PHY 101 Lecture 4 B 4 1 The

PHY 101: Lecture 4 B • • • 4. 1 The Concepts of Force and Mass 4. 2 Newton’s First Law of Motion 4. 3 Newton’s Second Law od Motion 4. 4 The Vector Nature of Newton’s Second Law of Motion 4. 5 Newton’ Third Law of Motion 4. 6 Types of Forces: An Overview 4. 7 The Gravitational Force 4. 8 The Normal Force 4. 9 Static and Kinetic Frictional Forces 4. 10 The Tension Force 4. 11 Equilibrium Applications of Newton’s Laws of Motion 4. 12 Nonequilibrium Applications of Newton’s Laws of Motion

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4. 6 Types of Forces: An Overview

Force • Push or pull that could potentially cause motion • Force is a vector quantity • Two types of forces: Ø Contact forces Ø Noncontact forces or force-at-a-distance • Contact Forces: Ø Arise from physical contact between two objects Ø Normal, Friction, Tension, Push / Pull • Noncontact Forces: Ø Gravity

Basic Forces of the Universe • • Gravity Electromagnetic Weak (nuclear) Strong (nuclear)

Forces • In the course we will consider five forces ØBasic Force • Gravity ØElectromagnetism • • Normal Friction Tension Push / Pull

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4. 7 The Gravitational Force

Newton’s Law of Universal Gravity The Story - 1 • Newton observes apple falling from a tree • Apple is accelerated, since its velocity changes from zero as it moves toward the ground • By Newton’s 2 nd Law there must be a force acting on the apple to cause acceleration • Call the force “gravity” • Imagine tree is twice as high • The apple is still accelerated towards the ground • Suggests that this force reaches to top of the tallest tree

Newton’s Law of Universal Gravity The Story - 2 • Newton has a brilliant insight • If the force of gravity reaches the top of the highest tree, might it not reach even further • Might it not reach all the way to the orbit of the Moon • The orbit of the Moon about the Earth could be a consequence of the gravitational force • Acceleration due to gravity could change the velocity of the Moon in such a way that it follows an orbit around the earth

Newton’s Law of Universal Gravity The Story - 3 • Suppose we fire a cannon ball horizontally from a high mountain • The projectile will eventually fall to Earth • As we increase the muzzle velocity, the projectile will travel further and further before returning to Earth • If the cannon fired the cannon ball with the right velocity, the projectile would travel completely around the Earth

Newton’s Law of Universal Gravity The Story - 4 • The projectile is always falling because of gravity but never reaching the Earth, which is curving away at the same rate that the projectile falls • That is, the cannon ball would have been put into orbit around the Earth • Newton concluded that the orbit of the Moon was of exactly the same nature • The moon continuously “fell” in its path around the Earth due to gravity, thus producing an orbit

Newton’s Law of Universal Gravity The Story - 5 • Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other

Dependence on Distance - 1 • Acceleration on earth’s surface is 9. 81 m/s 2 • Center of moon is 3. 84 x 108 meters from center of earth • Moon’s orbital period is 27. 3 days = 2. 36 x 106 seconds

Dependence on Distance - 2 • Centripetal acceleration of moon due to earth’s gravitational attraction is • ac = 4 p 2 r/T 2 • (we will soon learn formula) Ø T is orbital period of moon • 4 p 2 r/T 2 • 4 p 2(3. 84 x 108)/(2. 36 x 106)2 • 0. 00272 m/s 2

Dependence on Distance - 3 • ratio of earth’s acceleration to moon’s acceleration • 9. 81/0. 0272 = 3607 • ratio of earth’s radius to distance of moon’s center from earth’s center is 60 • 3607 is approximately 602 • bigger the radius the smaller the acceleration • therefore, dependence on distance is 1/r 2

Mass Dependence – 1 • Galileo showed that different masses fall with the same acceleration • Let m 1 be mass of falling object and m 2 mass of earth • Therefore, the formula for the force of gravity must be like m 1 h(m 2)/r 2 = m 1 a • h is some formula containing only m 2 • Cancellation of m 1 from both sides of the equation means that a does not depend on the mass m 1

Mass Dependence - 2 • If Newton’s Third Law applies to gravity • Then earth attracts moon and moon attracts earth with forces of the same magnitude • It should not matter which mass is called m 1 and which is called m 2 • Dependence on mass must be m 1 h(m 2) • h is a function

Mass Dependence - 3 • h is a functions • For example: Øh is m 2, (m 2)2, sin 2(m 2), e 2(m 2), … • If the interchange of m 1 and m 2 produces the same force • Then h must be m 2 • Force of gravity is proportional to m 1 m 2/r 2

