Lecture 6. Applying Newton s Laws Free body diagrams Friction
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1 Lecture 6 Applying Newton s Laws Free body diagrams Friction
2 ACT: Bowling on the Moon An astronaut on Earth kicks a bowling ball horizontally and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. His foot hurts: A. More B. Less C. The same
3 EXAMPLE: Box hanging from the ceiling. A box of mass m hangs from the ceiling. Determine the tension on the string.
4 EXAMPLE: Box hanging from the ceiling. A box of mass m hangs from the ceiling. Determine the tension on the string. T box,string T W = ma = 0 T = W = mg W box,earth
5 ACT: Fishing A fish is being yanked upward out of the water with a line that can stand a maximum tension of 180 N. The string snaps when the acceleration of the fish is 8 m/s 2. What is the mass of the fish? A. 8 kg snap! B. 10 kg C. 18 kg a = 8 m/s 2
6 A little more difficult: two boxes. String 1 String 1 T s1,ceiling T s1,u String 2 String 2 T s2,u T s2,l
7 A little more difficult: two boxes. String 1 Upper box: String 1 T s1,ceiling T 1 T 2 W U = m U a = 0 T U,s2 String 2 T U,s1 W U,E T s1,u String 2 T 1 = W U + T 2 = W U + W L = (m U + m L ) g Lower box: T L,s2 T s2,u T 2 W L = m L a = 0 T 2 = W L = m L g W L,E T s2,l
8 What if the boxes hang from the ceiling of an accelerated elevator? String 1 T U,s1 T U,s2 String 2 W U,E a T L,s2 W L,E
9 What if the boxes hang from the ceiling of an accelerated elevator? String 1 Upper box: T U,s1 T 1 T 2 W U = m U a T U,s2 T L,s2 String 2 W U,E a Lower box: T 2 W L = m L a Simply keep the a in the equations! W L,E
10 ACT: Sharing the weight. In both of the cases depicted below, a 2-kg weight is supported by two strings. In which case are the students holding the strings exerting a force of greater magnitude? 1. Case A 2. Case B A 3. Both the same B
11 Cable example Find the tension in the cables. θ θ m
12 Cable example Find the tension in the cables. θ T 1 m T 2 θ y mg x x : T 1 cosθ T 2 cosθ = 0 T 1 = T 2 T y : T 1 sinθ + T 2 sinθ mg = 0 2T sinθ = mg T mg 2sin Small θ, large T It is impossible for a real cable (m > 0) to be completely horizontal (it would require infinite tension, and then the cable snaps).
13 N by pivot Atwood s Machine Reasonable direction of motion: Take + in this direction T T 2mg T 2ma T mg ma T = m (a + g) (Mg = 0) T T m 2m mg 2mg 2 mg m( a g ) 2ma g 4 a ; T mg 3 3 mg T 2mg Note: Net force is in the right direction for each box!
14 ACT: A weight vs. a hand In figure 1, a 10-kg mass hangs from a string and pulls on a box of mass m. In figure 2, a hand provides a constant downward force of 98.1 N and pulls on another box of mass m. The pulleys and strings are all ideal and massless. Where does the box experience a larger acceleration? m m 10 kg Fig. 1 T 2 = 98.1 N Fig. 2 A. In figure 1. B. In figure 2. C. It s the same in both.
15 A. Mg B. Mg/2 Example: Pulley How much force does the worker have to exert in order to support the mass M at constant height h off the ground? C. Mgh D. Mg/h E. Mg/(2h) h M
16 A. Mg B. Mg/2 Example: Pulley How much force does the worker have to exert in order to support the mass M at constant height h off the ground? C. Mgh D. Mg/h E. Mg/(2h) h M
17 orange pulley: T 2T green pulley: T 2T mass M: T Mg 0 3 T 1 T 2 T 1 T T Mg 2 T T 1 T 3 T 1 T 3 T 1 Mg 2 Mg
18 Optional: Example: Pulleys A sack of weight w hangs motionless from a system of pulleys. All ropes and pulleys are massless. What is the magnitude of the force is exerted by the worker?
