2. Vectors Vector Operations Components Trig Applications

3. Objectives I can add vector using the head to tail method. I can subtract vectors using the head to head method. I can use trigonometric functions to find the components of a vector. I add components of vectors to find a resultant vector.

4. Vector Addition Head to tail method Parallelogram method 8 N 4 N 3 N Suppose 3 forces act on an object at the same time. F net is not 15 N because these forces aren’t working together. But they’re not completely opposing each either. So how do find F net ? The answer is to add the vectors... not their magnitudes, but the vectors themselves. There are two basic ways to add vectors w/ pictures:

5. Tip to Tail Method in-line examples Place the tail of one vector at the head of the other. The vector sum (also called the resultant) is shown in red. It starts where the black vector began and goes to the head of the blue one. In these cases, the vector sum represents the net force. You can only add or subtract magnitudes when the vectors are in-line! 16 N 20 N 4 N 20 N 16 N 12 N 9 N 12 N 21 N

6. Head to Tail – 2 Vectors 5 m 2 m To add the red and blue displacement vectors first note: Vectors can only be added if they are of the same quantity—in this case, displacement. 5 m 2 m blue + red Interpretation: Walking 5 m in the direction of the blue vector and then 2 m in the direction of the red one is equivalent to walking in the direction of the black vector. The displacement walked this way is the black vector’s magnitude. Place the vectors head to tail and draw a vector from the tail of the first to the head of the second.

7. Commutative Property blue + red red + blue As with scalars quantities and ordinary numbers, the order of addition is irrelevant with vectors. Note that the resultant (black vector) is the same magnitude and direction in each case. (We’ll learn how to find the resultant’s magnitude soon.)

8. Head to Tail – 3 Vectors We can add 3 or more vectors by placing them head to tail in any order, so long as they are of the same type (force, velocity, displacement, etc.). blue + green + red

9. Parallelogram Method This time we’ll add red & blue by placing the tails together and drawing a parallelogram with dotted lines. The resultant’s tail is at the same point as the other tails. It’s tip is at the intersection of the dotted lines. Note: Opposite sides of a parallelogram are congruent.

10. Comparison of Methods red + blue Tip to tail method Parallelogram method The resultant has the same magnitude and direction regardless of the method used.

11. Opposite of a Vector v - v If v is 17 m/s up and to the right, then -v is 17 m/s down and to the left. The directions are opposite; the magnitudes are the same.

12. Vector Subtraction red - blue blue - red Put vector tails together and complete the triangle, pointing to the vector that “comes first in the subtraction.” Why it works: In the first diagram, blue and black are tip to tail, so blue + black = red  red – blue = black. Note that red - blue is the opposite of blue - red.

13. Vector Components A 150 N force is exerted up and to the right. This force can be thought of as two separate forces working together, one to the right, and the other up. These components are perpendicular to each other. Note that the vector sum of the components is the original vector (green + red = black). The components can also be drawn like this: 150 N Horizontal component Vertical component

14. Finding Components with Trig  v v cos  v sin  Multiply the magnitude of the original vector by the sine & cosine of the angle made with the given. The units of the components are the same as the units for the original vector. Here’s the correspondence: cosine  adjacent side sine  opposite side

15. Component Example 34 m/s 30.814 m/s 25  14.369 m/s A helicopter is flying at 34 m/s at 25  S of W (south of west). The magnitude of the horizontal component is 34 cos 25   30.814 m/s. This is how fast the copter is traveling to the west. The magnitude of the vertical component is 34 sin 25   14.369 m/s. This is how fast it’s moving to the south.

16. Pythagorean Theorem 34 m/s 30.814 m/s 25  14.369 m/s Since components always form a right triangle, the Pythagorean theorem holds: (14.369) 2 + (30.814) 2 = (34) 2. Note that a component can be as long, but no longer than, the vector itself. This is because the sides of a right triangle can’t be longer than the hypotenuse.

17. Component Form Instead of a magnitude and an angle, vectors are often specified by listing their horizontal and vertical components. For example, consider this acceleration vector: a = 10 m/s 2 at 53.13  N of W 53.13  a = 10 m/s 2 3 m/s 2 4 m/s 2 In component form: a =  -3, 4  m/s 2

18. Component Form Some books use parentheses rather than angle brackets. The vector F =  2, -1, 3  N indicates a force that is a combination of 2 N to the east, 1 N south, and 3 N up. Its magnitude is found w/ the Pythag. theorem: F = [2 2 + (-1) 2 + 3 2 ] 1/2 = 3.742 N

19. Finding the direction of a vector x =  5, -2  meters is clearly a position to the southeast of a given reference point. If the reference pt. is the origin, then x is in the 4 th quadrant. The tangent of the angle relative to the east is given by: 5 m 2 m  tan  = 2 m / 5 m   = tan -1 (0.4) = 21.801  The magnitude of x is (25 + 4) 1/2 = 5.385 m. Thus, the vector is 5.385 m at 21.801  S of E.

