Exam 1 Review IEA Section 2 2/5/2018 Section 3 2/5/2018
ALAC review session ALAC will have a review session in preparation for IEA Exam 1 Monday (Today) February 5th 8PM-10PM DCC 330
Test 1 Wednesday 2/7/2018 8:00 9:50 am Not 9:00 as stated in your SIS schedule Students arriving late will have to turn in their tests at 9:50 Covers lectures 1-5
Test 1 Rooms Sec-01 DARRIN 308 Sec-02 Darrin 318 This is not a typo Sec-03 DARRIN 308 Sec-04 DARRIN 308 Sec-05 DARRIN 308
Test 1 Four problems equally weighted 1. Vectors (2D/3D) 2. Forces (2D/3D) 3. 2D particle equilibrium 4. Linear Algebra (Gauss-Jordan Elimination Method)
Test 1 Closed book and closed notes. For maximum credit, show all set-ups and all details necessary (see syllabus). You can use only a hand-held calculator for calculations (no smartphone, laptop, etc.). Cell phones, audio devices, headphones and laptops must be off and stored away. Any student found deviating from this policy will be asked to leave the exam room, given a grade of zero, and no retest will be allowed.
Test 1 Students entitled to extra time should: get a note from the Dean of Students and give (or email) a copy to their instructor by 2/02/2018. remain in the test room until escorted by a TA to a different room (DCC 239) to finish their test.
Grade-Challenging Sessions Monday 2/12/18 in DARRIN 236 Tuesday 2/13/18 in DARRIN 236 6:00 8:00 pm
Makeup Test 1 Wednesday 2/14/2018 5:00 6:50 pm Room: JONSSN 5119 Only students with a valid reason and a note from the office of the Dean of Students will be allowed to take it No makeup for the makeup test
The sum of two vectors analytic representation (two dimensional ) y (v 1 +w 1,v 2 +w 2 ) v 2 (w 1,w 2 ) w 2 w v (v 1,v 2 ) v 1 w 1 x v + w = (v 1 + w 1, v 2 + w 2 ) v + w = (v 1 + w 1 )i + (v 2 + w 2 ) j
The sum of two vectors rectangular components (Three dimensional ) z a (a 1,a 2,a 3 ) y b (b 1,b 2,b 3 ) x a + b = (a 1 + b 1, a 2 + b 2, a 3 + b 3 ) a + b = (a 1 + b 1 )i + (a 2 + b 2 ) j + (a 3 + b 3 ) k
Example Find the components of the vector having initial point P 1 and terminal point P 2 Solution: P 1 (-1,0,2), P 2 (0,-1,0) V = (0 - (-1), -1-0, 0-2) = (1,-1,-2) Final minus initial
EXAMPLE G Given: Two forces F 1 and F 2 are applied to a hook. Find: The resultant force in Cartesian vector form. Plan: 1) Using geometry and trigonometry, write F 1 and F 2 in Cartesian vector form. 2) Then add the two forces (by adding x and y-components).
Solution: First, resolve force F 1. F x = 0 = 0 lb F y = 500 (4/5) = 400 lb F z = 500 (3/5) = 300 lb EXAMPLE (continued) Now, write F 1 in Cartesian vector form (don t forget the units!). F 1 = {0 i + 400 j + 300 k} lb
EXAMPLE (continued) Now, resolve force F 2. F 2z = -800 sin 45 = 565.7 lb F 2 = 800 cos 45 = 565.7 lb F 2 can be further resolved as, F 2x F 2y = 565.7 cos 30 = 489.9 lb = 565.7 sin 30 = 282.8 lb F 2 F 2z Thus, we can write: F 2 = {489.9 i + 282.8 j 565.7 k } lb
EXAMPLE (continued) So F R = F 1 + F 2 and F 1 = {0 i + 400 j + 300 k} lb F 2 = {489.9 i + 282.8 j 565.7 k } lb F R = { 490 i + 683 j 266 k } lb
Orthogonal versus Perpendicular You want to start to move away from using the word Perpendicular Perpendicular is a word that has meaning in 2- Dimensional geometry Orthogonal is a word that generalizes the concept into higher dimensions
Orthogonal defined The line that is normal to this plane is orthogonal to the plane This line is perpendicular to any line that one can draw in the plane (in a new plane that contains both lines)
Orthogonal vectors and the dot product If two vectors are orthogonal, then their dot product is zero. If the dot product of two vectors is zero,then they are orthogonal. You can dot vectors with different units to find this!
ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4) We resolve vectors into components using the x and y-axis coordinate 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.
For example, F = F x i + F y 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.
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.
POSITION VECTOR 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. ALWAYS subtract the tail coordinates from the tip coordinates!
DEFINITION The dot product of vectors A and B is defined as A B = A B cos θ. The angle θ 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.
DOT PRODUCT DEFINITON (continued) Examples: By definition, 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
USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS For these two vectors in Cartesian form, one can find the angle by a) Find the dot product, A B = (A x B x + A y B y + A z B z ), b) Find the magnitudes (A & B) of the vectors A & B, and c) Use the definition of dot product and solve for θ, i.e., θ = cos -1 [(A B)/(A B)], where 0 º θ 180 º.
USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS For these two vectors in Cartesian form, one can find the angle by a) Find the dot product, A B = (A x B x + A y B y + A z B z ), b) Find the magnitudes (A & B) of the vectors A & B, and c) Use the definition of dot product and solve for θ, i.e., θ = cos -1 [(A B)/(A B)], where 0 º θ 180 º.
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 aa = A x U x + A y U y + A z U z
DETERMINING THE PROJECTION OF A VECTOR (continued) 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 )
FBD at A F D A EQUATIONS OF 2-D EQUILIBRIUM y A A F B 30 x F C = 392.4 N FBD at A Or, written in a scalar form, ΣF x = 0 and Σ F y = 0 Since particle A is in equilibrium, the net force at A is zero. So F B + F C + F D = 0 or ΣF = 0 In general, for a particle in equilibrium, Σ F = 0 or Σ F x i + Σ F y j = 0 = 0 i + 0 j (a vector equation) These are two scalar equations of equilibrium (E-of-E). They can be used to solve for up to two unknowns.
SIMPLE SPRINGS Spring Force = spring constant * deformation of spring or F = k * s
Augmented matrix x 1 +3 x 2 + 4x 3 = 8 2x 1 + 5x 2-8x 3 = 1 3x 1 + 7x 2-9x 3 = 0 1 3 4 8 2 5 8 1 3 7 9 0
The rules of Matrix operations You can multiply any row by a constant You can add two rows together and replace one of the rows with the resultant You can exchange two rows You can subtract two rows and replace one of the rows with the resultant You can combine addition/subtraction with multiplying a row.
Gauss-Jordan elimination A systematic procedure for solving system of linear equations by transforming the augmented matrix to a reduced row-echelon form x = 2 y = 3 z = 1 1 0 0 2 0 1 0 3 0 0 1 1 Reduced row-echelon form
Reduced row-echelon form 1) If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1 (leading 1). 2) If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix. 3) In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row. 4) Each column that contains a leading 1 has zeros everywhere else. x = 2 y = 3 z= 1 1 0 0 2 0 1 0 3 0 0 1 1 Reduced row-echelon form