical components
Then resolve the horizontal components Fxy into x- and y-components
Fx, Fy, and Fz are the desired scalar components of F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

61. Rectangular Components
Dot Product
The dot product of two vectors P and Q is defined as the product of their magnitudes times the cosine of the angle α between them
If n is a unit vector in a specified direction, the projection of F in the n-direction has the magnitude Fn = F.n
The projection of F in the n-direction can be expressed as a vector quantity by multiplying its scalar component (F.n) with the unit vector n to give Fn = (F.n)n = F.nn
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

62. Rectangular Components
Dot Product
Statics Notes 2013 [Force Systems]
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63. Rectangular Components
Angle between two Vectors
Statics Notes 2013 [Force Systems]
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64. Example 10
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65. Example 10 (Solution) Statics Notes 2013 [Force Systems]
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66. Example 10 (Solution) Statics Notes 2013 [Force Systems]
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67. Example 10 (Solution) Statics Notes 2013 [Force Systems]
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68. Example 10 (Solution) Statics Notes 2013 [Force Systems]
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69. Three Dimensional Force System
Moment and Couple
Moments in 3D can be calculated using scalar (2D) approach but it can be difficult and time consuming
Thus, it is often easier to use a mathematical approach called the vector cross product
Using the vector cross product, MO = r  F
Here r is the position vector from point O to any point on the line of action of F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

70. 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 
The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

71. Evaluating the Cross Product
Of even more utility, the cross product can be written as
Each component can be determined using 22 determinants
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

72. Cartesian Vector Formulation
What is ?
Magnitude:
= (1)(1)sin90o = 1
Direction:
Here is a simple way of remembering this
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

73. Moment In 3D
So, the cross-product expression for the moment MO may be written in the determinant form as:
By expanding the above equation using 22 determinants, we get
MO = (ryFZ − rZ Fy)i − (rxFz − rzFx )j + (rxFy − ryFx )k
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

74. Moment In 3D
In the shown figure, the three components of a force F acting at a point located relative to O by the vector r is illustrated
The scalar magnitude of the moments of these forces about the positive x-, y-, and z-axes through O can be obtained from the moment-arm rule, and are
Which agree with the relative terms in the determinant expresion for the cross product r × F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

75. Moment about an Arbitrary Axis
Our goal is to find the moment of F about any axis λ through O
First compute the moment of F about any arbitrary point O that lies on the λ axis using the cross product
MO = r  F
Now, find the component of MO along the axis λ using the dot product
Mλ = MO •n
Here n is a unit vector in the λ-direction
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

76. Moment about an Arbitrary Axis
To obtain the vector expression for the moment Mλ of F about λ, multiply the magnitude by the direction unit vector n to obtain
The expression r × F•n is known as a triple scalar product
The triple scalar product may be represented by the determinant
where α, β, and γ are the direction cosines of the unit vector n
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

77. Varignon’s Theorem in 3D
In the shown figure, a system of concurrent forces F1, F2, F3, .....
The sum of the moments about O of these forces is
Using the symbol Mo to represent the sum of the moments on the left side of the a bove equation, we have
Mo = ∑ (r × F) = r × R
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

78. Couples in 3D M = rA × F + rB × (–F) = (rA – rB) × F M = r × F
A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d
With the cross-product notation, the combined moment about point O of the forces forming the couple is
M = rA × F + rB × (–F) = (rA – rB) × F
where rA and rB are position vectors which run from point O to to arbitrary points A and B on the lines of action of F and –F
Because rA – rB = r, M can be express as
M = r × F
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

79. Couple vectors obey all of the rules which govern vector quantities
Couples in 3D
The moment of a couple is free vector (The moment of a couple is the same about all points)
whereas the moment of a force about a point (which is also the same about a defined axis through the point) is a sliding vector
Couple vectors obey all of the rules which govern vector quantities
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

80. Example 11
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81. Example 11 (Solution) Statics Notes 2013 [Force Systems]
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82. Example 11 (Solution) Statics Notes 2013 [Force Systems]
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83. Example 12
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84. Example 12 (Solution) Statics Notes 2013 [Force Systems]
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85. Example 13
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86. Example 13 (Solution) Statics Notes 2013 [Force Systems]
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87. Example 13 (Solution) Statics Notes 2013 [Force Systems]
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88. Example 14
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89. Example 14 (Solution) Statics Notes 2013 [Force Systems]
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90. Example 14 (Solution) Statics Notes 2013 [Force Systems]
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91. Three Dimensional Force System
Resultants
The Resultant of a system of forces is the simplest force combination which can replace the original forces without altering the external effect on the rigid body to which the forces are applied
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

92. Resultants of A force System
When several forces act on a body, you can move each force and its associated couple moment to a common point O
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

93. Resultants of A force System
Now you can add all the forces and couple moments together and find one resultant force-couple moment pair
In general, the force system can be expressed as
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

94. Resultants for Several Special Force Systems
Concurrent Forces: When forces are concurrent at a point, there are no moments about the point of concurrency
Parallel Forces: For a system of parallel forces not all in the same plane, the magnitude of the parallel resultant force R is simply the magnitude of algebraic sum of the given forces. The position of its line of action is obtained from the principle of moments by requiring that r × F = Mo
Coplanar Forces: Are discussed earlier
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

95. Resultants for Several Special Force Systems
Wrench Resultants: When the resultant couple vector M is parallel to the resultant force R
A wrench is positive if the couple and force vectors points in the same direction and negative if they points in opposite directions
Statics Notes 2013 [Force Systems]
Dr. Sameer Shadeed

96. Example 15
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97. Example 15 (Solution) Statics Notes 2013 [Force Systems]
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98. Example 16
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99. Example 16 (Solution) Statics Notes 2013 [Force Systems]
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100. Example 16 (Solution) Statics Notes 2013 [Force Systems]
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101. Example 17
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102. Example 17 (Solution) Statics Notes 2013 [Force Systems]
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103. Example 17 (Solution) Statics Notes 2013 [Force Systems]
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104. Example 17 (Solution) Statics Notes 2013 [Force Systems]
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