46 30° W W || W  FAFA F || FF FNFN Equations from Diagram: W || = F || F N = F  + W  m = 5 W = 49 W  = 42.4 W || = 24.5 cos30 = F || F A cos30 =

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Presentation transcript:

46 30° W W || W  FAFA F || FF FNFN Equations from Diagram: W || = F || F N = F  + W  m = 5 W = 49 W  = 42.4 W || = 24.5 cos30 = F || F A cos30 = 24.5 F A Same as W || F A = 28.3 N F N = F  + W  F N = F N = 56.6 N

50 T W T - W = F NET T - 5(9.8) = 5(3) T = 64 N

53 FNFN W FfFf F N = W F f = F NET Find d: v i = 7 v f = 0 d = ? a = Find a:Find F f : same as F NET F NET = m a 32.3 = 65.8a a =.491 F f =  F N F f =.05(645) F f = v f 2 = v i 2 + 2ad 0 2 = (.491)d d = 49.9 m

55 30° W W || W  FNFN FfFf W || - F f = F NET F N = W  m = 3 W=29.4 W  =25.5 W || = 14.7 A. v i =0 d = 2 t = 1.5 a = ? d = v i t + ½ at 2 2 = ½ a(1.5) 2 a = 1.78 m/s 2 B. W || - F f = F NET 14.7 – F f = 3(1.78) F f = 9.36 N F f =  F N 9.36 =  (25.5)  =.367 C. Already found this ! 9.36 N D. v f 2 = v i 2 + 2ad v f 2 = 2(1.78)2 v f = 2.67 m/s

56 FfFf W W – F f = F NET Find v f :Find a :Find F NET : v i = 0 v f = ? d = 25 a = F NET = ma W – F f = F NET 75(9.8) – 95 = F NET F NET = = 75a a = v f 2 = v i 2 + 2ad v f 2 = 2(8.53)(25) v f = 20.7 m/s

59 35° W W || W  FNFN FAFA FfFf W || = F f W  + F A = F N m = 3 W = 29.4 W  = 24.1 W || = 16.9 Find F A : W  + F A = F N Find F A : F f =  F N 16.9 =.300F N F N = F A = 56.3 F A = 32.2 N

61 °° W W || W  F N FAFA W || = F A W  = F N sin  = W || W sin  = Same as F A  = 1.4°

A. Whole: W FNFN FAFA F NET = ma 180 = 9a a = 20 m/s 2 For ALL blocks B. Resultant force is NET force F NET = ma = 2(20) = 40 N For m 1 F NET = ma = 3(20) = 60 N For m 2 F NET = ma = 4(20) = 80 N For m 3 C. Between m 1 and m 2 : W FNFN F from m1 F NET = ma = 7(20) = 140 N D. Between m 2 and m 3 : 80 N…..it is the NET force on m 3