20. Adding vectors in component form If F 1 =  3, 7  N andF 2 =  2, -4  N, then the net force is simply given by: F net =  5, 3  N. Just add the horizontal and vertical components separately. F1F1 F2F2 F net F1F1 F2F2 3 N 7 N 2 N 4 N

22. Other component pairs There are an infinite number of component pairs into which a vector can be split. Note that green + red = black in all 3 diagrams, and that green and red are always perpendicular. The angle is different in each diagram, as well as the lengths of the components, but the Pythagorean theorem holds for each. The pair of components used depends on the geometry of the problem.  v v cos  v sin  v v   v cos  v sin  v cos  v sin 

24. Scalar Multiplication x -2x 3x3x Scalar multiplication means multiplying a vector by a real number, such as 8.6. The result is a parallel vector of a different length. If the scalar is positive, the direction doesn’t change. If it’s negative, the direction is exactly opposite. Blue is 3 times longer than red in the same direction. Black is half as long as red. Green is twice as long as red in the opposite direction. ½ x

25. Other Operations Vectors are not multiplied, at least not the way numbers are, but there are two types of vector products that will be explained later. –Cross product –Dot product –These products are different than scalar mult. There is no such thing as division of vectors –Vectors can be divided by scalars. –Dividing by a scalar is the same as multiplying by its reciprocal.

26. Comparison of Vectors 15 N 43 m 0.056 km 27 m/s Which vector is bigger? The question of size here doesn’t make sense. It’s like asking, “What’s bigger, an hour or a gallon?” You can only compare vectors if they are of the same quantity. Here, red’s magnitude is greater than blue’s, since 0.056 km = 56 m > 43 m, so red must be drawn longer than blue, but these are the only two we can compare.

29. Vectors 2-D Force & Motion Problems Trig Applications Relative Velocities Free Body Diagrams Vector Operations Components Inclined Planes Equilibrium

30. Inclined Plane m  mg perpendicular component parallel component A crate of chop suey of mass m is setting on a ramp with angle of inclination . The weight vector is straight down. The parallel component (blue) acts parallel to the ramp and is the component of the weight pulling the crate down the ramp. The perpendicular component (red) acts perpendicular to the ramp and is the component of the weight that tries to crush the ramp. Note: red + blue = black continued on next slide

31. Inclined Plane (continued) m  mg mg cos  mg sin   The diagram contains two right triangles.  is the angle between black and blue.  +  = 90  since they are both angles of the right triangle on the right. Since blue and red are perpendicular, the angle between red and black must also be . Imagine the parallel component sliding down (dotted blue) to form a right triangle. Being opposite , we use sine. Red is adjacent to , so we use cosine.  continued on next slide mg sin 

32. Inclined Plane (continued) m  mg mg cos  mg sin  The diagram does not represent 3 different forces are acting on the chop suey at the same time. All 3 acting together at one time would double the weight, since the components add up to another weight vector. Either work with mg alone or work with both components together.

33. How the incline affects the components m mg mg cos  mg sin  The steeper the incline, the greater  is, and the greater sin  is. Thus, a steep incline means a large parallel component and a small horizontal one. Conversely, a gradual incline means a large horizontal component and a small vertical one. m mg mg cos  mg sin  Extreme cases: When  = 0, the ramp is flat; red = mg; blue = 0. When  = 90 , the ramp is vertical; red = 0; blue = mg.

34. Inclined Plane - Pythagorean Theorem m  mg mg cos  mg sin  (mg sin  ) 2 + (mg cos  ) 2 = (mg) 2 (sin 2  + cos 2  ) = (mg) 2 (1) = (mg) 2 Let’s show that the Pythagorean theorem holds for components on the inclined plane:

35. Inclined Plane: Normal Force m  mg mg cos  mg sin  N = mg cos  Recall normal force is perpen- dicular to the contact surface. As long as the ramp itself isn’t accelerating and no other forces are lifting the box off the ramp or pushing it into the ramp, N matches the perpendicular component of the weight. This must be the case, otherwise the box would be accelerating in the direction of red or green. N > mg cos  would mean the box is jumping off the ramp. N < mg cos  would mean that the ramp is being crushed.

