Presentation on theme: "6-2 Using Newton’s Laws. Mass and Weight Objects at different weights will still fall at the same rate Weight force The force of gravity acting upon an."— Presentation transcript:
Mass and Weight Objects at different weights will still fall at the same rate Weight force The force of gravity acting upon an object (F g ) Ball falling through air (neglecting air resistance) Only force acting on it is F g. Acceleration is g Newton’s 2 nd Law becomes F g = mg Both force and acceleration are downward
Scale What does a scale actually measure? Scale exerts an upward force on you Because you are not accelerating, F net = 0 Ergo, the magnitude of the spring force (F sp ) is equal to your weight, F g.
Weighing self in an elevator Your mass is 75 kg. You stand on a bathroom scale in an elevator. Going up!! Starting from rest the elevator accelerates at 2.0 m/s 2 for 2.0 sec then continues at constant speed. What is the scale reading during the acceleration? m = 75 kga = +2.0 m/s 2 t = 2.0 sec Unknown = F scale
Strategy Fnet = Fscale – Fg Rearrange: Fscale = Fnet + Fg = ma + mg F scale = ma + mg or F scale = m(a + g) = (75 kg)(2.0 m/s 2 + 9.80 m/s 2 ) = 890 N
Lifting a Bucket A 50 kg bucket is being lifted by a rope. The rope is guaranteed not to break is the tension is 500 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. Assuming the acceleration is constant, is the rope in danger of breaking? m = 50 kg,v = 3.0 m/s,vo = 0.0 m/s,d = 3.0 m Unknown = F T = ?
Strategy Net force is the vector sum of F T (positive) and F g (negative) F net = F T – Fg Rearrange: F T = F net + F g = ma + mg Because v o is zero, a = v 2 / 2d F T = m(a + g) = m(v 2 /2d + g) = (50 kg)((9.0m 2 /s 2 )/(2x3.0m) + 9.80 m/s 2 ) F T = 570 N
Friction Force Friction is minimized in solving force and motion problems But friction is everywhere!!!!! Static Friction Force exerted on one surface by the other when there is no relative motion between the two Static FF acts in response to other forces Kinetic Friction Force exerted on one surface by the other when the surfaces are in motion
Model Friction Forces Friction forces put into an equation Kinetic Friction F f,kinetic = µ k F N Usually called the coefficient of kinetic friction Static Friction 0 ≤ F f,static ≤ µ s F N Equation tells you that Static friction force can be anywhere from 0 to the maximum force is before movement
Typical Coefficients of Friction Surfaceµsµs µkµk Rubber on concrete0.800.65 Rubber on Wet concrete0.600.40 Wood on Wood0.500.20 Steel on Steel (dry)0.780.58 Steel on Steel (oil)0.150.06 Teflon on Steel0.04
Balanced Friction Forces Problems You push a 25 kg wooden box across the wooden floor at a constant speed of 1.0 m/s. How much force do you exert on the box?
Unbalanced Friction forces If the force you exerted on the box is doubled, what is the resulting acceleration on the box? m = 25 kg v = 1.0 m/s µ k = 0.20 F p = 2(49 N) = 98 N Unknown a = ???
If the force you exerted on the box is doubled, what is the resulting acceleration on the box? y-direction: F N = F g = mg; F N = mg x-direction: F P – F f = ma F f = µ k F N = µ k mg F net = ma a = F net / m = F P - µ k mg/ m = (F P /m) - µ k g a = (98 N / 25 kg) – (0.20)(9.8 m/s 2 ) a = 2.0 m/s 2
A boy exerts a 36 N horizontal force as he pulls a 52 N sled across a cement sidewalk at a constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? (Ignore air resistance)
Suppose the sled runs on packed snow. The coefficient of friction is now only 0.12. If a person weighting 650 N sits on the sled, what force is needed to pull the sled across the snow at a constant speed?
Air drag and terminal velocity This force depends upon… speed of the motion, becoming larger as speed increases Size and Shape of the object Density Kind of fluid Terminal Velocity Velocity of an object in free fall becomes equal to the drag force on the object
Periodic Motion Playground swing, guitar string, pendulum Each object has one position where F Net = 0 In equilibrium When object is pulled away from equilibrium, the net force becomes nonzero and pulls it back toward equilibrium If the force that restores the object to its eq. position is directly proportional to displaecment, motion resulted is Simple Harmonic Motion Period: time to complete one cycle of motion (T) Amplitude: maximum distance object moves from equilibrium
Pendulum Bob and suspended string (l) String exerts a tension force (FT) Gravity exerts the weight force (Fg) Vector sum of the two forces produces the net force Period of a pendulum of length l T = 2 л√ (l/g) Notice period is only determined by length of string and gravity not mass or amplitude