6: Applications of Newton's Laws


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1 6: Applications of Newton's Laws Friction opposes motion due to surfaces sticking together Kinetic Friction: surfaces are moving relative to each other a.k.a. Sliding Friction Static Friction: surfaces are not moving relative to each other. Static Friction prevents stationary objects from moving until sufficient force has been applied. Friction Applied Force Coefficient of Friction Frictional forces depend upon how hard the surfaces are being pressed together > force perpendicular to the surface = normal force the types of surfaces that are in contact > coefficient of friction Kinetic Friction Static Friction f k = k N f s, max = s N f s s N p250c6:1
2 Typical coefficients of friction Materials kinetic, m k static, m S Rubber on concrete (dry) Rubber on concrete (wet) Steel on Steel Glass on glass Wood on leather Copper on steel Steel on ice Waxed ski on snow Teflon on Teflon Example: A 50.0 g saltshaker is slid down a counter top with an initial speed of 1.15 m/s whereupon it comes to rest in a distance of m. What is the coefficient of kinetic friction between the shaker and the table? p250c6:2
3 Example: A sea lion slides from rest down an 3.00 m long ramp into a pool of water which is inclined at an angle of 23. If the coefficient of kinetic friction is 0.26, how long does the sea lion's ride last? p250c6:3
4 Example: A flatbed truck slowly tilts back. At an angle of 23.2 a crate on the flatbed begins to slide. What is the coefficient of static friction between the bed and the truck? p250c6:4
5 From the previous example, what is the force of static friction when the truck bed is level? When it is at 20? Thought Question: What type of friction occurs between the tires of a car and the road when operating normally? p250c6:5
6 Strings and springs Tension: the force necessary to hold a string (rope, cable, etc) taut. (frictionless, low mass) pulleys do not reduce tension. T T both sides of rope contribute to tension on pulley T What are the readings? add forces as vectors! T T 1.00 kg 1.00 kg 1.00 kg 1.00 kg p250c6:6
7 Force: Hooke s Law for elastic deformations stretch or compress F = kx F x = k x unstretched half stretched stretched x F = 0 F = k x/2 F = k x Example: A 1.50 kg object is hung from a spring with spring constant k = 250 N/m. How far is the spring stretched from its equilibrium length? p250c6:7
8 Equilibrium Static Equilibrium: no acceleration with bodies at rest. Dynamic Equilibrium: no acceleration with moving bodies Equilibrium implies F i = 0 so that F xi = 0 and F yi = 0 Connected Objects use actionreaction objects will have related changes in motion chose coordinates so that velocity and acceleration are the same p250c6:8
9 Example: Find expressions for the acceleration of the system and the tension in the string. Find values if m 1 is 1.50 kg and m 2 is kg. m 1 T=F 1 1 (+) m 2 2 (+) T w w 1 T=F 2 =m 1 gt p250c6:9
10 T (+) (+) w 2 T=F 2 w 1 1 m 1 = 3.1kg 2 m 2 = 4.4kg p250c6:10
11 A curved path requires an inward force Center seeking = Centripetal v v v Centripetal force is the force perpendicular to the velocity of an object moving along a curved path. The centripetal force is directed toward the center of curvature of the path. v = a t = (F/m)dt examples: ball on a string, car rounding a corner. Centrifugal Effect: the fictitious force felt by an object when the frame of reference moves along (and therefore accelerates) along a curved path. This effect is simply inertia. Stop the force and the object will undergo straight line motion. p250c6:11
12 Uniform Circular Motion motion in a circle at constant speed centripetal force F c and centripetal acceleration a c is always directed towards the center centripetal force and acceleration have constant magnitudes a cp v r 2 F cp ma cp mv r 2 p250c6:12
13 A 1200kg car rounds a corner with radius r = 45m. If the static friction between the tires and the road is m s = 0.82, what is the greatest speed the car can have in the corner without skidding? phys p250c6:13
14 A roadway is banked at the proper angle so that a car, without any assistance from friction, can round the corner safely. Find the appropriate banking for a 900 kg car traveling at 20.5 m/s in a turn of radius 85.0 m/s. phys p250c6:14
15 While driving along at constant speed of 17.0 m/s you encounter a dip in the road. The dip can be approximated as a circular arc, 65.0 m in radius. What is the normal force exerted by a car seat on a 80.0 kg passenger at the bottom of the dip? How does this compare to the weight of the passenger? p250c6:15
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