36. Net Force on a Frictionless Inclined Plane m  mg mg cos  mg sin  N = mg cos  With no friction, F net = mg + N = mg cos  + mg sin  + N = mg sin . (mg cos  + N = 0 since their magnitudes are equal but they’re in directions opposite. That is, the perpendicular component of the weight and the normal cancel out.) Therefore, the net force is the parallel force in this case.

37. Acceleration on a Frictionless Ramp m  mg mg cos  mg sin  Here F net = mg sin  = m a. So, a = g sin . Since sin  has no units, a has the same units as g, as they should. Both the net force and the acceleration are down the ramp.

38. Incline with friction at equilibrium m  mg mg cos  mg sin  f s = mg sin  N = mg cos  At equilibrium F net = 0, so all forces must cancel out. Here, the normal force cancels the perpendicular component of the weight, and the static frictional force cancels the parallel component of the weight. continued on next slide

39. Incline with friction at equilibrium (cont.) m  mg mg cos  mg sin  f s = mg sin  N = mg cos  f s   s N =  s mg cos . Also, f s = mg sin  (only because we have equilibrium). So, mg sin    s mg cos . Since the mg’s cancel and tan  = sin  / cos , we have  s  tan . continued on next slide

40. Incline with friction at equilibrium (cont.) mg mg cos  mg sin  f s = mg sin  N = mg cos  Suppose we slowly crank up the angle, gradually making the ramp steeper and steeper, until the box is just about to budge. At this angle, f s = f s, max =  s N =  s mg cos . So now we have mg sin  =  s mg cos , and  s = tan .  An adjustable ramp is a convenient way to find the coefficient of static friction between two materials. (Neither of these quantities have units.)

41. Acceleration on a ramp with friction mg mg cos  mg sin  f k =  k mg cos  N = mg cos   In order for the box to budge, mg sin  must be greater than f s, max which means tan  must be greater than  s. If this is the case, forget about  s and use  k. f k =  k N =  k mg cos . F net = mg sin  - f k = m a. So, mg sin  -  k mg cos  = m a. The m’s cancel, which means a is independent of the size of the box. Solving for a we get: a = g sin  -  k g cos . Once again, the units work out right.

42. Parallel applied force on ramp mg mg cos  mg sin  fkfk N  F A In this case F A and mg sin  are working together against friction. Assuming F A + mg sin  > f s, max the box budges and the 2 nd Law tells us F A + mg sin  - f k = m a. Mass does not cancel out this time. If F A were directed up the ramp, we’d have acceleration up or down the ramp depending on the size of F A compared to mg sin . If F A were bigger, friction acts down the ramp and a is up the ramp.

43. Non-parallel applied force on ramp mg mg cos  mg sin  fkfk N  Suppose the applied force acts on the box, at an angle  above the horizontal, rather than parallel to the ramp. We must resolve F A into parallel and perpendicular components (orange and gray) using the angle  + . F A serves to increase acceleration directly and indirectly: directly by orange pulling the box down the ramp, and indirectly by gray lightening the contact force with the ramp (thereby reducing friction). F A  F A cos(  +  ) F A sin(  +  )  continued on next slide

44. Non-parallel applied force on ramp (cont.) mg mg cos  mg sin  fkfk N  F A  F A cos(  +  ) F A sin (  +  )  Because of the perp. comp. of F A, N < mg cos . Assuming F A sin(  +  ) is not big enough to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, F A sin(  +  ) + N = mg cos . Remember, N is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, N = mg cos  - F A sin(  +  ), which means f k =  k N =  k [mg cos  - F A sin(  +  )]. continued on next slide

45. Non-parallel applied force on ramp (cont.) mg mg cos  mg sin  fkfk N  F A  F A cos(  +  ) F A sin(  +  )  Assuming the combined force of orange and blue is enough to budge the box, we have F net = orange + blue - brown = m a. Substituting, we have F A cos(  +  ) + mg sin  -  k [mg cos  - F A sin(  +  )] = m a.

46. Support Beam mg 1 1 T1T1 2 2 T2T2 continued on next slide Hanging Sign Problem

47. mg 1 1 T1T1 2 2 T2T2 Since the sign is not accelerating in any direction, it’s in equilibrium. Since it’s not moving either, we call it Static Equilibrium. Thus, red + green + black = 0. continued on next slide Hanging sign f.b.d. Free Body Diagram

48. mg T 1 T 2 Vector Equation: T 1 + T 2 + m g = 0 continued on next slide As long as F net = 0, this is true no matter many forces are involved. Hanging sign force triangle F net = 0 means a closed vector polygon !

