Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Presentasi Message Confidentiality ... by Uliel Azmie 291 views
- чил презентация by Doltz 185 views
- 4 sem pathology - 2005 to 2010 1 by Siam Weng Loong 117 views
- Best Lab Furniture and Instruments ... by Best Lab Furnitur... 898 views
- Presentati Sisteam Antrian Real-Time by Uliel Azmie 361 views
- Wikis (3) by Elizabeth Rodrigu... 150 views

1,633 views

Published on

- 1. Chapter 3 Kinematics in TwoDimensions; Vectors
- 2. VectorsAn Introduction
- 3. There are two kinds of quantities…• Scalars are quantities that have magnitude only, such as – position – speed – time – mass• Vectors are quantities that have both magnitude and direction, such as – displacement – velocity – acceleration
- 4. Notating vectors• This is how you notate a vector… R R• This is how you draw a vector… R headtail
- 5. Direction of Vectors• Vector direction is the direction of the arrow, given by an angle.• This vector has an angle that is between 0o and 90o. A θ x
- 6. Vector angle ranges y Quadrant II Quadrant I90o < θ < 180o 0 < θ < 90o θ θ x θ θ Quadrant III Quadrant IV180o < θ < 270o 270o < θ < 360o
- 7. Direction of Vectors• What angle range would this vector have?• What would be the exact angle, and how would you determine it? Between 180o and 270o θ x θ B or between - 90o and -180o
- 8. Magnitude of Vectors• The best way to determine the magnitude (or size) of a vector is to measure its length.• The length of the vector is proportional to the magnitude (or size) of the quantity it represents.
- 9. Sample Problem• If vector A represents a displacement of three miles to the north, then what does vector B represent? Vector C? B A C
- 10. Equal Vectors• Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).• Draw several equal vectors.
- 11. Inverse Vectors• Inverse vectors have the same A length, but opposite direction.• Draw a set of -A inverse vectors.
- 12. The Right Triangle hy poopposite te nu se θ adjacent
- 13. Pythagorean Theorem• hypotenuse2 = opposite2 + adjacent2• c2 = a2 + b2 hy po opposite te nu se θ adjacent
- 14. Basic Trigonometry functions• sin θ = opposite/hypotenuse• cos θ = adjacent/hypotenuse• tan θ = opposite/adjacent hy po opposite te nu se SOHCAHTOA θ adjacent
- 15. Inverse functions• θ = sin-1(opposite/hypotenuse)• θ = cos-1(adjacent/hypotenuse)• θ = tan-1(opposite/adjacent) hy po opposite te nu se SOHCAHTOA θ adjacent
- 16. Sample problem• A surveyor stands on a riverbank directly across the river from a tree on the opposite bank. She then walks 100 m downstream, and determines that the angle from her new position to the tree on the opposite bank is 50o. How wide is the river, and how far is she from the tree in her new location?
- 17. Sample problem• You are standing at the very top of a tower and notice that in order to see a manhole cover on the ground 50 meters from the base of the tower, you must look down at an angle 75o below the horizontal. If you are 1.80 m tall, how high is the tower?
- 18. Vectors: x-component• The x-component of a vector is the “shadow” it casts on the x-axis.• cos θ = adjacent ∕ hypotenuse• cos θ = Ax ∕ A• Ax = A cos θ A θ x Ax
- 19. Vectors: y-component• The y-component of a vector is the “shadow” it casts on the y-axis.• sin θ = opposite ∕ hypotenuse• sin θ = Ay ∕ A• Ay = A sin θ y Ay A Ay θ x
- 20. Vectors: angle• The angle a vector makes with the x- axis can be y determined by the components. Ry• It is calculated by the inverse tangent θ x Rx function• θ = tan-1 (Ay/Ax)
- 21. Vectors: magnitude• The magnitude of a vector can be determined by the y components. R• It is calculated using the Pythagorean Ry Theorem. x• R2 = Rx2 + Ry2 Rx
- 22. Practice Problem• You are driving up a long inclined road. After 1.5 miles you notice that signs along the roadside indicate that your elevation has increased by 520 feet. a) What is the angle of the road above the horizontal?
