Mechanics 2012


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Mechanics 2012

  1. 1. Mechanics
  2. 2. Contents  Kinematic Concepts: Displacement Speed vs. Velocity Acceleration Changing Units Instantaneous vs. Average Frames of Reference Graphical Representation of Position Graphical Representation of Velocity Graphical Representation of Acceleration  Equations for Constant Acceleration  Vertical Motion of Objects  Air Resistance  Force and Mass: Springs
  3. 3. Mechanics Mechanics is the study of:  motion,  force and,  energy.
  4. 4. Kinematic Concepts Kinematics is the part of mechanics that describes how objects move. The why objects move as they do is called dynamics.
  5. 5. Kinematic Concepts Displacement When a person moves over a time interval,  they change their position in space. This change in position is called the:  displacement. It is a vector quantity.
  6. 6. Kinematic Concepts The further it moves from its starting position,  the greater it’s displacement.
  7. 7. Kinematic Concepts direction; If the object moves in the opposite  as defined,  its displacement will be negative.
  8. 8. Kinematic Concepts Displacement is given the symbol s;  or sometimes x,  it’s S.I. unit is the (m)etre.
  9. 9. Kinematic Concepts The term distance is a scalar:  the symbol is d. Distance is more useful when purchasing a car.  The distance travelled is important;  while the direction it travelled in is not.
  10. 10. Kinematic Concepts A person who walks 100 m east and then;  100 m west has travelled a distance of,  200 m. Their displacement however is 0 m. They have ended up at their starting point. Distance vs Displacement
  11. 11. Kinematic Concepts Speed vs Velocity Although speed and velocity are used interchangeably in everyday life,  both terms have specific meanings.
  12. 12. Kinematic Concepts Speed is defined as;  the distance travelled by an object in a given time interval. This will give us the average speed. Mathematically, it can be represented by: average speed distancetravelled time taken
  13. 13. Kinematic Concepts As distance is one of the variables,  speed must be a scalar. Velocity is used to signify both magnitude and direction hence;  it is a vector.
  14. 14. Kinematic Concepts The average velocity is defined as:  the change in position of the object;  in a given time interval. Mathematically, it can be described as: average velocity displacement time taken v s t
  15. 15. Kinematic Concepts The S.I. unit for both speed and velocity is m s-1. Always include a direction when giving the value of the velocity.
  16. 16. Kinematic Concepts Acceleration. An object whose velocity is changing is:  accelerating. Acceleration is defined as:  the rate of change in velocity. a v t
  17. 17. Kinematic Concepts  Which cars below are acceleration and which cars are traveling at constant velocity?  Which car accelerates fastest?
  18. 18. Kinematic Concepts The red car is traveling at constant velocity. The blue car is accelerating fastest.  Its rate of change of velocity is greatest.
  19. 19. Kinematic Concepts Since an object always accelerates in a given direction,  acceleration is a vector quantity. The S.I. units are m s-2
  20. 20. Kinematic Concepts Changing Units When using kinematic equations,  the data given is not always given in S.I. units. The data needs to be converted to S.I. Units;  before they can be substituted into an equation.
  21. 21. Kinematic Concepts An example is speed. Very often speed is given in km h-1. The kilometres need to be converted to metres and the hours into seconds. There are 1000 m in 1 km  3600 s in 1 hr (60 x 60). 1 100 km 100 x 1000 100 km h 27.7 m s -1 1h 1 x 60 x 60
  22. 22. Kinematic Concepts There is doubt over whether the correct answer should be:  30 m s-1,  28 m s-1 or,  27.7 m s-1  due to the number of significant figures given.
  23. 23. Kinematic Concepts To avoid this problem, always give your data using scientific notation. In the above example,  1.00 x 102 m s-1 would eliminate the problem  hence the answer would be 27.7 m s-1.
  24. 24. Kinematic Concepts Instantaneous vs Average If you drive a car for 240 km in 3 hrs,  your average speed is 80 km h-1. It is unlikely that for every part of the journey,  you would be travelling at 80 km h-1. At each instant, your speed would change.
