1. Applications involving Friction

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, μ k or μ s, for kinetic and static coefficients.  Friction depends very little on surface area.

4. Friction Coefficients  The friction force, F fr, is always perpendicular to the normal force, F N.  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.  F max = μ s F N and since static friction can vary from 0 to max, we write  μ s is generally more than μ k as its harder to start an object than keep it moving.

6. Example 1  A 10.0kg box rests on the floor. μ s = 0.4 and μ k =0.3  Determine the force of friction, F fr, acting on the box if the horizontal applied force, F A = 20N. If F A = 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, Σf y = ma = 0 yields F N – mg = 0.  In all cases, the normal force, F N = mg = (10.0kg)(9.80m/s/s)= 98 N.  The force of static friction will oppose any applied force up to the maximum of  F fr = μ s F N = (0.40)(98N)=39N  If F A = 20 N, the box won’t move so F fr = -20N to balance the applied force.

8. Continued  If F A = 40N, which is more than the maximum static friction force, the box will accelerate and we have kinetic friction, F fr = μ k F N  F fr = (0.30)(98N) = 29 N.  ΣF x = F A + F fr = 40 N + (-29N) = 11N, so the box will accelerate at a = F net / mass.  a = 11N / 10.0 kg = 1.1 m/s/s in a direction of the applied force. (positive horizontal). FAFA FfFf FgFg FNFN

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?  F N is always perpendicular to the surface and F fr 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.