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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. Distinguish static and kinetic coefficients of friction Differentiate between Newton’s 1 st , 2 nd and 3 rd Laws.

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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. • Distinguish static and kinetic coefficients of friction • Differentiate between Newton’s 1st, 2nd and 3rd Laws Assignment: HW4, (Chapters 6 & 7 (1st part) due 2/17, 9 am, Wednesday) For Thursday, Read Chapter 7 1st Exam Thursday, Feb. 17 from 7:15-8:45 PM Chapters 1-7

  2. Identifying Forces: Non-contact FB,G All objects having mass exhibit a mutually attractive force (i.e., gravity) that is distance dependent At the Earth’s surface this variation is small so little “g” (the associated acceleration) is typically set to 9.80 or 10. m/s2

  3. Contact (i.e., normal) Forces FB,T Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point).

  4. No net force  No acceleration FB,G (Force vectors are not always drawn at contact points) y FB,T Normal force is always  to a surface

  5. No net force  No acceleration • If zero velocity then “static equilibrium” • If non-zero velocity then “dynamic equilibrium” • This label, static vs. dynamic, is observer dependent • Forces are vectors

  6. A special contact force: Friction • 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”)! j N FAPPLIED i ma fFRICTION mg

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

  8. Friction... N F ma fF mg • Force of friction acts to oppose motion: • Parallel to a surface • Perpendicular to a Normal force. j i

  9. Static Friction with a bicycle wheel • 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?

  10. Important notes • Many contact forces are conditional and, more importantly, they are not necessarily constant • We have a general notion of forces is from everyday life. • In physics the definition must be precise. • Aforce is an action which causes a body to accelerate. (Newton’s Second Law) • On a microscopic level, all forces are non-contact

  11. Pushing and Pulling Forces • A rope is an example of something that can pull • You arm is an example of an object that can push or push

  12. Examples of Contact Forces:A spring can push

  13. A spring can pull

  14. Ropes provide tension (a pull) In physics we often use a “massless” rope with opposing tensions of equal magnitude

  15. 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 ff F F N N N F ff ff mg mg mg

  16. 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 ?

  17. 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 2nd Law (Resolve vectors into appropriate components)

  18. 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

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

  20. 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!

  21. 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

  22. 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

  23. 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

  24. 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 |

  25. 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

  26. 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

  27. 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”

  28. 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

  29. 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

  30. 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

  31. 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

  32. Coefficients of Friction

  33. 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

  34. 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

  35. 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

  36. Recap • Assignment: HW4, (due 2/17, 9 am, Wednesday) • Read Chapter 7 • 1st Exam Wednesday, Feb. 17 from 7:15-8:45 PM

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