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**Vector Operation and Force Analysis**

Lecture #2 (ref Ch 2) Vector Operation and Force Analysis R. Michael PE 8/14/2012

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**Ch. 2 Key Concepts A little on vectors (and scalars)**

Finding Resultant Force (vector addition) Graphical Approach (sec 2.3) Triangle method (or successive triangle method) Parallelogram method Polygon method (good if more than one force) Finding resultants by resolving forces into components!! (sec 2.4-Addition of a System of Coplanar Forces)

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**Scalar vs. Vector Scalar Quantity**

A mathematical expression possessing only magnitude characterized by a positive or negative number The following are classified as Scalar Quantities Mass Volume Length

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**Scalar vs. Vector Vector**

Physical quantity that requires both a magnitude and a direction for its complete description. possessing magnitude and direction and must be added using Vector Operations The following are classified as Vectors Displacements Velocities Accelerations Moments

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**SCALARS AND VECTORS (Section 2.1)**

Scalars Vectors Examples: Mass, Volume Force, Velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font, a line, an arrow or a “carrot” Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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Vector Notation In Slides and handouts Vectors will be denoted as a BOLD letter. Example; (a+b) will denote a scalar addition (A+B) will denote a vector addition When hand writing a Vector use an arrow over the letter to denote it is a Vector. [ ] A

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Vector Notation Vector Notation for Rectangular Components of a Vector (Force) - Because the directional sense of the axes of the rectangular coordinate system are known, Rectangular Vector Components can be written in a couple of different ways. Cartesian Vector Notation – Cartesian unit vectors (i and j) are used to designate the x-axis and y-axis respectively where F=Fxi+Fyj. Magnitude and Direction – Define the Vector by magnitude, units, and angle it makes with respect to the x-axis - F= 45N ° 135 vs 45

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**Components of a Vector A Vector Magnitude Vector Designation 500 N**

Head Tail Vector Designation 500 N

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**Vector Operations Multiplication and Division of a Vector by a Scalar**

Product of Vector (A) and Scalar (b) = bA = a vector with the same direction as A but with the magnitude multiplied by the scalar (b). Example – If a 500 lb force acting along the x-axis is doubled, it becomes a 1000 lb force acting along the x-axis.

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Vector Addition Vectors can be compared to giving directions. Go north 4 steps, Go east 3 steps. The vector would be defined as the arrow pointing from where you started to where you are now. The magnitude would be defined by how far are you from where you started (not how far you traveled to get there)

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**VECTOR OPERATIONS (Section 2.2)**

Scalar Multiplication and Division Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**2.3 APPLICATION OF VECTOR ADDITION**

There are three concurrent forces acting on the hook due to the chains. We need to decide if the hook will fail (bend or break)? To do this, we need to know the resultant force acting on the hook. FR Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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Vector Addition The addition of two vectors results in a resultant vector (P+Q=R) where R is a vector pointing from the starting point of P to the ending point of Q. Resultant Vector

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Parallelogram Law By drawing construction lines parallel to the vectors, the resultant vector goes from the point of origin to the intersection of the construction lines

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Triangle Method Place the tail of B to the head of A. The Resultant (R) can be found by connecting the Tail of A to the Head of B. This forms the third leg of the triangle and the resultant vector. B A A + B Resultant

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**Summary: VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE**

Triangle method (always ‘tip to tail’): Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**Example #1 Triangle Method – Find Resultant (mag and direction) 30 N**

y 30 N 50 N x 45° 30° See solution in notes

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**Successive Triangle Method**

If there are multiple vectors to be added together, add the first two vectors to find the first resultant. Once the first Resultant (R1) is found, add the next vector to the resultant to find (R2). Can be repeated as many times as necessary to add all the vectors (it also does not matter what order they are added in, the end resultant will be the same). See HO, normally just resolve into components

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Polygon Method Polygon method is similar to the Successive Triangle Method but no intermediate resultants are calculated Graphically measure length and direction of R!! See HO, do example on board

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**We ‘resolve’ vectors into components using the x and y axis system.**

ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4) – Basically finding resultant vectors by breaking forces up into components and adding! We ‘resolve’ vectors into components using the x and y axis system. Each component of the vector is shown as a magnitude and a direction. 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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**F = Fx i + Fy j or F' = F'x i + ( F'y ) j**

For example, F = Fx i + Fy j or F' = F'x i + ( F'y ) j The x and y axis are always perpendicular to each other. Together, they can be directed at any inclination. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**ADDITION OF SEVERAL VECTORS**

Step 1 is to resolve each force into its components. Step 2 is to add all the x-components together, followed by adding all the y components together. These two totals are the x and y components of the resultant vector. Step 3 is to find the magnitude and angle of the resultant vector. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**An example of the process:**

Break the three vectors into components, then add them. FR = F1 + F2 + F3 = F1x i + F1y j F2x i + F2y j + F3x i F3y j = (F1x F2x + F3x) i + (F1y + F2y F3y) j = (FRx) i + (FRy) j Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**Remember! You can also represent a 2-D vector with a magnitude and angle:**

Or, Cartesian Vector Notation: FR = (FRx) i + (FRy) j Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**F2-8: Determine the magnitude and direction of the resultant force.**

Side question: What additional force would you have to apply so the net force acting on the hook was zero (think, same magnitude but opposite direction for the resultant force found in step 1).

