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CHAPTER 4 The Laws of Motion

Newton’s First Law: An object at rest remains at rest and an object in motion continues in motion with constant velocity (constant speed in straight line) unless acted on by a net external force. CHAPTER 4 The Laws of Motion. “in motion” or

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CHAPTER 4 The Laws of Motion

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  1. Newton’s First Law: An object at rest remains at rest and an object in motion continues in motion with constant velocity (constant speed in straight line) unless acted on by a net external force. CHAPTER 4The Laws of Motion • “in motion” • or • “at rest” – with respect to the chosen frame of reference • “net force” – vector sum of all the external forces acting on the object • – FNet,x and FNet,y calculated separately • Forces: Contact Forces • *Applied Forces (push or pull) • *Normal Force (supporting force) • *Frictional Force (opposes motion) • Field Forces • *Gravitational • ·Magnetic • ·Electrostatic • *The typical four forces analyzed in our study of classical mechanics

  2. Newton’s Second Law: The acceleration of an object is directly proportional to the net force acting on it FNet = ma • Mass – The measurement of inertia (“inertial mass”) • Inertia – The tendency of an object to resist any attempt to change its motion • Book Example: • Strike golf ball w/golf club • Strike bowling ball w/golf club • Which has greatest inertia? • Which has greatest mass? Dimensional Analysis F = ma = kg x m/s2 = newton = N 1 newton = 1 kg · m/s2

  3. Weight and the Gravitational Force Mass – an amount of matter (“gravitational mass”) “Your mass on the Moon equals your mass on Earth.” Weight – the magnitude of the force of gravity acting on an amount of matter F = ma Fg = mg w = mg NOTE: Your text treats weight (w) as a scalar rather than as a vector. Example Your mass is 80kg. What is your weight? w = 80kg · 9.8m/s2 w = 780 kg·m/s2 w = 780 N

  4. Example: (Contact Force) Book Table Example: (Field Force) Earth Moon FMoon FEarth Newton’s Third Law: If two objects interact, the force exerted on object 1 by object 2 is equal in magnitude but opposite in direction to the force exerted on object 2 by object 1 Book pushes down on table with force of 9.8.N Table pushes up on book with force of 9.8.N Net Force on book =9.8N – 9.8N = 0N Hence, book does not accelerate up or down. Earth pulls on Moon equal to the force the Moon pulls on Earth.

  5. FN Fa Ff Fg Problem Solving Strategy • Remember: We are working now with only 4 forces. • Applied Force Fa • Normal Force FN • Frictional Force Ff • Gravitational Force Fg Draw a Sketch Determine the Magnitude of Forces in “x” and in “y” Direction FN often equals Fg (object does not accelerate up off surface or accelerate downward through surface) FNet,y = FN – Fg = 0 N FNet,x = Fa – Ff = ma Ff< Fa Label forces on Sketch Solve Problem

  6. FN=540N Fa=25N Ff=15N Fg=540N Example 1: Sliding “Box” Problem (Horizontal Fa) “Box” = hockey puck = shopping cart = tire = dead cat = etc. A 55 kg shopping cart is pulled horizontally with a force of 25N. The frictional force opposing the motion is 15N. How fast does the cart accelerate? Fa = 25N Ff = 15N FNet,x = 25N – 15N = ma = 10.N = 55kg·a Fg = mg = 55kg · 9.8m/s2 = 540N FN = Fg = 540N a = .18m/s2

  7. Fa FN Fa,y =15N Fa,x =20N Ff Fg Example 2: Sliding “Box” Problem (Pulled at an Angle) A dead cat with a mass of 7.5kg is pulled off the road by a passing motorist. The motorist pulls the cat by its tail which is at an angle of 37° to the horizontal. A force of 25N is applied. The force of friction opposing motion is 18N. How fast does the cat accelerate? =25N 59N= 18N= =74N m = 7.5kg Fa = 25N Fa,x = Fa cos37 = 20.N Fa,y = Fa sin 37 = 15N FN + Fa,y = Fg (up forces equal down forces) Fg = mg = 74N FN = 74N – 15N FN = 59N Ff = 18N FNet,x = Fa,x – Ff = ma 20.N – 18N = 7.5kg · a a = .27m/s2

