1. Scalars and Vectors A scalar quantity is one that can be described by a single number: temperature, speed, mass A vector quantity deals inherently with both magnitude and direction: velocity, force, displacement, acceleration

2. Scalars and Vectors By convention, the length of a vector arrow is proportional to the magnitude of the vector. 8 lb 4 lb Arrows are used to represent vectors. The direction of the arrow gives the direction of the vector.

3. Vectors The plane travels with a velocity relative to the ground which is vector sum of the plane’s velocity (relative to the air) plus the wind velocity The resultant is a combination of both motions

4. Vector Procedures These are the rules you must play by to add vectors and find the resultant: 1.Draw each vector to scale and label 2.Vectors should be drawn in order as they occurred and labeled (I.e.v 1, v 2 …..) 3.Draw each successive vector starting from the tip of the preceding vector. (This is called the “tip to tail” method 4.The vector sum is found by drawing the resultant vector arrow from the tail of the first vector to the tip of the last vector Vector v 1 is 1 unit east or +1 unit Vector v 2 is 3 units west or - 3 units v1v1 v2v2 v 1 + v 2 = R v1v1 v2v2 Vector (v 1 + v 2 ) is 2 units west or - 2 units A1A1 5.For two-vector operations in two dimensions the resultant forms the diagonal of a parallelogram with the component vectors forming the sides. R = A 1 + A 2  A2A2 R = Runits @  0 (  measured from East counterclockwise)

5. Use a ruler and vector rules to find the resultant of the following vector additions. Record your answer using vector convention. F1F1 F2F2 F1F1 F2F2 R = F 1 + F 2 P1P1 P2P2 P2P2 P1P1 R = P 1 + P 2 P1P1 R = P 2 + P 1

6. Vector Addition and Subtraction When a vector is multiplied by -1, the magnitude of the vector remains the same, but the direction of the vector is reversed.

7. Vector Addition and Subtraction

8. Finding the components of a vector To find the magnitude of a vector quantity in a particular direction you can use the parallelogram rule 1.Draw the vector and treat it as the resultant 2.Form a parallelogram around the resultant in the two directions of interest. In physics problems these are usually perpendicular directions to simplify the problem 3.Draw the component vectors along the sides of the parallelogram a ahah avav E E paralel E perpendicular E a Pythagorean Theorem: a 2 = a h 2 + a v 2 Pythagorean Theorem: E 2 = E paralel 2 + E perpendicular 2

9. Using Trig to solve 2D-Vector Problems Right-Angled Trig Relationships Trig functions for right-angled triangles are just geometric ratios. You probably memorized these ratios in your math classes using So/h..Ca/h..To/a. Here they are once more!  o h a sin  = o / h cos  = a / h tan  = sin  / cos  = o / a 1.Sketch vectors on a coordinate system. (Two perpendicular axes) 2.Find (resolve) the components of all vectors in the two directions of interest using the parallelogram rule. Make sure that you use the normal conventions of + and – arithmetic quantities. 3.Sum the components in the two directions of interest. 4.Find the magnitude of R using Pythagorean theorem 5.Use the tangent rule to determine the direction R as measured from the East by convention or a specified direction Follow these rules!

10. Practice, Practice…..Practice R v h = 35 m/s 60 0 Use trig relationships to determine i) the other component and ii) the resultant R i) tan  = v v / v h v v = v h tan  v v = (35 m/s) tan 60 0 v v = 60.6 m/s ii) R =  (v v 2 + v h 2 ) R =  ((60.6 m/s) 2 + (35 m/s) 2 ) R = 70 m/s  = 180 0 - 60 0 = 120 0  R = 70 m/s @ 120 0 v

11. Determine the resultant R of the following vectors by using the horizontal and vertical components of A and B A = 80 Newtons B = 60 Newtons 60 0 30 0 + +- HV A B A cos 60 0 -80 N cos 60 0 A sin 60 0 +80 N sin 60 0 - 40 N+ 69.3 N B cos 30 0 +60 N cos 30 0 B sin 30 0 60 N sin 30 0 + 52 N+ 30 N R = A + B + 12 N+ 99.3 N R = 100 Newtons  R =  (R H 2 + R v 2 ) R =  ((12 N) 2 + (99.3 N) 2 ) R = 100 N  = tan -1 (R v / R H )  = tan -1 (99.3 N / 12 N)  = 83 0 R = 100 N @ 83 0 BHBH BvBv AHAH AvAv