  25. 25. Kinematic Concepts The speedometer in the car gives;  the instantaneous speed. The instantaneous speed is defined as:  the average speed over an indefinitely short time interval.
  26. 26. Kinematic Concepts The same definition can be used for velocity. The formula for velocity needs to be changed to:  accommodate the difference between average,  and instantaneous velocity.
  27. 27. Kinematic Concepts Average velocity: _ s s v av t v t Instantaneous velocity: s v t
  28. 28. Kinematic Concepts As t becomes very small,  approaching zero,  s approaches zero as well. However, the ratio approaches a definite value.
  29. 29. Kinematic Concepts This definite value is known as the instantaneous velocity. The symbol for instantaneous velocity is:  v no av or bar above the v. s v t
  30. 30. Kinematic Concepts The same procedure can be used for acceleration and so the equations become: Average acceleration: _ v v a av t a t Instantaneous acceleration: v a t
  31. 31. Kinematic Concepts Relationship between Velocity & Acceleration
  32. 32. Kinematic Concepts Frames of Reference The simple question ‘how fast are you moving’ is quite complex. In your seat now, you are moving at:  0 m s-1 relative to the room. If the room were on the equator,  you would be moving at 1700 km h-1.
  33. 33. Kinematic Concepts At the pole you would be spinning on the spot. This however is not true relative to the Sun. Two planes moving at 1000 km h-1 relative to the ground:  may be stationary relative to each other.
  34. 34. Kinematic Concepts All motion is relative. We must choose our frame of reference in each description of motion we give. Relative Motion (frame of reference) Frames of Reference Movie (Part 1) Frames of Reference Movie (Part 2) Frames of Reference Movie (Part 3) Frames of Reference Movie (Part 4)
  35. 35. Kinematic Concepts Usually the motion of objects is considered relative to the earth. This is called:  the inertial frame of reference. Unless otherwise stated,  this will be assumed. Inertial Frame
  36. 36. Graphical Representation ofMotion  Graphical Representation of Position  Multiflash photographs of an object in motion;  can be taken and,  data collected from it.  An example of a multiflash photo is the toy car on the next slide.
  37. 37. Graphical Representation ofMotion To take this photo;  the film needs to be exposed for a period of time in darkness,  with a strobe flashing at a known rate.
  38. 38. Graphical Representation ofMotion From the data collected,  graphs can be drawn. By plotting position on the vertical axis and;  time on the horizontal,  the graph will look like:
  39. 39. Graphical Representation ofMotion
  40. 40. Graphical Representation ofMotion A number of facts can be ascertained from this graph.1. As the line is straight,  the change in position per unit time, is constant.  This means the velocity is constant.
  41. 41. Graphical Representation ofMotion2. The magnitude of the velocity can be obtained by the slope; changein position v changein tim e
  42. 42. Graphical Representation ofMotion If the graph is steep, it means:  there is a greater change in position,  per unit time and the object,  is moving relatively fast.
  43. 43. Graphical Representation ofMotion
  44. 44. Graphical Representation ofMotion  What happens when two cars traveling at different speeds but with constant velocity?
  45. 45. Graphical Representation ofMotion If the graph is horizontal:  the object is stationary. If the slope is negative:  the object is moving back towards its starting position with,  constant velocity.
  46. 46. Graphical Representation ofMotion
  47. 47. Graphical Representation ofMotion  In graph a); 1. The person has moved at constant velocity over the interval t1 to point p 2. They then remained stationary for the period t2 3. The person then returned to the original starting position at a constant, but slower velocity over the interval t3.
  48. 48. Graphical Representation ofMotion  In graph b); 1. The person has moved at constant velocity over the interval t1 to point p 2. They then remained stationary for the period t2 3. The person then continued in the original direction for the same distance p as in interval t1 at a constant, but slower velocity over the interval t3.
  49. 49. Graphical Representation ofMotion  Graphical Representation of Velocity  Consider graph a) from above.  From the information from the graph, a graph of the person’s velocity can be drawn.
  50. 50. Graphical Representation ofMotion
  51. 51. Graphical Representation ofMotion  The graph is slightly idealised as:  the person could not travel at the constant velocity,  at every instant of the journey.
