CHAPTER 4 The Laws of Motion Newton’s First Law: 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” “at rest” – with respect to the chosen frame of reference “net force” “net force” – vector sum of all the external forces acting on the object – F Net,x and F Net,y calculated separately Forces: 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: Newton’s Second Law:The acceleration of an object is directly proportional to the net force acting on it F Net = 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/s 2 = newton = N 1 newton = 1 kg · m/s 2

Weight and the Gravitational Force Mass Mass – an amount of matter (“gravitational mass”) “Your mass on the Moon equals your mass on Earth.” Weight Weight – the magnitude of the force of gravity acting on an amount of matter F = ma F g = 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/s 2 w = 780 kg·m/s 2 w = 780 N

Newton’s Third Law: 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: 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: Example: (Field Force) Earth F Moon Moon F Earth 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 F a Normal ForceF N Frictional ForceF f Gravitational Force F g Draw a Sketch FNFN FaFa FgFg FfFf Determine the Magnitude of Forces in “x” and in “y” Direction F N often equals F g (object does not accelerate up off surface or accelerate downward through surface) F Net,y = F N – F g = 0 N F Net,x = F a – F f = ma F f < F a Label forces on Sketch Solve Problem

Example 1: Example 1: Sliding “Box” Problem (Horizontal F a ) “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? F a =25N F N =540N F g =540N F f =15N F a = 25N F f = 15N F Net,x = 25N – 15N = ma = 10.N = 55kg·a a =.18m/s 2 F g = mg = 55kg · 9.8m/s 2 = 540N F N = F g = 540N

Example 2: 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? m = 7.5kg F a = 25N F a,x = F a cos37 = 20.N F a,y = F a sin 37 = 15N F f = 18N F Net,x = F a,x – F f = ma 20.N – 18N = 7.5kg · a a =.27m/s 2 F N + F a,y = F g (up forces equal down forces) F g = mg = 74N F N = 74N – 15N F N = 59N =74N =15N =20N 59N= 18N= F a,x F a,y FNFN FgFg FfFf FaFa =25N

Friction Friction opposes motion. Kinetic Friction opposes motion of a moving object. Static Friction opposes motion of a stationary object. F f =  F N  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  F f,s > F f,k Applied Physics Example: Anti-lock Brakes

Example 3: 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? FNFN FgFg FfFf FaFa F a,x F a,y 37 ° F a = F a,x = F a,y = F g = F N + F a,y = F g F N = 120N F f,s = At the point where box started to move F f,s =  s F N = F a,x =  s · 120N = 40N  s = N F a cos 37 = 40.N F a sin 37 = 30.N mg = 150N F a,x

Forces on an Inclined Plane F g 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. F g,x = F g sin  F g,y = F g cos  F N = F g,y (in opposite direction) F a and F f will be along our new x-axis  = 30°  FgFg F gy F gx x y

Example Problem 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 F g · sin  = 245N · sin30 = 123N (to left along x-axis) Label Forces on your sketch Solve the Problem  k F N =.567 · 212N = 120N (to left along x-axis) Determine the Magnitude of the forces in x and y directions  = 30°  FgFg F gy F gx xy m = F g = F g,x = F g,y = mg = 25.0kg · 9.80m/s 2 = 245N (down) 25.0 kg F g · cos  = 245N · cos30 = 212N (down along y-axis) F a = F N = F f = 245N (to right along x-axis) F g cos  = 212N (up along y-axis)

Solve the Problem F a – F f – F g,x = 245N – 120.N – 123N = 2N F Net,x = ma x 2N = 25.0kg · a x a x =.08 m/s 2 NOTE: The box may be moving up the incline at any velocity. However, at the specified conditions it will be accelerating. F Net,x =

Example Problem (Connected Objects – Flat Surface) Two similar objects are pulled across a horizontal surface at constant velocity. The required F a 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: F T = 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). FTFT FaFa

F g,1 = F g,2 = F N,1 = F N,2 = F f,1 = F f,2 = F Net,x = F a =  k =.20 F T = F T = 110N FTFT F f,1 m 1 = 125kg F g,1 F N,1 F a =350.N F f,2 m 2 = 55kg F g,2 F N,2 m 1 g = 125kg · 9.80m/s 2 = 1230N (down) m 2 g = 55kg · 9.80m/s 2 = 540N (down) 1230N (up) 540N (up)  k · 1230N (left)  k · 540N (left) 0NConstant Velocity a = 0m/s2 = ma F f,1 + F f,2 =  k · 1230N +  k · 540N F a = 350.N =  k (1230N + 540N)  k · 540N

Example Problem Example Problem (Elevators) m1m1 m2m2 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. m 1 g = 25.0kg · 9.80m/s 2 = 245N (down on right) m 2 g = 18.0kg · 9.80m/s 2 = 176N (down on left) 245N – 176N = 69N (down on right) a = 1.60 m/s 2 m 1 accelerates down m 2 accelerates up 69N = ma = (m 1 + m 2 ) · a 69N = (25.0kg kg) · a F g,1 = F g,2 = F g,net =