The document discusses friction and its relationship to static and kinetic forces. Static friction acts before motion, while kinetic friction acts during motion. Both are related to the normal force and can be expressed using coefficients of friction. Common friction problems involve determining if motion will occur based on applied forces and coefficients, or finding coefficients given force conditions. Sample problems demonstrate setting up force diagrams and equations to solve these types of problems.

1. Chapter 8 - Friction Sections 8.1 - 8.4

19. Sample Problem 8.1 A 300 lb block is on the inclined plane.  s = .25 and  k = .2. Determine whether the block is in equilibrium and the value of F. FBD 300 lb 100 lb y x F N 3 3 4 4 5 5

20. Sample Problem 8.1 cont. ΣF x = 0 = 100(4/5) - F(4/5) - N(3/5) N = 100(4/5 x 5/3) - F(4/5 x 5/3) N = 133.3 - 1.33 F ΣF y = 0 = 100(3/5) - 300 + N(4/5) - F(3/5) F = -240(5/3) + N(4/5 x 5/3) F = -400 + 1.33 N F = -400 + 1.33(133.3 - 1.33F) F + 1.72 F = -400 + 173 2.73 F = -227 F = -83.1 lb

21. Sample Problem 8.1 cont. N = 133.3 - 1.33(-83.1) N = 133.3 + 110.6 N = 243.8 lb F m =  s N = .25(2438) = 62 lb since F > F m the block will slide and F actual =  k N = .2(243.8) = 48.8 lb

22. Sample Problem 8.2 If  s = 0.35 and  k = 0.25, determine the force P needed to (a) start the block moving up the incline, (b) to keep it moving, and (c) to prevent it from sliding down. F m =  s N F m = .35N 25° 25° N F P 25° 800 N y x

23. Sample Problem 8.2 cont. ΣF x = 0 = - P + F cos 25° + N sin 25° P = .35 N(cos 25°) + N sin 25° P = .317 N + .423 N P = .74 N ΣF y = 0 = -800 + N cos 25° - F sin 25° 800 LB = .906 N - (.35 N) x .42 N = 800/.758 = 1055 N P = .74(1055) = 781 N So when P exceeds 781 N the block will start to move.

24. Sample Problem 8.2 cont. Once the block starts moving: Use the same equations, but F =  k N = .25 N ΣF x = 0 = - P + F cos 25° + N sin 25° P = .25 N cos 25° + N sin 25° P = .65 N ΣF y = 0 = -800 N + N cos 25° - F sin 25° 800 = N cos 25° - .25 N sin 25° 800 = .8 N N = 1000 N P = .65 x 1000 = 650 N

25. Sample Problem 8.2 cont. To prevent the block from sliding down: F m =  s N ΣF x = 0 = - P - F cos 25° + N sin 25° P = -(.35 N)cos 25° + N sin 25° P = -.317 N + .423 N P = .106 N ΣF y = 0 = -800 + N cos 25° + F sin 25° 800 = .906 N + .423(.35 N) 800 = 1.05 N N = 759 N P = .106(759) P = 80.5 N N P 25° w F

26. Solution 8.1 Given:  = 30°, P = 50 lb Find: Friction Force acting on block.  P 250 lb  s = 0.30  k = 0.20

27. Solution 8.1 cont. 250 lb x y F N 50 lb 30° 30° Assume Equilibrium + ΣF y = 0: N - (250 lb)cos 30° - (50 lb)sin 30° = 0 N = +241.5 lb N = 241.5 lb + ΣF x = 0: F - (250 lb)sin 30° + (50 lb)cos 30° = 0 F = +81.7 lb F = 81.7 lb

28. Solution 8.1 cont. Maximum Friction Force: F m =  s N = (0.30)(241.5 lb) = 72.5 lb Since F > F m , Block moves down Friction Force: F =  k N = (0.20)(241.5 lb) = 48.3 lb F = 48.3 lb

29. Solution 8.2 Given:  = 35°, P = 100 lb Find: Friction Force acting on block.  P 250 lb  s = 0.30  k = 0.20

30. Solution 8.2 cont. 250 lb x y F N 100 lb 35° 35° Assume Equilibrium + ΣF y = 0: N - (250 lb)cos 35° - (100 lb)sin 35° = 0 N = +262.15 lb N = 262.15 lb + ΣF x = 0: F - (250 lb)sin 35° + (100 lb)cos 35° = 0 F = +61.48 lb F = 61.48 lb

31. Solution 8.2 cont. Maximum Friction Force: F m =  s N = (0.30)(262.15 lb) = 78.64 lb Since F < F m , Block is in equilibrium Friction Force: F = 61.5 lb

32. Solution 8.3 Given:  = 40°, P = 400 N Find: Friction Force acting on block. 800 N  25°  s = 0.20  k = 0.15 P

33. Solution 8.3 cont. 400 N 800 N N F x y 40° 25° 15° 25° Assume Equilibrium + ΣF y = 0: N - (800 N)cos 25° + (400 N)sin 15° = 0 N = +621.5 N N = 621.5 N + ΣF x = 0: - F + (800 N)sin 25° - (400 N)cos 15° = 0 F = -48.28 N F = 48.28 N

34. Solution 8.3 cont. Maximum Friction Force: F m =  s N = (0.20)(621.5 N) = 124.3 N Since F < F m , Block is in equilibrium F = 48.3 N