Galileo & Friction 2000 yrs prior to inertia idea, the popular belief was that all objects want to come to a rest. BUT 1600's: Galileo reasoned that

Galileo & Friction 2000 yrs prior to inertia idea, the popular belief was that all objects want to come to a rest. BUT 1600's: Galileo reasoned that moving objects eventually stop only because of a force called friction.

From friction discovery to inertia: If the opposite incline was elevated at nearly a 0-degree angle, then the ball would roll almost forever in an effort to reach the original height. AND If the opposing incline was not even inclined at all, then... an object in motion would continue in motion forever

Newton's First Law : The Law of Inertia Isaac Newton generalized Galileo's findings in a single statement: An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced (net) force.

Newton s Second Law Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced.

Newton s Second Law The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass F a 2F 2a

Newton s Second Law Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law can be stated mathematically by F = ma or F NET = ma UNITS The unit of m x a is kg x (m/s 2 ) So F has units of kg x (m/s 2 ) Which has been defined as the NEWTON (abbrev. N)

Newton s Third Law For every action, there is an equal and opposite reaction.

What are action and reaction? Action and Reaction are just names of forces.

Recap Newton s three laws of motion 1 st Law: Law of Inertia 2 nd Law: F NET = ma 3 rd Law: For every force there is an equal and opposite force.

Newton s Laws of motion deal with FORCES

Force: a push or a pull Vector Unit: Newton (N) 1N = 1 kg m/s 2 U.S. Customary unit for force: 1 lb = slug ft/ s 2 1N ~ ¼ lb

Categories of Forces Contact Forces (Tension, Friction) Field Forces (Electrical, Gravitational)

Friction: a force acting when two surfaces rub against each other. Friction opposes motion. Is caused by the irregularities in surfaces.

Weight: the gravitational force with which the earth pulls an object vector unit: N

Weight and Mass Weight is the force due to gravity. It is directed downward and it varies from location to location. Mass is a universal constant which is a measure of the inertia of a body. Mass is not a force! F = m a so that: W = mg and m = W g

F g = mg Weight = mass acceleration due to gravity. This follows directly from F = ma. Weight is the force of gravity on a body. Near the surface of the Earth, g = 9.8 m/s 2.

Normal Force: a force acting perpendicular to the interface of contact between two objects.

Normal Force The normal force is the force exerted by a surface on an object. 22/31

Normal Force Normal Force The normal force always acts outward and perpendicular to the surface of the object that is touching it. The symbol for the normal force is n.

Normal forces aren t always up Normal means perpendicular. A normal force is always perpendicular to the contact surface. N mg For example, if a flower pot is sitting on an incline, N is not vertical; it s at a right angle to the incline.

Up Normal force directions You re standing on level ground. You re at the bottom of a circle while flying a loopthe-loop in a plane. Sideways A ladder leans up against a wall. You re against the wall on a spinning ride at an amusement park when the floor drops out. At an angle A race car takes a turn on a banked track. Down You re in a roller coaster at the top of a loop.

Normal Force The normal force is always perpendicular to the surface that produces it.

Normal forces aren t always up N mg mg > N

Normal force When an object lies on a table or on the ground, the table or ground must exert an upward force on it, otherwise gravity would accelerate it down. N m mg In this particular case, N = mg. So, F net = 0; hence a = 0.

Another example where N < mg N is less than the weight because N + F y = Weight

Normal Force The normal force may be equal to, greater than, or less than the weight.

Cases in which N ¹ mg 1. Mass on incline 2. Applied force acting on the mass 3. Nonzero acceleration, as in an elevator or launching space shuttle N N F A N a mg mg mg

When does N = mg? If the following conditions are satisfied, then N = mg: The object is on a level surface. There s nothing pushing it down or pulling it up. The object is not accelerating vertically.

Free Body Diagram:

Using Newton s Second Law Example 1. A resultant force of 40 N gives a block an acceleration of 8 m/s 2. What is the weight of the block near the surface of the Earth? a F = 40 N To find weight, we must first 8 m/s 2 find the mass of the block: W=? F F = ma; m= a 40 N m = = 5 kg 8 m/s 2 Now find weight of a 5-kg mass on earth. W = mg = (5 kg)(9.8 m/s 2 ) W = 49.0 N = 50 N

More about Newton s 2nd Law You must be certain about which body we are applying it to

More about Newton s 2nd Law You must be certain about which body we are applying it to F net must be the vector sum of all the forces that act on that body Only forces that act on that body are to be included in the vector sum Net force component along an axis is related to the acceleration along that same axis

Vector Nature of Force Vector force: has magnitude and direction Net Force: a resultant force acting on object!!!!! F = å net F = F + F + F... 1 2 3 + You must use the rules of vector addition to obtain the net force on an object

Resultant of forces is the Net Force A resultant force is a single force which can replace a set of forces acting on an object. It has exactly the same effect on the object as the original set. Parallel forces 4 N 7 N Resultant force = 7 + 4 = 11 N 5 N 12 N Resultant force = 12-5 = 7 N

Example 2 The man pushes on a 400. N box with a force of 200. N at 53.1 o. (a) Find the vertical and horizontal components of the force he exerts on the box. (b) Find the normal force on the box. F x = 200. (sin 53.1)= 160. N F y = 200. (cos 53.1)= -120. N F norm + F y + F g = 0 F norm 120.N 400. N = 0 F norm = 520. N

Force and acceleration a. When did the object in the figure have constant velocity? 4-5 sec b. Which interval corresponds to the greatest change in speed of the body? 2-3 sec c. During which time interval(s) did the body accelerate? 0-4 sec d. During which time interval(s) did the body decelerate? 5-6 sec The following graph shows the force on an object moving in the positive direction.