49. T 2 mg T 1 11 22 T 1 cos  1 T 2 cos  2 T 1 sin  1 T 2 sin  2 T 1 cos  1 = T 2 cos  2 Horizontal: Vertical: T 1 sin  1 + T 2 sin  2 = mg We use Newton’s 2 nd Law twice, once in each dimension: Hanging sign equations Components & Scalar Equations

50. Support Beam 35  T1T1 62  T2T2 75 kg Answers: Accurately draw all vectors and find T 1 & T 2. T 1 = 347.65 N T 2 = 606.60 N Hanging sign sample

51. Vector Force Lab Simulation Go to the link below. This is not exactly the same as the hanging sign problem, but it is static equilibrium with three forces. Equilibrium linkEquilibrium link 1.Change the strengths of the three forces (left, right, and below) to any values you choose. (The program won’t allow a change that is physically impossible.) 2.Record the angles that are displayed below the forces. They are measured from the vertical. 3.Using the angles given and the blue and red tensions, do the math to prove that the computer program really is displaying a system in equilibrium. 4.Now click on the Parallelogram of Forces box and write a clear explanation of what is being displayed and why.

52. 3 - Way Tug-o-War Bugs Bunny, Yosemite Sam, and the Tweety Bird are fighting over a giant 450 g Acme super ball. If their forces remain constant, how far, and in what direction, will the ball move in 3 s, assuming the super ball is initially at rest ? Bugs: 95 N Tweety: 64 N Sam: 111 N To answer this question, we must find a, so we can do kinematics. But in order to find a, we must first find F net. 38°43° continued on next slide

53. 3 - Way Tug-o-War (continued) Sam: 111 N Bugs: 95 N Tweety: 64 N 38°43° 87.4692 N 68.3384 N 46.8066 N 43.6479 N First, all vectors are split into horiz. & vert. comps. Sam’s are purple, Tweety’s orange. Bugs is already done since he’s purely vertical. The vector sum of all components is the same as the sum of the original three vectors. Avoid much rounding until the end. continued on next slide

54. 95 N 87.4692 N 68.3384 N 46.8066 N 43.6479 N continued on next slide 3 - Way Tug-o-War (continued) 16.9863 N 40.6626 N Next we combine all parallel vectors by adding or subtracting: 68.3384 + 43.6479 - 95 = 16.9863, and 87.4692 - 46.8066 = 40.6626. A new picture shows the net vertical and horizontal forces on the super ball. Interpretation: Sam & Tweety together slightly overpower Bugs vertically by about 17 N. But Sam & Tweety oppose each other horizontally, where Sam overpowers Tweety by about 41 N.

55. 3 - Way Tug-o-War (continued) 16.9863 N 40.6626 N F net = 44.0679 N  Find F net using the Pythagorean theorem. Find  using trig: tan  = 16.9863 N / 40.6626 N. The newtons cancel out, so  = tan -1 (16.9863 / 40.6626) = 22.6689 . (tan -1 is the same as arctan.) Therefore, the superball experiences a net force of about 44 N in the direction of about 23  north of west. This is the combined effect of all three cartoon characters. continued on next slide

56. 3 - Way Tug-o-War (final) a = F net / m = 44.0679 N / 0.45 kg = 97.9287 m/s 2. Note the conversion from grams to kilograms, which is necessary since 1 m/s 2 = 1 N / kg. As always, a is in the same direction as F net.. a is constant for the full 3 s, since the forces are constant. 22.6689  97.9287 m/s 2 Now it’s kinematics time: Using the fact  x = v 0 t + 0.5 a t 2 = 0 + 0.5 (97.9287)(3) 2 = 440.6792 m  441 m, rounding at the end. So the super ball will move about 441 m at about 23  N of W. To find out how far north or west, use trig and find the components of the displacement vector.

57. 3 - Way Tug-o-War Practice Problem The 3 Stooges are fighting over a 10 000 g (10 thousand gram) Snickers Bar. The fight lasts 9.6 s, and their forces are constant. The floor on which they’re standing has a huge coordinate system painted on it, and the candy bar is at the origin. What are its final coordinates? 78  Curly: 1000 N Moe: 500 N 93  Larry: 150 N Hint: Find this angle first. Answer: ( -203.66, 2246.22 ) in meters

58. How to budge a stubborn mule Big Force Little Force It would be pretty tough to budge this mule by pulling directly on his collar. But it would be relatively easy to budge him using this set-up. (explanation on next slide)

59. How to budge a stubborn mule (cont.) overhead view treemule little force Just before the mule budges, we have static equilibrium. This means the tension forces in the rope segments must cancel out the little applied force. But because of the small angle, the tension is huge, enough to budge the mule! treemule little force T T (more explanation on next slide)

60. How to budge a stubborn mule (final) tree mule little force T T Because  is so small, the tensions must be large to have vertical components (orange) big enough to team up and cancel the little force. Since the tension is the same throughout the rope, the big tension forces shown acting at the middle are the same as the forces acting on the tree and mule. So the mule is pulled in the direction of the rope with a force equal to the tension. This set-up magnifies your force greatly. 

