2. Midterm Review  Five Problems 2-D/3-D Vectors, 2-D/3-D equilibrium, Dot Product, EoE, Cross Product, Moments  Closed Book & Note  Allowed to bring one half-sheet of notes  Calculator OK  Exam will start promptly at 7:30pm and end at 8:50pm

3. Midterm Review 1 ft = 0.3048 m 1 lb = 4.4482 N 1 slug = 14.5938 kg

4. VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law: Triangle method (always ‘tip to tail’): How do you subtract a vector? How can you add more than two concurrent vectors graphically ?

5. Vector Subtraction  R’ = A – B = A + (-B)  Example, pg. 19

6. “Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse. RESOLUTION OF A VECTOR

7. CARTESIAN VECTOR NOTATION (Section 2.4) Each component of the vector is shown as a magnitude and a direction. 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.

8. For example, F = F x i + F y j or F' = F' x i + F' y j The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination.

9. ADDITION OF SEVERAL VECTORS  Step 3 is to find the magnitude and angle of the resultant vector. Step 1 is to resolve each force into its 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.

10. Example of this process,

11. You can also represent a 2-D vector with a magnitude and angle.

12. A UNIT VECTOR 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. For a vector A with a magnitude of A, a unit vector is defined as U A = A / A.

13. 3-D CARTESIAN VECTOR TERMINOLOGY Consider a box with sides A X, A Y, and A Z meters long. The vector A can be defined as A = (A X i + A Y j + A Z 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´ = (A X 2 + A Y 2 ) 1/2. The magnitude of the position vector A can now be obtained as A = ((A´) 2 + A Z 2 ) ½ = (A X 2 + A Y 2 + A Z 2 ) ½

14. The direction or orientation of vector A is defined by the angles , , and . These angles are measured between the vector and the positive X, Y and Z axes, respectively. Their range of values are from 0° to 180° Using trigonometry, “direction cosines” are found using the formulas These angles are not independent. They must satisfy the following equation. cos ²  + cos ²  + cos ²  = 1 This result can be derived from the definition of a coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector of any position vector: or written another way, u A = cos  i + cos  j + cos  k. 3-D CARTESIAN VECTOR TERMINOLOGY (continued)

15. ADDITION/SUBTRACTION OF 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 = A X i + A Y j + A Z k and B = B X i + B Y j + B Z k, then A + B = (A X + B X ) i + (A Y + B Y ) j + (A Z + B Z ) k or A – B = (A X - B X ) i + (A Y - B Y ) j + (A Z - B Z ) k.

16. POSITION VECTOR 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 (X A, Y A, Z A ) and ( X B, Y B, Z B ), respectively. The position vector directed from A to B, r AB, is defined as r AB = {( X B – X A ) i + ( Y B – Y A ) j + ( Z B – Z A ) 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!

17. FORCE VECTOR DIRECTED ALONG A LINE (Section 2.8) 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: a) Find the position vector, r AB, along two points on that line. b) Find the unit vector describing the line’s direction, u AB = (r AB /r AB ). c) Multiply the unit vector by the magnitude of the force, F = F u AB.

18. DEFINITION The dot product of vectors A and B is defined as AB = A B cos . Angle  is the smallest angle between the two vectors and is always in a range of 0 º to 180 º. 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.

19. DOT PRODUCT DEFINITON (continued) Examples:i j = 0 i i = 1 A B= (A x i + A y j + A z k) (B x i + B y j + B z k) = A x B x +A y B y + A z B z

20. USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS For the given two vectors in the Cartesian form, one can find the angle by a) Finding the dot product, A B = (A x B x + A y B y + A z B z ), b) Finding the magnitudes (A & B) of the vectors A & B, and c) Using the definition of dot product and solving for , i.e.,  = cos -1 [(A B)/(A B)], where 0 º    180 º.

21. DETERMINING THE PROJECTION OF A VECTOR You can determine the components of a vector parallel and perpendicular to a line using the dot product. Steps: 1. Find the unit vector, U aa´ along line aa´ 2. Find the scalar projection of A along line aa´ by A || = A U = A x U x + A y U y + A z U z

22. 3.If needed, the projection can be written as a vector, A ||, by using the unit vector U aa´ and the magnitude found in step 2. A || = A || U aa´ 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 || ) DETERMINING THE PROJECTION OF A VECTOR (continued)

23. EQUATIONS OF 2-D EQUILIBRIUM Or, written in a scalar form,  F x = 0 and  F y = 0 These are two scalar equations of equilibrium (EofE). They can be used to solve for up to two unknowns. Since particle A is in equilibrium, the net force at A is zero. So F AB + F AC + F AD = 0 or  F = 0 FBD at A A In general, for a particle in equilibrium,  F = 0 or  F x i +  F y j = 0 = 0 i + 0 j (A vector equation)

24. EXAMPLE Write the scalar EofE: +   F x = T B cos 30º – T D = 0 +   F y = T B sin 30º – 2.452 kN = 0 Solving the second equation gives: T B = 4.90 kN From the first equation, we get: T D = 4.25 kN Note : Engine mass = 250 KgFBD at A

25. SPRINGS, CABLES, AND PULLEYS Spring Force = spring constant * deformation, or F = k * S With a frictionless pulley, T 1 = T 2.

26. THE EQUATIONS OF 3-D EQUILIBRIUM When a particle is in equilibrium, the vector sum of all the forces acting on it must be zero (  F = 0 ). This equation can be written in terms of its x, y and z components. This form is written as follows. (  F x ) i + (  F y ) j + (  F z ) k = 0 This vector equation will be satisfied only when  F x = 0  F y = 0  F z = 0 These equations are the three scalar equations of equilibrium. They are valid at any point in equilibrium and allow you to solve for up to three unknowns.

27. MOMENT OF A FORCE - SCALAR FORMULATION (continued) In the 2-D case, the magnitude of the moment is M o = F d As shown, d is the perpendicular distance from point O to the line of action of the force. In 2-D, the direction of M O is either clockwise or counter-clockwise depending on the tendency for rotation.

28. MOMENT OF A FORCE - SCALAR FORMULATION (continued) For example, M O = F d and the direction is counter-clockwise. Often it is easier to determine M O by using the components of F as shown. Using this approach, M O = (F Y a) – (F X b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force. F a b d O a b O F F x F y

29. CROSS PRODUCT (Section 4.2) 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  U C Here U C is the unit vector perpendicular to both A and B vectors as shown (or to the plane containing the A and B vectors).

30. CROSS PRODUCT (continued) The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product. For example: i  j = k Note that a vector crossed into itself is zero, e.g., i  i = 0

31. CROSS PRODUCT (continued) Of even more utility, the cross product can be written as Each component can be determined using 2  2 determinants.

32. MOMENT OF A FORCE – VECTOR FORMULATION (Section 4.3) Moments in 3-D can be calculated using scalar (2-D) 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, M O = r  F. Here r is the position vector from point O to any point on the line of action of F.

33. MOMENT OF A FORCE – VECTOR FORMULATION (continued) So, using the cross product, a moment can be expressed as By expanding the above equation using 2  2 determinants (see Section 4.2), we get (sample units are N - m or lb - ft) M O = ( r y F Z - r Z F y ) i - ( r x F z - r z F x ) j + ( r x F y - r y F x ) k The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D formulation.