  52. 52. Graphical Representation ofMotion  In a real situation;  velocity does not change in zero time and a more likely description is,  the one shown above.
  53. 53. Graphical Representation ofMotion  What about v/t graphs for the two cars previously?
  54. 54. Graphical Representation ofMotion V vs t graph animation
  55. 55. Graphical Representation ofMotion Multiflash photography can be used to obtain:  direct values of average velocity. This is because time intervals are:  very short and,  constant.
  56. 56. Graphical Representation ofMotion The distance from one image to the next is:  the change in position of the object in,  a specific interval of time. Changes in position in equal intervals of time are:  direct measures of the average velocity,  over those intervals.
  57. 57. Graphical Representation ofMotion The area under the Velocity vs. Time graph can:  be used to obtain further information. This information relates to:  the change in position of the object.
  58. 58. Graphical Representation ofMotion _ changein position v changein tim e Rearranging this equation:  change in position is given by the product; average velocity x time taken The graph on the next slide deals with constant velocity.
  59. 59. Graphical Representation ofMotion
  60. 60. Graphical Representation ofMotion As the velocity is constant;  the average velocity must be v. Hence, the change in position:  will be vt. This is also equal to the:  area under graph in time interval t. This relationship holds for all v/t graphs;  as in the example on the next slide:
  61. 61. Graphical Representation ofMotion
  62. 62. Graphical Representation ofMotion  The motion in the graph above describes:  motion under non-zero constant acceleration  eg motion under the force of gravity.
  63. 63. Graphical Representation ofMotion  Since velocity changes in a regular way;  the average velocity is,  the average of the initial velocity and,  the final velocity.  The horizontal line through P represents this.
  64. 64. Graphical Representation ofMotion The areas a and b are equal:  one being under the graph and,  the other, above the line.
  65. 65. Graphical Representation ofMotion The product of vavt is:  the area under the graph and,  gives the change in position.
  66. 66. Graphical Representation ofMotion Another way to calculate the change in position is:  to divide the area under the graph into,  a rectangle and triangle and,  add the two solutions.
  67. 67. Graphical Representation ofMotion Any area below the axis corresponds to:  motion in the opposite direction where,  the change in position is opposite in sign.
  68. 68. Graphical Representation ofMotion Representation of Acceleration Graphical Velocity (ms -1 ) Time (sec)
  69. 69. Graphical Representation ofMotion In the example, the slope, Velocity (ms -1 )  and therefore the rate at which velocity is changing,  is constant. Constant Acceleration Graph Time (sec)
  70. 70. Graphical Representation ofMotion Acceleration is defined as:  the rate of change of velocity. Mathematically, this can be written as: v a t
  71. 71. Graphical Representation ofMotion The slope of the graph is given by: rise/run. In the example, the rise = v and the run as t. Velocity (ms -1 ) Time (sec)
  72. 72. Graphical Representation ofMotion  This gives the formula: v t  This corresponds to our formula for acceleration.
  73. 73. Graphical Representation ofMotion This corresponds to our formula for acceleration. The slope of a v/t graph gives:  acceleration. Acceleration is measured in m s-2 and;  is a vector quantity so,  direction must always be included.
  74. 74. Graphical Representation ofMotion A car is stationary at the lights when the lights change to green. Another car is moving when the lights turn green. What is the displacement of each car after 3 seconds?
  75. 75. Graphical Representation ofMotion
  76. 76. Graphical Representation ofMotion Red car Area of triangle = ½ b x h Displacement = ½ x 3 x 12 Displacement = 18 m
  77. 77. Graphical Representation ofMotion Blue car Area of rectangle = b x h Displacement = 3 x 10 Displacement = 30 m
  78. 78. Graphical Representation ofMotion  Watch the animation again.  What is the acceleration of the red car after 3 seconds?
  79. 79. Graphical Representation ofMotion Slope = ryse/run Acceleration = 12/3 Acceleration = 4 m s-1 Watch the animation again and determine the displacement of both cars after 9 seconds.