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**Given: Three concurrent forces acting on a tent post. **

EXAMPLE Given: Three concurrent forces acting on a tent post. Find: The magnitude and angle of the resultant force. Plan: a) Resolve the forces into their x-y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**F2 = {– 450 cos (45°) i + 450 sin (45°) j } N **

EXAMPLE (continued) F1 = {0 i j } N F2 = {– 450 cos (45°) i sin (45°) j } N = {– i j } N F3 = { (3/5) 600 i + (4/5) 600 j } N = { 360 i j } N Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**Summing up all the i and j components respectively, we get, **

EXAMPLE (continued) Summing up all the i and j components respectively, we get, FR = { (0 – ) i + ( ) j } N = { i j } N Using magnitude and direction: FR = ((41.80)2 + (1098)2)1/2 = 1099 N = tan-1(1098/41.80) = 87.8° x y FR Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**Given: Three concurrent forces acting on a bracket**

GROUP PROBLEM SOLVING Given: Three concurrent forces acting on a bracket Find: The magnitude and angle of the resultant force. Plan: a) Resolve the forces into their x and y components. b) Add the respective components to get the resultant vector. c) Find magnitude and angle from the resultant components. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**GROUP PROBLEM SOLVING (continued)**

F1 = { (5/13) 300 i + (12/13) 300 j } N = { i j } N F2 = {500 cos (30°) i sin (30°) j } N = { i j } N F3 = { 600 cos (45°) i 600 sin (45°) j } N { i j } N Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**GROUP PROBLEM SOLVING (continued)**

Summing up all the i and j components respectively, we get, FR = { ( ) i + ( – 424.3) j }N = { i j } N x y FR Now find the magnitude and angle, FR = ((972.7)2 + (102.7)2) ½ = N = tan–1( / ) = 6.03° From Positive x axis, = 6.03° Do example on board Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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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 cos (30°) j C) 80 sin (30°) i – 80 cos (30°) j D) 80 cos (30°) i sin (30°) j 30° x y F = 80 N Answers: 1. C 2. C 2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i j } N and F2 = { 20 i j } N . A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.1,2.2,2.4

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**Rectangular Components of 3D Forces**

3D Force Vector – Vector defining a Force in more than one Cartesian Plane defined by its location and rectangular components Rectangular Components - Components that fall along the Cartesian coordinate system axes Coordinate Angles (θx, θy, θz)– The angle a vector makes with the individual axes of the Cartesian Coordinate System

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**Many structures and machines involve 3-Dimensional Space.**

APPLICATIONS Many structures and machines involve 3-Dimensional Space. In this case, the power pole has guy wires helping to keep it upright in high winds. How would you represent the forces in the cables using Cartesian vector form? We’ll solve this later Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**APPLICATIONS (continued)**

In the case of this radio tower, if you know the forces in the three cables, how would you determine the resultant force acting at D, the top of the tower? We’ll solve this later Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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Coordinate Angles b g a Note, book uses a, b, g:

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Coordinate Angles The values of the three angles are not independent, they are related by the identity: cos2(θx) + cos2(θy) + cos2(θz) = 1

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**Resolving a 3D Force Vector into its Rectangular Components**

Given the magnitude of a force vector (F) and its Coordinate angles (θx, θy, θz): Fx = Fcos(θx) Fy = Fcos(θy) Fz = Fcos(θz) Note, book uses a, b, g:

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**Resultant of a 3D Force Vector from its Rectangular Components**

If given the components of a 3D force (Fx, Fy, Fz), the force can be determined by: Magnitude (F) = √(Fx2+Fy2+Fz2) The Coordinate Angles of the Force Vector can be found by cos(θx) = Fx/F cos(θy) = Fy/F cos(θz) = Fz/F Do Example on board, then HO examples

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**Addition of 3D Force Vectors**

Forces are easy to add once they are broken down into their rectangular components. The components of the resultant force can be found as follows: Rx=ΣFx Ry=ΣFy Rz=ΣFz

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3D Resultant Force The magnitude of the resultant force is equal to the square root of the addition of the scalar quantity of each leg squared: R = √(Rx2+Ry2+Rz2) The Coordinate Angles of the resultant can be found by: cos(θx) = Rx/R cos(θy) = Ry/R cos(θz) = Rz/R See HO’s

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**ADDITION OF CARTESIAN VECTORS (Section 2.6)**

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 . Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**IMPORTANT NOTES Sometimes 3-D vector information is given as:**

a) Magnitude and the coordinate direction angles, or, b) Magnitude and projection angles. You should be able to use both these types of information to change the representation of the vector into the Cartesian form, i.e., F = {10 i – 20 j k} N . Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**Given: Two forces F1 and F2 are applied to a hook.**

EXAMPLE Given: Two forces F1 and F2 are applied to a hook. Find: The resultant force in Cartesian vector form. Plan: G Using geometry and trigonometry, write F1 and F2 in Cartesian vector form. 2) Then add the two forces (by adding x and y components). Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**Solution : First, resolve force F1.**

Fx = 0 = 0 lb Fy = 500 (4/5) = 400 lb Fz = 500 (3/5) = 300 lb Now, write F1 in Cartesian vector form (don’t forget the units!). F1 = {0 i j k} lb Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**We are given only two direction angles, and . **

Now resolve force F2. We are given only two direction angles, and . So we need to find the value of . Recall that cos ² () + cos ² () + cos ² () = 1. Now substitute what we know: cos ² (30°) + cos ² () + cos ² (45) = 1. Solving, = 75.5° or 104.5°. Since the vector is pointing in the positive direction, = 75.5° Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**So, using u A = cos i + cos j + cos k .**

Now that we have the coordinate direction angles, we can find uG and use it to determine F2 = 800 uG lb. So, using u A = cos i + cos j + cos k . F2 = {800 cos (30°) i cos (75.5°) j 800 cos (45°) k )} lb F2 = {712.8 i j k } lb Now, R = F1 + F2 or R = {713 i j 308 k} lb Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 2.5,2.6

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**Good 3D problem. Know resultant mag and direction, F1 mag and direction, F2 mag. Find F2 direction.**

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