1. Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals:  Solve 1D & 2D motion with friction  Utilize Newton’s 2 nd Law  Differentiate between Newton’s 1 st, 2 nd and 3 rd Laws  Begin to use Newton’s 3 rd Law in problem solving

2. Physics 207: Lecture 8, Pg 2 Friction with no acceleration l No net force l So frictional force just cancels applied force F F APPLIED f f FRICTION gmggmg N i j

3. Physics 207: Lecture 8, Pg 3 Friction... l Friction is caused by the “microscopic” interactions between the two surfaces:

4. Physics 207: Lecture 8, Pg 4 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, As F increases so does f s F m FBD fsfs N mg

5. Physics 207: Lecture 8, Pg 5 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, F m FBD fsfs N mg  F x = 0 = -F + f s f s = F  F y = 0 = - N + mg N = mg Force applied FrictionForce

6. Physics 207: Lecture 8, Pg 6 Static friction, at maximum (just before slipping) f S is proportional to the magnitude of N f s =  s N F m fsfs N mg Force applied FrictionForce

7. Physics 207: Lecture 8, Pg 7 Model of Static Friction Magnitude: f is proportional to the applied forces such that f s ≤  s N    s called the “coefficient of static friction” Direction:  If just a single “applied” force, friction is in opposite direction

8. Physics 207: Lecture 8, Pg 8 Kinetic (f k < f s ) Dynamic equilibrium, moving but acceleration is still zero F m 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

9. Physics 207: Lecture 8, Pg 9 Model of Sliding Friction N l Direction:  to the normal force vector N and opposite to the velocity. fN l Magnitude: f k is proportional to the magnitude of N fN  f k =  k N The constant  k is called the “coefficient of kinetic friction” Logic dictates that  S >  K for any system

10. Physics 207: Lecture 8, Pg 10 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

11. Physics 207: Lecture 8, Pg 11 Sliding friction (f k 0 A change in velocity As F increases f k remains nearly constant (but now there is acceleration) F m FBD fkfk N mg  F x = -F + f k = net Force  F y = 0 = - N + mg  N = mg v f k =  k N

12. Physics 207: Lecture 8, Pg 12 Acceleration, Inertia and Mass l The tendency of an object to resist any attempt to change its velocity is called Inertia l Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity (acceleration) If mass is constant then If force constant  l Mass is an inherent property of an object l Mass is independent of the method used to measure it l Mass is a scalar quantity l The SI unit of mass is kg |a| m

13. Physics 207: Lecture 8, Pg 13 Mass l We have an idea of what mass is from everyday life. l In physics:  Mass (in Phys 207) is a quantity that specifies how much inertia an object has (i.e. a scalar that relates force to acceleration) (Newton’s Second Law) l Mass is an inherent property of an object. l Mass and weight are different quantities; weight is usually the magnitude of a gravitational (non-contact) force. “Pound” (lb) is a definition of weight (i.e., a force), not a mass!

14. Physics 207: Lecture 8, Pg 14 Exercise Newton’s 2 nd Law A. increasing B. decreasing C. constant in time D. Not enough information to decide l An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time (read this text carefully). l The speed of the object is

15. Physics 207: Lecture 8, Pg 15 1 st : Frictionless experiment (with a ≠ 0) Two blocks are connected on the table as shown. The table is frictionless. Find the acceleration of mass 2. Requires two FBDs T Mass 2  F x = m 2 a x = -T  F y = 0 = N – m 2 g m1m1 m2m2 m2gm2g N m1gm1g T Mass 1  F y = m 1 a y = T – m 1 g Notice a y = a x = a Eliminate T m 1 a + m 2 a = m 1 a = m 1 / (m 2 +m 1 )g

16. Physics 207: Lecture 8, Pg 16 Experiment with friction (with a ≠ 0) Two blocks, of m 1 & m 2, are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Given an a, find  K. Similar but now with friction. T Mass 2  F x = m 2 a = -T + f k = -T +  k N  F y = 0 = N – m 2 g m1m1 m2m2 m2gm2g N m1gm1g T fkfk Mass 1  F y = m 1 a = T – m 1 g T = m 1 g + m 1 a =  k m 2 g – m 2 a   k = (m 1 (g+a)+m 2 a)/m 2 g

17. Physics 207: Lecture 8, Pg 17 Experiment with friction (with a = 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find  K. Dynamic equilibrium: Set m 2 and adjust m 1 to find place when a = 0 and v ≠ 0 Requires two FBDs m1m1 m2m2 m2gm2g N m1gm1g T T Mass 2  F x = 0 = -T + f f = -T +  k N  F y = 0 = N – m 2 g fkfk Mass 1  F y = 0 = T – m 1 g T = m 1 g =  k m 2 g   k = m 1 /m 2