61. Relative Velocities in 1 D Schmedrick and his dog, Rover, are goofing around on a train. Schmed can throw a fast ball at 23 m/s. Rover can run at 9 m/s. The train goes 15 m/s. continued on next slide Question 1: If Rover is sitting beside the tracks with a radar gun as the train goes by, and Schmedrick is on the train throwing a fastball in the direction of the train, how fast does Rover clock the ball? v BT = velocity of the ball with respect to the train = 23 m/s v TG = velocity of the train with respect to the ground = 15 m/s v BG = velocity of the ball with respect to ground = 38 m/s This is a simple example, but in general, to get the answer we add vectors: v BG = v BT + v TG (In this case we can simply add magnitudes since the vectors are parallel.)

62. Relative Velocities in 1 D (cont.) Velocities are not absolute; they depend on the motion of the person who is doing the measuring. Write a vector sum so that the inner subscripts match. The outer subscripts give the subscripts for the resultant. This trick works even when vectors don’t line up. Vector diagrams help (especially when we move to 2-D). v BG = v BT + v TG v BT = 23 m/s v TG = 15 m/s v BG = 38 m/s continued on next slide

63. Question 2: Let’s choose the positive direction to be to the right. If Schmedrick is standing still on the ground and Rover is running to the right, then the velocity of Rover with respect to Schmedrick = v RS = +9 m/s. From Rover’s perspective, though, he is the one who is still and Schmedrick (and the rest of the landscape) is moving to the left at 9 m/s. This means the velocity of Schmedrick with respect to Rover = v SR = -9 m/s. Therefore, v RS = - v SR The moral of the story is that you get the opposite of a vector if you reverse the subscripts. Relative Velocities in 1 D (cont.) continued on next slide v SR v RS

64. Relative Velocities in 1 D (cont.) Question 3: If Rover is chasing the train as Schmed goes by throwing a fastball, at what speed does Rover clock the ball now? v BT = 23 m/s v TG = 15 m/s v BG = 29 m/s Note, because Rover is chasing the train, he will measure a slower speed. (In fact, if Rover could run at 38 m/s he’d say the fastball is at rest.) This time we need the velocity of the ball with respect to Rover: v BR = v BT + v TG + v GR = v BT + v TG - v RG = 23 + 15 - 9 = 29 m/s. Note how the inner subscripts match up again and the outer most give the subscripts of the resultant. Also, we make use of the fact that v GR = - v RG. v RG = 9 m/s

65. River Crossing Current 0.3 m/s campsite boat You’re directly across a 20 m wide river from your buddies’ campsite. Your only means of crossing is your trusty rowboat, which you can row at 0.5 m/s in still water. If you “aim” your boat directly at the camp, you’ll end up to the right of it because of the current. At what angle should you row in order to trying to land right at the campsite, and how long will it take you to get there? river continued on next slide

66. River Crossing (cont.) Current 0.3 m/s campsite boat river 0.3 m/s 0.5 m/s Because of the current, your boat points in the direction of red but moves in the direction of green. The Pythagorean theorem tells us that green’s magnitude is 0.4 m/s. This is the speed you’re moving with respect to the campsite. Thus, t = d / v = (20 m) / (0.4 m/s) = 50 s.  = tan -1 (0.3 / 0.4)  36.9 .  0.4 m/s continued on next slide

67. River Crossing: Relative Velocities Current 0.3 m/s campsite river 0.3 m/s 0.5 m/s  0.4 m/s The red vector is the velocity of the boat with respect to the water, v BW, which is what your speedometer would read. Blue is the velocity of the water w/ resp. to the camp, v WC. Green is the velocity of the boat with respect to the camp, v BC. The only thing that could vary in our problem was . It had to be determined so that red + blue gave a vector pointing directly across the river, which is the way you wanted to go. continued on next slide

68. River Crossing: Relative Velocities (cont.) v WC v BW  v BC v BW = vel. of boat w/ respect to water v WC = vel. of water w/ respect to camp v BC = vel. of boat w/ respect to camp v BW + v WC = v BC Look how they add up: The inner subscripts match; the out ones give subscripts of the resultant. This technique works in 1, 2, or 3 dimensions w/ any number or vectors.