  80. 80. Graphical Representation ofMotion
  81. 81. Graphical Representation ofMotion Red Car Displacement = area of triangle + area of rectangle. Displacement = (½ x 3 x 12) + (9 x 12) Displacement =18 + 72 Displacement = 90 m
  82. 82. Graphical Representation ofMotion Blue Car Displacement = area of rectangle Displacement = 9 x 10 Displacement = 90 m Watch the animation again. When do the two cars pass each other? Does it agree with your calculations?
  83. 83. Graphical Representation ofMotion
  84. 84. Graphical Representation ofMotion Graphs of position, velocity and acceleration can be drawn for the same object.
  85. 85. Graphical Representation ofMotion What happens if it is traveling backwards?
  86. 86. Graphical Representation ofMotion What happens when it is being pushed forward?
  87. 87. Graphical Representation ofMotion Traveling in the opposite direction?
  88. 88. Graphical Representation ofMotion What happens when it is pulled backward?
  89. 89. Graphical Representation ofMotion Traveling in the opposite direction?
  90. 90. Graphical Representation ofMotion Very complicated motion can be studied using graphs. Watch the two stage rocket as it is launched, run out of fuel and returns to Earth.
  91. 91. Graphical Representation ofMotion
  92. 92. Graphical Representation ofMotion The Moving Man Motion Graphs If you would like more practice at drawing motion graphs diagrams, try this web site: aph.html
  93. 93. Equations for UniformlyAccelerated Motion The previous equations used can:  be applied to all types of motion. However, when acceleration is constant:  our mathematical description can be taken further.
  94. 94. Equations for UniformlyAccelerated Motion Velocity -1 (ms ) v . v av u . t Time (sec)
  95. 95. Equations for UniformlyAccelerated Motion In the graph on the previous slide, as the slope is constant;  the acceleration is also constant. If we let u and v be the velocity at the start and end of the time interval t;  average velocity can be described as;
  96. 96. Equations for UniformlyAccelerated Motion u v vav 2 and from previously, s vav t Comparing these equations, we can say: u v s  2 t
  97. 97. Equations for UniformlyAccelerated Motion  The average acceleration can also be given as: v u a t  Rearranging the equation gives: v u at 
  98. 98. Equations for UniformlyAccelerated Motion These two equations describe motion:  at constant acceleration,  in terms of five variables;  u, v, t, s, a.
  99. 99. Equations for UniformlyAccelerated Motion Equation u does not use a and;  v does not use s. Using algebra, we can derive 3 more equations,  each one not using one variable listed above.
  100. 100. Equations for UniformlyAccelerated Motion u v s  2 t  rearranging gives; v u t s 2
  101. 101. Equations for UniformlyAccelerated Motion  Equation 2: v = u + at rearranged becomes; v u t a  substituting for t we have:
  102. 102. Equations for UniformlyAccelerated Motion v u v u s 2 a v2 u 2 s 2a  v2 - u2 = 2as w
  103. 103. Equations for UniformlyAccelerated Motion u v s  2 t  Equation v v = u + at  Substituting for v from v we get; u (u at ) s 2 t
  104. 104. Equations for UniformlyAccelerated Motion 2u at s 2 2 t s=ut + ½at2 
  105. 105. Equations for UniformlyAccelerated Motion Exercise: Try and derive y, which is independent of u.
  106. 106. Equations for UniformlyAccelerated Motion VARIABLES EQUATIONTOK u v sPhysicists reduce concepts to u v t s amathematical equations to reduce 2 tambiguity. Does mathematics and u v t s a v = u + atequations reduce ambiguity due tolanguage? u v t s a v2 - u2 = 2as u v t s a 1 2 s ut at 2 1 2 u v t s a s vt at 2
  107. 107. Vertical Motion of Objects The Earth, near its surface, has a uniform gravitational field. Objects that move perpendicular to the Earth’s surface,  move parallel to the gravitational field.
  108. 108. Vertical Motion of Objects The acceleration experienced by the object will be constant. a = 9.81 m s-2 down. If the object is moving towards the Earth,  the object’s speed will increase at the rate of,  9.81 m s-1 every second. This is assuming that air resistance is negligible.
