Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: (Finish Chap. 6 and begin Ch. 7)  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 Assignment: HW4, (Chap. 6 & 7, due 10/5) Finish reading Chapter 7 1 st Exam Wed., Oct. 7 th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall

Physics 207: Lecture 8, Pg 2 Net force gives rise to acceleration! In physics:  A force is an action which causes an object to accelerate (translational & rotational) This is Newton’s Second Law and mass plays a role

Physics 207: Lecture 8, Pg 3 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!

Physics 207: Lecture 8, Pg 4 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 object’s surroundings 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

Physics 207: Lecture 8, Pg 5 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

Physics 207: Lecture 8, Pg 6 Exercise Newton’s 2 nd Law A. A B. B C. C D. D E. G A 10 kg 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?

Physics 207: Lecture 8, Pg 7 Now: Back to forces that oppose motion

Physics 207: Lecture 8, Pg 8 Static and Kinetic Friction l Friction is a model force that exists between objects & it is conditional At Static Equilibrium: A block, mass m, with a horizontal force F applied, Direction: N 1. Force vector  to the normal force vector N Opposite to the direction of acceleration if  were 0. Magnitude: f is proportional to the applied forces such that f s ≤  s N   s called the “coefficient of static friction”

Physics 207: Lecture 8, Pg 9 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

Physics 207: Lecture 8, Pg 10 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

Physics 207: Lecture 8, Pg 11 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

Physics 207: Lecture 8, Pg 12 Sliding Friction: Modeling 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

Physics 207: Lecture 8, Pg 13 Case study... big F l Dynamics: x-axis i :ma x = F  K N y-axis j : ma y = 0 = N – mg or N = mg soF  K mg = m a x axmaxaxmax F gmggmg N i j  K mg v fkfk fkfk

Physics 207: Lecture 8, Pg 14 Case study... little F l Dynamics: x-axis i :ma x = F  K N y-axis j : ma y = 0 = N – mg or N = mg soF  K mg = m a x axmaxaxmax F gmggmg N i j  K mg v fkfk fkfk

Physics 207: Lecture 8, Pg 15 Coefficients of Friction Material on Material  s = static friction  k = kinetic friction steel / steel add grease to steel metal / ice brake lining / iron tire / dry pavement tire / wet pavement0.80.7

Physics 207: Lecture 8, Pg 16 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

Physics 207: Lecture 8, Pg 17 A 2 nd experiment 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

Physics 207: Lecture 8, Pg 18 An experiment (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. Non-equilibrium: Set m 2 and adjust m 1 to find regime where a ≠ 0 Requires two FBDs 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

Physics 207: Lecture 8, Pg 19 Home Exercise 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?

Physics 207: Lecture 8, Pg 20 Home exercise  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 = ( · 0.40 ) / (0.20 ·10.) = 0.46

Physics 207: Lecture 8, Pg 21 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

Physics 207: Lecture 8, Pg 22 Inclined plane with “Normal” and Frictional Forces 1.Static Equilibrium Case 2.Dynamic Equilibrium (see 1) 3.Dynamic case with non-zero acceleration Block weight is mg Normal Force Friction Force “Normal” means perpendicular   mg cos  f  x y mg sin   F = 0 F x = 0 = mg sin  – f F y = 0 = –mg cos  + N with mg sin  = f ≤  S N if mg sin  >  S N, must slide Critical angle  s = tan  c

Physics 207: Lecture 8, Pg 23 Inclined plane with “Normal” and Frictional Forces 1.Static Equilibrium Case 2.Dynamic Equilibrium Friction opposite velocity (down the incline) mg Normal Force Friction Force “Normal” means perpendicular   mg cos  fKfK  x y mg sin   F = 0 F x = 0 = mg sin  – f k F y = 0 = –mg cos  + N f k =  k N =  k mg cos  F x = 0 = mg sin  –  k mg cos    k = tan   (only one angle) v

Physics 207: Lecture 8, Pg 24 Inclined plane with “Normal” and Frictional Forces 3. Dynamic case with non-zero acceleration Result depends on direction of velocity Weight of block is mg Normal Force Friction Force Sliding Down   v mg sin  f k Sliding Up F x = ma x = mg sin  ± f k F y = 0 = –mg cos  + N f k =  k N =  k mg cos  F x = ma x = mg sin  ±  k mg cos  a x = g sin  ±  k g cos 

Physics 207: Lecture 8, Pg 25 The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction Putting it all together gives two different accelerations, a x = g sin  ± u k g cos . A tidy result but ultimately it is the process of applying Newton’s Laws that is key.  F x = ma x = mg sin  ± u k N  F y = ma y = 0 = -mg cos  + N

Physics 207: Lecture 8, Pg 26 Velocity and acceleration plots: Notice that the acceleration is always down the slide and that, even at the turnaround point, the block is always motion although there is an infinitesimal point at which the velocity of the block passes through zero. At this moment, depending on the static friction the block may become stuck.

Physics 207: Lecture 8, Pg 27 Recap Assignment: HW4, (Chapter 6 & 7 due 10/5) Finish Chapter 7