69. Law of Sines The river problem involved a right triangle. If it hadn’t we would have had to use either component techniques or the two laws you’ll also do in trig class: Law of Sines & Law of Cosines. Law of Sines: sin A sin B sin C a b c = = Side a is opposite angle A, b is opposite B, and c is opposite C. AB C c b a

70. Law of Cosines Law of Cosines: a 2 = b 2 + c 2 - 2 b c cosA This side is always opposite this angle. These two sides are repeated. It doesn’t matter which side is called a, b, and c, so long as the two rules above are followed. This law is like the Pythagorean theorem with a built in correction term of -2 b c cos A. This term allows us to work with non-right triangles. Note if A = 90 , this term drops out (cos 90  = 0), and we have the normal Pythagorean theorem. A B C c b a

71. v WA = vel. of Wonder Woman w/ resp. to the air v AG = vel. of the air w/ resp. to the ground (and Aqua Man) v WG = vel. of Wonder Woman w/ resp. to the ground (Aqua Man) Wonder Woman Jet Problem Suppose Wonder Woman is flying her invisible jet. Her onboard controls display a velocity of 304 mph 10  E of N. A wind blows at 195 mph in the direction of 32  N of E. What is her velocity with respect to Aqua Man, who is resting poolside down on the ground? We know the first two vectors; we need to find the third. First we’ll find it using the laws of sines & cosines, then we’ll check the result using components. Either way, we need to make a vector diagram. continued on next slide

72. The 80  angle at the lower right is the complement of the 10  angle. The two 80  angles are alternate interior. The 100  angle is the supplement of the 80  angle. Now we know the angle between red and blue is 132 . Wonder Woman Jet Problem (cont.) continued on next slide 10  32  v WA v AG v WG v WA + v AG = v WG 80  195 mph 304 mph v WG 80  32  100 

73. Wonder Woman Jet Problem (cont.) 195 mph 304 mph v 132  By the law of cosines v 2 = (304) 2 + (195) 2 - 2 (304) (195) cos 132 . So, v = 458 mph. Note that the last term above appears negative, but it’s actually positive, since cos 132  < 0. The law of sines says:  sin 132  sin  v 195 = So, sin  = 195 sin 132  / 458, and   18.45  80  This mean the angle between green and the horizontal is 80  - 18.45   61.6  Therefore, from Aqua Man’s perspective, Wonder Woman is flying at 458 mph at 61.6  N of E.

74. Wonder Woman Problem: Component Method 32  v WA = 304 mph v AG = 195 mph 10  This time we’ll add vectors via components as we’ve done before. Note that because of the angles given here, we use cosine for the vertical comp. of red but sine for the vertical comp. of blue. All units are mph. 304 195 103.3343 165.3694 52.789 299.3816 continued on next slide

75. Wonder Woman: Component Method (cont.) 304 195 103.3343 165.3694 52.789 299.3816 103.3343 52.789 165.3694 299.3816 402.7159 mph 218.1584 mph 458.0100 mph Combine vertical & horiz. comps. separately and use Pythag. theorem.  = tan -1 (218.1584 / 402.7159) = 28.4452 .  is measured from the vertical, which is why it’s 10  more than . 

76. Comparison of Methods We ended up with same result for Wonder Woman doing it in two different ways. Each way requires some work. You will only want to use the laws of sines & cosines if: the vectors form a triangle. you’re dealing with exactly 3 vectors. (If you’re adding 3 vectors, the resultant makes a total of 4, and this method would require using 2 separate triangles.) Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick.

77. floor Free body diagrams # 1 m F1F1 F 2 Two applied forces; F 2 < mg; coef. of kinetic friction =  k For the next several slides, draw a free body diagram for each mass in the set-up and find a (or write a system of 2 nd Law equations from which you could find a.) v F1F1 F 2 fkfk mg m a = F 1 - f k = F 1 -  k N = F 1 -  k (mg - F 2 ) (to the right). There is not enough info to determine whether or not N is bigger than F 2. N answer:

78. Free body diagrams # 2 Bodies start at rest; m 3 > m 1 + m 2 ; frictionless pulley with negligible mass. answer : T1T1 m3 gm3 g T1T1 m1 gm1 gT2T2 T2T2 m2 gm2 g Let’s choose clockwise as the + direction. m 1 : T 1 - m 1 g -T 2 = m 1 a m 2 : T 2 - m 2 g = m 2 a m 3 : m 3 g - T 1 = m 3 a system: m 3 g - m 1 g - m 2 g = (m 1 + m 2 + m 3 ) a (Tensions are internal and cancel out.) So, a = (m 3 - m 1 - m 2 ) g / (m 1 + m 2 + m 3 ) If masses are given, find a first with last equation and substitute to find the T ’s. m 1 m 3 m 2