18. Physics 207: Lecture 8, Pg 18 To repeat, net force  acceleration In physics:  A force is an action which causes an object to accelerate (translational & rotational) This is Newton’s Second Law

19. Physics 207: Lecture 8, Pg 19 Home Exercise Newton’s 2 nd Law A. A B. B C. D D. F E. G A mass undergoes motion along a line with velocities as given in the figure below. In regards to the stated letters for each region, in which is the magnitude of the force on the mass at its greatest?

20. Physics 207: Lecture 8, Pg 20 Remember: Forces are Conditional l Notice what happens if we change the direction of the applied force l The normal force can increase or decrease l Here the normal force exceeds mg Let a=0 F fFfF mgmg N  F sin  +mg i j  F sin 

21. Physics 207: Lecture 8, Pg 21 Home Exercise Newton’s 2 nd Law A. 4 x as long B. 2 x as long C. 1/2 as long D. 1/4 as long A constant force is exerted on a cart that is initially at rest on an air table. The force acts for a short period of time and gives the cart a certain final speed s. Air Track Cart Force In a second trial, we apply a force only half as large. To reach the same final speed, how long must the same force be applied (recall acceleration is proportional to force if mass fixed)?

22. Physics 207: Lecture 8, Pg 22 Home Exercise Newton’s 2 nd Law Solution Air Track Cart Force (B) 2 x as long F = ma Since F 2 = 1/2 F 1 a 2 = 1/2 a 1 We know that under constant acceleration, v = a  t So, a 2  t 2 = a 1  t 1 we want equal final velocities 1/2 a 1 /  t 2 = a 1 /  t 1  t 2 = 2  t 1

23. Physics 207: Lecture 8, Pg 23 Home Exercise Newton’s 2 nd Law A. 8 x as far B. 4 x as far C. 2 x as far D. 1/4 x as far A force of 2 Newtons acts on a cart that is initially at rest on an air track with no air and pushed for 1 second. Because there is friction (no air), the cart stops immediately after I finish pushing. It has traveled a distance, D. Air Track Cart Force Next, the force of 2 Newtons acts again but is applied for 2 seconds. The new distance the cart moves relative to D is:

24. Physics 207: Lecture 8, Pg 24 Home Exercise Solution Air Track Cart Force (B) 4 x as long We know that under constant acceleration,  x = a (  t) 2 /2 (when v 0 =0) Here  t 2 =2  t 1, F 2 = F 1  a 2 = a 1

25. Physics 207: Lecture 8, Pg 25 Sample Problem l You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. l What is the value of μ k for jelloium on steel?

26. Physics 207: Lecture 8, Pg 26 Sample Problem  F x =ma = F - f f = F -  k N = F -  k mg  F y = 0 = N – mg  k = (F - ma) / mg & x = ½ a t 2  0.80 m = ½ a 4 s 2 a = 0.40 m/s 2  k = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46

27. Physics 207: Lecture 8, Pg 27 Forces at different angles Case 1 Case 2 F mg N Case1: Downward angled force with friction Case 2: Upwards angled force with friction Cases 3,4: Up against the wall Questions: Does it slide? What happens to the normal force? What happens to the frictional force? mg Cases 3, 4 mg N N F F f f f

28. Physics 207: Lecture 8, Pg 28 Example (non-contact) Consider the forces on an object undergoing projectile motion F B,E = - m B g EARTH F E,B = m B g F B,E = - m B g F E,B = m B g

29. Physics 207: Lecture 8, Pg 29 Gravity Newton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m 1 and m 2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:

30. Physics 207: Lecture 8, Pg 30 Cavendish’s Experiment F = m 1 g = G m 1 m 2 / r 2 g = G m 2 / r 2 If we know big G, little g and r then will can find m 2 the mass of the Earth!!!

31. Physics 207: Lecture 8, Pg 31 Example (non-contact) F B,E = - m B g EARTH F E,B = m B g F B,E = - m B g F E,B = m B g Compare: g = G m 2 / 4000 2 g ’ = G m 2 / (4000+40) 2 g / g ’ = / (4000+40) 2 / 4000 2 = 0.98 Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)

32. Physics 207: Lecture 8, Pg 32 Recap Assignment: HW4, (Chapter 6 & 7 due 10/4) For Monday finish Chapter 7