1. Physics 201: Lecture 9, Pg 1 Lecture 8 l Goals:  Solve 1D & 2D problems introducing forces with/without friction  Utilize Newton’s 1 st & 2 nd Laws  Begin to use Newton’s 3 rd Law in problem solving

2. Physics 201: Lecture 9, Pg 2 Newton’s Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, F NET =   F = ma Law 3: Forces occur in pairs: F A, B = - F B, A (For every action there is an equal and opposite reaction.) Read: Force on A by B

3. Physics 201: Lecture 9, Pg 3 Exercise Newton’s 3rd Law A. 2 B. 4 C. 6 D. Something else a b l Two blocks are being pushed by a massless finger on a horizontal frictionless floor. l How many action-reaction force pairs are present in this exercise?

4. Physics 201: Lecture 9, Pg 4 Newton’s Law The acceleration of an object is directly proportional to the net external force acting upon it. The constant of proportionality is the mass. l This is a vector expression

5. Physics 201: Lecture 9, Pg 5 Analyzing Forces: Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. Eat at Bucky’s A hanging sign is an example of static equilibrium (depends on observer) What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ?

6. Physics 201: Lecture 9, Pg 6 Free Body Diagram Step two: Sketch in force vectors Step three: Apply Newton’s 1 st & 2 nd Laws (Resolve vectors into appropriate components)  mgmg  T1T1 T2T2 Eat at Bucky’s   T1T1 T2T2 mgmg Step one: Define the system

7. Physics 201: Lecture 9, Pg 7 Free Body Diagram Eat at Bucky’s   T1T1 T2T2 mgmg Vertical : y-direction 0 = -mg + T 1 sin  1 + T 2 sin  2 Horizontal : x-direction 0 = -T 1 cos  1 + T 2 cos  2

8. Physics 201: Lecture 9, Pg 8 Pushing and Pulling Forces l String or ropes are examples of things that can pull l You arm is an example of an object that can push or push

9. Physics 201: Lecture 9, Pg 9 Scale Problem l You are given a 5.0 kg mass and you hang it directly on a fish scale and it reads 50 N (g is 10 m/s 2 ). l Now you use this mass in a second experiment in which the 5.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale. l What force does the fish scale now read? 5.0 kg 50 N ? 5.0 kg

10. Physics 201: Lecture 9, Pg 10 Scale Problem l Step 1: Identify the system(s). In this case it is probably best to treat each object as a distinct element and draw three force body diagrams.  One around the scale  One around the massless pulley (even though massless we can treat is as an “object”)  One around the hanging mass l Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.) ? 5.0 kg

11. Physics 201: Lecture 9, Pg 11 Scale Problem   F y = 0 in all cases 1: 0 = -2T + T ’ 2: 0 = T – mg  T = mg 3: 0 = T ” – W – T ’ (not useful here) Substituting 2 into 1 yields T ’ = 2mg = 100 N (We start with 50 N but end with 100 N) ? 5.0 kg ? 1.0 kg -mg T -T -T ’ T”T” W 1: 2: 3: T ’

12. Physics 201: Lecture 9, Pg 12 Newton’s 2 nd Law, Forces are conditional A. P + C < W B. P + C > W C. P = C D. P + C = W A woman is straining to lift a large crate, without success. It is too heavy. We denote the forces on the crate as follows: P is the upward force being exerted on the crate by the person C is the contact or normal force on the crate by the floor, and W is the weight (force of the earth on the crate). Which of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: force up is positive & down is negative)

13. Physics 201: Lecture 9, Pg 13 Another experiment A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant force of 10 N. What is the acceleration of the block if the table top is frictionless?  F x = ma x = -T  F y = 0 = N - mg a x = -T/m = 5 m/s 2 m1m1 T mg FBD N

14. Physics 201: Lecture 9, Pg 14 Version 2 A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant horizontal force of 10 N. The surface is frictionless and it is inclined 30 ° from horizontal. What is the acceleration of the block down the slide?  F x = ma x = T x +W x  F y = 0 = N + T y +W y ma x = -T cos 30 ° +W sin 30 ° a x = -T/m 3 ½ /2 - g / 2 a x = -5(1.732)/2- 5 = -7.2 m/s 2 What is the apparent weight of the block on the slide? N = T y +W y = T sin 30 ° -W cos 30 ° N = -10(0.5) + 20(0.866) = 12.3 N m T mg = W FBD N i j 30 °

