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Lecture 8

Lecture 8. Goals: Solve 1D & 2D motion with friction Utilize Newton’s 2 nd Law Differentiate between Newton’s 1 st , 2 nd and 3 rd Laws Begin to use Newton’s 3 rd Law in problem solving. Friction with no acceleration. No net force So frictional force just cancels applied force.

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Lecture 8

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  1. Lecture 8 • Goals: • 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

  2. Friction with no acceleration • No net force • So frictional force just cancels applied force j N FAPPLIED i fFRICTION mg

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

  4. Friction: Static friction Static equilibrium: A block with a horizontal force F applied, As F increases so does fs FBD N F m fs mg

  5. Friction: Static friction FrictionForce Force applied Static equilibrium: A block with a horizontal force F applied, S Fx = 0 = -F + fs fs = F S Fy = 0 = - N + mg N = mg FBD N F m fs mg

  6. Static friction, at maximum (just before slipping) fSis proportional to the magnitude of N fs = ms N N F fs m mg FrictionForce Force applied

  7. Model of Static Friction Magnitude: fis proportional to the applied forces such that fs≤ms N • ms called the “coefficient of static friction” Direction: • If just a single “applied” force, friction is in opposite direction

  8. Kinetic (fk < fs) Dynamic equilibrium, moving but acceleration is still zero FBD S Fx = 0 = -F + fk fk = F S Fy = 0 = - N + mg  N = mg v N F m fk mg fk = mk N

  9. Model of Sliding Friction • Direction:  to the normal force vector N and opposite to the velocity. • Magnitude: fkis proportional to the magnitude of N • fk= kN • The constant k is called the “coefficient of kinetic friction” • Logic dictates that S > Kfor any system

  10. Coefficients of Friction

  11. Sliding friction (fk < fs) but now |a| > 0 A change in velocity As F increases fk remains nearly constant (but now there is acceleration) FBD S Fx = -F + fk = net Force S Fy = 0 = - N + mg  N = mg v N F m fk mg fk = mk N

  12. Acceleration, 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 method used to measure it • Mass is a scalar quantity • The SI unit of mass is kg

  13. 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

  14. 1st: Frictionless experiment (with a ≠ 0) N m2 m2g Two blocks are connected on the table as shown. The table is frictionless. Find the acceleration of mass 2. T Requires two FBDs T m1 m1g Mass 1 S Fy = m1ay = T – m1g Notice ay = ax = a Eliminate T m1a + m2a = m1 a =m1 / (m2+m1)g Mass 2 S Fx = m2ax = -T S Fy = 0 = N – m2g

  15. Experiment with friction (with a ≠ 0) Two blocks, of m1 & m2 , are connected on the table as shown. The table has unknown static and kinetic friction coefficients. Given an a, find mK. N T Similar but now with friction. m2 fk T m1 m2g 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

  16. To repeat, net force  acceleration In physics: • Aforce is an action which causes an object to accelerate (translational & rotational) This is Newton’s Second Law

  17. Home Exercise Newton’s 2nd Law A 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 • D • F • G

  18. Remember: Forces are Conditional j i • Notice what happens if we change the direction of the applied force • The normal force can increase or decrease • Here the normal force exceeds mg F N  F sinq +mg q F sin q Let a=0 fF mg

  19. 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

  20. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Consider the forces on an object undergoing projectile motion

  21. Gravity Newton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m1and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:

  22. Cavendish’s Experiment F = m1 g = G m1 m2 / r2 g = G m2 / r2 If we know big G, little g and r then will can find m2 the mass of the Earth!!!

  23. Example (non-contact) FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi) Compare: g= G m2 / 40002 g’ = G m2 / (4000+40)2 g / g’ = / (4000+40)2 / 40002 = 0.98

  24. Recap Assignment: HW4, (Chapter 6 & 7 due 10/4) For Monday finish Chapter 7

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