79. Free body diagrams # 3 m2m2 m1m1 m3m3 v kk m 1 > m 3 m1 gm1 g T1 T1 m3 g m3 g T2 T2 Note: T 1 must be > T 2 otherwise m 2 couldn’t accelerate. T 2 - m 3 g = m 3 a T 1 - T 2 -  k m 2 g = m 2 a m 1 g - T 1 = m 1 a system: m 1 g -  k m 2 g - m 3 g = (m 1 + m 2 + m 3 ) a T1 T1 T2 T2 fkfk m2 gm2 g N answer:

80. Free body diagrams #4 m v answer: Rock falling down in a pool of water mg - D = m a. So, a = (mg - D) / m. Note: the longer the rock falls, the faster it goes and the greater D becomes, which is proportional to v. Eventually, D = mg and a becomes zero, as our equation shows, and the rock reaches terminal velocity. D mg m

81. Free body diagrams #5 answer: cotton candy Fe A large crate of cotton candy and a small iron block of the same mass are falling in air at the same speed, accelerating down. R mg R Since the masses are the same, a = (mg - R) / m for each one, but R is bigger for the cotton candy since it has more surface area and they are moving at the same speed (just for now). So the iron has a greater accelera- tion and will be moving faster than the candy hereafter. The cotton candy will reach terminal vel. sooner and its terminal vel. will be less than the iron’s.

82. Free body diagrams #6 a The boxes are not sliding; coefficients of static friction are given. answer: m1m1 m3m3 m2m2 11 22 m2m2 There is no friction acting on m 2. It would not be in equilibrium otherwise. T = m 3 g = f 1   1 N 1 =  1 (m 1 + m 2 ) g f 1 ’s reaction pair acting on table is not shown. m3 g m3 g m3m3 N1 N1 m1 gm1 g T m2 gm2 g m2 gm2 g N2N2 f1f1 T m1m1  2 is extraneous info in this problem, but not in the next slide.

83. Free body diagrams #6 b Boxes accelerating (clockwise); m 1 & m 2 are sliding; coef’s of kinetic friction given. answer: m1m1 m3m3 m2m2 11 22 v m2m2 There is friction acting on m 2 now. It would not be accelerating otherwise. m 3 g - T = m 3 a ; f 2 = m 2 a ; T - f 1 - f 2 = m 1 a, where f 1 =  1 N 1 =  1 (m 1 + m 2 ) g and f 2 =  2 N 2 =  2 m 2 g. m3 g m3 g N1 N1 m1 gm1 g T m2 gm2 g m2 gm2 g N2N2 f1f1 T m1m1 f2f2 f2f2 Note: f 2 appears twice; they’re reaction pairs.

84. Free body diagrams # 7 m 1  m2 m2 kk v Boxes moving clockwise at constant speed. answer: m2 g m2 g T m 1 g T m 1 g sin  f k m 1 g cos  N Since a = 0, m 2 g = T = m 1 g sin  + f k = m 1 g sin  +  k m 1 g cos   m 2 = m 1 (sin  +  k cos  ). This is the relationship between the masses that must exist for equilibrium. Constant velocity is the same as no velocity when it comes to the 2 nd Law. Note: sin , cos , and  k are all dimensionless quantities, so we have kg as units on both sides of the last equation.

85. Free body diagrams # 8 Mr. Stickman is out for a walk. He’s moseying along but picking up speed with each step. The coef. of static friction between the grass and his stick sneakers is  s. v answer: mg N fsfs Here’s a case where friction is a good thing. Without it we couldn’t walk. (It’s difficult to walk on ice since  s is so small.) We use f s here since we assume he’s not slipping. Note: friction is in the direction of motion in this case. His pushing force does not appear in the free body diag. since it acts on the ground, not him. The reaction to his push is friction. F net = f s So, m a = f s  f s, max =  s m g Thus, a   s g.

86. Free body diagrams #9 ground m  F kk v f k mg F sin  F cos  N Note:  is measured with respect to the vertical here. Box does not get lifted up off the ground as long as F cos   mg. If F cos   mg, then N = 0. Box budges if F sin  > f s, max =  s N =  s (mg - F cos  ). While sliding, F sin  -  k (mg - F cos  ) = m a. answer:

87. Dot Products First recall vector addition in component form:  x 1, y 1, z 1  x 2, y 2, z 2  + =  x1 + x2, y1 + y2, z1 + z2  x1 + x2, y1 + y2, z1 + z2  It’s just component-wise addition. Note that the sum of two vectors is a vector. For a dot product we do component-wise multiplication and add up the results:  x 1, y 1, z 1  x 2, y 2, z 2   = x 1 x 2 + y 1 y 2 + z 1 z 2 Note that the dot product of two vectors is a scalar! Ex:  -2, 3, 10  N  1, 6, -5  m = -2 + 18 - 50 = -34 N m  Dot products are used to find the work done by a force applied over a distance, as we’ll see in the future.