  109. 109. Vertical Motion of Objects
  110. 110. Vertical Motion of Objects True or False?1. The elephant encounters a smaller force of air resistance than the feather and therefore falls faster.2. The elephant has a greater acceleration of gravity than the feather and therefore falls faster.
  111. 111. Vertical Motion of Objects3. Both elephant and feather have the same force of gravity, yet the acceleration of gravity is greatest for the elephant.4. Both elephant and feather have the same force of gravity, yet the feather experiences a greater air resistance.
  112. 112. Vertical Motion of Objects5. Each object experiences the same amount of air resistance, yet the elephant experiences the greatest force of gravity.6. Each object experiences the same amount of air resistance, yet the feather experiences the greatest force of gravity.
  113. 113. Vertical Motion of Objects7. The feather weighs more than the elephant, and therefore will not accelerate as rapidly as the elephant.8. Both elephant and feather weigh the same amount, yet the greater mass of the feather leads to a smaller acceleration.
  114. 114. Vertical Motion of Objects9. The elephant experiences less air resistance and than the feather and thus reaches a larger terminal velocity.10.The feather experiences more air resistance than the elephant and thus reaches a smaller terminal velocity.
  115. 115. Vertical Motion of Objects11.The elephant and the feather encounter the same amount of air resistance, yet the elephant has a greater terminal velocity. If you answered True to any of these questions, you need to review your understanding.
  116. 116. Vertical Motion of Objects
  117. 117. Vertical Motion of Objects Objects which move vertically upwards:  will slow down at the rate of 9.81 m s-1,  every second until it is stationary. It will then start to accelerate towards the earth at:  the rate of 9.81 m s-1 every second. Vertical Motion
  118. 118. Air Resistance Objects falling in the Earth’s uniform gravitational field have two opposing forces acting on it.  Gravity acts towards the Earth, pulling the object downward. Any resistance force opposes motion.
  119. 119. Air Resistance This means, in this case;  air resistance acts upwards. The faster the object falls;  the greater the air resistance. As the object accelerates under the force of gravity;  the greater the air resistance.
  120. 120. Air Resistance This slows the rate at which the object;  accelerates towards the earth. Eventually, the two forces cancel each other out. This means there is no acceleration and;  the velocity becomes constant. This is known as terminal velocity.
  121. 121. Air Resistance The terminal velocity is different:  for different objects. Sky diver’s have a terminal velocity of :  about 150 to 200 km h-1 depending on;  the mass of the sky diver and,  their orientation.
  122. 122. Air Resistance Sky divers tend to try to:  increase the air resistance thereby,  reducing the terminal velocity. This gives them a longer free fall. The parachute:  greatly increases air resistance and,  cuts the terminal velocity to,  between 15 and 20 km h-1.
  123. 123. Air Resistance
  124. 124. Forces and Dynamics To maintain the velocity of an object such as a wheelbarrow:  a push or pull is required. To maintain acceleration:  either by changing the speed of an object or changing its direction,  a push or pull is required.
  125. 125. Forces and Dynamics The name we use for this push or pull is:  force. A force is required to:  keep things moving or,  change their motion. A force can also be used to:  deform an object. This happens when a tennis racquet hits a tennis ball.
  126. 126. Forces and Dynamics Using an air track, we can eliminate (or very nearly);  the force of friction on a glider. This leaves the force of gravity and:  the force of the hand that pushes a glider.
  127. 127. Forces and Dynamics How can we determine the effect of a single force when;  there are two forces acting? We must make an assumption. Gravitational forces being vertical:  have no effect on motion,  in the horizontal plane.
  128. 128. Forces and Dynamics Recall the air track practical:  once the hand released the glider,  it moved with a constant velocity. This indicates that the assumption is true.
  129. 129. Forces and Dynamics  There are cases when more than one force acts on a object.  These forces may oppose each other:  Pushing an object forward and,  Friction backward  Or a tug of war
  130. 130. Forces and Dynamics Or they may act at some other angle:  Running for the football and,  being hip and shouldered. A diagram can be drawn to show these forces. It is called a Free-body Diagram.
