Engineering Mechanics Statics Zheng Tinghui Ph. D DEPT. OF Applied Mechanics SICHUAN UNIVERSITY

Mechanics Study of what happens to a “thing” (the technical name is “BODY”) when FORCES are applied to it. The subject is subdivided into three branches:

Units of Measurement Four fundamental physical quantities: Length, Time, Mass, and Force. Newton’s 2nd Law relates them: F = m * a We use this equation to develop systems of units. Three of four (base units) are independent and the fourth unit is derived from the equation. N=

Chapter 2 Force Vector These communication towers are stabilized by cables that exert forces at the points of connection. In this chapter, we will show how to express these forces as Cartesian vectors.

2.1-2.2 Vectors and Scalars 1. Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity 2. For vector addition you have to use ______ law. A) Newton’s Second B) the arithmetic C) Pascal’s D) the parallelogram A vector is a quantity that has both a magnitude and a direction. Vector operations include multiplication and division by a scalar; addition; subtraction and resolution.

Scalar Multiplication and Division Vector Addition: Parallelogram Law Triangle method (always ‘tip to tail’) Vector Subtraction: R’ = A – B = A + (-B)

Vector Resolution “Resolution” of a vector is breaking up a vector into components having known lines of a action. It is kind of like using the parallelogram law in reverse.

2.3 Vector Addition of Force Two common problems in Statics involve either finding the resultant force, knowing its components, or resolving a known force into two components. There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket? by using parallelogram law; 2) by using the rectangular-component method (algebra menthod).

2.4 Cartesian Vector Notation We ‘resolve’ vectors into components using the x and y axes system. The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes. Cartesian Vector

Coplanar Force Resultants Step 1 is to resolve each force into its rectangular components Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector. Step 3 is to find the magnitude and angle of the resultant vector.

Example Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force. FR

F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN Solution FR F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30° i – 36 sin 30° j } kN = { 31.18 i – 18 j } kN

Solution FR Summing up all the i and j components respectively, FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN = { 16.82 i + 3.49 j } kN FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN  = tan-1(3.49/16.82) = 11.7°

Attention Quiz 1. Resolve F along x and y axes and write it in vector form. F = { ___________ } N A) 80 cos (30°) i - 80 sin (30°) j B) 80 sin (30°) i + 80 cos (30°) j C) 80 sin (30°) i - 80 cos (30°) j D) 80 cos (30°) i + 80 sin (30°) j 30° x y F = 80 N 2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N . A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N

2.5 Cartesian Vectors How will you represent each of the cable forces in Cartesian vector form? Given the forces in the cables, how will you determine the resultant force acting at D, the top of the tower?

A Unit Vector For a vector A with a magnitude of A, an unit vector is defined as UA = A / A . Characteristics of a unit vector: a) Its magnitude is 1. b) It is dimensionless. c) It points in the same direction as the original vector (A). The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and z axes respectively.

3-D Cartesian Vector Consider a box with sides AX, AY, and AZ meters long. The vector A can be defined as A = (AX i + AY j + AZ k) m The projection of the vector A in the x-y plane is A´. The magnitude of this projection, A´, is found by using the same approach as a 2-D vector: A´ = (AX2 + AY2)1/2 . The magnitude of the vector A can now be obtained as A = ((A´)2 + AZ2) ½ = (AX2 + AY2 + AZ2) ½ The direction of the vector A can be definite by angles They must satisfy the following equation. cos ²  + cos ²  + cos ²  = 1

3-D Cartesian Vector Consider a box with sides AX, AY, and AZ meters long. The vector A can be defined as A = (AX i + AY j + AZ k) m The projection of the vector A in the x-y plane is A´. The magnitude of this projection, A´, is found by using the same approach as a 2-D vector: A´ = (AX2 + AY2)1/2 .

The Second Expression of a vector Recall, the formula for finding the unit vector of any vector: So a unit vector can be written in another way u A = cos  i + cos  j + cos  k Finally, Vector A may be expressed in unit vector form as A= AuA =Acos  i + Acos  j + Acos  k =Axi + Ayj + Azk

2.6 Addition/Subtraction of Cartesian Vectors Once individual vectors are written in Cartesian form, it is easy to add or subtract them. The process is essentially the same as when 2-D vectors are added. For example, if A = AX i + AY j + AZ k and B = BX i + BY j + BZ k , then A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k or A – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k .