15. Physics 201: Lecture 9, Pg 15 Version 3 A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant horizontal force of 34.6 N. The surface is frictionless and it is inclined 30 ° from horizontal. What is the acceleration of the block down the slide?  F x = ma x = T x +W x  F y = 0 = N + T y +W y ma x = -T cos 30 ° +W sin 30 ° a x = -T/m 3 ½ /2 - g / 2 a x = -17.3(1.732)/ 2- 5 = -35 m/s 2 What is the apparent weight of the block on the slide? N = T y +W y = T sin 30 ° -W cos 30 ° N = -34.6(0.5) + 20(0.866) = 0.0 N !!! m T mg = W FBD N i j 30 °

16. Physics 201: Lecture 9, Pg 16 Another experiment: A modified Atwood’s machine Two blocks, m 1 & m 2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward. What is the acceleration of the horizontal block? Requires two FBDs and Newton’s 3 rd Law. m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2  F x = m 2 a 2x = -T  F y = 0 = N – m 2 g Mass 1  F y = m 1 a 1y = T – m 1 g Correlated motion: |a 1y | = | a 2x | ≡ a If m 1 moves up moves m 2 right

17. Physics 201: Lecture 9, Pg 17 Another experiment: A modified Atwood’s machine Two blocks, m 1 & m 2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward. What is the acceleration of the horizontal block. Requires two FBDs and Newton’s 3 rd Law. m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2 B.  F x = m 2 a = -T  F y = 0 = N – m 2 g Mass 1 A.  F y = m 1 a= T – m 1 g Subbing in T from B into A m 1 a= - m 2 a – m 1 g m 1 a + m 2 a = m 1 g a = m 1 g / (m 1 + m 2 )

18. Physics 201: Lecture 9, Pg 18 A “special” contact force: Friction l What does it do?  It opposes motion (velocity, actual or that which would occur if friction were absent!) l How do we characterize this in terms we have learned?  Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)! mama F F APPLIED f f FRICTION gmggmg N i j

19. Physics 201: Lecture 9, Pg 19 If no acceleration l No net force l So frictional force just equals applied force l Key point: It is conditional! F F APPLIED f f FRICTION gmggmg N i j

20. Physics 201: Lecture 9, Pg 20 Friction... l Friction is caused by the “microscopic” interactions between the two surfaces:

21. Physics 201: Lecture 9, Pg 21 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, As F increases so does f s F m1m1 FBD fsfs N mg  F x = 0 = -F + f s  f s = F  F y = 0 = - N + mg  N = mg

22. Physics 201: Lecture 9, Pg 22 Static friction, at maximum (just before slipping) Equilibrium: A block, mass m, with a horizontal force F applied, N Direction: A force vector  to the normal force vector N and the vector is opposite to the direction of acceleration if  were 0. Magnitude: f S is proportional to the magnitude of N f s =  s N F m fsfs N mg

23. Physics 201: Lecture 9, Pg 23 Kinetic or Sliding friction (f k < f s ) Dynamic equilibrium, moving but acceleration is still zero As F increases f k remains nearly constant (but now there acceleration is acceleration) F m1m1 FBD fkfk N mg  F x = 0 = -F + f k  f k = F  F y = 0 = - N + mg  N = mg v f k =  k N

24. Physics 201: Lecture 9, Pg 24 Model of Static Friction (simple case) Magnitude: f is proportional to the applied forces such that f s ≤  s N    s called the “coefficient of static friction” Direction:  Opposite to the direction of system acceleration if  were 0

25. Physics 201: Lecture 9, Pg 25 Sliding Friction N l Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity. fN l Magnitude: f k is proportional to the magnitude of N fN g  f k =  k N ( =  K  mg in the previous example) The constant  k is called the “coefficient of kinetic friction” Logic dictates that  S >  K for any system

26. Physics 201: Lecture 9, Pg 26 Coefficients of Friction Material on Material  s = static friction  k = kinetic friction steel / steel0.60.4 add grease to steel0.10.05 metal / ice0.0220.02 brake lining / iron0.40.3 tire / dry pavement0.90.8 tire / wet pavement0.80.7

27. Physics 201: Lecture 9, Pg 27 Other Forces are Conditional l Notice what happens if we change the direction of the applied force l The normal force can increase or decrease Let a=0 F fFfF mgmg N i j

28. Physics 201: Lecture 9, Pg 28 An experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find  S Static equilibrium: Set m 2 and add mass to m 1 to reach the breaking point. Requires two FBDs m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2  F x = 0 = -T + f s = -T +  S N  F y = 0 = N – m 2 g fSfS Mass 1  F y = 0 = T – m 1 g T = m 1 g =  S m 2 g   S = m 1 /m 2

29. Physics 201: Lecture 9, Pg 29 Static Friction with a bicycle wheel (not simple) l You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? l You are breaking and the bicycle is slowing down What is the direction of the frictional force?

30. Physics 201: Lecture 9, Pg 30 For Thursday l Read Chapter 6