Example 4. A 54-g tennis ball is in contact with the racket for a distance of 40. cm as it leaves with a velocity of 48 m/s. What is the average force on the ball? First, draw sketch and list given quantities: Given: v o = 0; v f = 48 m/s x = 40. cm; m = 54 g a =? Consistent units require converting grams to kilograms and centimeters to meters: Given: v o = 0; v f = 48 m/s x = 0.40 m; m = 0.054 kg; a =?

Example 4 (Cont). A 54-g tennis ball is in contact with the racket for a distance of 40. cm as it leaves with a velocity of 48 m/s. What is the average force on the ball? Knowing that F = m a, we need first to find acceleration a: v f2 = v 02 + 2ax a 2 2 f 0 2 ax = v - v ; 2 (48 m/s) ; 2880 m/s 2 = a= 2(0.40 m) 0 2 v f 2x F = ma F= (0.054 kg)(2880 m/s 2 ); F = 156 N a = = 160 N

5. A 4.58 kg crate is at rest on a level icy surface. A cord suddenly exerts a force F = 13.0 N at an angle of 15.5 above the horizontal. At 3.3 seconds what is the crate's speed? F net = F x = 13.0 N (cos 15.5 o ) = 12.5 N a = F/m = 12.5 N / 4.58kg = 2.74 m/s 2 For final velocity v f = v i + at = 9.0 m/s

Newton s Third Law For every action, there is an equal and opposite reaction. Action and Reaction are just names of forces.

Examples: Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left.

Example Action-Reaction Pairs The rocket exerts a downward force on the exhaust gases. The gases exert an equal but opposite upward force on the rocket. F G F R

Example: Walking Forward When weight is on back foot it acts by pushing back on the floor. Reaction is the force of the floor pushing your body forward. Action Reaction

Two Logical Difficulties with Newton s Third Law 1. 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.)

The First Difficulty Addressed If the action and reaction forces are the same size, how can two objects push on each other and get different accelerations? Newton s Second Law says that the acceleration of an object depends not only on the force on it, but on the object s mass.

Earth / Apple apple EARTH 0.40 kg 3.92 N 3.92 N m = 5.98 10 24 kg A 0.40 kg apple weighs 3.92 N (W = mg). The apple s weight is Earth s force on it. The apple pulls back just as hard. So, the same force acts on both bodies. Since their masses are different, so are their accelerations (2 nd Law). The Earth s mass is so big, it s acceleration is negligible.

Earth / Apple (cont.) The products are the same, since the forces are the same m a = m a Apple s little mass Apple s big acceleration Earth s big mass Earth s little acceleration

And the same force acting on objects of different mass will produce different accelerations! Same Force F net m a F = m net = a

Newton s three laws all work together

Example 6: A 60.-kg athlete exerts a force on a 10.-kg skateboard. If she receives an acceleration of 4.0 m/s 2, what is the acceleration of the skateboard? Force on runner = -(Force on board) m r a r = -m b a b Force on Board Force on Runner (60. kg)(4.0 m/s 2 ) = -(10 kg) a b (60 kg)(4 m/s) a = =-24 m/s -(10 kg) a = - 24 m/s 2 2

The Other Logical Problem 2. If Newton s Third Law action & reaction forces are equal and opposite, how come they don t always cancel, making net force and acceleration impossible?

Classic Problem: How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force? Aren t these balanced forces resulting in no acceleration? NO!!!

The Keys to Understanding Only forces pushing or pulling on an object affect the object s motion. Only forces that act on the same object can cancel. Newton s Third Law action and reaction forces push/pull on different objects, so they don t cancel.

The ground on horse/ horse on ground reaction pair must be larger than the wagon on horse/ horse on wagon reaction pair in order for the horse and cart to accelerate forward.

NORMAL AND GRAVITATION FORCES ACTION REACTION PAIR??

N and mg are NOT an Action-Reaction Pair! N m mg F s The dot represents the man. mg, his weight, is the force on the man due to the Earth. F E is the force on the Earth due to the man. Earth F E N, the normal force, is the force on the man due to the surface. F s is the force on the surface (ground) due to the man. The red vectors are an action-reaction pair. So are the blue vectors. Action-reaction pairs always act on two different bodies!