- 23. Practice Problem• You are driving up a long inclined road. After 1.5 miles you notice that signs along the roadside indicate that your elevation has increased by 520 feet. b) How far do you have to drive to gain an additional 150 feet of elevation?
- 24. Practice Problem• Find the x- and y-components of the following vectors a) R = 175 meters @ 95o
- 25. Practice Problem• Find the x- and y-components of the following vectors b) v = 25 m/s @ -78o
- 26. Practice Problem• Find the x- and y-components of the following vectors c) a = 2.23 m/s2 @ 150o
- 27. Graphical Addition of Vectors Day 2
- 28. Graphical Addition of Vectors1) Add vectors A and B graphically by drawing them together in a head to tail arrangement.2) Draw vector A first, and then draw vector B such that its tail is on the head of vector A.3) Then draw the sum, or resultant vector, by drawing a vector from the tail of A to the head of B.4) Measure the magnitude and direction of the resultant vector.
- 29. Practice Graphical Addition B B A R A + B = RR is called the resultant vector!
- 30. The Resultant and the Equilibrant• The sum of two or more vectors is called the resultant vector.• The resultant vector can replace the vectors from which it is derived.• The resultant is completely canceled out by adding it to its inverse, which is called the equilibrant.
- 31. The Equilibrant Vector B A -R R A + B =The vector -R is calledRtheequilibrant.If you add R and -R you get anull (or zero) vector.
- 32. Graphical Subtraction of Vectors1) Subtract vectors A and B graphically by adding vector A with the inverse of vector B (-B).2) First draw vector A, then draw -B such that its tail is on the head of vector A.3) The difference is the vector drawn from the tail of vector A to the head of -B.
- 33. Practice Graphical Subtraction -B B C A A - B = C
- 34. Practice Problem• Vector A points in the +x direction and has a magnitude of 75 m. Vector B has a magnitude of 30 m and has a direction of 30o relative to the x axis. Vector C has a magnitude of 50 m and points in a direction of -60o relative to the x axis.a) Find A + Bb) Find A + B + Cc) Find A – B.
- 35. a)
- 36. b)
- 37. c)
- 38. Vector Addition Laboratory
- 39. Vector Addition Lab 1.Attach spring scales to force board such that they all have different 2.Slip graph paper between scales and board and carefully trace your setIn Class 3.Record readings of all three spring scales. 4.Detach scales from board and remove graph paper. 5.On top of your tracing, draw a force diagram by constructing vectorsHomework 6.On a separate sheet of graph paper, add the three vectors together 7.Did you get a resultant? Did you expect one? 8.You must have a separate set of drawings for each member of
- 40. Vector Addition by Component
- 41. Component Addition of Vectors1) Resolve each vector into its x- and y- components. Ax = Acosθ Ay = Asinθ Bx = Bcosθ By = Bsinθ Cx = Ccosθ Cy = Csinθ etc.1) Add the x-components (Ax, Bx, etc.) together to get Rx and the y-components (Ay, By, etc.) to get Ry.
- 42. Component Addition of Vectors3) Calculate the magnitude of the resultant with the Pythagorean Theorem (R = √Rx2 + Ry2).4) Determine the angle with the equation θ = tan-1 Ry/Rx.
- 43. Practice Problem• In a daily prowl through the neighborhood, a cat makes a displacement of 120 m due north, followed by a displacement of 72 m due west. Find the magnitude and displacement required if the cat is to return home.
- 44. Practice Problem• If the cat in the previous problem takes 45 minutes to complete the first displacement and 17 minutes to complete the second displacement, what is the magnitude and direction of its average velocity during this 62-minute period of time?
- 45. Relative Motion Day 3
- 46. Relative Motion• Relative motion problems are difficult to do unless one applies vector addition concepts.• Define a vector for a swimmer’s velocity relative to the water, and another vector for the velocity of the water relative to the ground. Adding those two vectors will give you the velocity of the swimmer relative to the ground.
