 # An-Najah National University College of Engineering

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Example 1 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 1 (Solution) Statics Notes 2013 [Force Systems]

Example 1 (Solution) Statics Notes 2013 [Force Systems]

Example 2 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 2 (Solution) Statics Notes 2013 [Force Systems]

Example 2 (Solution) Statics Notes 2013 [Force Systems]

Example 3 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 3 (Solution) Statics Notes 2013 [Force Systems]

Example 4 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 4 (Solution) Statics Notes 2013 [Force Systems]

Example 4 (Solution) Statics Notes 2013 [Force Systems]

Example 4 (Solution) Statics Notes 2013 [Force Systems]

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

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

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

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

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

Example 5 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 5 (Solution) Statics Notes 2013 [Force Systems]

Example 5 (Solution) Statics Notes 2013 [Force Systems]

Example 5 (Solution) Statics Notes 2013 [Force Systems]

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

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

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

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

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

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

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

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

Example 7 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 7 (Solution) Statics Notes 2013 [Force Systems]

Example 8 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 8 (Solution) Statics Notes 2013 [Force Systems]

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

Resultants Algebraic Method Statics Notes 2013 [Force Systems]

Resultants Algebraic Method Statics Notes 2013 [Force Systems]

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

Example 9 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 9 (Solution) Statics Notes 2013 [Force Systems]

Example 9 (Solution) Statics Notes 2013 [Force Systems]

Example 9 (Solution) Statics Notes 2013 [Force Systems]

Example 9 (Solution) Statics Notes 2013 [Force Systems]

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

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

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

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

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

Rectangular Components
Dot Product Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Rectangular Components
Angle between two Vectors Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 10 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 10 (Solution) Statics Notes 2013 [Force Systems]

Example 10 (Solution) Statics Notes 2013 [Force Systems]

Example 10 (Solution) Statics Notes 2013 [Force Systems]

Example 10 (Solution) Statics Notes 2013 [Force Systems]

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

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

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

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

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

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

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

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

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

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

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

Example 11 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 11 (Solution) Statics Notes 2013 [Force Systems]

Example 11 (Solution) Statics Notes 2013 [Force Systems]

Example 12 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 12 (Solution) Statics Notes 2013 [Force Systems]

Example 13 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 13 (Solution) Statics Notes 2013 [Force Systems]

Example 13 (Solution) Statics Notes 2013 [Force Systems]

Example 14 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 14 (Solution) Statics Notes 2013 [Force Systems]

Example 14 (Solution) Statics Notes 2013 [Force Systems]

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

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

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

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

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

Example 15 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 15 (Solution) Statics Notes 2013 [Force Systems]

Example 16 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 16 (Solution) Statics Notes 2013 [Force Systems]

Example 16 (Solution) Statics Notes 2013 [Force Systems]

Example 17 Statics Notes 2013 [Force Systems] Dr. Sameer Shadeed

Example 17 (Solution) Statics Notes 2013 [Force Systems]

Example 17 (Solution) Statics Notes 2013 [Force Systems]