Lecture 7. Forces: Newton s Laws. Problem-Solving Tactics: Friction and Centripetal Motion. Physics 105; Summer How do we jump?
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1 ecture 7 Problem-Solving Tactics: Friction and Centripetal Motion (H&W, Chapters 5-6) Newton s aws I. If no net force acts on a body, then the body s velocity cannot change. II. The net force on a body is equal to the product of the body s mass and acceleration. III. When two bodies interact, the force on the bodies from each other are always equal in magnitude and opposite in direction (F 12 = -F 21 ) Force is a vector Force has direction and magnitude Mass connects Force and acceleration; Physics 105; Summer 2006 ecture 7a Andrei Sirenko, NJIT 1 ecture 7a Andrei Sirenko, NJIT 2 How do we jump? Forces: Gravitational Force: F g = mg down to the ground Tension Force: T along the string Spring Force: F s = -kx Normal Force: N perpendicular to the support Friction Force Static; maximum value f s = µ st N opposite to the component of other forces parallel to the support Kinetic; value f k = µ kin N opposite to the velocity, parallel to the support µ st > µ kin ecture 7a Andrei Sirenko, NJIT 3 ecture 7a Andrei Sirenko, NJIT 4
2 elative Motion/eference Frames elative Motion/eference Frames ecture 7a Andrei Sirenko, NJIT 5 ecture 7a Andrei Sirenko, NJIT 6 elative Motion/eference Frames Inertial Frames: T T = mg mg 48.2 degrees; v boat = 2.2 m/s ecture 7a Andrei Sirenko, NJIT 7 ecture 7a Andrei Sirenko, NJIT 8 v
3 Inertial Frame; There are no Pseudo Forces: T cos = mg T sin = ma a =g tan Non-inertial Frame Pseudo Forces a Inertial Frame a ecture 7a Andrei Sirenko, NJIT 9 ecture 7a Andrei Sirenko, NJIT 10 Non-Inertial Frame; There is a Pseudo Force: ma Newton s aws do not work!!! Combination of Forces: Net Force ma =T sin F net = F 1 + F 2 F net = ma ecture 7a Andrei Sirenko, NJIT 11 ecture 7a Andrei Sirenko, NJIT 12
4 Uniform Circular Motion Centripetal acceleration Top view: Net Force and Centripetal Force Period Centripetal force : F = ma ecture 7a Andrei Sirenko, NJIT 13 ecture 7a Andrei Sirenko, NJIT 14 Centripetal Force is a combination of: Centripetal Force and Tension Force: Gravitational Force: down to the ground Tension Force: along the string mg T ma = T ma = mv 2 / = T Normal Force: N perpendicular to the support ma = N mg ma = mv 2 / Static Friction Force maximum value F fr max = µ st N ecture 7a Andrei Sirenko, NJIT 15 ecture 7a Andrei Sirenko, NJIT 16
5 Centripetal Force and Kinetic Friction Force: Uniform Circular Motion Centripetal acceleration Period Kinetic friction does not affect Centripetal acceleration directly ecture 7a Andrei Sirenko, NJIT 17 ecture 7a Andrei Sirenko, NJIT 18 Uniform Circular Motion Centripetal acceleration Period: Problem #1 = 2 m Find v, T, and a v = 100 m /12 s = 8.33 s; = 100/π = 31.8 m a = (8.33) 2 /31.8 m/s 2 = 2.2 m/s 2 ecture 7a Andrei Sirenko, NJIT 19 ecture 7a Andrei Sirenko, NJIT 20
6 Problem solving tactics: = 2 m Find v,t, and a Problem solving tactics: = 2 m Find v,t, and a sin = / = 0.4; tan = (/)/ (1-(/) 2 ) ½ = 0.44 X: ma = T * sin Y: ma = 0 = - mg + T * cos ecture 7a Andrei Sirenko, NJIT 21 ecture 7a Andrei Sirenko, NJIT 22 Problem solving tactics: sin = / = 0.4; tan = (/)/ (1-(/) 2 ) ½ = 0.44 X: ma = T * sin Y: ma = 0 = - mg + T * cos ma = mg*sin / cos = mg*tan T = mg/ cos a = g*tan = 2 m Find v,t, and a Centripetal Force originates from the tension force! Circular motion: ma = mv 2 / a = v 2 / v = (a) ½ ecture 7a Andrei Sirenko, NJIT 23 Problem is solved: sin = / = 0.4; tan = (/)/ (1-(/) 2 ) ½ = 0.44 X: ma = T * sin Y: ma = 0 = - mg + T * cos ma = mg*sin / cos = mg*tan T = mg/ cos a = g*tan T = 5 kg *9.8 m/s 2 /(1-(2m/5m) 2 ) ½ = 53 N a = 4.3 m/s 2 ; ma = 5 kg * 4.3 m/s^2 = 21 N; = 2 m Find v,t, and a Circular motion: ma = mv 2 / a = v 2 / v = (a) ½ v = (4.3*2) ½ m/s = 2.9 m/s ecture 7a Andrei Sirenko, NJIT 24
7 Net Force and Centripetal Force Problem #2 = 20 m; µ st = 0.5 ecture 7a Andrei Sirenko, NJIT 25 ecture 7a Andrei Sirenko, NJIT 26 Problem #2 = 20 m; µ st = 0.5 Problem #2 = 20 m; µ st = 0.5 X: ma = N sin + µ st N cos Y: 0 = N cos -mg-µ st N sin ecture 7a Andrei Sirenko, NJIT 27 ecture 7a Andrei Sirenko, NJIT 28
8 Problem #2 X: ma = N *(sin + µ st cos ) Y: N = mg / (cos - µ st sin ) ma = mv 2 / = 20 m; µ st = 0.5 Problem #2 ma = N *(sin + µ st cos ) N = mg / (cos - µ st sin ) ma = m v max 2 / m v max 2 / = mg *(sin + µ st cos )/ (cos - µ st sin ) v max2 = g* *(sin + µ st cos )/ (cos - µ st sin ) Mass m disappeared!!! v max 12.8 m/s 29 ml/h = 20 m; µ st = 0.5 ecture 7a Andrei Sirenko, NJIT 29 ecture 7a Andrei Sirenko, NJIT 30 QZ # 7 Analyze the previous problem 1. = 20 m; µ st = 0.5; What is going to happen to the static friction force for the case when the velocity of the track is doubled: v = v max * 2 What is going to happen to the track? (describe) 2. = 20 m; µ st = 0.5, v=5 m/s; m=3000 kg, and = 0 What is the value and direction of the static friction force? ecture 7a Andrei Sirenko, NJIT 31 ecture 7a Andrei Sirenko, NJIT 32
9 ecture 7a Andrei Sirenko, NJIT 33 ecture 7a Andrei Sirenko, NJIT 34
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