  131. 131. Forces and Dynamics Consider a statue resting on a table. What forces are acting on it? As it is at rest:  there are no unbalanced forces. Gravity (Fg) is acting downwards. What force balances gravity?
  132. 132. Forces and Dynamics The table exerts an upward force. The table is compressed by the statue  Due to its elasticity it pushes upwards. This is called a contact force.  As it occurs when two objects are in contact.
  133. 133. Forces and Dynamics When the contact force acts:  perpendicular to the common surface of contact, it is called,  the Normal Force (FN). A diagram of the objects and their forces can be drawn.
  134. 134. Forces and Dynamics
  135. 135. Forces and Dynamics We are interested in the forces and so:  we will only include vectors for the forces. FN Fg
  136. 136. Forces and Dynamics Notice: The vectors are of the same length:  Indicating same magnitude of force. The vectors are in opposite directions:  Indicating the forces oppose each other. The vectors have been labeled:  with appropriate symbols.
  137. 137. Forces and Dynamics What happens when you add a third force? Consider a box at rest on a table. What forces are acting on it?  Fg and:  FN
  138. 138. Forces and Dynamics FN Fg
  139. 139. Forces and Dynamics  What if you were to push down on the box with a force of 40 N? FN 40 N Fg
  140. 140. Forces and Dynamics Notice that the combined length of:  the Fg and 40 N vectors,  equal the FN vector. This indicates the box is stationary. What happens if you pull up with a force of 40 N?
  141. 141. Forces and Dynamics FN 40 N Fg
  142. 142. Forces and Dynamics Notice that the combined length of:  the FN and 40 N vectors,  equal the Fg vector. This indicates the box is stationary.
  143. 143. Forces and Dynamics What forces act on a shopping trolley?
  144. 144. Forces and Dynamics
  145. 145. Forces and Dynamics As FN = Fg The force making the trolley move is equal to: Fp = Force supplied by the person. This is an unbalanced force and:  it will cause an acceleration.
  146. 146. Forces and Dynamics What happens when the ( 10 kg) box in the earlier example is:  connected to another box (12 kg),  by a string and both,  are pulled along a table? What do the Free Body Diagrams look like?
  147. 147. Forces and Dynamics
  148. 148. Forces and Dynamics Box 1 FN T FP Fg
  149. 149. Forces and Dynamics Notice: The vectors Fg and FN are of the same length:  Indicating same magnitude of force.  No vertical motion The vector Fp is longer than T indicating:  acceleration to the right
  150. 150. Forces and Dynamics Box 2 FN T Fg
  151. 151. Forces and Dynamics Free Body Diagram Another one If you would like more practice at drawing free body diagrams, try this web site bd.html
  152. 152. Forces and Dynamics Suppose you wanted to see what was in the box in the earlier example. You pull it towards you:  using a string attached with,  a force of 40 N at,  an angle of 30o above the horizontal..
  153. 153. Forces and Dynamics
  154. 154. Forces and Dynamics This becomes more difficult to analyse. Notice FN Fg This is because some of the force supplied by the person is:  acting in an upward direction. Forces can be resolved into components.
  155. 155. Forces and Dynamics Forces can be resolved into two components:  horizontal and vertical. FORCE F F vert F horiz
  156. 156. Forces and Dynamics  This is useful when forces are applied at an angle but:  the effective force is in a particular direction,  such as pushing a roller Applied force F Effective Force = F . horiz
  157. 157. Forces and Dynamics Forces can combine to give very different results. A 10 N block hanging vertically from one horizontal string:  when measured by a spring balance would read,  10N.
  158. 158. Forces and Dynamics 10 N 10 N
  159. 159. Forces and Dynamics As the block is stationary:  the force of the scale pulling up balances,  the force downwards supplied by,  the weight. If 2 spring balances support the block:  the total weight will be the same.
  160. 160. Forces and Dynamics However each spring balance will support equal amounts:  ie 5N. 5N 5N 10 N
  161. 161. Forces and Dynamics Again the weight down is balanced by:  the two springs supplying a force up. In all situations the two forces vectors:  one vertically downward and,  the other vertically upwards,  must balance.