- 47. Relative MotionVs VwVw Vt = V s + V w
- 48. Relative Motion Vs Vw VwVt = V s + V w
- 49. Relative Motion Vw Vs VwVt = V s + V w
- 50. Practice Problem• You are paddling a canoe in a river that is flowing at 4.0 mph east. You are capable of paddling at 5.0 mph.a) If you paddle east, what is your velocity relative to the shore?b) If you paddle west, what is your velocity relative to the shore?c) You want to paddle straight across the river, from the south to the north.At what angle to you aim your boat relative to the shore? Assume east is 0o.
- 51. Practice Problem• You are flying a plane with an airspeed of 400 mph. If you are flying in a region with a 80 mph west wind, what must your heading be to fly due north?
- 52. Solving 2-D Problems• Resolve all vectors into components – x-component – Y-component• Work the problem as two one-dimensional problems. – Each dimension can obey different equations of motion.• Re-combine the results for the two components at the end of the problem.
- 53. Sample Problem• You run in a straight line at a speed of 5.0 m/s in a direction that is 40o south of west. a) How far west have you traveled in 2.5 minutes? b) How far south have you traveled in 2.5 minutes?
- 54. Sample Problem• A roller coaster rolls down a 20o incline with an acceleration of 5.0 m/s2. a) How far horizontally has the coaster traveled in 10 seconds? b) How far vertically has the coaster traveled in 10 seconds?
- 55. Sample ProblemA particle passes through the origin with a speed of 6.2 m/straveling along the y axis. If the particle accelerates in thenegative x direction at 4.4 m/s2. a) What are the x and y positions at 5.0 seconds?
- 56. Sample ProblemA particle passes through the origin with a speed of 6.2 m/straveling along the y axis. If the particle accelerates in thenegative x direction at 4.4 m/s2. b) What are the x and y components of velocity at this time?
- 57. Projectile Motion• Day 4
- 58. 3-5 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earths gravity; its path is a parabola.
- 59. Projectile Motion• Something is fired, thrown, shot, or hurled near the earth’s surface.• Horizontal velocity is constant.• Vertical velocity is accelerated.• Air resistance is ignored.
- 60. 1-Dimensional Projectile• Definition: A projectile that moves in a vertical direction only, subject to acceleration by gravity.• Examples: – Drop something off a cliff. – Throw something straight up and catch it.• You calculate vertical motion only.• The motion has no horizontal component.
- 61. 2-Dimensional Projectile• Definition: A projectile that moves both horizontally and vertically, subject to acceleration by gravity in vertical direction.• Examples: – Throw a softball to someone else. – Fire a cannon horizontally off a cliff. – Shoot a monkey with a blowgun.• You calculate vertical and horizontal motion.
- 62. Horizontal Component of Velocity• Is constant• Not accelerated• Not influence by gravity• Follows equation:• x = Vo,xt
- 63. Horizontal Component of Velocity
- 64. Vertical Component of Velocity• Undergoes accelerated motion• Accelerated by gravity (9.8 m/s2 down)• Vy = Vo,y - gt• y = yo + Vo,yt - 1/2gt2• Vy2 = Vo,y2 - 2g(y – yo)
- 65. Horizontal and Vertical
- 66. Horizontal and Vertical
- 67. Zero Launch Angle Projectiles
- 68. Launch angle• Definition: The angle at which a projectile is launched.• The launch angle determines what the trajectory of the projectile will be.• Launch angles can range from -90 o (throwing something straight down) to +90 o (throwing something straight up) and everything in between.
- 69. Zero Launch angle vo• A zero launch angle implies a perfectly horizontal launch.
- 70. Sample Problem• The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.
- 71. Sample Problem• An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?
- 72. Sample Problem• Playing shortstop, you throw a ball horizontally to the second baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later. a) How far were you from the second baseman? b) What is the distance of the vertical drop?
- 73. General Launch Angle Projectiles Day 5
- 74. General launch angle vo θ• Projectile motion is more complicated when the launch angle is not straight up or down (90 o or – 90o), or perfectly horizontal (0o).
- 75. General launch angle vo θ• You must begin problems like this by resolving the velocity vector into its components.
