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Transcript
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. “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 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: 1. Strike golf ball w/golf club 2. 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 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 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 Example: (Contact Force) Book Table 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. Example: (Field Force) Earth F Moon FEarth Moon Earth pulls on Moon equal to the force the Moon pulls on Earth. 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 FN Fa Ff Fg 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 Label forces on Sketch Ff < Fa Solve Problem 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? FN=540N Ff=15N Fa=25N Fg=540N Fa = 25N Ff = 15N FNet,x = 25N – 15N = ma = 10.N = 55kg·a a = .18m/s2 Fg = mg = 55kg · 9.8m/s2 = 540N FN = Fg = 540N 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? 59N= FN 18N= Ff Fa,y =15N Fa,x =20N Fg =74N m = 7.5kg Fa = 25N Fa,x = Fa cos37 = 20.N Fa,y = Fa sin 37 = 15N FN + Fa,y = Fg Ff = 18N (up forces equal down forces) FNet,x = Fa,x – Ff = ma Fg = mg = 74N 20.N – 18N = 7.5kg · a FN = 74N – 15N a = .27m/s2 FN = 59N 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 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? FN 37 ° Fa,y Fa,x Ff Fg Fa = 50.N Fa,x = Fa cos 37 = 40.N Fa,y = Fa sin 37 = 30.N Fg = mg = 150N FN + Fa,y = Fg FN = 120N Ff,s = Fa,x At the point where box started to move Ff,s = s FN = Fa,x = s· 120N = 40N s = .33 Forces on an Inclined Plane y Fgx x Fgy = 30° Fg 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 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. y x Draw a Sketch Fgx Fgy = 30° Fg Determine the Magnitude of the forces in x and y directions m = 25.0 kg Fa = 245N (to right along x-axis) 2 Fg = mg = 25.0kg · 9.80m/s FN = Fg cos = 212N = 245N (down) (up along y-axis) Fg,x = Fg · sin = 245N · sin30 Ff = kFN = .567 · 212N = 120N = 123N (to left along x-axis) (to left along x-axis) Fg,y = Fg · cos = 245N · cos30 = 212N (down along y-axis) Label Forces on your sketch Solve the Problem Solve the Problem FNet,x = Fa – Ff – Fg,x = 245N – 120.N – 123N = 2N FNet,x = max 2N = 25.0kg · ax 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. Example Problem (Connected Objects – Flat Surface) FT Fa 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). FN,2 m2 = 55kg Ff,2 FN,1 FT m1 = 125kg Ff,1 Fg,1 Fg,2 Fg,1 = m1g = 125kg · 9.80m/s2 = 1230N (down) Fg,2 = m2g = 55kg · 9.80m/s2 = 540N (down) FN,1 = 1230N (up) FN,2 = 540N (up) Ff,1 = k · 1230N (left) Ff,2 = k · 540N (left) FNet,x = 0N Constant Velocity a = 0m/s2 = ma Fa = Ff,1 + Ff,2 = k · 1230N + k · 540N Fa = 350.N = k (1230N + 540N) FT = k · 540N FT = 110N k = .20 Fa=350.N Example Problem (Elevators) m1 m2 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. Fg,1 = m1g = 25.0kg · 9.80m/s2 = 245N (down on right) Fg,2 = m2g = 18.0kg · 9.80m/s2 = 176N (down on left) Fg,net = 245N – 176N 2 a = 1.60 m/s = 69N (down on right) m1 accelerates down 69N = ma = (m1 + m2) · a m2 accelerates up 69N = (25.0kg + 18.0kg) · a