Lect07 - Physics 211: Lecture 7, Pg 1 Physics 211: Lecture...

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Unformatted text preview: Physics 211: Lecture 7, Pg 1 Physics 211: Lecture 7 Physics 211: Lecture 7 Today Today ’ ’ s Agenda s Agenda z Friction Î What is it? Î How do we characterize it? Î Model of friction Î Static & Kinetic friction (kinetic = dynamic in some languages) z Some problems involving friction Physics 211: Lecture 7, Pg 2 New Topic: New Topic: Friction Friction z What does it do? Î It opposes relative motion of two objects that touch ! m a F APPLIED m g N i j j f FRICTION some roughness here Physics 211: Lecture 7, Pg 3 Surface Friction... Surface Friction... z Friction is caused by the “microscopic” interactions between the two surfaces: Friction causes heat ! Physics 211: Lecture 7, Pg 4 Surface Friction... Surface Friction... z Force of friction acts to oppose relative motion: Î Parallel to surface. Î Perpendicular to N ormal force. m a F f F m g N i j j Physics 211: Lecture 7, Pg 5 Model for Sliding Model for Sliding (kinetic) (kinetic) Friction Friction z The direction of the frictional force vector is perpendicular to the normal force vector N in the direction that opposes relative motion between the two surfaces. z The magnitude of the frictional force vector | f F | is proportional to the magnitude of the normal force | N N | . Î | f F | = μ K | N N | ( = μ K | m g g | in the previous example ) Î The “heavier” something is, the greater the friction will be...makes sense! z The constant μ K is called the “coefficient of kinetic friction.” Suitcase Physics 211: Lecture 7, Pg 6 Model... Model... z Dynamics: i : F − μ K N = ma j : N = mg so F − μ K mg = ma m a F m g N i j j μ K mg Physics 211: Lecture 7, Pg 7 Lecture 7, Lecture 7, Act 0 Act 0 Friction Friction A small box of mass is being pulled to the right by a horizontal string. It slides with friction on top of a bigger box. Assume both boxes were initially at rest. In which direction does the frictional force act on the bottom box? A) To the right B) To the left C) Friction does not affect the bottom box, only the top box. M T m friction Physics 211: Lecture 7, Pg 8 Lecture 7, Lecture 7, Act 0 Act 0 Solution Solution z Draw FBD of the top box: m N mg T f = μ K N = μ K mg Physics 211: Lecture 7, Pg 9 Lecture 7, Lecture 7, Act 0 Act 0 Solution Solution z Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2 ....
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This note was uploaded on 02/10/2009 for the course PHYS 211 taught by Professor Tonyliss during the Spring '09 term at University of Illinois, Urbana Champaign.

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Lect07 - Physics 211: Lecture 7, Pg 1 Physics 211: Lecture...

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