19 T 1 T T T 3 T T T T T 4 T 2 T 4 T T w T 2 T 1 = T 2 = T 3 = T 4 = 2T w = 4T T 4 w T 2 + T 4 = w T?
20 In practice, just count the number of ropes providing support to hanging pulleys. T 4 w DEMO: Pulleys
21 Kinetic and Static Friction A friction is a force that resists relative motion of two bodies in contact with each other. The origin of friction is microscopic irregularities of a surface. Force of friction is always parallel to the surface. Static friction exits between two objects which have a contact and do not move relative one another. Kinetic friction occurs between two objects which have a contact and are moving relative one another.
22 Static Friction Static friction on an object by a surface is opposite in direction to tendancy of motion of the object relative to the surface. The kinetic friction force on an object by can take any value between zero and a maximu value, f s,max. f s μ s N Examples: Allows us to walk, accelerates and decelerates cars (when no skidding), A box on an incline not sliding, etc.
23 Friction f k = μ k N f s μ s N For any two material: μ k μ s Materials Kinetic, μ k Static, μ s Rubber on concrete (dry) Steel on Steel Copper on Steel Rubber on Concete (wet) Steel on ice Teflon on Teflon
24 Example: Trying to move a trunk friction μ s N μ k N F by you
25 Example: Trying to move a trunk friction μ s N F by you f S by floor μ k N F by you
26 Example: Trying to move a trunk friction μ s N F by you f S by floor μ k N F by you
27 Example: Trying to move a trunk friction μ s N F by you f S by floor μ k N F by you
28 Example: Trying to move a trunk friction μ s N F by you f S by floor μ k N F by you
29 Example: Trying to move a trunk friction μ s N F by you f k by floor μ k N F by you
30 Example: Trying to move a trunk friction μ s N F by you f k by floor μ k N F by you Static friction Kinetic friction
31 Car The static friction between the tire and the road pushes the car forward when speeding up without slipping. Rotation of the tire DEMO: Tire Static friction on tire by road
32 Car Speeding up too much so tires slip, then kinetic friction pushes the car forward. Rotation of the tire kinetic friction on tire by road
33 EXAMPLE: Box on incline with friction A box slides down an incline with angle θ = 45. The coefficient of kinetic friction between the box and the plane is μ k = 0.2. Find the acceleration of the box. θ
34 x: W x f k = ma Y: N W y = 0 mg sinθ μ k N = ma N mg cosθ = 0 mg sinθ μ k mg cosθ = ma y a = g (sinθ μ k cosθ ) = 5.5 m/s 2 N B,I x f k B,I W x W y θ W B,E θ
35 Example: Incline and pulley Find the acceleration of the boxes when the system below is released. Friction is negligible. 2m m 35
36 Example: Incline and pulley Find the acceleration of the boxes when the system below is released. Friction is negligible. 2m Same acceleration for both (they move together) m 35
37 1. Draw free body diagram for both boxes. 2. Select axes 3. Write Newton s 2 nd law T N y T mg ma { 2mg sinθ T =2ma N 2mg cosθ = 0 x T m mg 2m 2mg 35 x
38 4. Solve equations { 2 mg sin θ T =2 ma T mg =ma 2 mg sinθ mg = 3 ma g (2 sinθ 1) = 3 a a g 3 2sin 1 = g 3 (2 sin 35 1) = g = 0.48 m/s2 a If θ <30, a < 0 x m 2m θ y x
39 EXAMPLE: Incline and pulley, with friction Same system, but μ s = 0.2 and μ k = 0.1 What happens when the system is released? 2m m 35
40 Does the system move at all? Net force, along the direction of motion, without friction: F net = 2 mg sin 35 mg =0.15 mg Maximum static friction force: f S,MAX S N = μ s 2 mg cos mg This equals to the needed friction that keeps the box from sliding (setting a = 0 in the previous example)! Motion without friction f S f s < f s,max N y It does not move! T 2m x T m 2mg 35 x μ S = 0.2 μ k = 0.1 mg
41 ACT: Magnitude of friction What is the magnitude of the static force in the system we just studied? A mg B mg C mg f S N F f net, no friction S,max T 2m T m 2mg 35 mg 0.33mg 0.15mg
42 ACT: Book against wall When you push a book against a wall, the static friction between the wall and the book can prevent it from falling. If you press harder, the friction force will be: A. Larger than before B. The same C. Smaller than before.
