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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.
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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.
0 Force Concept Contact Forces Ex: car on road, ball bounce Non-Contact Ex: magnetism, gravity / 2
units Force units (SI): newton, N 1N ≈ ¼ lb. 1N = (1kg)(1m/s/s) N/kg = m/s/s 3
Inertia is ‘resistance’ to change in velocity Measurement: Mass SI Unit: Kilogram (Kg) / 0 4
Universal Law of Gravity all matter is weakly attracted attraction is inverse-square with distance G = 6.67x10-11 N·m2/kg2 Example: Two 100kg persons stand 1.0m apart 5
g vs G G is universal g ~ Mass and Radius / 6
Normal Force 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.
Normal forces are? Always vertically upward. Always vertically downward. Can point in any direction. 0 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.
Force Label Notation • F = general force • FN = normal force • f = frictional force • w = mg = Fg = weight • T = tension force • / 10
0 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 0 constant velocity Force Diagram table force Fnet = 0 a = 0 weight force 12
Example: Net-force on 0.5kg • Net-force = 4N: Acceleration = 4N/0.5kg = 8m/s/s • 5N, Right; 3N Left; Net-force = 2NAcceleration = 2N/0.5kg = 4m/s/s • Falling; Net-force = mgAcceleration = mg/m = g = 9.8m/s/s • /
0 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
0 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 0 net force equals the mass times its acceleration. 16
g’s • one “g” of acceleration = 9.8m/s/s • “two g’s” = 19.6m/s/s, etc. • Example: What is the net force on a 2100kg SUV that is accelerating at 0.75g? • Net-force = ma = m(0.75g) = 0.75mg = ¾ weight of car. • / 17
Block on Frictionless Incline a = wx/m =mgsinq/m = gsinq. Fn = wy. 18
Newton’s 3rd Law of Motion • equal-sized oppositely-directed forces • Independent of mass • Pair-notation x x 19
Newton’s 3rd Law Pair Notation • use “x” marks on forces that are 3rd Law pairs. • Use “xx” for a different interaction, etc. 20
Force Diagram each object. Which has greater acceleration when released? Spring Force Spring Force x x Acceleration = F/m Acceleration = F/(2m) 21
0 Friction • Static Friction “sticking force” • Kinetic Friction “sliding force” • Coefficients: 0 = min, 1 ~ max • e.g. teflon around 0.05 • Rubber on concretearound 1.0 22
0 Using Coefficients of Friction • Ex. 10kg block. FN = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30. 23
A 3kg object sits on a frictionless table. Two horizontal forces act, one is 2N in the y-direction, the other 4N in the x-direction. A top-view diagram will be shown. What is the magnitude of the net-force acting? Fnet 2 2 4 25
Fnet 2 2 4 What direction does the 3kg mass accelerate? parallel to Fnet by Newton’s 2nd Law. We are in Quadrant I since x and y are both +
Fnet = F = (2m)a = (2kg)(1m/s/s) = 2N Fnet = T = ma = (1kg)(1m/s/s) = 1N / F Two 1kg Blocks; a = 1m/s/s 28
F Two 1kg Blocks; F = 10N • a = F/(2m) = 10N/2kg = 5 m/s/s • T = ma = (1kg)(5m/s/s) = 5N • / 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
60 30 30 90 Mg, 300 deg.
Inclined Plane Forces • Fxnet = FNcos90 + mgcos300 = (0.02)(a) • = 0 + (0.02)(9.8)(0.5) = (0.02)a • accel = 4.9 m/s/s • Fynet = FNsin90 + mgsin300 = (0.02)(0) • FN + (0.02)(9.8)(-.866) = 0 • FN = 0.17N
Ex: Newton’s 2nd Law 0 acceleration Fnet 33
Coefficients of Friction Ex: Block&Load = 580grams If it takes 2.4N to get it moving and 2.0N to keep it moving 34
Example: 0 1. 3kg box on level frictionless surface. F=86N acts 60° below horizontal. 35
1.(cont) 0 36
0 Q1. 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. • F12 = “force of object 1 on object 2” • F21 = “force of object 2 on object 1” • F34 = “force of object 3 on object 4” • Etc. 38
Interaction Notation Symbols • F12 – general force, 1 on 2 • N12 – normal contact force, 1 on 2 • f12 – frictional force, 1 on 2 • W12 – gravitational force, 1 on 2 • T12 – tension force, 1 on 2 • m12 – magnetic force, 1 on 2 • e12 – 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
0 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 f21 exists, then f12 also exists, and is opposite in direction to f21. • f21 and f12 act on different objects. velocity 1 1 f21 2 f12 42
Example: A 10kg box is being pushed along a horizontal surface by a force of 15N. A frictional force of 5N 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 10N horizontal (0 vertical) The x-acceleration is: 43