1. Chapter 4 Forces and the Laws of Motion

2. Aristotle’s view on motion

3. Two types of Motion Natural motion – either straight up or down Violent motion – was imposed motion, result of a force.

4. Force A push or a pull exerted on some object.

5. Galileo’s view dispelled Aristotle’s A force is not necessary to keep an object moving. Introduced ‘friction’,

6. Friction Force between materials that are moving past one another. Force that opposes motion.

9. The cause of an acceleration, or the change in an object’s speed.

10. Newton (N) SI/Metric unit of force. Is the amount of force that, when acting on a 1 kg mass, produces an acceleration of 1 m/s 2.

11. A ‘dyne’ is the measurement of force when using grams and centimeters.

12. Weight Is the measure of the magnitude of the gravitational force exerted on an object.

13. Conversion Factor 1 N = 0.225 lb 1 lb = 4.448 N

14. Figure out what your weight is in Newton’s.

15. 190 lb4.448 N 1 lb 845.12 N

16. Forces can act through: 1. Contact 2. At a distance

17. Contact Forces A force that arises from the physical contact of two objects. Ex: Throwing a ball or pulling a sled.

18. Field Forces A force that can exist between objects, even in the absence of physical contact between the objects.

19. Ex: force of gravity or attraction/repulsion of electrical charges.

20. The effect of a force depends upon the magnitude of the force and the direction of the force. Therefore, a force is a vector.

21. Force Diagram Is a diagram, in which, all forces acting on an object are represented by vector arrows. Sometimes these are called ‘free-body diagrams”.

22. Look at figures 4 – 3 & 4 – 4 on pages 126 – 128.

23. Section 4 – 2: Newton’s First Law

24. An object at rest remains at rest, and an object in motion will continue in motion with a constant velocity (in a straight line),

25. unless the object experiences a net external force.

26. Inertia Is the tendency of an object to maintain its state of motion.

27. Galileo said that every material object has a resistance to the change of its state of motion.

28. Newton’s first law is sometimes called the ‘Law of Inertia’ Meaning that in the absence of force a body will preserve its state of motion.

32. What is “mass”? The amount of matter an object contains. Mass measures the inertia of an object.

33. All objects made of matter have inertia - that is, they resist accelerations (Newton’s First Law), but some objects resist more than others.

34. Mass is not Volume. Mass is a scalar quantity. SI unit of mass is the kilogram (kg).

35. Mass is not Weight Mass is a property of an object that measures how much it resists accelerating. An object is difficult to accelerate because it has mass

36. Net External Force (F net ) Is the total force resulting from a combination of external forces on an object; sometimes called the ‘resultant force’.

38. The downward force due to gravity is the object/body’s weight. The upward force is due to the force exerted by the contact surface on the object.

39. Ex 1: A man is pulling on his dog with a force of 70 N directed at an angle of 30 degrees to the horizontal. Find the x and y components of the force.

40. Force diagram 70 N

41. Force diagram 70 N x y 

42. X component: since we have a right triangle and know the angle and the hypotenuse. We will use the cosine trig function.

43. adj cos  = ----- hyp F x = hyp cos  F x = 70 cos 30 F x = 60.6 N

44. y component: since we have a right triangle and know the angle and the hypotenuse. We will use the sine trig function.

45. opp sin  = ----- hyp F y = hyp sin  F y = 70 sin 30 F y = 35 N

46. Ex 2: A crate is pulled to the right with a force of 82 N, to the left with a force 115 N, upward with a force of 565 N, and downward with a force of 236 N.

47. Force Diagram 565 N 82 N 236 N 115 N

48. A) Find the net external force in the x direction. Remember: right is (+) and left is (-) x = 82 N + (- 115 N) x = - 33 N

49. B) Find the net external force in the y direction. Remember: up is (+) and down (-) y = 565 N + (- 236 N) y = 329 N

50. C) Find magnitude and direction of the F net on the crate. F net 329 N - 33 N

51. Use the Pythagorean Theorem to find F net and the inverse tangent function to find the direction.

60. Since we are on the negative x-axis, this is the same as an angle of 95.7 o from the positive x-axis.

61. In your notes do problems 3 & 4 on HPg 133.

62. Mass is a measurement of inertia. Ex: A golf ball and a basketball, if the same F net acts on both, the golf ball will accelerate more.

63. Equilibrium The state in which there is no change in the body’s motion.

64. Equilibrium Means "Zero Acceleration" Forces in balance: Equilibrium

65. An object is in equilibrium when the vector sum of the forces acting on it is equal to zero.

66. Is the object moving? Yes, but at constant velocity.

67. Section 4-3: Newton’s Second and Third Laws

68. Force is proportional to mass and acceleration Think about pushing a stalled car. If you try to push it by yourself, you can move it, but not very fast.