43 Example: Box on truck A box with mass m = 50 kg sits on the back of a truck. The coefficients of friction between the box and the truck are μ k = 0.2 and μ s = 0.4. What is the maximum acceleration that the truck can have without the box slipping? A. 2.0 m/s 2 B. 3.1 m/s 2 C. 3.9 m/s 2 D. 4.9 m/s 2 E. 9.8 m/s 2 a m
44 a f s B,T N B,T Direction of motion relative to the truck in the absence of friction Not slipping: static W B,E f s = m B a N-W = 0 F s,max = m B a MAX a max f S,MAX m B SN m B SmB g m B S g 0.4g 3.9 m/s Answer C 2
45 EXAMPLE: Is N = mg always? A box of mass m = 1.5 kg is being pulled by a string with tension T = 45 N. The string makes an angle θ = 15 with the horizontal. What is the normal force on the box by the floor? θ T m
46 EXAMPLE: Is N = mg always? A box of mass m = 1.5 kg is being pulled by a string with tension T = 45 N. The string makes an angle θ = 15 with the horizontal. What is the normal force on the box by the floor? N T T θ T y = T sinθ θ m T x = T cosθ FBD mg Y direction: F y = m a y N +T sin θ mg = 0 N = mg T sin θ ( <mg) =(1.5 kg)(9.80 m/s 2 ) (45 N sin 15 ) = 3.05 N mg = (1.5 kg)(9.80 m/s 2 ) = 14.7 N
47 EXAMPLE: Box on another box A box of mass m 1 = 1.5 kg is being pulled by a horizontal string with tension T = 45 N. It slides with friction (μ k = 0.50, μ s = 0.70) on top of a second box of mass m 2 = 3.0 kg, which in turn sits on a frictionless floor. Find the acceleration of box 2. T a 1 m 1 μ s = 0.5 a 2? m 2 frictionless
48 T m 1 N f k m 2 f k m 1 g For box 2: f K = μ K N = m 2 a 2 From box 1, we know that N - m 1 g = 0 a 2 N m 1.5 kg k k g 0.50 (9.8 m/s ) 2.5 m/s m2 m2 3.0 kg
49 The magnitude of the tension did not play any role! EXCEPT that the tension just needs to be large enough so the boxes cannot move together. If they moved together, the acceleration of both blocks would be: T 45 N 10 m/s 2 a m 1 m m 4.5 kg T 1 2 The static friction would be the only horizontal force on m 2 : F on m2 = f s = m 2 a = (3.0 kg)(10 m/s 2 ) = 30 N But static friction has a maximum value: f s,max =μ s N = μ s m 1 g ( where N =m 1 g ) m 2 f s = 0.7 (1.5 kg) (9.8 m/s 2 ) = 10.3 N f s N m 1 g f s > f s,max!!!! Since f s required for locking blocks together is larger than the Maximum value friction can provide, Blocks will not stick together in motion for the given value of T
50 ACT: Force and acceleration (2) Two blocks of masses m and 2m are pushed together along a horizontal, frictionless surface by a force F. The magnitude of the net force on block B is: A. 1/3 F B. 2/3 F C. F F m A 2m B
51 Newton s Third Law For every force, or action there is an equal but opposite force, or reaction. Forces ALWAYS happen in pairs. F F AB BA
52 Newton s third law: Action and Reaction Example: Gravitation. You attract the Earth! F g you,earth F g you,earth = F g Earth, you F g Earth, you But the acceleration that this produces on the Earth is a 2 your weight (70 kg)(10 m/s ) m/s, 24 M 6 10 kg Earth (nothing to be too proud of )
53 Leaning on a wall N wall, person N person,wall
54 Rocket Ship Newton s third law implies that if a rocket accelerates forwards, something must be pushed backwards. In outer space, there isn t much else around besides its own fuel. rocket f
