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Lecture 7

Lecture 7. "Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools." New York Times editorial, 1921 ,

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Lecture 7

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  1. Lecture 7 "Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools." New York Times editorial, 1921, about Robert Goddard's revolutionary rocket work. "Correction: It is now definitely established that a rocket can function in a vacuum. The 'Times' regrets the error." New York Times editorial, July 1969.

  2. Lecture 7 • Goals: • Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws. • Distinguish static and kinetic coefficients of friction • Differentiate between Newton’s 1st, 2nd and 3rd Laws Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am, Wednesday) Read Chapter 7 1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7

  3. Exercise,Newton’s 2nd Law • P + C < W • P + C > W • P = C • P + C = W

  4. 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!

  5. Inertia and Mass |a| m • The tendency of an object to resist any attempt to change its velocity is called Inertia • Massis 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  • Mass is an inherent property of an object • 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

  6. ExerciseNewton’s 2nd Law • increasing • decreasing • constant in time • Not enough information to decide • 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

  7. Exercise Newton’s 2nd Law A 10 kg mass undergoes motion along a line with a 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? • B • C • D • F • G

  8. Moving forces around string T1 -T1 • Massless strings: Translate forces and reverse their direction but do not change their magnitude (we really need Newton’s 3rd of action/reaction to justify) • Massless, frictionless pulleys: Reorient force direction but do not change their magnitude T2 T1 -T1 -T2 | T1 | = | -T1 | = | T2 | = | T2 |

  9. Scale Problem ? 1.0 kg • You are given a 1.0 kg mass and you hang it directly on a fish scale and it reads 10 N (g is 10 m/s2). • Now you use this mass in a second experiment in which the 1.0 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. • What force does the fish scale now read? 10 N 1.0 kg

  10. Scale Problem ? • 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 • Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.) 1.0 kg

  11. Scale Problem ? ? 1.0 kg 1.0 kg T ’ 3: 1: 2: • SFy = 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 = 20 N (We start with 10 N but end with 20 N) T” T -T -T W -T ’ -mg

  12. No Net Force, No acceleration…a demo exercise • In this demonstration we have a ball tied to a string undergoing horizontal UCM (i.e. the ball has only radial acceleration) 1 Assuming you are looking from above, draw the orbit with the tangential velocity and the radial acceleration vectors sketched out. 2 Suddenly the string brakes. 3 Now sketch the trajectory with the velocity and acceleration vectors drawn again.

  13. Static and Kinetic Friction • Friction exists between objects and its behavior has been modeled. At Static 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 m were 0. Magnitude:fis proportional to the applied forces such that fs≤ms N ms called the “coefficient of static friction”

  14. Friction: Static friction FBD N F m1 fs mg Static equilibrium: A block with a horizontal force F applied, As F increases so does fs S Fx = 0 = -F + fs fs = F S Fy = 0 = - N + mg  N = mg

  15. 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 m were 0. Magnitude:fSis proportional to the magnitude of N fs = ms N N F fs m mg

  16. Kinetic or Sliding friction (fk < fs) Dynamic equilibrium, moving but acceleration is still zero As F increases fk remains nearly constant (but now there acceleration is acceleration) FBD S Fx = 0 = -F + fk fk = F S Fy = 0 = - N + mg  N = mg v N F m1 fk mg fk = mk N

  17. Sliding Friction: Quantitatively • Direction: A force vector  to the normal force vector N and the vector is opposite to the velocity. • Magnitude: fkis proportional to the magnitude of N • fk= kN ( = Kmg in the previous example) • The constant k is called the “coefficient of kinetic friction” • Logic dictates that S > Kfor any system

  18. Coefficients of Friction

  19. An experiment N m2 m2g Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find mS T Static equilibrium: Set m2 and add mass to m1 to reach the breaking point. Requires two FBDs fS T m1 m1g Mass 1 S Fy = 0 = T – m1g T = m1g = mSm2g  mS=m1/m2 Mass 2 S Fx = 0 = -T + fs= -T + mSN S Fy = 0 = N – m2g

  20. A 2nd experiment N m2 m2g Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find mK. T Dynamic equilibrium: Set m2 and adjust m1 to find place when a = 0 and v ≠ 0 Requires two FBDs fk T m1 m1g Mass 1 S Fy = 0 = T – m1g T = m1g = mkm2g  mk=m1/m2 Mass 2 S Fx = 0 = -T + ff= -T + mkN S Fy = 0 = N – m2g

  21. An experiment (with a ≠ 0) N m2 m2g Two blocks are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Design an experiment to find mK. T Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0 Requires two FBDs fk T m1 m1g Mass 1 S Fy = m1a = T – m1g T = m1g + m1a = mkm2g – m2a mk= (m1(g+a)+m2a)/m2g Mass 2 S Fx = m2a = -T + fk= -T + mkN S Fy = 0 = N – m2g

  22. Inclined plane with “Normal” and Frictional Forces At first the velocity is v up along the slide Can we draw a velocity time plot? What the accelerationversus time? “Normal” means perpendicular Normal Force Friction Force Sliding Down fk Sliding Up v q q mg sin q Weight of block ismg Note: If frictional Force = Normal Force  (coefficient of friction) Ffriction =  Fnormal = m mg sin q then zero acceleration

  23. Recap • Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am, Wednesday) • Read Chapter 7 • 1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7

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