- 76. Resolving the velocity• Use speed and the launch angle to find horizontal and vertical velocity components Vo Vo,y = Vo sin θ θVo,x = Vo cos θ
- 77. Resolving the velocity• Then proceed to work problems just like you did with the zero launch angle problems. Vo Vo,y = Vo sin θ θVo,x = Vo cos θ
- 78. Sample problem• A soccer ball is kicked with a speed of 9.50 m/s at an angle of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?
- 79. Sample problem• Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25o above the horizontal. When the snowballs land, is the speed of A greater than, less than, or the same speed of B? Verify your answer by calculation of the landing speed of both snowballs.
- 80. Projectiles launched over level ground• These projectiles have highly symmetric characteristics of motion.• It is handy to know these characteristics, since a knowledge of the symmetry can help in working problems and predicting the motion.• Lets take a look at projectiles launched over level ground.
- 81. Trajectory of a 2-D Projectile y x• Definition: The trajectory is the path traveled by any projectile. It is plotted on an x-y graph.
- 82. Trajectory of a 2-D Projectile y x• Mathematically, the path is defined by a parabola.
- 83. Trajectory of a 2-D Projectile y x• For a projectile launched over level ground, the symmetry is apparent.
- 84. Range of a 2-D Projectile y x Range• Definition: The RANGE of the projectile is how far it travels horizontally.
- 85. Maximum height of a projectile y Maximum Height x Range• The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward.
- 86. Maximum height of a projectile y Maximum Height x Range• The vertical velocity component is zero at maximum height.
- 87. Maximum height of a projectile y Maximum Height x Range• For a projectile launched over level ground, the maximum height occurs halfway through the flight of the projectile.
- 88. Acceleration of a projectile y g g g g g x• Acceleration points down at 9.8 m/s 2 for the entire trajectory of all projectiles.
- 89. Velocity of a projectile y v v v vo vf x• Velocity is tangent to the path for the entire trajectory.
- 90. Velocity of a projectile y vx vy vx vx vy vy vx vx vy x• The velocity can be resolved into components all along its path.
- 91. Velocity of a projectile y vx vy vx vx vy vy vx vx vy x• Notice how the vertical velocity changes while the horizontal velocity remains constant.
- 92. Velocity of a projectile y vx vy vx vx vy vy vx vx vy x• Maximum speed is attained at the beginning, and again at the end, of the trajectory if the projectile is launched over level ground.
- 93. Velocity of a projectile vo θ θ- vo• Launch angle is symmetric with landing angle for a projectile launched over level ground.
- 94. Time of flight for a projectile tto = 0 • The projectile spends half its time traveling upward…
- 95. Time of flight for a projectile tto = 0 2t • … and the other half traveling down.
- 96. Position graphs for 2-D projectilesy y x x t t
- 97. Velocity graphs for 2-D projectiles Vy Vx t t
- 98. Acceleration graphs for 2-D projectiles ay ax t t
- 99. Projectile Lab
- 100. Projectile LabThe purpose is to collect data to plot a trajectory for aprojectile launched horizontally, and to calculate thelaunch velocity of the projectile. Equipment is provided,you figure out how to use it.• What you turn in: 1. a table of data 2. a graph of the trajectory 3. a calculation of the launch velocity of the ball obtained from the data• Hints and tips: 1. The thin paper strip is pressure sensitive. Striking the paper produces a mark. 2. You might like to hang a sheet of your own graph paper on the brown board.
- 101. More on Projectile Motion
- 102. The Range Equation• Derivation is an important part of physics.• Your book has many more equations than your formula sheet.• The Range Equation is in your textbook, but not on your formula sheet. You can use it if you can memorize it or derive it!
- 103. The Range Equation• R = vo2sin(2θ)/g. – R: range of projectile fired over level ground – vo: initial velocity – g: acceleration due to gravity θ: launch angle
- 104. Deriving the Range Equation
- 105. Review Day
- 106. Sample problem• A golfer tees off on level ground, giving the ball an initial speed of 42.0 m/s and an initial direction of 35o above the horizontal. a) How far from the golfer does the ball land?
