Chapter 4 Newton’s Laws of Motion

 Aristotle – (4 th Century B.C.) Studied Motion Motion due to Natural (up/down) & Violent Forces (push/pull) * external causes  Copernicus (1473 – 1543) Sun Centered System.  Galileo – Supported Copernicus Only when friction is present… is force needed to keep an object moving

 Newton – ( ) Born on the day Galileo died Dec. 25, Laws of Motion 1. Objects at Rest (or Motion) will remain at Rest (or in motion) 2. F = ma 3. Action – Reaction

Forces  Forces – are the cause of motion (Push or Pull)  Dynamics -> the study of motion considering its causes; the study of the effects of forces on matter Historically: Kinematics were 1 st done by Galileo and Sir Isaac Newton developed dynamics

Force: Push or Pull on an object Measured in Newtons (N) 1N= 1kg x 1 m/s 2 F = m x a = ma

 Force Diagram (Free – Body Diagram) FfFf FNFN FAFA F w (force due to gravity) F N = Normal Force F w = Force of Weight F A = Applied Force F f = Frictional Force Or F r * Net Force = total of All forces acting on an object * Equilibrium is reached when the Net Force equals ___________.

Newton’s First Law  Every object remains at rest or in motion in a straight line at constant speed unless it is acted on by another force

Newton’s Second Law  The acceleration of an object is directly Proportional to the new external Force acting on the object and inversely proportional to the object’s mass : a = F m Hence the equation: F = ma

Newton’s Third Law: W hen 2 objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1 ( ACTION-REACTION )  *Force of Friction is proportional to the Normal Force and opposes the Applied Force * Friction depends on the Surface*

1) Gravitational (Gravity) 2) Electromagnetic 3) Nuclear Forces A) Weak B) Strong Basic Kinds of Forces

 Force Diagram (Free – Body Diagram) On an Inclined Plane θ1θ1 θ2θ2 θ 1 = θ 2 FAFA FNFN FfFf FwFw

Coefficient of Friction: μ ( mu) The ratio of the forces of friction to the normal force acting between μ = F f F N Coefficient of kinetic friction μ k = F k F N Coefficient of static friction μ s = F s max F N

Things to Know about μ  μ = Greek letter “mu”  Units of μ? NONE  0 < μ < 1  There are actually 2 different μ’s 1. μ s – coefficient of static friction 2. μ k – coefficient of kinetic friction  Generally we use μ k F f = μF N μ s > μ k

Things to Study and Understand!!  Table 4 – 2  Samples to Study (Add to notes) 4B pg 137 4C pg 145 4D pg 146 Add these to your notes!

Example: A 40N wooden block moving on a wooden table due to a 14N force What is the coefficient of friction? F A = 14 NF w = 40Nμ = ? Hint: constant velocity -> F A = F f with no vertical motion F w = F N

Example F A = 14 NF w = 40 Nμ = ? constant velocity -> F A = F f a) F A = F f = μF N = μF w ~ F N = F w F f /F N = F A / F w = μ = 14N / 40N = b) F w = 40N + 20N = 60N at cont. V is Horizontal F A = F f = μF N = (.35)(60N) = N

Review--- Generally, for a vector F at θ o F x = Fcosθ F y = Fsinθ FxFx FyFy F θ Y X FyFy

Reminder: When resolving a vector into its components: Let F = o FxFx FyFy F 62N 75 o Y X FyFy Note: cos 75 o = F x F sinθ = F y F F x = F cos75 o = (62N)(.2588) F x = 16N F y = Fsin75 o = (62N)(.9659) F y = 59.9N

Review of quadrant signs and standard position (always measure from 0 o ) F x : + F y : + F x : - F y : + F x : + F y : - F x : - F y : - 0 o STD I II III IV * Note Sign Patterns per quadrant 180 o

Hmwk. Chapter 4 #1 Book 4A pg 133 1,2,32. a. -33N b. 329N c. 330 N 4B pg 138 1,2, m/s N 4C pg 145 1,2,32. a. 1.5 b D pg 147 1