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Friction

Friction. Definition and Cause. A reaction force present when one object is in contact with another and moves (or tries to) across their mutual surface Due only to two factors: The nature of the surfaces in contact (smooth, rough, etc)

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Friction

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  1. Friction

  2. Definition and Cause • A reaction force present when one object is in contact with another and moves (or tries to) across their mutual surface • Due only to two factors: • The nature of the surfaces in contact (smooth, rough, etc) • The amount of normal force exerted by each surface on the other (Fn) • Surfaces: even the smoothest surfaces have imperfections (ridges, depressions) at the microscopic scale From http://www.pxleyes.com/blog/2010/03/5-things-you-need-to-know-about-raytracing/ • When these surfaces move across each other, the ridges and depressions bind and release, generating a force in the opposite direction to motion

  3. Causes (cont.) • Given the previous picture, explain why an object in contact with paper has more friction than the same object in contact with aluminum. • Explain the statement: the amount of friction between two surfaces is different than the amount of friction between any two other surfaces • Changing the normal force also changes the amount of friction: FN Ff Fapp FN Fg 2 x mass = 2 x FN = 2x Ff Fapp Ff Fg

  4. Example Fappx • The Fappshouldbe broken into its components: • Fnet= 0 (uniform motion) • Ff = -Fappx • Fappx= 150 (cos 15) =144N Ff = -1.4 x 102 N = • What is the force of friction if the crate is sliding at a constant speed when being pushed at an angle of 15 (down) with a force of 150N? 15 Fappy 150N FN Ff Fappx Fapp Fg+ Fappy

  5. Example 2 Fappx • Look at the vertical components now • Fnet= 0 (uniform motion) • FN = -(Fg+ Fappy) • FN = -(mg + -150sin15) =144N=-[(58kg)(-9.81m/s2)–150sin15]= 607N = 6.1x102 N • If the crate is 58kg, what normal force is the ground applying to it? 15 Fappy 150N FN Ff Fappx Fapp Fg+ Fappy

  6. Static and Kinetic Friction • Static friction occurs between two surfaces before motion begins – tries to prevent motion • Strongest Frictional Force • Kinetic Friction occurs between two surfaces after an objet is in motion…tries to slow it down • Weaker than static friction • Explain Why?

  7. Coefficient of Friction • Since the amount of friction is different for every 2 surfaces, and proportional to Fn, we can define an equation for Ff • Ff = Fn , where  (pronounced mue) is called the coefficient of friction ( •  is determined empirically (experimentally) for every 2 surfaces in contact • Every two surfaces have a static  and a kinetic  • A simple set-up to measure : • Using Fnet= Fapp+ Ff = 0 if the block is not moving (static) or moving at constant speed (kinetic) then:Ff = -Fapp(measured by scale) • Ff = Fn so  = Ff/ Fn • Fn = -mg (in this case) •  = Ff/(mg) [note: drop all‘-’ signs…just interested in number not direction] Picture Reference

  8. Coefficient of Friction example • A 75.0kg sled needs 250N of force to get pushed into motion. What is the coefficient of static friction between the runners and the snow? • Since Fnet= 0 (looking at the point of first motion) • Ff = Fapp= -250N • Ff = Fn so  = Ff/ Fn • Fn = -mg (in this case) •  = Ff/ (mg)= 250N/[(75.0kg)(9.81m/s2)]= 0.334 FN Ff Fapp Fg

  9. Coefficient of Friction example 2 • The coefficient of kinetic friction for the same sled is 0.25. Calculate how much force it would take to accelerate it at 3.2 m/s2 after it was already in motion. • Fnet= Fapp+ Ff = ma • Fapp= ma – Ff • Ff = -Fn (direction is important here!) • Fapp= ma –(-Fn) • Fn= -mg • Fapp= ma –[- (-mg)]= (75.0kg)(3.2m/s2)+[0.25(-75.0kg)(-9.81m/s2)] = 424N = 4.2x102N FN Ff Fapp Fg

  10. Assignment • Read pg 172-173, 182-183 • Do pg 172 pracprob 1 • Do pg 185 pp 1-2 • Do pg 188 pp1 • Work on assignment (can do coefficient of friction questions)

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