  8. Friction Friction opposes motion. Kinetic Friction opposes motion of a moving object. Static Friction opposes motion of a stationary object. Ff = FN static = coefficient of static friction kinetic = coefficient of kinetic friction s > k Why? Static condition: peaks and valleys of the two surfaces overlap each other. Kinetic Condition: surfaces slide over each other touching only at their peaks s > kFf,s> Ff,k Applied Physics Example: Anti-lock Brakes

  9. Fa FN Fa,y 37 ° Fa,x Ff Fg Example 3: Sliding “Box” (Pulled at Angle: advanced) A box is pulled at a 37° angle with increasingly applied force. The box which has a mass of 15kg begins to move when the applied force reaches 50.N. What is the coefficient of static friction between the box and the surface? Fa = Fa,x = Fa,y = Fg = FN + Fa,y = Fg FN = 120N 50.N Fa cos 37 = 40.N Fa sin 37 = 30.N mg = 150N Fa,x Ff,s = At the point where box started to move Ff,s = s FN = Fa,x = s· 120N = 40N s = .33

  10. y x Fgx  Fgy  = 30° Fg Forces on an Inclined Plane Fg is always directed straight down. We then choose a Frame of Reference where the x-axis is parallel to the incline and the y-axis perpendicular to the incline. Fg,x = Fg sin Fg,y = Fg cos FN = Fg,y (in opposite direction) Fa and Ff will be along our new x-axis

  11. y x Fgx  Fgy  = 30° Fg Example Problem(Inclined Plane) A 25.0kg box is being pulled up a 30° incline with a force of 245N. The coefficient of kinetic friction between the box and the surface is .567. Calculate the acceleration of the box. Draw a Sketch Determine the Magnitude of the forces in x and y directions 25.0 kg m = Fg = Fg,x = Fg,y = 245N (to right along x-axis) Fa = FN = Ff = mg = 25.0kg · 9.80m/s2 = 245N (down) Fg cos = 212N (up along y-axis) Fg · sin = 245N · sin30 = 123N (to left along x-axis) kFN = .567 · 212N = 120N (to left along x-axis) Fg · cos = 245N · cos30 = 212N (down along y-axis) Label Forces on your sketch Solve the Problem

  12. Solve the Problem Fa – Ff – Fg,x = 245N – 120.N – 123N = 2N FNet,x = max 2N = 25.0kg · ax FNet,x = ax = .08 m/s2 NOTE: The box may be moving up the incline at any velocity. However, at the specified conditions it will be accelerating.

  13. FT Fa Example Problem (Connected Objects – Flat Surface) Two similar objects are pulled across a horizontal surface at constant velocity. The required Fa is 350.N. The mass of the leading object is 125kg while the mass of the trailing object is 55kg. The values for k are the same for each object. Calculate k and calculate the Force of “Tension” in the connecting rope. NOTE: FT = Force of Tension is not a new type of force. It is just a specific type of applied force. • Label the forces. • Calculate the magnitude of the forces. • Solve the problem(s).

  14. FN,2 FN,1 FT m2 = 55kg m1 = 125kg Fa=350.N Ff,2 Ff,1 Fg,1 Fg,2 Fg,1 = Fg,2 = FN,1 = FN,2 = Ff,1 = Ff,2 = FNet,x = Fa = m1g = 125kg · 9.80m/s2 = 1230N (down) m2g = 55kg · 9.80m/s2 = 540N (down) 1230N (up) 540N (up) k · 1230N (left) k · 540N (left) 0N Constant Velocity a = 0m/s2 = ma Ff,1 + Ff,2 = k · 1230N + k · 540N k = .20 Fa = 350.N = k (1230N + 540N) FT = k · 540N FT = 110N

  15. Example Problem (Elevators) Two weights are connected across a frictionless pulley by weightless string. Mass of object 1 is 25.0kg. The mass of object 2 is 18.0kg. Determine the acceleration of the two objects. m1 m2 a = 1.60 m/s2 m1 accelerates down m2 accelerates up Fg,1 = Fg,2 = Fg,net = m1g = 25.0kg · 9.80m/s2 = 245N (down on right) m2g = 18.0kg · 9.80m/s2 = 176N (down on left) 245N – 176N = 69N (down on right) 69N = ma = (m1 + m2) · a 69N = (25.0kg + 18.0kg) · a

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