Lecture 8

1. Lecture 8 • Goals: • Solve 1D & 2D problems introducing forces with/without friction • Utilize Newton’s 1st & 2nd Laws • Begin to use Newton’s 3rd Law in problem solving

2. Newton’s Laws Read: Force on A by B Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, FNET = F = ma Law 3: Forces occur in pairs: FA , B = - FB , A (For every action there is an equal and opposite reaction.)

3. Exercise Newton’s 3rd Law a b • 2 • 4 • 6 • Something else • Two blocks are being pushed by a massless finger on a horizontal frictionless floor. • How many action-reaction force pairs are present in this exercise?

4. Newton’s Law The acceleration of an object is directly proportional to the net external force acting upon it. The constant of proportionality is the mass. • This is a vector expression

5. Analyzing Forces: Free Body Diagram Eat at Bucky’s A heavy sign is hung between two poles by a rope at each corner extending to the poles. A hanging sign is an example of static equilibrium (depends on observer) What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ?

6. Free Body Diagram T2 T1 Eat at Bucky’s Step one: Define the system q2 q1 T2 T1 q2 q1 mg mg Step two: Sketch in force vectors Step three: Apply Newton’s 1st & 2nd Laws (Resolve vectors into appropriate components)

7. Free Body Diagram T1 T2 q2 q1 Eat at Bucky’s mg Vertical : y-direction 0 = -mg + T1 sinq1 + T2 sinq2 Horizontal : x-direction 0 = -T1 cosq1 + T2 cosq2

8. Pushing and Pulling Forces • String or ropes are examples of things that can pull • You arm is an example of an object that can push or push

9. Scale Problem ? 5.0 kg • You are given a 5.0 kg mass and you hang it directly on a fish scale and it reads 50 N (g is 10 m/s2). • Now you use this mass in a second experiment in which the 5.0 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? 50 N 5.0 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.) 5.0 kg

11. Scale Problem ? ? 1.0 kg 5.0 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 = 100 N (We start with 50 N but end with 100 N) T” T -T -T W -T ’ -mg

12. Newton’s 2nd Law, Forces are conditional • P + C < W • P + C > W • P = C • P + C = W

13. Another experiment m1 A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant force of 10 N. What is the acceleration of the block if the table top is frictionless? N FBD S Fx = max = -T S Fy = 0 = N - mg ax = -T/m = 5 m/s2 T mg

14. Version 2 j i A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant horizontal force of 10 N. The surface is frictionless and it is inclined 30° from horizontal. What is the acceleration of the block down the slide? N S Fx = max = Tx+Wx S Fy = 0 = N + Ty+Wy max = -T cos 30°+W sin 30° ax = -T/m 3½/2 - g / 2 ax = -5(1.732)/2- 5 = -7.2 m/s2 What is the apparent weight of the block on the slide? N = Ty+Wy = T sin 30° -W cos 30° N = -10(0.5) + 20(0.866) = 12.3 N FBD T m 30° 30° 30° mg = W

15. Version 3 j i A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant horizontal force of 34.6 N. The surface is frictionless and it is inclined 30° from horizontal. What is the acceleration of the block down the slide? N S Fx = max = Tx+Wx S Fy = 0 = N + Ty+Wy max = -T cos 30°+W sin 30° ax = -T/m 3½/2 - g / 2 ax = -17.3(1.732)/ 2- 5 = -35 m/s2 What is the apparent weight of the block on the slide? N = Ty+Wy = T sin 30° -W cos 30° N = -34.6(0.5) + 20(0.866) = 0.0 N !!! FBD T m 30° 30° 30° mg = W

16. Another experiment: A modified Atwood’s machine N m2 m2g Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward. What is the acceleration of the horizontal block? T Requires two FBDs and Newton’s 3rd Law. T m1 Mass 1 S Fy = m1a1y= T – m1g m1g Correlated motion: |a1y| = | a2x| ≡ a If m1 moves up moves m2 right Mass 2 S Fx = m2a2x = -T S Fy = 0 = N – m2g

17. Another experiment: A modified Atwood’s machine N m2 m2g Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward. What is the acceleration of the horizontal block. T Requires two FBDs and Newton’s 3rd Law. T m1 Mass 1 A. S Fy = m1 a= T – m1g m1g Subbing in T from B into A m1 a= - m2 a – m1g m1 a + m2 a = m1g a = m1g / (m1 + m2 ) Mass 2 B. S Fx = m2 a = -T S Fy = 0 = N – m2g

18. A “special” contact force: Friction j i • What does it do? • It opposes motion (velocity, actual or that which would occur if friction were absent!) • How do we characterize this in terms we have learned? • Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)! N FAPPLIED ma fFRICTION mg

19. If no acceleration • No net force • So frictional force just equals applied force • Key point: It is conditional! j N FAPPLIED i fFRICTION mg

20. Friction... • Friction is caused by the “microscopic” interactions between the two surfaces:

21. 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

22. 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

23. 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

24. Model of Static Friction (simple case) Magnitude: fis proportional to the applied forces such that fs≤ms N • ms called the “coefficient of static friction” Direction: • Opposite to the direction of system acceleration if m were 0

25. Sliding Friction • 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

26. Coefficients of Friction

27. Other Forces are Conditional • Notice what happens if we change the direction of the applied force • The normal force can increase or decrease F j N i Let a=0 fF mg

28. 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

29. Static Friction with a bicycle wheel (not simple) • You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? • You are breaking and the bicycle is slowing down What is the direction of the frictional force?

30. For Thursday • Read Chapter 6