88. Dot Product Properties The dot product of two vectors is a scalar. It can be proven that a  b = a b cos , where  is the angle between a and b. The dot product of perpendicular vectors is zero. The dot product of parallel vectors is simply the product of their magnitudes. A dot product is commutative: A dot product can be performed on two vectors of the same dimension, no matter how big the dimension. a  b = b  ab  a

89. Unit Vectors in 2-D The vector v =  -3, 4  indicates 3 units left and 4 units up, which is the sum of its components: v =  -3, 4  =  -3, 0  +  0, 4  Any vector can be written as the sum of its components. Let’s factor out what we can from each vector in the sum: v =  -3, 4  = -3  1, 0  + 4  0, 1  The vectors on the right side are each of magnitude one. For this reason they are called unit vectors. A shorthand for the unit vector  1, 0  is i. A shorthand for the unit vector  0, 1  is j. Thus, v =  -3, 4  = -3 i + 4 j

90. Unit Vectors in 3-D v =  7, -5, 9  =  7, 0, 0  +  0, -5, 0  +  0, 0, 9  One way to interpret the vector v =  7, -5, 9  is that it indicates 7 units east, 5 units south, and 9 units up. v can be written as the sum components as follows: = 7  1, 0, 0  - 5  0, 1, 0  + 9  0, 0, 1  = 7 i - 5 j + 9 k In 3-D we define these unit vectors: i =  1, 0, 0 , j =  0, 1, 0 , and k =  0, 0, 1  (continued on next slide)

91. Unit Vectors in 3-D (cont.) x y z 1 i j 1 k 1 The x-, y-, and z-axes are mutually perpendicular, as are i, j, and k. The yellow plane is the x-y plane. i and j are in this plane. Any point in space can be reached from the origin using a linear combination of these 3 unit vectors. Ex: P = (-1.8, -1.4, 1.2) so the vector P -1.8 i – 1.4 j + 1.2 k will extend from the origin to P.

92. Determinants To take a determinant of a 2  2 matrix, multiply diagonals and subtract. The determinant of A is written | A | and it equals 3 (11) - 4 (-2) = 33 + 8 = 41. A = In order to do cross products we will need to find determinants of 3  3 matrices. One way to do this is to expand about the 1 st row using minors, which are smaller determinants within a determinant. To find the minor of an element, cross out its row and column and keep what remains. Minor of a: Minor of b: Minor of c: cont. on next slide

93. Determinants (cont.) (Minor of a) - b (Minor of b ) + c (Minor of c ) By definition, = a - b + c = a (e i - h f ) - b (d i - g f ) + c (d h - g e) Determinants can be expanded about any row or column. Besides cross products, determinants have many other purposes, such as solving systems of linear equations.

94. Cross Products Let v 1 =  x 1, y 1, z 1  and v 2 =  x 2, y 2, z 2 . By definition, the cross product of these vectors (pronounced “v 1 cross v 2 ”) is given by the following determinant. v1  v2 =v1  v2 = x 1 y 1 z 1 x 2 y 2 z 2 i j k = (y 1 z 2 - y 2 z 1 ) i - (x 1 z 2 - x 2 z 1 ) j + (x 1 y 2 - x 2 y 1 ) k Note that the cross product of two vectors is another vector! Cross products are used a lot in physics, e.g., torque is a vector defined as the cross product of a displacement vector and a force vector. We’ll learn about torque in another unit.

95. a  ba  b a  b. Right hand rule  b a a  ba  b A quick way to determine the direction of a cross product is to use the right hand rule. To find a  b, place the knife edge of your right hand (pinky side) along a and curl your hand toward b, making a fist. Your thumb then points in the direction of It can be proven that the magnitude of is given by: a b sin  | a  b | =| a  b | = where  is the angle between a and b.

96. Dot Product vs. Cross Product 1. The dot product of two vectors is a scalar; the cross product is another vector (perpendicular to each of the original). 2. A dot product is commutative; a cross product is not. In fact, a  b = - b  a.- b  a.  x 1, y 1, z 1  x 2, y 2, z 2   = x 1 x 2 + y 1 y 2 + z 1 z 2 3. Dot product definition: Cross product definition: v1  v2 =v1  v2 = x 1 y 1 z 1 x 2 y 2 z 2 i j k 4. a  b = a b cos , and a b sin  | a  b | =| a  b | =