Constant of Proportionality • meter and kilogram were defined without any concern for the force of gravity • The constant of proportionality, G, adjusts for the fact the meter and kilogram are not appropriate to gravity

Newton’s Law of Universal Gravity • Every particle in the universe exerts an attractive force on every other particle • A particle is a piece of matter, small enough in size to be regarded as a mathematical point • For two particles that have mass m 1 and m 2 and are separated by a distance r, the force that each exerts on the other is directed along the line joining the particles and has a magnitude given by • G is the universal gravitational constant, whose value is found experimentally • G = 6. 673 x 10 -11 Nm 2/kg 2

Newton’s Law of Universal Gravity Special Requirements • Newton’s law of gravitation applies only to particles • Newton proved that an object of finite size can be considered a particle for purposes of the gravitation law, provided the mass of the object is distributed with spherical symmetry about its center • In this case, r is the distance between the centers of the spheres and not between the outer surfaces • The gravitational forces that the spheres exert on each other are the same as if the entire mass of each were concentrated at its center

Newton’s Law of Universal Gravity Example • Calculate the gravitational force between the Earth and the Moon Ø F = GMe. Mm/rem 2 Ø F = 6. 673 x 10 -11 x 5. 98 x 1024 x 7. 35 x 1022/(3. 85 x 108)2 Ø F = 2. 0 x 1020 N

Newton’s Law of Universal Gravity Weight - 1 • Weight of an object on or above the earth is the gravitational force that the earth exerts on the object • Weight always acts downward, toward the center of the earth • On or above another astronomical body, the weight is the gravitational force exerted on the object by the body • SI Unit of Weight: newton (N) • Symbol for Weight: W

Newton’s Law of Universal Gravity Weight - 2 • m is the mass of the object • ME is the mass of the Earth • r is the distance from the center of the earth to the object • r must be equal to or greater than the radius of the Earth

Newton’s Law of Universal Gravity Relation between Mass and Weight • Mass ØA quantitative measure of inertia ØAn intrinsic property of matter ØDoes not change as an object is moved from one location to another • Weight ØGravitational force acting on the object ØCan vary, depending on how far the object is above the earth’s surface or whether it is near another body such as the moon

Newton’s Law of Universal Gravity Weight at Earth’s Surface • m is the mass of the object • ME is the mass of the Earth • Re is radius of the Earth

Newton’s Law of Universal Gravity Mass / Weight - Example • What is the mass of a person weighing 740 N on the Earth? ØW = mg Øm = W/g Øm = 740 / 9. 8 = 75. 5 kg

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4. 8 The Normal Force

Normal Force • Normal force is one component of the force that a surface exerts on an object with which it is in contact • This is component that is perpendicular to the surface • Depending on the physical situation, the Normal Force can be vertical, horizontal, or some other direction • For example: – Block exerts a force on table by pressing down on it – Consistent with the third law, table exerts an oppositely directed force of equal magnitude on block – This reaction force is the normal force

Normal Force Apparent Weight • There are situations in which a scale does not give correct weight • In such situations, the reading on the scale gives only the “apparent” weight, rather than the gravitational force or “true” weight • The apparent weight is the force that the object exerts on the scale with which it is in contact • Consider a person on a scale in an elevator: Ø If the elevator is not moving or moving with constant velocity, the scale registers the true weight Ø If the elevator is accelerating, apparent weight and the true weight are not equal Ø When the elevator accelerates upward, the apparent weight is greater than the true weight

Apparent Weight – Example 1 • Consider a person on a scale in an elevator • Normal force of scale on person is N in up direction • Weight of gravity, mg, on person is in down direction • Newton’s second law gives Ø F = N – mg = ma Ø N = ma + mg • “a” can be either + for up acceleration or – for down acceleration • N is the force the scale exerts on the person • By Newton’s third law, N is the force the person exerts on the scale, the apparent weight

Apparent Weight – Example 2 • A man of mass 80 kg stands in an elevator accelerating up at 4 m/s 2 • Calculate the Normal force on the man ØWeight down = -mg = -80(9. 8) = -784 N ØF = ma ØN – 784 = 80(4) ØN = 784 + 320 = 1004 N

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4. 9 Static and Kinetic Friction Forces

Friction Force • When an object is in contact with a surface, there is a force acting on the object • Component of this force that is perpendicular to the surface is called the normal force • When the object moves or attempts to move along the surface, there is also a component of force that is parallel to the surface • This parallel component is called the frictional force, or friction