More Examples Involving Normal Force

F app is 25 N upwards ((50 N) sin 30 degrees). So F norm is 75 N

(a) Find the normal force on the object. (b) Determine the acceleration of the box. (c) Determine its velocity after being pulled by the applied force for 2.0 seconds.

F grav = m g = (20 kg) (9.8 m/s 2 ) = 196 N F y = (50 N) sin(45 degrees) = 35.4 N In order for the vertical forces to balance, F norm + F y = F grav F norm = F grav - F y = 196 N - 35.4 N = 160.6 N F x = (50 N) cos (45 degrees) = 35.4 N F net is 35.4 N, right since the only force which is not balanced is F x. The acceleration is: a = F net / m = (35.4 N) / (20 kg) = 1.77 m/s 2 The velocity after 2.0 seconds can be calculated using a kinematic equation: v f = v i + a t = 0 m/s + (1.77 m/s 2 ) (2.0 s) = 3.54 m/s

1. A 100-kg crate moves down an incline which exerts a friction force of 255 N. Fill in the blanks below in order to determine the net force on the crate and the acceleration down the incline.

Since this angle is 60 o, This angle must be 30 o

The force of gravity is 980 N and the components of this force are: F parallel = 980 N sin 30 degrees = 490 N F perpendicular = 980 N cos 30 degrees = 849 N Now the normal force can be determined to be 849 N The net force is 235 N (i.e., 490 N - 255 N). The acceleration is 2.35 m/s 2 (F net /m = 235 N/100 kg).

Applications of Newton s Laws Involving Friction When a body is in motion along rough a surface, the force of kinetic friction acts opposite to the direction of the body s velocity. The magnitude of the force of kinetic friction depends on the nature of the two sliding surfaces. For a given surface, experiment shows that the force is approximately proportional to the normal force between the two surfaces.

Coefficient of Kinetic Friction The force of friction between hard surfaces depends very little on the total surface area in contact. We can write the proportionality as an equation by inserting a constant of proportionality, µ k : F fr = µ k F N

Coefficient of Kinetic Friction This relation is not a fundamental law; it is an experimental relation. It is not a vector equation since the two forces are perpendicular to one another. The term µ k is called the coefficient of kinetic friction, and its value depends on the nature of the two surfaces.

Coefficient of Static Friction There is also static friction, which refers to a force parallel to the two surfaces that can arise even when they are not sliding.

Coefficient of Static Friction This is the force of static friction exerted by the floor on the desk. If you push with greater force without moving the desk, the force of static friction also has increased. If you push hard enough, the desk will eventually start to move, and kinetic friction takes over. At this point, you have exceeded that maximum force of static friction, which is given by F max = µ s F N

Coefficient of Static Friction The constant µ s is the coefficient of static friction. Since the force of static friction can vary from zero to this maximum value, we write F fr < µ s F N

Example 2 Friction: static and kinetic A 10-kg mystery box rests on a horizontal floor. The coefficient of static friction is µ s = 0.40 and the coefficient of kinetic friction µ k = 0.30. Determine the force of friction, F fr, acting on the box if a horizontal external applied force F A is exerted on it with a magnitude (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.

Conceptual Example To push or pull a sled? Your little sister wants a ride on her sled. If you are flat on the ground, will you exert less force if you push her or pull her? Assume the same angle q in both cases.

Conceptual Example 3 A box against a wall. You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically?

The horizontal force you apply produces a normal force on the box exerted by the wall. The force of gravity (mg) is balanced by the upward static friction force. Static friction force is proportional to normal force, so the harder you push, the greater the normal force, and the greater the force of static friction.

Example 3 Pulling against friction. A 10.0-kg box is pulled along a horizontal surface by a force F P of 40.0 N that is applied at a 30.0 o angle. We assume a coefficient of kinetic friction of 0.30. Calculate the acceleration.

F NET = F x - F f F NET = Fcos 30 o - µf N where F N = F g - F y F NET = 11.2 N a = F NET /m (Newton s 2 nd Law) a = 1.12 m/s 2

Example 4 The skier. The skier in the figure has just begun descending a 30 o slope. Assuming the coefficient of kinetic friction is 0.10, calculate (a) her acceleration, and (b) the speed she will reach after 4.0 s. use v f = v i + at a = 4.05 m/s 2 v f = 16.2 m/s

Example 5 Measuring µ k. Suppose that the snow is slushy and the skier moves down the 30 o slope at constant speed. What can you say about the coefficient of friction, µ k?

Coupled Motion 1. Two boxes and a pulley. Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box I and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the chord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn t stretch. As box II moves down, box I moves to the right. 5-5

100

Coupled Motion 2. A plane, a pulley, and two boxes. A box of mass m 1 = 10.0 kg rests on a surface inclined at q = 37 o to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box of mass m 2, which hangs freely. (a) If the coefficient of static friction µ s = 0.40, determine what range of values for m 2 will keep the system at rest. (b) If the coefficient of kinetic friction is µ k = 0.30, and m 2 = 10.0 kg, determine the acceleration of the system.

103

104