  162. 162. Forces and Dynamics This looks logical for a vertical orientation but how do they work for non-vertical situations? 10 N 10 N 10 N 10 N
  163. 163. Forces and Dynamics  The vertically up vector must still equal 10 N and so by vector addition:  the two scales will read more than 5 N each,  perhaps 10 N.  If the angle is increased to 60o from the vertical:  120o between the scales,  The reading is much higher.
  164. 164. Forces and Dynamics 10 N 20 N 20 N 10 N
  165. 165. Forces and Dynamics If we continue to do this until the angle is 90o,  the horizontal rope must support a force that is,  much greater than the original weight.
  166. 166. Forces and Dynamics Gymnasts who hold their arms out horizontally from the rings:  must supply a force that is considerably greater than their own body weight. This is an extreme test of strength.
  167. 167. Forces and Dynamics To determine exact values for the components:  Trigonometry must be used. FORCE F F vert F horiz
  168. 168. Forces and Dynamics Fvert = opposite side Fhoriz = adjacent sideF = applied force sin = opposite/hypotenuse Opposite = Fvert = F sin cos = adjacent/hypotenuse Adjacent = Fhoriz = F cos
  169. 169. Forces and Dynamics Springs If you hang a weight from a spring:  apply a force,  it stretches. Add more weight:  It stretches more Remove the weights:  It returns to its original length.
  170. 170. Forces and Dynamics  The spring is said to be elastic.  By stretching a spring:  or compressing it,  The amount it is stretched is:  Directly proportional to the force applied.  This was first noticed by the British Physicist:  Robert Hooke  In the 17th century
  171. 171. Forces and Dynamics It is now remembered as Hooke’s Law. The applied force F and:  The extension (or compression) x Can be represented mathematically.F x A graph of this can be shown.
  172. 172. Forces and Dynamics F x
  173. 173. Forces and Dynamics Compare the line to the equation of a straight line: y = mx + c y=F x=x c = 0 (as through the origin) To turn the proportionality into an equation:
  174. 174. Forces and Dynamics F = mx The slope is constant and so is given a special symbol:  k All constants in Physics have special symbols. The equation now becomes: F = kx
  175. 175. Forces and Dynamics The full form of the equation is: F = -kx This is because the force is a restoring force.  It always acts to try an return the spring to the original length. Hookes Law
  176. 176. Forces and Dynamics You will need to be able to: draw a graph  Given data Determine the spring constant k:  For a particular spring.
  177. 177. Forces and Dynamics Determine the force required to:  extend a spring by a certain amount Determine the amount a spring is stretched by:  for a given force.
  178. 178. Forces and Dynamics An object that moves with constant velocity:  requires no force. Although this is one experiment, it can be shown to be true:  in all situations  whether it is a person ice-skating or,  a person slipping on a banana peel.
  179. 179. Forces and Dynamics Newton investigated this phenomenon,  which he embodied in his first law: Provided no external force acts, the velocity of any object will remain constant unless an unbalanced force acts upon the object.
  180. 180. Forces and Dynamics A body is resistant to change. This resistance of a body to change is called inertia. Does this hold for a stationary object? A stationary object has zero velocity:  and so it will remain at zero.
  181. 181. Forces and Dynamics Plates on a tablecloth are at rest. If you pull the tablecloth quickly enough from under the plates:  the small and brief force of friction,  between the plates and the tablecloth is,  not significant enough to,  appreciably move the dishes.
  182. 182. Forces and Dynamics Does this hold for tug of war contest? When the two sides are even:  the forces are equal in magnitude and,  opposite direction. The resultant force will be zero. This implies that the two sides will not move.
  183. 183. Forces and Dynamics Does this apply to a car moving on a straight road;  at a constant velocity? The car will slow to a stop unless:  the force applied by the engine continues.
  184. 184. Forces and Dynamics This appears to disobey the law:  until friction is taken into account. We can ignore gravity because:  it is acting in a vertical direction. Friction however is a retarding force acting:  in the opposite direction to the motion.
  185. 185. Forces and Dynamics This is an example of translational equilibrium. Two forces are being applied in opposite directions. As the forces cancel each other out,  no acceleration occurs. The object continues to move at:  a constant velocity.