- 107. Sample problem• A golfer tees off on level ground, giving the ball an initial speed of 42.0 m/s and an initial direction of 35o above the horizontal. b) The next golfer hits a ball with the same initial speed, but at a greater angle than 45o. The ball travels the same horizontal distance. What was the initial direction of motion?
- 108. 3-5 Projectile Motion It can be understood by analyzing the horizontal and vertical motions separately. 4 monkey problem
- 109. 3-5 Projectile Motion The speed in the x-direction is constant; in the y- direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.
- 110. 3-5 Projectile MotionIf an object is launched at an initial angle of θ0with the horizontal, the analysis is similar exceptthat the initial velocity has a vertical component.
- 111. 3-6 Solving Problems Involving Projectile MotionProjectile motion is motion with constantacceleration in two dimensions, where theacceleration is g and is down.
- 112. 3-6 Solving Problems Involving Projectile Motion1. Read the problem carefully, and choose the object(s) you are going to analyze.2. Draw a diagram.3. Choose an origin and a coordinate system.4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.5. Examine the x and y motions separately.
- 113. 3-6 Solving Problems Involving Projectile Motion6. List known and unknown quantities.Remember that vx never changes, and thatvy = 0 at the highest point.7. Plan how you will proceed. Use theappropriate equations; you may have tocombine some of them.
- 114. 3-7 Projectile Motion Is Parabolic In order to demonstrate that projectile motion is parabolic, we need to write y as a function of x. When we do, we find that it has the form: This is indeed the equation for a parabola.
- 115. 3-8 Relative VelocityWe already considered relative speed in onedimension; it is similar in two dimensionsexcept that we must add and subtract velocitiesas vectors.Each velocity is labeled first with the object, andsecond with the reference frame in which it hasthis velocity. Therefore, vWS is the velocity of thewater in the shore frame, vBS is the velocity of theboat in the shore frame, and vBW is the velocity ofthe boat in the water frame.
- 116. 3-8 Relative VelocityIn this case, the relationship between thethree velocities is: (3-6)
- 117. Summary of Chapter 3• A quantity with magnitude and direction is avector.• A quantity with magnitude but no direction isa scalar.• Vector addition can be done either graphicallyor using components.• The sum is called the resultant vector.• Projectile motion is the motion of an objectnear the Earth’s surface under the influence ofgravity.
- 118. 3-1 Vectors and Scalars A vector has magnitude as well as direction. Some vector quantities: displacement, velocity, force, momentum A scalar has only a magnitude. Some scalar quantities: mass, time, temperature 1
- 119. 3-2 Addition of Vectors – Graphical Methods For vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs, as the figure indicates.
- 120. 3-2 Addition of Vectors – Graphical MethodsIf the motion is in two dimensions, the situation issomewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem.
- 121. 3-2 Addition of Vectors – Graphical MethodsAdding the vectors in the opposite order gives thesame result:
- 122. 3-2 Addition of Vectors – Graphical Methods Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip” method.
- 123. 3-2 Addition of Vectors – Graphical MethodsThe parallelogram method may also be used;here again the vectors must be “tail-to-tip.”
- 124. 3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction.Then we add the negative vector:
- 125. 3-3 Subtraction of Vectors, and Multiplication of a Vector by a ScalarA vector V can be multiplied by a scalar c; theresult is a vector cV that has the same directionbut a magnitude cV. If c is negative, the resultantvector points in the opposite direction.
- 126. 3-4 Adding Vectors by ComponentsAny vector can be expressed as the sumof two other vectors, which are called itscomponents. Usually the other vectors arechosen so that they are perpendicular toeach other.
- 127. 3-4 Adding Vectors by Components If the components are perpendicular, they can be found using trigonometric functions.
- 128. 3-4 Adding Vectors by ComponentsThe components are effectively one-dimensional,so they can be added arithmetically:
- 129. 3-4 Adding Vectors by ComponentsAdding vectors:1. Draw a diagram; add the vectors graphically.2. Choose x and y axes.3. Resolve each vector into x and y components.4. Calculate each component using sines and cosines.5. Add the components in each direction.6. To find the length and direction of the vector, use:

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment