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Physics 207 – Lecture 8 Lecture 8 Goals: (Finish Chap. 6 and begin Ch. 7) Solve 1D & 2D motion with friction Utilize Newton’s 2nd Law Differentiate between Newton’s 1st, 2nd and 3rd Laws Begin to use Newton’s 3rd Law in problem solving Assignment: HW4, (Chap. 6 & 7, due 10/5) Finish reading Chapter 7 1st Exam Wed., Oct. 7th from 7:15-8:45 PM Chapters 1-7 in room 2103 Chamberlin Hall Physics 207: Lecture 8, Pg 1 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 r r r ∑ F = Fnet = ma ≠ 0 ∑ Fx = max ∑ Fy = may ∑ Fz = maz and mass plays a role Physics 207: Lecture 8, Pg 2 Page 1 Physics 207 – Lecture 8 Mass We have an idea of what mass is from everyday life. 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) Mass is an inherent property of an object. 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 3 Inertia and Mass The tendency of an object to resist any attempt to change its velocity is called Inertia Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity (acceleration) r r a ∝ Fnet If mass is constant then If force constant r | a |∝ 1 m |a| Mass is an inherent property of an object m Mass is independent of the object’s surroundings Mass is independent of the method used to measure it Mass is a scalar quantity The SI unit of mass is kg Physics 207: Lecture 8, Pg 4 Page 2 Physics 207 – Lecture 8 Exercise Newton’s 2nd Law 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). The speed of the object is A. B. C. D. increasing decreasing constant in time Not enough information to decide Physics 207: Lecture 8, Pg 5 Exercise Newton’s 2nd Law 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? A. B. C. D. E. A B C D G Physics 207: Lecture 8, Pg 6 Page 3 Physics 207 – Lecture 8 Now: Back to forces that oppose motion Physics 207: Lecture 8, Pg 7 Static and Kinetic Friction 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: 1. Force vector ⊥ to the normal force vector N 2. Opposite to the direction of acceleration if µ were 0. Magnitude: f is proportional to the applied forces such that fs ≤ µs N µs called the “coefficient of static friction” Physics 207: Lecture 8, Pg 8 Page 4 Physics 207 – Lecture 8 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, Σ Fx = 0 = -F + fs fs = F FBD Σ Fy = 0 = - N + mg N = mg N As F increases so does fs F m fs 1 mg Physics 207: Lecture 8, Pg 9 Static friction, at maximum (just before slipping) Equilibrium: A block, mass m, with a horizontal force F applied, Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the direction of acceleration if µ were 0. Magnitude: fS is proportional to the magnitude of N fs = µs N N F m mg Physics 207: Lecture 8, Pg 10 Page 5 fs Physics 207 – Lecture 8 Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero Σ Fx = 0 = -F + fk FBD fk = F v Σ Fy = 0 = - N + mg N = mg N F As F increases fk remains nearly constant (but now there acceleration is acceleration) m fk 1 mg fk = µk N Physics 207: Lecture 8, Pg 11 Sliding Friction: Modeling Direction: A force vector ⊥ to the normal force vector N and the vector is opposite to the velocity. Magnitude: fk is proportional to the magnitude of N fk = µk N ( = µK mg g in the previous example) The constant µk is called the “coefficient of kinetic friction” Logic dictates that µS > µK for any system Physics 207: Lecture 8, Pg 12 Page 6 Physics 207 – Lecture 8 Case study ... big F Dynamics: x-axis i : y-axis j : max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax so fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 8, Pg 13 Case study ... little F Dynamics: x-axis i : y-axis j : so max = F − µKN may = 0 = N – mg or N = mg F − µKmg = m ax fk v j N F i ma ax fk µK mg mg g Physics 207: Lecture 8, Pg 14 Page 7 Physics 207 – Lecture 8 Coefficients of Friction Material on Material µs = static friction µk = kinetic friction steel / steel 0.6 0.4 add grease to steel 0.1 0.05 metal / ice 0.022 0.02 brake lining / iron 0.4 0.3 tire / dry pavement 0.9 0.8 tire / wet pavement 0.8 0.7 Physics 207: Lecture 8, Pg 15 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 µS N T Static equilibrium: Set m2 m2 and add mass to m1 to reach the breaking point. Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g fS T m1 m2g m1g Mass 2 Σ Fx = 0 = -T + fs = -T + µS N Σ Fy = 0 = N – m2g T = m1g = µS m2g µS = m1/m2 Page 8 Physics 207: Lecture 8, Pg 16 Physics 207 – Lecture 8 A 2nd experiment Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find µK. T Dynamic equilibrium: Set m2 and adjust m1 to find place when T a = 0 and v ≠ 0 fk m1 Requires two FBDs Mass 1 Σ Fy = 0 = T – m1g N m2 m2g m1g Mass 2 Σ Fx = 0 = -T + ff = -T + µk N Σ Fy = 0 = N – m2g T = m1g = µk m2g µk = m1/m2 Physics 207: Lecture 8, Pg 17 An experiment (with a ≠ 0) Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. N Design an experiment to find µK. T Non-equilibrium: Set m2 T and adjust m1 to find regime where a ≠ 0 m1 Requires two FBDs m1g Mass 1 Σ Fy = m1a = fk m2 m2g Mass 2 T – m1g Σ Fx = m2a = -T + fk Σ Fy = 0 = N – m2g = -T + µk N T = m1g + m1a = µk m2g – m2a µk = (m1(g+a)+m2a)/m2g Physics 207: Lecture 8, Pg 18 Page 9 Physics 207 – Lecture 8 Home Exercise 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. What is the value of µk for jelloium on steel? Physics 207: Lecture 8, Pg 19 Home exercise Σ Fx =ma = F - ff = F - µk N = F - µk mg Σ Fy = 0 = N – mg µk = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2 a = 0.40 m/s2 µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46 Physics 207: Lecture 8, Pg 20 Page 10 Physics 207 – Lecture 8 Forces at different angles 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? Cases 3, 4 Case 2 Case 1 F N N ff F F N ff ff mg mg mg Physics 207: Lecture 8, Pg 21 Inclined plane with “Normal” and Frictional Forces 1. Static Equilibrium Case “Normal” means perpendicular 2. Dynamic Equilibrium (see 1) Normal Force 3. Dynamic case with non-zero acceleration Friction ΣF=0 f Force Fx= 0 = mg sin θ – f mg sin θ Fy= 0 = –mg cos θ + N mg cos θ θ with mg sin θ = f ≤ µS N if mg sin θ > µS N, must slide Critical angle µs = tan θc θ θ Block weight is mg Page 11 Physics 207: Lecture 8, Pg 22 y x Physics 207 – Lecture 8 Inclined plane with “Normal” and Frictional Forces 1. Static Equilibrium Case “Normal” means perpendicular Normal Force 2. Dynamic Equilibrium Friction opposite velocity v (down the incline) ΣF=0 fK Friction Force mg sin θ Fx= 0 = mg sin θ – fk Fy= 0 = –mg cos θ + N Fx= 0 = mg sin θ – µk mg cos θ µk = tan θ (only one angle) y mg cos θ θ fk = µk N = µk mg cos θ θ θ x mg Physics 207: Lecture 8, Pg 23 Inclined plane with “Normal” and Frictional Forces 3. Dynamic case with non-zero acceleration Normal Result depends on direction of velocity Force Friction Force Sliding Down v mg sin θ θ Fx= max = mg sin θ ± fk θ fk Sliding Up Weight of block is mg Fy= 0 = –mg cos θ + N fk = µk N = µk mg cos θ Fx= max = mg sin θ ± µk mg cos θ ax = g sin θ ± µk g cos θ Page 12 Physics 207: Lecture 8, Pg 24 Physics 207 – Lecture 8 The inclined plane coming and going (not static): the component of mg along the surface > kinetic friction Σ Fx = max = mg sin θ ± uk N Σ Fy = may = 0 = -mg cos θ + N Putting it all together gives two different accelerations, ax = g sin θ ± uk g cos θ. A tidy result but ultimately it is the process of applying Newton’s Laws that is key. Physics 207: Lecture 8, Pg 25 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 26 Page 13 Physics 207 – Lecture 8 Recap Assignment: HW4, (Chapter 6 & 7 due 10/5) Finish Chapter 7 Physics 207: Lecture 8, Pg 27 Page 14