Chapter 5 Newton s Laws of Motion. What determines acceleration on objects?

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Chapter 5 Newton s Laws of Motion What determines acceleration on objects? 1

Units of Chapter 5 Force and Mass Newton s First Law of Motion Newton s Second Law of Motion Newton s Third Law of Motion The Vector Nature of Forces Normal Force, Tension, Free body diagram, Problem solving skills Weight, Apparent weight 2

Force: push or pull 5-1 Force and Mass Force is a vector it has magnitude and direction 3

Mass is the measure of how hard it is to change an object s velocity. Mass can also be thought of as a measure of the quantity of matter in an object. 5-1 Force and Mass The more matter one object has, the harder it is to change its velocity. 4

5-2 Newton s First Law of Motion If you stop pushing an object, does it stop moving? Only if there is friction! In the absence of any net external force, an object will keep moving at a constant speed in a straight line, or remain at rest. This is also known as the Law of Inertia. If the net force on an object is zero, it s velocity is constant (a=0). INERTIA (need safety belt) Attention: If there is friction, Air resistance net force is not zero.) 5

5-2 Newton s First Law of Motion In order to change the velocity of an object, magnitude or direction, a net force is required. Example: Object on desk. Net force = Objects want to keep doing what they are already doing. For example: in space Movie Unstoppable Your dog knows 6

5-3 Newton s Second Law of Motion If net force is not zero, velocity will not be constant. a 0 Object s acceleration will be in the same directions as the net force. Acceleration direction is not the velocity direction. Acceleration direction determines velocity change. F net = ma Acceleration is proportional to net force. The more net force, the larger the acceleration. 7

5-3 Newton s Second Law of Motion Acceleration is inversely proportional to mass: Same Force, less mass object gets bigger a. Same Force, more mass object gets smaller a. More mass, more inertia, harder to change v, less a Less mass, less inertia, easier to change v, more a F net = ma 8

5-3 Newton s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: (5-1) 9

SI unit for forces: Newton, (N) F= m a; 1 Newton of net force gives 1 kg mass 1m/s 2 acceleration. 1 N = 1 kg * m/s 2 Attention: when you solve problem, if you first convert every value to standard SI unit. (m, s, m/s, m/s2, N,.) you can plug the numbers and automatically get SI unit in your answer. In order to give object of mass m a downward acceleration of g=9.8 m/s 2, how much force is needed? Gravity force = m g 10

Weight The weight of an object on the Earth s surface is the gravitational force exerted on it by the Earth., also known as Gravity, G G=mg 1 kg mass weights 9.8 Newton on earth. 9.8 Newton gravity force gives 1kg mass an acceleration of 9.8m/s 2 11

5-3 Newton s Second Law of Motion 12

5-4 Newton s Third Law of Motion The 3rd law of Newton s law is about two objects! Not one object! Forces always come in pairs. If object A exerts a force F on object B (action) then object B must exert a force negative F on object A (reaction). Action and reaction forces are always equal in size and opposite in directions. Action /Reaction pair examples: I push you, I feel your resistant. These two forces are always equal and opposite (Will you push a nail tip or blade with large force? No, because the reaction resistant force from the blade is as big as the push you give to it) 13

1. Newton s 3 rd law applications: 2. Throw heavy object forward. You feel recoil backward. 3. Swim, push water back, so that you can move forward (water s reaction force on you is forward) 4. Walk/run, you need to press down and back, so that floor s reaction force is upward and forward. (That s why on ice and sand it s so hard to walk or drive) 5, Water hose sneak around. 6, Air balloon flies upward, while gas in it was squeezed downward. (demo) 7, Your gravity, you 100 kg earth act 980 N gravity force on you downward. 14 You act 980 N, gravity force on Earth upward.

5-4 Newton s Third Law of Motion Some action-reaction pairs: 15

These two forces are not action and reaction pair: I push you on left side. He push you on your right side. On one object bearing two forces. These two forces are not action and reaction. They can be equal or not. Nobody guarantee these two forces to be equal. I push you and you resist me, these two forces are action and reaction pair. Newton s 3 rd law guarantee these two forces to be equal in side and opposite in directions. These is very useful for calculations. 16

Free-body diagrams: A free-body diagram shows every force acting on an object. -Use a dot to represent the object of interest. -Use one arrow with correct length and direction to represent each force. These two forces are two forces acting on one object. 17 These two forces can be equal or not.

Exercise for yourself: Forces add as vectors. A hockey puck is acted on by one or more forces, as shown in Figure 5-19. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, starting with the smallest. Indicate ties with an equal sign. (Use only the symbols < and =, for example A < B=C.) A < D < B < C 18

5-5 The Vector Nature of Forces: Forces in Two Dimensions The easiest way to handle forces in two dimensions is to treat each dimension separately, as we did for kinematics. 19

5-7 Special forces, Normal Forces The normal force is the force exerted by a surface on an object. Very often it is a supporting force. 20

5-7 Special forces, Normal Forces Normal force is the force from a surface. Label N. Direction: Always perpendicular & away from surface. Size : depend on press of object. How much you press, how much It supports. The normal force may be equal to, greater than, or less than the weight. 21

Special forces, Tension, Lable T When you pull on a string or rope, it becomes taut. We say that there is tension in the string. Direction: Always along rope, always pointing away from Object. Size : Tension inside massless rope in all location has the same size.