55 A Book on a Table Normal on book by table Are forces shown Newton pair?
56 Normal on table by book Normal on book by table
57 Book on Table The full story W TE W ET N BT W BE N TE N TB W EB N ET Action-Reaction Pairs Normal force between book and table N BT = N TB Gravitational force between book and earth W BE = W EB Normal force between table and earth N TE = N ET Gravitational force between table and earth W TE = W ET The book does not accelerate W BE +N BT =0 The table does not accelerate W TE +N TB +N TE =0 Does the earth accelerate?
58 Free Body Diagram It is a diagram with all the forces acting on one object. one all N BT N TE W BE W TE N TB You should always draw a free-body diagram before attempting an application of Newton s second law!!! * * This instructor declines all responsibility for a failed question and will disregard any whining if a free-body diagram has not been drawn.
59 Example: Apparent weight John has a mass of 100 kg and is standing on a scale in an elevator which is accelerating upwards from rest at 2 m/s². What will the scale read? What does a scale measure? The magnitude of the normal force on the scale by John, N JS = N SJ N on John, by scale a (not part of John s free body diagram) W on John, by Earth Newton s 2nd law on John: JS JE J N W m a N on scale, by John John moves rigidly with the elevator
60 John has a mass of 100 kg and standing on a scale in an elevator which is accelerating upwards from rest at 2 m/s². What will the scale read? N W m a JS JE JS J J J N m g m a N m g m a JS J J J m g a 2 2 (100 kg)(9.8 m/s 2.0 m/s ) 1180 N N on John, by scale a W on John, by Earth If the scale is in kg, it will read: N 1180 N kg JS g 9.8 m/s Check: When the elevator is at rest (a = 0), the scale must read the correct weight, 100 kg (980N). Note: In Physics or at least in this course-, the word weight refers to mg, not to what a scale reads.
61 ACT: Force and acceleration (3) Two blocks of masses m and 2m are pushed together along a horizontal, frictionless surface by a force F. The magnitude of the force on block A by block B is: A. 1/3 F F m A 2m B B. 2/3 F C. F
62 ACT: Force and acceleration (3) Two blocks of masses m and 2m are pushed together along a horizontal, frictionless surface by a force F. The magnitude of the force on block A by block B is: A. 1/3 F F m A 2m B B. 2/3 F C. F
63 Example: Box on an incline A hand keeps a 35-kg box from sliding down a frictionless incline. The plane of the incline makes an angle θ = 25 with the horizontal. What is the magnitude of the force exerted by hand? A. 35 N B. 311 N C. 343 N D. 145 N E. 100 N θ
64 Example: Box on an incline A hand keeps a 35-kg box from sliding down a frictionless incline. The plane of the incline makes an angle θ = 25 with the horizontal. What is the magnitude of the force exerted by hand? Draw the free-body diagram A. 35 N B. 311 N C. 343 N D. 145 N Choose axes (draw them!) Use Newton s 2 nd law in the x and y-directions. E. 100 N θ
65 A hand keeps a 35-kg box from sliding down a frictionless incline. The plane of the incline makes an angle θ = 25 with the horizontal. What is the magnitude of the force exerted by hand? W x F = m a x = 0 N - W y = m a y = 0 y F B,hand N B,I F = W x = mg sinθ N = W y = mg cosθ F = mg sinθ = (35 kg)(9.8 m/s 2 )sin(25 ) = 145 N (Answer D) x W x = mg sinθ mg cosθ = W y θ θ W B,E
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