1 / 15

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

Download Presentation

CHAPTER 4 The Laws of Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related