6-2 Strings and Springs The tension in a real rope will vary along its length, due to the weight of the rope. Here, we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated. Tension inside massless rope in all location has the same size.

** problem solving strategy and steps This is the most important slides for this semester. 1, Identify and isolate the object of interest, a dot. 2, Identify all forces on the object acted by other objects. Label them using arrows of correct length and direction 3, Pick x y axis. ( based on acceleration direction for convenience) 4, Break up all forces into x-y components 5, Set equations: Add all force components in x directions = ma x Add all force components in y directions = ma y F net x = m a x, F net y = m a y (x & y are independent ) 6, Solve for any two unknowns, from the above two equations. (If know a, can solve force. If know forces 24 can find a. Again Math skills here. )

6-3 Translational Equilibrium The traffic light problem. When an object is in translational equilibrium, (Not moving or moving at constant velocity.) the net force on it is zero: This allows the calculation of unknown forces. Add all force components in x directions = 0 Add all force components in y directions = 0

Example 1 An object of mass m being dragged with force F at angle q. (know m, F, q) Find Normal force N, and acceleration 1- object m, use a dot 2- label all forces on it mg, N, F 3- pick axes 4- components. F both x & y direction x direction: Fx component: Positive + or -? y direction: Fy component: Positive Fx= +Fcosq Fy = +Fsinq 5- Total F=ma for both x & y directions F net x : Fcos θ =m a x, F net y : N+Fsinθ-mg =ma y = 0 6- solve a x : solve N : a x =Fcos θ / m N = mg - Fsinθ 26

Example 2 Incline plane. No friction. We know m and angle q, Find acceleration, and N. 1, object: dot 2, forces: mg, N, anything else? 3, pick x-y axes cleverly. (Along motion direction, So that we know for sure a y =0) 4, force components: weight along incline: mgsinq weight perpendicular to incline: mgcosq It s very important to do the drawing yourself. 5, Total F=ma; x direction : mgsinq =ma x y direction: N-mgcosq=0 6, solve equations: a x =gsinq; N=mgcosq; Check for special angles. If q=0, a=0, N=mg; If q=90, a=g, N=0, 27 Make sense. Large q, Larger sinq, Larger a, smaller q, Less N

28

5-6 Apparent Weight Apparent weight: Your perception of your weight is based on the contact forces between your body and your surroundings. It s N or T If you are accelerating your surroundings, your apparent weight may be more or less than your actual weight. If Accelerating downward, Define downward to be positive direction: mg-n=ma y ; so N= mg- ma y N<mg, you still has same mg, but no enough support. That s what frightens you badly in those terror rides. 29 Apparent weight =N <mg,

If acceleration is downward, Mass unchanged. Gravity unchanged. But Normal force is less. Apparent weight is less. This is what frightens you in the theme park rides. Even in complete dark, you can tell N<mg, feel downward a In an elevator, that is accelerating upward: N>mg, N - mg = m a y ; N= mg + m ay ) > mg Mass unchanged, gravity unchanged, But apparent weight is more! Look at the poor girl. T= mg + m ay, when a is upward 30

If no elevator, try quickly stand up or squat on scale. Acceleration is not equal to 0 (Try it at home) This is how Wii fit detect your motion (acceleration) by measuring the Normal forces between it and you. Question, in an elevator, going down at constant velocity N > = < mg???? a=???? Answer: a= 0; N = mg Q: Constant velocity elevator going up? Apparent weight, N=?? Before next class, at least Read Chapter 6-1 friction, 31

Summary of Chapter 5 Force: a push or pull Mass: measures the difficulty in accelerating an object Newton s first law: if the net force on an object is zero, it stay at rest or keep constant velocity. Newton s second law: for one object Free-body diagram: a sketch showing all the forces on an object. Remember to decompose in x and y directions. Solve F=ma in both x and y direction separately Add all forces in x directions = ma x Add all forces in y directions = ma y 32

Summary of Chapter 5 Newton s third law: If object 1 exerts a force F on object 2, then object 2 exerts a force F on object 1. Contact forces: an action-reaction pair of forces produced by two objects in physical contact Forces are vectors Newton s second laws can be applied to x and y each components of the forces separately 33

Summary of Chapter 5 Weight: gravitational force exerted by the Earth on an object, Always DOWNWARD. On the surface of the Earth, W = mg Apparent weight: force felt from contact with a floor or scale. It s determined by both mg and acceleration Normal force: Always perpendicular to the surface. (away from the surface) Normal force is determined by the amount of pressing again the surface. It may be equal to, lesser than, or greater than the object s weight. Tension force in a string is always ALONG the string It is equal everywhere if the string has no mass. 34