Example Given:Two forces F and G are applied to a hook. Force F is shown in the figure and it makes 60° angle with the X-Y plane. Force G is pointing up and has a magnitude of 80 N with  = 111° and  = 69.3°. Find: The resultant force in the Cartesian vector form. G G F=100N

Solution : First, resolve force F. Fz = 100 sin 60° = 86.60 N F' = 100 cos 60°= 50.00 N Fx = 50 cos 45° = 35.36 N Fy = 50 sin 45° = 35.36 N G Now, you can write: F = {35.36 i – 35.36 j + 86.60 k} N G F=100N

we need to find the value of . Now resolve force G. we need to find the value of . Recall : cos ² () + cos ² () + cos ² () = 1. Now substitute what we know. We have cos ² (111°) + cos ² (69.3°) + cos ² () = 1. = 30.22° or 120.2°. G F=100N Since the vector is pointing up,  = 30.22°

G Now using the coordinate direction angles, we can get UG, and determine G = 80 UG N. G = {80 ( cos (111°) i + cos (69.3°) j + cos (30.22°) k )} N G = {- 28.67 i + 28.28 j + 69.13 k } N G F=100N Now, R = F + G or R = {6.69 i – 7.08 j + 156 k} N

Discussion How to write these forces in the following figures in the Cartesian Vector Form? x y F = 80 N

2.7 Position Vectors A position vector is defined as a fixed vector that locates a point in space relative to another point. Consider two points, A & B, in 3-D space. Let their coordinates be (XA, YA, ZA) and ( XB, YB, ZB ), respectively. The position vector directed from A to B, r AB , is defined as r AB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m Please note that B is the ending point and A is the starting point. So ALWAYS subtract the “tail” coordinates from the “tip” coordinates!

2.8 Force Vector Directed along a Line How can we represent the force along the wing strut in a 3-D Cartesian vector form? If a force is directed along a line, then we can represent the force vector in Cartesian Coordinates by using a unit vector and the force magnitude. So we need to: Wing strut a) Find the position vector, r AB , along two points on that line. b) Find the unit vector describing the line’s direction, uAB = (rAB/rAB). c) Multiply the unit vector by the magnitude of the force, F = F uAB .

A(0,0,30) B(12,-8,6) Figure: 02-02Ex13a 28

Figure: 02-02Ex13b 29

Figure: 02-02Ex15a 30

Figure: 02-02Ex15b 31

2.9 Dot Product For this geometry, can you determine angles between the pole and the cables? For force F at Point A, what component of it (F1) acts along the pipe OA? What component (F2) acts perpendicular to the pipe?

Definition The dot product of vectors A and B is defined as A•B = A B cos . Angle  is the smallest angle between the two vectors and is always in a range of 0º to 180º. Dot Product Characteristics: 1. The result of the dot product is a scalar (a positive or negative number). 2. The units of the dot product will be the product of the units of the A and B vectors.

Cartesian Vector Formulation Dot product for each of the Cartesian unit vectors i • i = 1 j • j = 1 k • k = 1 i • j = 0 i • k = 1 k • j = 1 Therefore

Application 1: to determine the angle formed between two vectors or intersecting lines

Application 2: to determine the components of a vector parallel and perpendicular to a line Steps: 1. Find the unit vector, Uaa´ along line aa´ 2. Find the scalar projection of A along line aa´ by A|| = A • U = AxUx + AyUy + Az Uz

3. If needed, the projection can be written as a vector, A|| , by using the unit vector Uaa´ and the magnitude found in step 2. A|| = A|| Uaa´ 4. The scalar and vector forms of the perpendicular component can easily be obtained by A  = (A 2 - A|| 2) ½ and A  = A – A|| (rearranging the vector sum of A = A + A|| )

Example Given: The force acting on the pole Find: The angle between the force vector and the pole, and the magnitude of the projection of the force along the pole OA. A

 = cos-1{(F • rOA)/(F rOA)}  = cos-1 {2/(10.95 * 3)} = 86.5° Solution: rOA = {2 i + 2 j – 1 k} m rOA = (22 + 22 + 12)1/2 = 3 m F = {2 i + 4 j + 10 k}kN F = (22 +42 +102)1/2 =10.95 kN  = cos-1{(F • rOA)/(F rOA)}  = cos-1 {2/(10.95 * 3)} = 86.5° A F • rOA = (2)(2) + (4)(2) + (10)(-1) = 2 kN·m uOA = rOA/rOA = {(2/3) i + (2/3) j – (1/3) k} FOA = F • uOA = (2)(2/3) + (4)(2/3) + (10)(-1/3)=0.667 kN Or FOA = F cos  = 10.95 cos(86.51°) = 0.667 kN

Group Quize Given: The force acting on the pole. Find: The angle between the force vector and the pole, and the magnitude of the projection of the force along the pole AO.

End of the Lecture Let Learning Continue Statics:The Next Generation (2nd Ed.) Mehta, Danielson & Berg Lecture Notes for Sections 4.1-4.4 41