69. However, if a couple of friends help, then it is much easier to move and you can make it move faster quicker.

70. Newton’s 2 nd Law The acceleration of an object is directly proportional to the Net External Force (F net ) acting on the object and inversely proportional to the object’s mass.

71. From this law we get: Normally it is written as:

72. a = F / m

73. 2a = 2 F / m

74. a = F / m 2a = 2 F / m a = 2 F / 2m

75. a = F / m

76. a/2 = F /2m

77. a = F / m a/2 = F /2m a/3 = F / 3m

79. Note: remember that it is the F net that cause the object to move.

80. Note:

81. Ex 3: Nathan applies a force of 20 N, causing the book to accelerate at a rate of 1.25 m/s 2. What is the mass of the book?

82. G: a = 1.25 m/s 2, F net = 20 N U: m = ? E: F net = ma or m = F net /a S: m = 20 N / 1.25 m/s 2 S: m = 16 kg

83. Ex 4: The F net on the propeller of a 3.2 kg model airplane is 7.0 N forward. What is the acceleration of the airplane?

84. G: F net = 7.0 N, m = 3.2 kg U: a = ? E: F net = ma or a = F net /m S: a = 7.0 N / 3.2 kg S: a = 2.2 m/s 2 forward

85. Remember the from Ch. 2

88. Ex 5: A 2 kg otter starts from rest at the top of a muddy incline 0.85 m long and slides down to the bottom in 0.5 sec. What is the net external force?

89. G:  x =.85 m,  t = 0.5s v i = 0 m/s, m = 2 kg U: F net = ? We need to find the acceleration, but we need final speed first.

90. We need to find the average speed first, than use that to find the final speed.

93. Using the 2 nd average speed equation. Rearrange for final speed

97. Now we can find acceleration.

100. Now we can find F net E: F net = ma S: F net =(2 kg)(6.8 m/s 2 ) S: F net = 13.6 N

101. On pg 138 (HP), Do Practice 4B: # 3 & 5.

102. Forces always exist in pairs. Ex: When you push against the wall, the wall pushes back. The forces are equal, but opposite.

103. Newton’s 3 rd Law

104. If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal in magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction.

105. A simpler alternative statement for Newton’s 3 rd Law: For every action, there is an equal and opposite reaction.

107. Action-reaction pair A pair of simultaneously equal, but opposite forces resulting from the interaction of 2 objects.

108. Action-Reaction forces each act on different objects. They do not result in equilibrium. Reread pg 139

112. Recoil is an example of the 3 rd Law

113. Field Forces also exist in pairs. Newton’s 3 rd Law also applies to field forces.

114. When an object is falling, its falling due to the force exerted by the earth, the reaction force is that the body is exerting an equal and opposite force on the earth.

115. Why don’t we notice the this reaction force? Because of Newton’s 2 nd Law, the mass of the earth is so great that the acceleration is so small it is negligible.

116. Section 4-4: Everyday Forces

117. Weight (F g ) The magnitude of the force of gravity acting on an object. Weight is NOT mass!

118. The weight of an object depends on the object’s mass. In fact, an object’s weight is directly proportional to the object’s mass.

119. The weight of an object also depends on the object’s location.

120. In fact, an object’s weight is directly proportional to its free fall acceleration, g at its current location.

121. F g = mg Where g = 10 m/s 2, unless otherwise specified. Weight is dependent on the force of gravity. It depends upon location. And acts downward.

122. The farther away from the center of the earth, the less ‘g’ becomes.

123. Also, the gravitational pull of the moon is about 1/6 th of the earth’s. So if you way 180 lb or 900 N on earth you’d weigh 30 lb or 150 N on the moon.

124. EX 6: A rocket with a mass of 10 kg is launched vertically with a force of 150 N. What is the rocket’s net force? What is the acceleration produced by this net force?

125. What forces are acting on this object? Weight – downward Force of Engine - upward

126. G: m = 10 kg, g = 10 m/s 2 F engine = 150 N U: F net = ? E: F net = F engine - F g F net = F engine - mg S: F net = 150 N - (10)(10) S: F net = 50 N

127. Remember the net force is what accelerates the object. U: a = ? E: a = F net /m S: a = 50 N / 10 kg S: a = 5 m/s 2

128. EX 7: A 20 kg rocket is accelerated upwards (vertically) at 3 m/s 2. What is the Force that provides this acceleration, this is the net force? What is the force provided by the rocket’s engine?

129. Remember the Net Force causes the object to accelerate.

130. G: a = 3 m/s 2, m = 20 kg, g = 10 m/s 2 U: F net = ? E: F net = ma S: F net = (20 kg)(3 m/s 2 ) S: F net = 60 N

131. U: F net = F engine - F g F engine = F net + F g F engine = F net + mg F engine = 60+ (20)(10) F engine = 260 N

132. Normal Force (F N ) A force exerted by one object on another in a direction perpendicular to the surface of contact.

