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**JJ205 ENGINEERING MECHANICS**

COURSE LEARNING OUTCOMES : Upon completion of this course, students should be able to: CLO 1. apply the principles of statics and dynamics to solve engineering problems (C3) CLO 2. sketch related diagram to be used in problem solving (C3) CLO 3. study the theory of engineering mechanics to solve related engineering problems in group (A3)

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**JJ205 ENGINEERING MECHANICS**

CHAPTER 2: FORCE VECTORS CLO 1. apply the principles of statics and dynamics to solve engineering problems (C3) Prepared by: THEEBENRAJ

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**Objectives: At the end of this chapter, student should be able to:**

Understand scalars and vectors Differentiate between scalars and vectors. Distinguish free vectors, sliding vectors, fixed vectors. Understand rectangular components Explain two forces acting on a particle. Understand vectors and vector operations. Calculate addition of vectors. Calculate subtraction of vectors. Determine resolution of vectors. Understand the resultant force of coplanar forces by addition. Explain scalar notation. Explain cartesian vector notation. Determine coplanar forces and resultant force.

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**Objectives: Understand Cartesian vectors.**

Explain right handed coordinate system. Explain cartesian unit vector. Apply cartesian vector representation. Understand the magnitude of cartesian vector. Determine the direction of cartesian vector. Understand resultant cartesian vector by addition and substraction. Solve problems regarding concurrent force system. Understand position vectors and x, y, z coordinates. Explain position vectors and x, y, z coordinate.

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**Objectives: Understand the force vector directed along the line.**

Explain the force vector directed along the line. Determine the force vector directed along the line. Understand the dot product. Apply laws of operation. Commutative law Multiplication by scalar Distributive law Formulate cartesian vector formulation.

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**Force On A Particle. Resultant of Two Forces**

A force represents the action of one body on another and generally characterized by its point of application, its magnitude, and its direction. The magnitude of the force is characterized by a certain number of units. SI units to measure the magnitude of a force are the Newton (N). Multiple: kilonewton (kN) = 1000N

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**The direction of a force is defined by the line of action and the sense of the force.**

30˚ A Magnitude of a force on particle A is 100N at 30˚. 100N 30˚ A The different sense of the force but have same magnitude and same direction.

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**Two forces, P and Q acting on a particle A:**

Can be replaced by a single force R which has the same effect on the particle. R is the resultant of the forces P and Q. The method is known as the parallelogram law for the addition of two forces. P A Q R = R P A A Q

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Example 2.1: The two forces P and Q act on bolt A. Determine their resultant. Solution: Using graphical solution: A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant are measured and found to be The triangle rule may also be used. Forces P and Q are drawn in tip-to-tail fashion.

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**Trigonometric Solution.**

Continue… Trigonometric Solution. The triangle rule is again used; two sides and the included angle are known. We apply the law of cosines: Now, applying the law of sines, we write; Solving equation (1) for sin A, we have; Using a calculator, we first compute the quotient, then its are sines, and obtain; (1)

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**Alternative Trigonometric Solution.**

Continue… Alternative Trigonometric Solution. We construct the right triangle BCD and compute; Then, using triangle ACD, we obtain; Again,

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Example 2.2 The screw eye in Figure (a) is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force. Solutions: Parallelogram Law: The parallelogram law of addition is shown in Figure (b). The two unknowns are the magnitude of FR and the angle θ (theta).

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continue Trigonometry: The vector triangle as shown in Fig. (c), is constructed from Fig. (b). FR is determined by using the law of cosines; The angle θ is determined by applying the law of sines; Thus, the direction ø (phi) of FR, measured from the horizontal, is;

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Vectors Vectors is defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. It is represented by arrows. The magnitude of a vector defines the length of the arrow used to represent the vector.

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Types of vectors: 1) Fixed vectors: cannot be moved without modifying the conditions of the problem. 2) Free vectors: couples, which represented by vectors which may be freely moved in space. 3) Sliding vectors: forces acting on a rigid body, which represented by vectors which can be moved, or slid, along their lines of action.