Static Friction Force Magnitude • When two surfaces are not in relative motion, the force of friction is called static friction • The magnitude fs of the static frictional force can have any value from zero up to a maximum value of fs. Max , depending on the applied force • In this equation, ms is the coefficient of static friction, and N is the magnitude of the normal force

Kinetic Friction Force Magnitude • When two surfaces are in relative motion, the force of friction is called kinetic friction • Magnitude of the kinetic frictional force is given by fk • fk = mk. N • In this equation, mk is the coefficient of kinetic friction, and N is the magnitude of the normal force

Friction Force Direction • Static Friction ØThe direction is opposite to the direction in which the object would move if there was no friction • Kinetic Friction ØThe direction is opposite to the direction in which the object is actually moving

Friction Force – Example 1 • In moving a 35. 0 -kg desk from one side of a classroom to the other, a professor finds that a horizontal force of 275 N is necessary to set the desk in motion, and a force of 195 N is necessary to keep it in motion at a constant speed • What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor? • (a) Ø Static friction Ø f s = m s. N Ø ms = fs/N = fs/mg = 275/(35)(9. 8) = 0. 802 • (b) Ø Kinetic friction Ø f k = m k. N Ø mk = fk/N = fk/mg = 195/(35)(9. 8) = 0. 569

Friction Force – Example 2 • • • An object has an initial velocity of 20 m/s Object slides on surface with mk = 0. 8 How far does the object slide until it stops? Ø Vertical • F = N – mg = ma = 0 • N = mg Ø Horizontal • Friction = -mk. N = -mkmg • F = -mkmg = ma • a = -mkg Ø Stopping distance • vf 2 = vi 2 + 2 ax • (we will soon learn this formula) • 0 = vi 2 – 2 mkgx • x = vi 2/2 mkg = 202/2/(. 8)/9. 8 • x = 25. 5 m

Friction Force – Example 3 • mk = 0. 4 • No friction Ø F = 400 = ma = 50 a Ø a = 8 m/s 2 • With friction Vertical Ø F = ma = 0 Ø -mg + N = 0 Ø N = mg = 50(9. 8) = 490 N Horizontal Ø fk = mk. N = 490(0. 4) = 196 N Ø F = P – fk = ma Ø 400 – 196 = 50 a Ø a = 204/50 = 4. 08 m/s 2

Friction Force – Example 4 • mk = 0. 4 • No friction Horizontal Ø Px = 400 cos 36. 9 = 320 Ø Py = 400 sin 36. 9 = 240 Ø F = Px = 320 = ma = 50 a Ø a = 320/50 = 6. 4 m/s 2 • With friction Vertical Ø F = -mg + N + Py = -490 + N + 240 = 0 Ø N = 490 – 240 = 250 Horizontal Ø fk = mk. N = 0. 4(250) = 100 Ø F = -fk + Px = -100 + 320 = ma = 50 a Ø a = 220/50 = 4. 4 m/s 2

Friction Force – Example 5 • mk = 0. 4 • No friction Horizontal Ø Px = 400 cos 36. 9 = 320 Ø Py = -400 sin 36. 9 = -240 Ø F = Px = 320 = ma = 50 a Ø a = 320/50 = 6. 4 m/s 2 • With friction Vertical v F = -mg + N + Py = -490 + N - 240 = 0 v N = 490 + 240 = 730 Horizontal Ø fk = mk. N = 0. 4(730) = 292 Ø F = -fk + Px = -292 + 320 = ma = 50 a Ø a = 28/50 = 0. 56 m/s 2

Friction Force Summarize Examples 3, 4, 5 • Example 4 with friction has greater a than Example 3 because box lifted up decreasing friction • Example 5 with friction has lower a than Example 4 because box is pushed down increasing friction • Note: a is acceleration Ex 3 a m/s 2 Horz Ex 4 a m/s 2 Pull Up Ex 5 a m/s 2 Push Down No Friction 8. 00 6. 40 Friction 4. 08 4. 40 0. 56

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4. 10 The Tension Force

Tension Force • Forces are applied by means of cables or ropes that are used to pull an object • If the rope has no mass, the tension is the same along the length of the rope ØLet the tension on one end of the rope be T 1 and the tension on the other end be T 2 ØF = ma ØT 1 – T 2 = (m=0)a = 0 ØTherefore, T 1 = T 2

Tension Force - Example • Part a of drawing shows a bucket of water suspended from pulley of a well; tension in the rope is 92. 0 N. • (a) Ø F = ma = 0 Ø 2 T – W = 0 Ø W = 2 T = 2 x 92 = 184 N • Part b shows same bucket of water being pulled up from well at a constant velocity. What is the tension in the rope in part b? Ø F = ma = 0 ØT – W = 0 Ø T = W = 184 N