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Chapter 4 Sec 8

Applications involving Friction. Chapter 4 Sec 8. What is Friction?. Friction is a FORCE that opposes or impedes the motion of an object. Friction is caused by microscopic bumps between solid objects in contact.

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Chapter 4 Sec 8

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  1. Applications involving Friction Chapter 4 Sec 8

  2. What is Friction? • Friction is a FORCE that opposes or impedes the motion of an object. • Friction is caused by microscopic bumps between solid objects in contact. • Friction can exist from sliding objects or rolling objects (though rolling friction is less than sliding). • Friction depends on the types of materials AND on the normal force (or weight)of the object.

  3. Types of Friction • Friction between solid objects sliding against each other (or rolling) is called Kinetic Friction. (Greek for moving) • Friction between solid objects can exist parallel to the surface of them even when they are not moving. This is Static Friction. • Each type of material changes the amount of Friction present, so we have a coefficient of friction, μk or μs, for kinetic and static coefficients. • Friction depends very little on surface area.

  4. Friction Coefficients • The friction force, Ffr, is always perpendicular to the normal force, FN. • In calculating kinetic friction: • μ depends on whether the object is wet or dry, what type of finish is on them, but NOT on speed of the objects sliding.

  5. Static Friction • Static friction is a force that exists between objects that are in contact, but NOT moving when a force is applied. • Eventually with a hard enough push, the object will move and kinetic friction takes over as you exceed the MAXIMUM static friction. • Fmax = μsFN and since static friction can vary from 0 to max, we write • μs is generally more than μk as its harder to start an object than keep it moving.

  6. Example 1 • A 10.0kg box rests on the floor. μs = 0.4 and μk =0.3 • Determine the force of friction, Ffr, acting on the box if the horizontal applied force, FA = 20N. If FA = 40N. Draw a free body diagram and label all forces.

  7. Solution for Example 1 • In the vertical direction there is no motion, so the net force, Σfy = ma = 0 yields FN – mg = 0. • In all cases, the normal force, FN = mg = (10.0kg)(9.80m/s/s)= 98 N. • The force of static friction will oppose any applied force up to the maximum of • Ffr = μsFN = (0.40)(98N)=39N • If FA = 20 N, the box won’t move so Ffr = -20N to balance the applied force.

  8. FN Continued FA Ff Fg • If FA = 40N, which is more than the maximum static friction force, the box will accelerate and we have kinetic friction, Ffr = μkFN • Ffr = (0.30)(98N) = 29 N. • ΣFx= FA + Ffr = 40 N + (-29N) = 11N, so the box will accelerate at a = Fnet / mass. • a = 11N / 10.0 kg = 1.1 m/s/s in a direction of the applied force. (positive horizontal).

  9. More Examples • Look at the Example 4-16 on page 99. (two boxes on a pulley) • Now look at the Example 4-17 on page 100. (skier on a hill) • Recall how to obtain components of weight when not perpendicular to the surface? • FN is always perpendicular to the surface and Ffr is always parallel to the surface here.

  10. Your turn to Practice • Please do Chapter 4 Review p 107 #s 38 and 39 • Please do Chapter 4 Review p 108 #s 40, 42, 43, 44, 46.

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