Chapter 4 Forces and Newtons Laws of Motion

Chapter 4: Forces and Newton’s Laws of Motion • • • Forces Newton’s Three Laws of Motion The Gravitational Force Contact Forces (normal, friction, tension) Application of Newton’s Second Law Apparent Weight Air Resistance Fundamental Forces CQ: 16, 18. P: 3, 5, 21, 37, 41, 63, 73, 87, 97, 147. 1

Force Concept Contact Forces Ex: car on road, ball bounce Non-Contact Ex: magnetism, gravity / 2

units • • Force units (SI): newton, N 1 N ≈ ¼ lb. 1 N = (1 kg)(1 m/s/s) N/kg = m/s/s 3

Inertia • is ‘resistance’ to change in velocity • Measurement: Mass • SI Unit: Kilogram (Kg) • / 4

Universal Law of Gravity • all matter is weakly attracted • attraction is inverse-square with distance • G = 6. 67 x 10 -11 N·m 2/kg 2 • Example: Two 100 kg persons stand 1. 0 m apart 5

g vs G • G is universal • g ~ Mass and Radius • / 6

Contact Forces • Surfaces in contact are often under compression: each surface pushes against the other. The outward push of each object is called the Normal Force. • If the objects move (even slightly) parallel to their surface the resistance force experienced is called the frictional force. 7

Normal forces are? 1. Always vertically upward. 2. Always vertically downward. 3. Can point in any direction. 8

Tension & Compression • Compressed objects push outward away from their center (aka Normal Force). • Stretched objects pull toward their center. This is called the Tension Force. 9

Force Label Notation • • • F = general force FN = normal force f = frictional force w = mg = Fg = weight T = tension force / 10

Net Force = change of motion vector sum of all forces acting on an object 11

Example: Net Force = 0, Ball rolls along a smooth level surface constant velocity Force Diagram table force Fnet = 0 a=0 weight force 12

Example: Net-force on 0. 5 kg • Net-force = 4 N: Acceleration = 4 N/0. 5 kg = 8 m/s/s • 5 N, Right; 3 N Left; Net-force = 2 N Acceleration = 2 N/0. 5 kg = 4 m/s/s • Falling; Net-force = mg Acceleration = mg/m = g = 9. 8 m/s/s • / 13

Newton’s Laws of Motion 1. An object maintains constant velocity when the Net -Force on it is zero. 2. An object’s acceleration equals the Net-Force on it divided by its mass. 3. Forces always occur in pairs equal in size and opposite in direction. 14

Force Diagrams • Object is drawn as a “point” • Each force is drawn as a “pulling” vector • Each force is labeled • Relevant Angles are shown • x, y axes are written offset from diagram • Only forces which act ON the object are shown 15

Example of a Force Diagram for a Sled net force equals the mass times its acceleration. 16

g’s • one “g” of acceleration = 9. 8 m/s/s • “two g’s” = 19. 6 m/s/s, etc. • Example: What is the net force on a 2100 kg SUV that is accelerating at 0. 75 g? • Net-force = ma = m(0. 75 g) = 0. 75 mg = ¾ weight of car. • / 17

Block on Frictionless Incline • a = wx/m =mgsinq/m • = gsinq. • F n = w y. 18

Newton’s 3 rd Law of Motion • equal-sized oppositely-directed forces • Independent of mass • Pair-notation x x 19

Newton’s 3 rd Law Pair Notation • use “x” marks on forces that are 3 rd Law pairs. • Use “xx” for a different interaction, etc. 20

Force Diagram each object. Which has greater acceleration when released? Spring Force x Acceleration = F/m Spring Force x Acceleration = F/(2 m) 21

Friction • • • Static Friction “sticking force” Kinetic Friction “sliding force” Coefficients: 0 = min, 1 ~ max e. g. teflon around 0. 05 Rubber on concrete around 1. 0 22

Using Coefficients of Friction • Ex. 10 kg block. FN = weight = mg = 98 N. Static coef. = 0. 50; Kinetic coef. = 0. 30. 23

Applications 24

A 3 kg object sits on a frictionless table. Two horizontal forces act, one is 2 N in the y-direction, the other 4 N in the xdirection. A top-view diagram will be shown. What is the magnitude of the net-force acting? 2 Fnet 2 4 25

What direction does the 3 kg mass accelerate? parallel to Fnet by Newton’s 2 nd Law. We are in Quadrant I since x and y are both + 2 Fnet 2 4 26

The magnitude of the acceleration is: 27

Two 1 kg Blocks; a = 1 m/s/s F • Fnet = F = (2 m)a = (2 kg)(1 m/s/s) = 2 N • Fnet = T = ma = (1 kg)(1 m/s/s) = 1 N • / 28

Two 1 kg Blocks; F = 10 N F • a = F/(2 m) = 10 N/2 kg = 5 m/s/s • T = ma = (1 kg)(5 m/s/s) = 5 N • / 29

4 Summary • Fnet = ma (Fnet = 0, v = constant) • forces always occur in pairs of equal size and opposite direction • various force types (& symbols) • equilibrium problems (a = 0) • dynamic problems (a ≠ 0) 30

30 60 90 30 Mg, 300 deg. 31

Inclined Plane Forces • Fxnet = FNcos 90 + mgcos 300 = (0. 02)(a) • = 0 + (0. 02)(9. 8)(0. 5) = (0. 02)a • accel = 4. 9 m/s/s • Fynet = FNsin 90 + mgsin 300 = (0. 02)(0) • FN + (0. 02)(9. 8)(-. 866) = 0 • FN = 0. 17 N 32

Ex: Newton’s 2 nd Law Fnet acceleration 33

Coefficients of Friction Ex: Block&Load = 580 grams If it takes 2. 4 N to get it moving and 2. 0 N to keep it moving 34

Example: 1. 3 kg box on level frictionless surface. F=86 N acts 60° below horizontal. 35

1. (cont) 36

Q 1. What are ax and FN if angle is 30? 37

Interaction Notation • Since all forces are ‘pairs’, label as interactions, e. g. 1 on 2, 2 on 1, etc. • F 12 = “force of object 1 on object 2” • F 21 = “force of object 2 on object 1” • F 34 = “force of object 3 on object 4” • Etc. 38

Interaction Notation Symbols • • F 12 – general force, 1 on 2 N 12 – normal contact force, 1 on 2 f 12 – frictional force, 1 on 2 W 12 – gravitational force, 1 on 2 T 12 – tension force, 1 on 2 m 12 – magnetic force, 1 on 2 e 12 – electrical force, 1 on 2 39

Gravitational Force • All masses attract via gravitational force • Attraction is weak for two small objects • Ex: Attraction between two bowling balls is so small it is hard to measure. • Force is proportional to mass product • Force is inversely proportional to the square of the distance between objects 40

Example: Net Force = 0. Block on a surface inclined 30° from horizontal. Applied force F acts 40° below horizontal. Net Force = 0 velocity = constant 41

Diagrams with Interaction Notation • If f 21 exists, then f 12 also exists, and is opposite in direction to f 21. • f 21 and f 12 act on different objects. 42

Example: A 10 kg box is being pushed along a horizontal surface by a force of 15 N. A frictional force of 5 N acts against the motion. We will want to (a) Calculate the net-force acting and (b) calculate the acceleration of the box. The net-horizontal force determines its x-acceleration The y-acceleration is known to be zero because it remains in horizontal motion, thus The net-force is 10 N horizontal (0 vertical) The x-acceleration is: 43