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continue 4) Equal vectors: two vectors have the same magnitude and the same direction, whether or not they also have the same point of application. They are denoted by same letter. 5) Negative vectors: vector having the same magnitude but opposite direction. P P P -P

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**Addition of Vectors Vectors add according to the parallelogram law.**

The sum of two vectors P and Q is obtained by attaching the two vectors to the same point A and constructing a parallelogram, using P and Q as two sides of the parallelogram. The diagonal that passes through A represents the sum of the vectors P and Q which denoted as P + Q. However, the magnitude of the vector P+Q is NOT In general, equal to the sum (P+Q) of the magnitudes of the vectors P and Q. Since that, we conclude that the addition of two Vectors is commutative, write as: P + Q = Q + P P + Q P A Q

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**From the only half of the parallelogram;**

continue Triangle Rule is an alternative method for determining the sum of two vectors from the parallelogram law. From the only half of the parallelogram; From the figures shown above, it confirms the fact that vectors addition is commutative. Q P + Q OR P P P + Q A A Q

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**Subtraction of Vectors**

Defined as the addition of the corresponding negative vector. P-Q representing the difference between the vectors P and Q is obtained by adding to P the negative vector –Q. we write; P – Q = P + (-Q) -Q P - Q P A

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Coplanar Vectors The sum of three vectors P, Q, and S was obtained graphically. The triangle rule was first applied to obtain the sum P+Q of the vectors P and Q. It was applied again to obtain the sum of vectors P+Q and S. For addition of vectors, Polygon Rule is applied by arranging the given vectors in tip-to-tail fashion and connecting the tail of the first vector with the tip of the last one. Q S P + Q P P + Q + S A Q S P P + Q + S A

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**P + Q + S = (P + Q) + S = P + (Q + S)**

continue The result obtained would have been unchanged if the vectors Q and S had been replaced by their sum Q + S. We may thus write; P + Q + S = (P + Q) + S = P + (Q + S) which expresses the fact that vector addition is associative. Recalling that vector addition has been shown, in the case of two vectors, to be commutative, we write: P + Q + S = (P + Q) + S = S + (P + Q) = S + (Q + P) = S + Q + P Q S Q + S P P + Q + S A

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continue This expression, as well as others which may be obtained in the same way, shows that the order in which several vectors are added together is immaterial. Q S P P + Q + S = S + Q + P A P S Q

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**Resolution of A Force Into Components**

Components of the original force F, is a single force F acting on a particle may be replaced by two or more forces which, together, have the same effect on the particle. The process of substituting them for F is called resolving the force F into components. There are two cases of particular interest: One of the two components. P is known. Second component, Q, is obtained by applying the triangle rule and joining the tip of P to the tip of F Magnitude and direction of Q are determined graphically or by trigonometry. Q P F A

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**The line of action of each component is known. **

continue The line of action of each component is known. The magnitude and sense of the components are obtained by applying the parallelogram law and drawing lines, through the tip of F, parallel to the given lines of action. This process leads to two well-defined components, P and Q, which can be determined graphically or computed trigonometrycally by applying the law if sines. Q F A P

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**Addition of a System of Coplanar Forces**

It will be found desirable to resolve a force into two components which are perpendicular to each other. In figure below, the force F has been resolved into component Fx along the x axis and a component Fy along the y axis. The parallelogram drawn to obtain the two components is a rectangle, and Fx and Fy are called rectangular components. y y F x OR Fy Fy F θ Fx θ O Fx x O

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**Cartesian Unit vectors:**

continue Cartesian Unit vectors: Two vectors of unit magnitude, directed respectively along the positive x and y axes, will be introduced at this point. They are denoted by i and j. Scalar components: The scalars Fx and Fy of forces F. Vector components: The actual component forces Fx and Fy of F. y j x i

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**Cartesian Vector Notation**

continue Cartesian Vector Notation Note that the rectangular components Fx and Fy of a force F may be obtained by multiplying respectively the unit vectors i and j by appropriate scalars. We write; Fx = Fx i Fy = Fy j and express F as the Cartesian vector, F = Fx i + Fy j y Fy = Fy j j F Fx = Fx i θ i x O

