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Chapter 4 Newton’s Laws of Motion
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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
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Newton – (1642 - 1727) Born on the day Galileo died Dec. 25, 1642 3 Laws of Motion 1. Objects at Rest (or Motion) will remain at Rest (or in motion) 2. F = ma 3. Action – Reaction
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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
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Force: Push or Pull on an object Measured in Newtons (N) 1N= 1kg x 1 m/s 2 F = m x a = ma
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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 ___________.
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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
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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
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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*
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1) Gravitational (Gravity) 2) Electromagnetic 3) Nuclear Forces A) Weak B) Strong Basic Kinds of Forces
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Force Diagram (Free – Body Diagram) On an Inclined Plane θ1θ1 θ2θ2 θ 1 = θ 2 FAFA FNFN FfFf FwFw
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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
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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
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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!
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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
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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
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Review--- Generally, for a vector F at θ o F x = Fcosθ F y = Fsinθ FxFx FyFy F θ Y X FyFy
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Reminder: When resolving a vector into its components: Let F = 62 N @ 75 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
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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
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Hmwk. Chapter 4 #1 Book 4A pg 133 1,2,32. a. -33N b. 329N c. 330 N 4B pg 138 1,2,42. 1.4 m/s 2 4. 14N 4C pg 145 1,2,32. a. 1.5 b. 1.3 4D pg 147 1
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