Newtons laws of motion

Newtons laws of motion PowerPoint PPT Presentation


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3 .1 Force,weight,and gravitational mass. What is a Force?. There are two types of forces that we can seeContact ForcesAt a distance forces. Force is an action that can change motion.A force is what we call a push or a pull, or any action that has the ability to change an object's motion.

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Newtons laws of motion

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1. 1 Norah Ali Al- moneef 25 March 2012

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7. Balanced Force A force that produces no change in an object’s motion because it is balanced by an equal, opposite force. 7 Norah Ali Al- moneef 25 March 2012

8. Unbalanced Forces Are forces that results in an object’s motion being changed. 8 Norah Ali Al- moneef 25 March 2012 Add together to equal greater force.Add together to equal greater force.

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12. 12 In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view of the Fig. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 220 N, and Charles pulls with force of magnitude 170 N. The direction of is not given. What is the magnitude of Betty's force ? 25 March 2012 Norah Ali Al- moneef

13. 13 Force components in the x-axis : 25 March 2012 Norah Ali Al- moneef

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36. Calculate force Three people are each applying 250 N of force to try to move a heavy cart. The people are standing on a rug. Someone nearby notices that the rug is slipping. How much force must be applied to the rug to keep it from slipping? Sketch the action and reaction forces acting between the people and the cart and between the people and the rug. 25 March 2012 36 Norah Ali Al- moneef

37. Locomotion is the act of moving or the ability to move from one place to another. 25 March 2012 37 Norah Ali Al- moneef

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42. The Other Logical Problem If the Newton’s Third Law action and reaction forces are always equal and opposite, how do two objects of different sizes get different accelerations in the same interaction? (When a bug hits a windshield, different things happen to the bug and windshield.) 25 March 2012 42 Norah Ali Al- moneef

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54. Equilibrium When the net force acting on an object is zero, the forces on the object are balanced. We call this condition equilibrium. 25 March 2012 54 Norah Ali Al- moneef

55. Equilibrium A moving object continues to move with the same speed and direction. Newton’s second law states that for an object to be in equilibrium, the net force, or the sum of the forces, has to be zero. 25 March 2012 55 Norah Ali Al- moneef

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59. Use a = F ÷ m. First add the forces to find the net force. F = - 75N - 25N + 45N + 55N = 0 N, a = 0 Calculating the net force from four forces 25 March 2012 59 Norah Ali Al- moneef

60. Use: net force = zero, Fw = mg and g = 9.8 N/kg. Fw = mg = (150 kg)(9.8 N/kg) = 1,470 N. Let F be the force in the other chain, equilibrium requires: F + (600 N) = 1,470 N F = 1,470 N – 600 N F = 870 N. Using equilibrium to find an unknown force 25 March 2012 60 Norah Ali Al- moneef

61. Applications of equilibrium Real objects can move in three directions: up-down, right-left, and front-back. The three directions are called three dimensions and usually given the names x, y, and z. When an object is in equilibrium, forces must balance separately in each of the x, y, and z dimensions. 25 March 2012 61 Norah Ali Al- moneef

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66. Solving Force Problems A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000 m/s2. (a) What force acts on the bullet? (b) What force acts on the rifle? ( c) If the bullet is in the rifle for 0.007 seconds, what is its muzzle velocity? (a) F = ma, F = 0.009 kg x 30,000 m/s2 = 270 N(grams must be converted to kilograms in order to use the MKS system and get newtons as force unit answer) (b) Newton’s 3rd Law (Action / Reaction) – the rifle is “shot” in the opposite direction with the same force as the bullet or – 270 N (c) (muzzle velocity means the velocity at which the bullet leaves the rifle barrel) Using the equation V = Vo + at, V = o + 30,000 x 0.007 = 210 m/s (Vo = 0 since the bullet is not moving before it is fired) 25 March 2012 66 Norah Ali Al- moneef

67. Solving Force Problems (a) How much force is needed to reduce the velocity of a 6400 Kg truck from 20 m/s to 10 m/s in 5 seconds? (b) What is its stopping distance? (a) F = ma, we must first find mass. The weight is 6400 Kg. Wt = mass x gravity therefore, m = w/g, mass = 6400 Kg /9.8 m/s2 = 653.06 Kg. Now, to find acceleration, V = Vo + at or a = (V – Vo) / t a = (20 – 10)/ 5 = - 2 m/s2 and F = 200 x (-2) = - 400 N (negative means the force is opposing the motion) (b) ?x = Vot + ½ at2 we get ?x = (20 x 5) + ½ (-2) 52 = 75 m is the distance traveled during stopping. 25 March 2012 67 Norah Ali Al- moneef

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71. Where does the calculus fit in? 25 March 2012 71 Norah Ali Al- moneef

72. Example on Newton’s 2. and 3. Law Suppose you are an astronaut in outer space giving a brief push to a spacecraft whose mass is bigger than your own 1) Compare the magnitude of the force you exert on the spacecraft, FS, to the magnitude of the force exerted by the spacecraft on you, FA, while you are pushing: 1. FA = FS 2. FA > FS 3. FA < FS 25 March 2012 72 Norah Ali Al- moneef

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74. Example: Three Forces 25 March 2012 74 Norah Ali Al- moneef