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**This may express the scalar components of F as follows: **

continue Scalar Notation. Indicates positive and negative Fx : Positive Fx when the vector component Fx has the same sense as the unit vector i. (same sense as the positive x axis). Negative Fx when Fx has the opposite sense. The positive and negative Fy is same as Fx Denoting by F the magnitude of the force F and by θ the angle between F and the x axis, measured counterclockwise from the positive x axis. This may express the scalar components of F as follows: Fx = F cosθ Fy = F sinθ y Fy F θ Fx x O

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Example 2.3 A force of 800 N is exerted on a bolt A as shown in Figure (a). Determine the horizontal and vertical components of the force. (a) The vector components of F are: Fx = -(655 N) i Fy = +(459 N) j May write in Cartesian vector form: F = -(655 N) i + (459 N) j

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Example 2.4 Resolve the 1000 N (≈ 100 kg) force acting on the pipe Fig. a, into components in the (a) x and y directions, and (b) x’ and y directions. Solution: In each case the parallelogram law is used to resolve F into its two components, and then the vector triangle is constructed to determine the numerical results by trigonometry.

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**The vector addition F = Fx + Fy is shown in Fig. b. **

continue Part (a) The vector addition F = Fx + Fy is shown in Fig. b. In particular, note that the length of the components is scaled along the x and y axes by first constructing lines from the tip of F parallel to the axes in accordance with the parallelogram law. From the vector triangle, Fig. c,

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**The vector addition F = Fx + Fy is shown in Fig. d. **

continue Part (b) The vector addition F = Fx + Fy is shown in Fig. d. Note carefully how the parallelogram is constructed. Applying the law of sines and using the data listed on the vector triangle, Fig. e, yields:

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Example 2.5 The force F acting on the frame shown in Fig.2-12a has a magnitude of 500 N and is to be resolved into two components acting along members AB and AC. Determine the angle θ, measured below the horizontal, so that the component FAC is directed from A toward C and has a magnitude of 400 N. Solution: by using the parallelogram law, the vector addition of the two components yielding the resultant is shown in Fig.b. Note carefully how the resultant force is resolved into two components FAB and FAC, which have specified lines of action. The corresponding vector triangle is shown in Fig.c.

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**The angle ø can be determined by using the law of sines:**

continue The angle ø can be determined by using the law of sines: Hence;

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**Show that in this case θ = 16.1˚ and FAB = 161 N.**

continue Using this value for θ, apply the law of cosines or the law of sines and show that FAB has a magnitude of 561 N. Notice that F can also be directed at an angle θ above the horizontal, as shown in Fig. d, and still produce the required component FAC. Show that in this case θ = 16.1˚ and FAB = 161 N.

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Example 2.6 The ring shown in Figure a is subjected to two forces, F1 and F2. if it is required that the resultant force have a magnitude of 1 kN and be directed vertically downward, determine (a) the magnitudes of F1 and F2 provided θ = 30˚, and (b) the magnitudes of F1 and F2 if F2 is to be a minimum.

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continue Solution: Part (a): A sketch of the vector addition according to the parallelogram law is shown in Fig. b. From the vector triangle constructed in Fig. c, the unknown magnitudes F1 and F2 are determined by using the law of sines:

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continue Part (b): If θ is not specified, then by the vector triangle, Fig. d, F2 may be added to F1 in various ways to yield the resultant 1000 N force. The minimum length or magnitude of F2 will occur when its line of action is perpendicular to F1. Any other direction, such as OA or OB, yields a larger value for F2. Hence, when θ = 90˚ - 20˚ = 70˚, F2 is minimum. From the triangle shown in Fig. e, it is seen that; F1 = 1000 sin 70˚ N = 940 N F2 = 1000 cos 70˚ N = 342 N

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**Coplanar Force Resultants.**

To determine the resultant of several coplanar forces Resolved each force into its x and y components. The respective components are added using scalar algebra since they are colinear. The resultant force is then formed by adding the resultants of the x and y components using parallelogram law. Example: Given three concurrent forces below: To solve this problem using Cartesian vector notation, each force is first represented as a Cartesian vector, i.e; F1 = F1x i + F1y j F2 = - F2x i + F2y j F3 = F3x i - F3y j The vector resultant is therefore; FR = F1 + F2+ F3 = F1x i + F1y j - F2x i + F2y j + F3x i - F3y j = (F1x - F2x i + F3x )i + (F1y + F2y - F3y ) j = (FRx )i + (FRy )j y F2 F1 x F3