133. N

134. The normal force is always to the surface of contact, but is not always opposite the direction of gravity.

136. F N = F g, but opposite if The F N can be calculated by the equation F N = mgcos , if the object is on an inclination.

137. FgFg FNFN   The value for  is the same for both.

139. Force of Friction Is the force that opposes motion. There are 2 types of friction:

140. Static Friction (F s ) The resistive force that opposes relative motion of 2 contacting surfaces that are at rest with respect with one another.

141. As long as an object does not move when a force is applied: As the applied force increase, the force of static friction increases. It increases until it reaches its max value F s,max.

142. Once you exceed the max value the object begins to move.

143. Kinetic Friction (F k ) The resistive force that opposes the relative motion of two contacting surfaces that are moving past one another.

144. Since it is easier to keep an object moving than it is to start it moving, the kinetic friction is less than static friction.

147. Frictional forces arise from the complex interaction of the surfaces, at a microscopic level.

148. It’s easier to push a chair than it is to move a desk at the same speed. Since the desk is heavier, it has a greater normal force.

149. Therefore, the force of friction is proportional to the normal force. Also, friction depends upon the nature of the surfaces in contact.

151. The frictional force between a chair and tile floor is less than the the force between a chair and carpet.

152. Coefficient of Friction  Is the ratio of the force of friction and the normal force acting between the objects.

153. F friction  = ---------- F normal

154. Note: The coefficient of friction has no units. From the equation the units cancel out.

155. Coefficient of Kinetic Friction

156. Coefficient of Static Friction

157. Remember that on a horizontal surface the normal force is equal to the object’s weight.

158. On pg 144, are values for the coefficient of friction, for both static and kinetic.

160. Ex 8: A 19 kg crate initially at rest, on a horizontal floor, requires a 75 N forces to set it in motion. Find the coefficient of static friction.

161. G: m =19 kg, F =F s =75 N U:  s = ? E: F s F s  s = -----= ------ F N mg

162. S: 75 N  s = --------------------- (19 kg)(10 m/s 2 ) S:  s = 0.39

163. From the Coefficient of friction what are the two surfaces in contact? Wood on Wood

164. Overcoming Friction Example

165. Ex 9: Andy pulls a wagon with a force of 90 N at an angle of 30 o to a horizontal surface. The mass of the wagon is 20 kg, and the coefficient of kinetic friction between the wagon and the side walk is 0.50. Find the acceleration of the wagon due to the net force.

166. FgFg FNFN FkFk F applied

167. Which way does the object accelerate? Only in the ‘x’ direction

168. In order to find the acceleration, we will use Newton’s 2 nd Law. F net,x = ma x We don’t know the F net,x, so we need to find it.

169. Since the object doesn’t move in the ‘Y’ direction the F net,y = 0 N. We need to find the net force in the ‘x’ direction.

170. F net,x = F applied,x + F k Since the applied force is at an angle, we only want the x- component.

171. F app,x = F app cos  F app,x = 90 N cos(30) F app,x = 77.9 N

172. Next find the F k F k =  F N We need to find the Normal Force. Does it equal the weight?

173. NO, since the F app has a y-component. We know the net force in the y-direction is zero, so: F net,y = F N + F app,y - F g =0

174. F N + F app,y - F g = 0 F N = F g - F app,y

175. F g = mg F g =(20)(10) F g = 200 N

176. F app,y = F app sin  F app,y = 90N sin(30) F app,y = 45N

177. F N + F app,y - F g = 0 F N = F g - F app,y F N = 200 N - 45N F N = 155 N

178. F k =  k  F N F k = (0.50)(155 N) F k = 77.5 N to the left So F k = 77.5 N

179. Now, find the F net,x F net,x = F app,x - F k F net,x =(77.9 N - 77.5 N) F net,x = 0.4 N

180. Now, find the horizontal acceleration. a x = F net,x /m a x = 0.4 N / 20 kg a x = 0.02 m/s 2

181. Air Resistance is a form of friction (F R ) When an object moves through a fluid, that fluid provides resistance in the direction opposite of the object’s motion.

182. Terminal Velocity When a free falling body accelerates, the object’s velocity increases. As the velocity increases, the object’s air resistance increases.

183. When the air resistance balances the force of gravity, the net force is zero. The objects continues to move downward at a constant maximum speed.

184. The maximum speed a free-falling body reaches due to air resistance equaling the force of gravity.

185. Eventually, the force R of air resistance becomes equal to the force exerted by the earth, and the object reaches equilibrium.

186. Why does the heavier person fall faster?

187. What does Pressure mean to you? Under Pressure Peer Pressure Feeling Pressured Air/Atmospheric Tire

188. Pressure The amount of force per unit area.

189. Force Pressure = ---------- Area

191. Pressure Units N/m 2 1 N/m 2 = 1 Pascal (Pa)

192. Ex: