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Lecture 7

Lecture 7. Goals: Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1 st and 2 nd laws. Differentiate between Newton’s 1 st , 2 nd and 3 rd Laws Use Newton’s 3 rd Law in problem solving.

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Lecture 7

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  1. Lecture 7 • Goals: • Solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws. • Differentiate between Newton’s 1st, 2nd and 3rd Laws • Use Newton’s 3rd Law in problem solving Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday) Finish Chapter 7 1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7

  2. 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.

  3. Friction revisited: 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

  4. Static friction, at maximum (just before slipping) Still 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 velocity. Magnitude:fSis proportional to the magnitude of N fs = ms N ms called the “coefficient of static friction” N F m fs mg

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

  6. Case study ... big F j i • Dynamics: x-axis i :max= F KN y-axis j :may = 0 = N–mgor N = mg soFKmg=m ax fk v N F max fk Kmg mg

  7. Case study ... little F • Dynamics: x-axis i :max= F KN y-axis j :may = 0 = N–mgor N = mg soFKmg=m ax fk v j N F i max fk Kmg mg

  8. 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” • As the normal force varies so does the frictional force

  9. Additional comments on Friction: • The force of friction does not depend on the area of the surfaces in contact (a relatively good approximation if there is little surface deformation) • Logic dictates that S > Kfor any system

  10. Coefficients of Friction

  11. 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 & mK. 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

  12. 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 & 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

  13. 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 mS & 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

  14. Sample Problem • 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?

  15. Sample Problem S Fx =ma = F - ff = F - mk N = F - mk mg S Fy = 0 = N – mg mk = (F - ma) / mg & x = ½ a t2  0.80 m = ½ a 4 s2 a = 0.40 m/s2 mk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46

  16. Another experiment m1 A block is connected to a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force T (by pulling on the rope) as shown in the plot. T (N) 50 T 40 30 (A) On the next slide are tables and plots of velocity vs. time (B) Can you deduce the various coefficients of friction and the mass ? 20 10 40 30 20 10 t (sec)

  17. The Experimental Data T (N) 50 25 20 40 30 15 20 10 5 10 speed (m/s) • t < 30 s puts constraints on ms (Static equilibrium) • t > 30 s reflects mk (Non-equilibrium) Const. accel. at 30-40 (T= 40 N) and 40-50 (T=50 N) second times 40 20 30 10 t (sec)

  18. Another experiment m1 A block is connected by a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force as shown in the plot. N FBD Static case (30 N & less) S Fx = 0 = -T + f = -T + mN S Fy = 0 = - N + mg T = m m g , 2 unknowns f T mg

  19. Another experiment N FBD f m1 A block is connected by a horizontal massless string. The table has coefficients of kinetic & static friction (mK & mS). There is a unknown mass m and you apply a variable force as shown in the plot. (B) Non-equilibrium S Fx = max = -T + f = -T + mK N S Fy = 0 = N – mg max = -T + mKm g Using information at right you can identify 2 equations and 2 unknowns T mg Notice that at 30 s < t < 40 s T= 40 N ax= Dv/Dt = -5/10 m/s2 and at 40 s < t < 50 s T= 50 N ax= Dv/Dt =-15/10 m/s2

  20. 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

  21. The inclined plane coming and going (not static):the component of mg along the surface > kinetic friction • Fx = max = mg sin q ± ukN > 0 • Fy = may = -mg cos q + N Putting it all together gives two different accelerations, ax = g sin q± uk g cos q. A tidy result but ultimately it is the process of applying Newton’s Laws that is key.

  22. Lecture 6 Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday) Finish Chapter 7 1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7

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