Transcript Friction
Chapter Friction and Equilibrium Friction Forces When two surfaces are in contact, friction forces oppose relative motion or impending motion. F Friction forces are parallel to the surfaces in contact and oppose motion or impending motion. Static Friction: No relative motion. Kinetic Friction: Relative motion. Friction and the Normal Force FN 8N 4N 2N 12 N FN 4N FN 6N The force required to overcome static or kinetic friction is proportional to the normal force, FN. Fs = msFN Fk = mkFN Friction forces are independent of area. 4N 4N If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact. Friction forces are independent of speed. 5 m/s 2 N 20 m/s 2 N The force of kinetic friction is the same at 5 m/s as it is for 20 m/s. Again, we must assume that there are no chemical or mechanical changes due to speed. The Static Friction Force When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value. FN Fs F Fw In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write: Fs = msFN Constant or Impending Motion For motion that is impending and for motion at constant speed, the resultant force is zero and SF = 0. (Equilibrium) Fs Fa Fk Fa Rest Constant Speed Fa – Fs = 0 Fa – Fk = 0 Here the weight and normal forces are balanced and do not affect motion. Friction and Acceleration When Fa is greater than the maximum fs the resultant force produces acceleration. Fk a Fa Constant Speed Fk = mkFN Note that the kinetic friction force remains constant even as the velocity increases. Case 1 EXAMPLE 1: If mk = 0.3 and ms = 0.5, what horizontal force Fa is required to just start a 250-N block moving? 1. Draw sketch and freebody diagram as shown. FN Fs Fa + Fw 2. List givens and label what is to be found: mk = 0.3; ms = 0.5; Fw = 250 N Find: Fa = ? to just start 3. Recognize for impending motion: Fa– Fs = 0 EXAMPLE 1(Cont.): ms = 0.5, Fw = 250 N. Find Fa to overcome Fs (max). Static friction applies. For this case: Fa – fs = 0 FN Fa Fs + 4. To find Fa we need to know fs , which is: fs = msFN 250 N 5. To find FN: SFy = 0 Fw = 250 N FN = ? FN – Fw = 0 FN = 250 N (Continued) EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find F to overcome fs (max). Now we know FN = 250 N. 6. Next we find fs from: fs = msFN = 0.5 (250 N) 7. For this case: Fa – fs = 0 Fa = fs = 0.5 (250 N) Fa = 125 N FN fs Fa + 250 N ms = 0.5 This force (125 N) is needed to just start motion. Next we consider F needed for constant speed. EXAMPLE 1(Cont.): If mk = 0.3 and ms = 0.5, what horizontal force Fa is required to move with constant speed? (Overcoming kinetic friction) SFy = may = 0 mk = 0.3 fk FN FN - F w = 0 Fa + mg F N = Fw Now: fk = mkFN = mkFw SFx = 0; Fa = (0.3)(250 N) Fa - fk = 0 Fa = fk = mkFw Fa = 75.0 N The Normal Force and Weight The normal force is NOT always equal to the weight. The following are examples: FN m F Here the normal force is less than weight due to upward component of F. 300 Fw F FN Fw Here the normal force is equal to only the component of weight perpendicular to the plane. Case 2 pushing Case 2 pulling Case3 on a ramp Example 3: What push Fa up the incline is needed to move a 230-N block up the incline at constant speed if mk = 0.3? Fa Step 1: Draw free-body including forces, angles and components.