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76. 76 example In the Figs. one or two forces act on a puck that moves over frictionless ice and along an x axis, in one-dimensional motion. The puck's mass is m = 0.20 kg. Forces and are directed along the axis and have magnitudes F1 = 4.0 N and F2 = 2.0 N. Force is directed at angle = 30° and has magnitude F3 = 1.0 N. In each situation, what is the acceleration of the puck? 25 March 2012 Norah Ali Al- moneef

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83. Free-body diagrams To keep track of the number and direction of all the forces in a system, it is useful to draw a free-body diagram. A free-body diagram makes it possible to focus on all forces and where they act 25 March 2012 83 Norah Ali Al- moneef

84. Free-body diagrams Forces due to weight or acceleration may be assumed to act directly on an object, often at its center. A reaction force is usually present at any point an object is in contact with another object or the floor. If a force comes out negative, it means the opposes another force. 25 March 2012 84 Norah Ali Al- moneef

85. Applications of equilibrium If an object is not moving, then you know it is in equilibrium and the net force must be zero. You know the total upward force from the cables must equal the downward force of the sign’s weight because the sign is in equilibrium. 25 March 2012 85 Norah Ali Al- moneef

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87. Rules for Ropes and Pulleys Force from rope points AWAY from object (Rope can only pull) Magnitude of the force is Tension Tension is same everywhere in the rope Tension does not change when going over pulley 25 March 2012 87 Norah Ali Al- moneef

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89. 89 The tension force The tension force pulls on a body. The cord is often assumed to be massless. We usually assume pulleys to be frictionless A string has a single tension force (magnitude). The direction depends on the body on which this force acts upon. The tension forces on two sides of a frictionless pulley are the same in magnitude. 25 March 2012 Norah Ali Al- moneef

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91. 91 example the Figure shows a block S (the sliding block) with mass M = 3.3 kg. The block is free to move along a horizontal frictionless surface such as an air table. 25 March 2012 Norah Ali Al- moneef

92. 92 example In Fig. a, a block B of mass M = 15.0 kg hangs by a cord from a knot K of mass mK, which hangs from a ceiling by means of two other cords. The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravitational force on the block. What are the tensions in the three cords? 25 March 2012 Norah Ali Al- moneef

93. 93 At a knot, the tensions are different. The tensions on both sides of a frictionless pulley are the same. 25 March 2012 Norah Ali Al- moneef

94. 94 example In Fig. a, a constant horizontal force of magnitude 20 N is applied to block A of mass mA = 4.0 kg, which pushes against block B of mass mB = 6.0 kg. The blocks slide over a frictionless surface, along an x axis. (a)  What is the acceleration of the blocks? 25 March 2012 Norah Ali Al- moneef

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99. 99 example In Fig. a, a cord holds stationary a block of mass m = 15 kg, on a frictionless plane that is inclined at angle = 27°. (a)  What are the magnitudes of the force on the block from the cord and the normal force on the block from the plane? 25 March 2012 Norah Ali Al- moneef

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103. 103 a 2.0 kg cookie tin is accelerated at 3.0 m/s2 in the direction shown by , over a frictionless horizontal surface. The acceleration is caused by three horizontal forces, only two of which are shown: of magnitude 10 N and of magnitude 20 N. What is the third force in unit-vector notation and as a magnitude and an angle? 25 March 2012 Norah Ali Al- moneef

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105. 105 Ali pulled a railroad cars (with his teeth) on his end of the rope with a constant force that was 2.5 times his body weight, at an angle of 30° from the horizontal. His mass m was 80 kg. The weight W of the cars was 700 kN, and he moved them 1.0 m along the rails. Assume that the rolling wheels encountered no retarding force from the rails. What was the speed of the cars at the end of the pull? 25 March 2012 Norah Ali Al- moneef

106. 106 SOLUTION:  We first find the acceleration of the cars in the x direction : 25 March 2012 Norah Ali Al- moneef

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124. Equilibrium and Forces It is much more difficult for a gymnast to hold his arms out at a 450 angle. To see why, consider that each arm must still support 350 N vertically to balance the force of gravity. 25 March 2012 124 Norah Ali Al- moneef

125. Forces in Two Dimensions The force in the right arm must also be 495 N because it also has a vertical component of 350 N. 25 March 2012 125 Norah Ali Al- moneef

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134. The maximum load that can safely be supported by a rope in an overhead hoist is 400 N. What is the maximum acceleration that can safely be given to a 25-kilogram object being hoisted vertically upward? Free Body Diagram 25 March 2012 134 Norah Ali Al- moneef

135. An object with a mass of 4.0 kg travels with a constant velocity of 4.8 m/s northward. It is then acted on by a force of 6.5 N in the direction of motion and a force of 9.5 N to the south, both of which continue even after the mass comes momentarily to rest. (a) How far will the object travel before coming to rest? (b) What will be its position 2.5 s after the object comes momentarily to rest? M = 4 kg, vo = 4.8 m/s, F1 = 6.5 N north, F2 = 9.5 N south. 25 March 2012 135 Norah Ali Al- moneef

136. ’ when v = 0. 0 = (4.8 m/s)2 + 2(-0.75 m/s2)x X = 15 m (b) Where will the object be 2.5 s later. We know its at x =15 with no velocity and -0.75 m/s2 acceleration. x = vot + ½ at2 = 0 m/s x 2.5 s + ½ (-0.75 m/s2)(2.5 s)2 = -2.3 m, So relative to its starting point its at x = 13 m 25 March 2012 136 Norah Ali Al- moneef

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