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**To solve this problem using Scalar notation, **

continue To solve this problem using Scalar notation, from figure shown, since x is positive to the right and y is positive upward, we have; The vector resultant is therefore; FR = = (FRx )i + (FRy )j In the general case, the x and y components of the resultant of any number of coplanar forces can be represented symbolically by the algebraic sum of the x and y components of all the forces; y F2y F1y F2x F1x x F3x F3y

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continue When applying these equations, it is important to use the sign convention establish for the components. Components having a directional sense along the positive coordinate axes are considered positive scalars. Components having a directional sense along the negative coordinate axes are considered negative scalars. If this convention is followed, then the signs of the resultant components will specify the sense of these components. For example; positive result indicates that the component has a directional sense which is in positive coordinate direction. y FRy FR θ x FRx

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continue Once the resultant components are determined, they may be sketched along the x and y axes in their proper direction, and the resultant force can be determined from vector addition. Then, the magnitude of FR can be found from the Pythagorean Theorem; which is: Also, the direction angle θ, which specifies the orientation of the force is determined from trigonometry: y FRy FR θ x FRx

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Example 2.7 Four forces act on bolt A as shown. Determine the resultant of the forces on the bolt. Force Magnitude, N x Component, N y Component, N F1 150 + 75.0 F2 80 27.4 + 75.2 F3 110 110.0 F4 100 25.9 Rx = Ry= Thus, the resultant R of the four forces is;

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continue The magnitude and direction of the resultant may now be determined. From the triangle shown, we have:

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Example 2.8 Determine the x and y components of F1 and F2 acting on the boom shown in Fig. (a). Express each force as a Cartesian vector.

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**The magnitude of each component is determined by trigonometry.**

continue Solution: Scalar Notation: F1 is resolved into x and y components using parallelogram law as shown in Fig. (b). The magnitude of each component is determined by trigonometry. Then, we have;

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**F2 is resolved into x and y components as shown in Fig. (c).**

continue Solution: F2 is resolved into x and y components as shown in Fig. (c). The slope of the line of action is indicated and could obtain the angle θ: Then, determine the magnitudes of the components;

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**Hence, using scalar notation;**

continue Solution: The magnitude of the horizontal component, F2x, was obtained by multiplying the force magnitude by the ratio of the horizontal leg of the slope triangle divided by the hypotenuse. The magnitude of the vertical component, F2y, was obtained by multiplying the force magnitude by the ratio of the vertical leg divided by the hypotenuse. Hence, using scalar notation;

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**Cartesian Vector Notation: **

continue Solution: Cartesian Vector Notation: The magnitudes and directions of the components of each force is determined. Thus, express each force as a Cartesian vector;

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Example 2.9 The link in Figure (a) below is subjected to two forces F1 and F2. determine the magnitude and orientation of the resultant force.

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**Scalar Notation: (*Can be solved using parallelogram law.) **

continue Solution 1: Scalar Notation: (*Can be solved using parallelogram law.) Each force is resolved into its x and y components, Figure (b). These components are summed algebraically. The “positive” sense of the x and y force components alongside each equation is indicated;

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**Scalar Notation: (*Can be solved using parallelogram law.) **

continue Solution 1: Scalar Notation: (*Can be solved using parallelogram law.) Magnitude of the resultant force, shown in Figure (c); From the vector addition, Figure (c), the direction angle θ is;

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**Cartesian Vector Notation: **

continue Solution 2: Cartesian Vector Notation: From Figure (b), each force is expressed as a Cartesian vector; Thus, The magnitude and direction of FR are determined in the same manner in solution 1.

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Example 2.10 The end of the boom O in Figure (a) below is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force.

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**Each force is resolved into its x and y components, Figure (b). **

continue Solution : Each force is resolved into its x and y components, Figure (b). Summing the x components; The “negative” sign indicates that FRx acts to the left, as noted by the small arrow. Summing the y components yields;

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**Magnitude of the resultant force, shown in Figure (c);**

continue Solution : Magnitude of the resultant force, shown in Figure (c); From the vector addition, Figure (c), the direction angle θ is;

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Exercises 2.31. Determine the x and y components of the 800 N force Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

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**Cartesian Vectors Right handed coordinate system.**

A right handed coordinate system will be used for developing the theory of vector algebra that follows. A rectangular or Cartesian coordinate system is said to be right-handed provided the thumb of the right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x toward the positive y axis.

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**Rectangular Components of a Vector**

continue Rectangular Components of a Vector A vector A may have one, two, or three rectangular components along the x, y, and z coordinate axes, depending on how the vector is oriented relative to the axes. When A is directed within an octant of the x, y, z frame, then, by two successive applications of the parallelogram law, we may resolve the vector into components as; A = A’ + Az and A = Ax + Ay Combining these equations, A represented by the vector sum of its three rectangular components; A = Ax + Ay + Az z Az A Ay y Ax A’ x

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continue Unit vector Specified as the direction of A since it has a magnitude of 1. If A is a vector having a magnitude A ≠ 0, then the unit vector having the same direction as A is represented by; From equation 2.1, the unit vector will be dimensionless since the unit will cancel out. Equation 2.2 therefore indicates that vector A may be expressed in terms of both its magnitude and direction separately Eg: A positive scalar defines the magnitude of A. uA (a dimensionless vector) defines the direction and sence of A. 2.1 2.2

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**Cartesian unit vectors.**

continue Cartesian unit vectors. In 3D, the set of Cartesian unit vectors i, j, k, is used to designate the directions of the x, y, z axes respectively. z The sense (or arrowhead) of these vectors will be described analytically by a plus or minus sign, depending on whether they are pointing along the positive or negative x, y, or z axes. k y j i x Figure shows the positive Cartesian unit vectors.

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**Cartesian Vector Representation.**

continue Cartesian Vector Representation. Magnitude of Cartesian Vector z Az k A k↑ Ay j j → y i ↙ Ax i x

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**Direction of a Cartesian Vector**

continue Direction of a Cartesian Vector The orientation of A is defined by the coordinate direction angle α (alpha), β (beta), and γ (gamma), measured between the tail of A and the positive x, y, z axes located at the tail of A. Each of the angles will be between 0˚ and 180˚. z A Az y Ax Ay x

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continue Direction cosine of A z z z Az A A A y y y Ax Ay x x x

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**A expressed in Cartesian vector form as:**

continue Direction angles: Since the magnitude of a vector is equal to the positive square root of the sum of the squares of the magnitudes of its components, and uA has a magnitude of 1. A expressed in Cartesian vector form as:

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Example 2.11 Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring in Fig. a.

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continue Solution: Since each force is represented in Cartesian vector form, the resultant force, shown in Fig. b, is: The magnitude of FR is found from equation above;

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continue The coordinate direction angles α, β, ɣ are determined from the components of unit vector acting in the direction of FR. So that, These angles are shown in Figure b. in particular, note that β > 90˚ since the j component of uFR is negative.

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**Position Vectors x, y, z coordinates.**

Right-handed coordinate system is used to reference the location of points in space. In many technical books, to require the positive z axis to be directed upward (the zenith direction) so that is measures the height of an object or the altitude of the point. The x and y axes then lie in the horizontal plane. Points in space are located relative to the origin of coordinates, O, by successive measurements along the x, y, z axes.

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**From the figure above, coordinate at point A: **

z 2 m B C y 4 m 4 m 2 m 1 m 6 m x A From the figure above, coordinate at point A: xA = +4 m along the x-axis yA = +2 m along the y-axis zA = -6 m along the z-axis thus, A (4,2,-6) B (0,2,0) C (6,-1,4)

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continue Position vector. The position vector, r, is defined as a fixed vector which locates a point in space relative to another point. For example, if r extends from the origin of coordinates, O, to point P(x,y,z), then, r can be expressed in Cartesian vector form as: r = xi + yj + zk z z k A r y j O y x i x

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**By the head-to-tail vector addition, we require: rA + r = rB **

continue z B (xB, yB, zB) r A (xA, yA, zA) rB rA y x By the head-to-tail vector addition, we require: rA + r = rB Solving for r and expressing rA and rB in cartesian vector form as: r = rB - rA = (xB i +yB j + zB k) – (xA i + yA j + zA k)

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continue z B r (zB – zA) A rB (xB – xA) rA y (yB – yA) x In other way in solving for r and expressing rA and rB in Cartesian vector: [+ i direction] [+ j direction] [+ k direction]

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Example 2.12 An elastic rubber band is attached to points A and B as shown in Fig. (a). Determine its length and its direction measured from A toward B.

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**First, establish a position vector A to B, Figure (b). **

continue Solution: First, establish a position vector A to B, Figure (b). The coordinates of the tail A (1 m, 0, -3 m) are substracted from the coordinates of the head B (-2 m, 2 m, 3 m), which yields; r = B – A = [-2 m – 1 m] i + [2 m – 0] j + [3 m – (-3 m)] k = { -3i + 2j + 6k} m

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**The magnitude of r represents the length of the rubber band.**

continue Solution: The components of r can be determined directly by realizing from fig. (a) that they represent the direction and distance one must go along each axis in order to move from A to B. The magnitude of r represents the length of the rubber band. Formulating a unit vector in the direction of r, we have;

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continue Solution: The components of this unit vector yield the coordinate direction angles: These angles are measured from the positive axes of a localized coordinate system placed at the tail of r, point A as shown in Fig. (c).

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**Force vector directed along a line**

The direction of a force is specified by two points through which its line of action passes. We can formulate F as a Cartesian vector by realizing that it has the same direction and sense as the position vector r directed from point A to point B on the cord. This common direction is specified by the unit vector u=r/r. Hence, F = Fu = F (r/r) z F B Force F is directed along the cord AB. A y x

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Example 2.13 The man shown in Figure (a) pulls on chord with a force of 350 N. Represent this force, acting on the support A, as a Cartesian vector and determine its direction.

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**The coordinates of the end points of the cord are: A (0, 0, 7.5 m) **

continue Solution: Force F is shown in Figure (b). The direction of this vector, u, is determined from the position vector r, which extends from A to B. The coordinates of the end points of the cord are: A (0, 0, 7.5 m) B (3 m, -2 m, 1.5 m) Forming the position vector, we have; r = B – A = (3 – 0) i + (-2 – 0) j + (1.5 – 7.5) k = { 3i – 2j – 6k} m

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**Magnitude of r, (represent the length of cord AB):**

continue Magnitude of r, (represent the length of cord AB): Forming the unit vector that defines the direction and sense of both r and F yields: Since F has a magnitude of 350 N and a direction specified by u, then,

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**Coordinate direction angles: **

continue Coordinate direction angles: Measured between r (or F) and the positive axes of a localized coordinate system with origin placed at A. From the components of the unit vector:

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Dot Product Dot product defines a particular method for “multiplying” two vectors and is used to solve the 3 Dimensional problems. Dot product of vectors A and B is A . B Defined as the product of the magnitudes of A and B and the cosine of the angle θ between their tails. Equation form: A . B = AB cos θ Where 0˚ ≤ θ ≤ 180˚ Dot product is often referred to as the scalar product of vectors; since the result is a scalar, NOT a vector. A B (Eqn. 2.3)

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**Laws of Operation Commutative Law Multiplication by a scalar**

continue Laws of Operation Commutative Law Multiplication by a scalar Distributive Law It is easy to prove the first and second laws by using Eqn. 2.3

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**Cartesian Vector Formulation**

continue Cartesian Vector Formulation Equation 2.1 may be used to find the dot product for each of the Cartesian unit vectors. Example: In similar manner: Should not be memorized, but understood.

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**Carrying out the dot-product operations, the final result becomes:**

continue Consider now the dot product of two general vectors A and B which are expressed in Cartesian vector form. We have: Carrying out the dot-product operations, the final result becomes: Thus, to determine the dot product of two Cartesian vectors, multiply their corresponding x,y,z components and sum their products algebraically. Since the result is a scalar, NOT to include any unit vector in final result.

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References: R.C. Hibbeler. (2004). Engineering Mechanics Statics – Third Edition. F.P. Beer, E.R. Johnston, Jr, E.R. Eisenberg. (2004). Vector Mechanics for Engineers. Statics. Seventh Edition in SI Units.

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