Example Problem Find the acceleration of each block in the following frictionless system: m1m1 m2m2 +y +x Coupled system: mass m 2 moves same distance in same time as mass m 1  v 1 = v 2  a 1 = a 2 = a Free–body diagrams: m1m1 m2m2 N m1gm1g m2gm2g T T Apply Newton’s 2 nd Law to block m 1 :  F x = m 1 a x  T = m 1 a x = m 1 a (no accel. in y–direction) (1)  F y = 0  m 1 g – N = 0  m 1 g = N

Example Problem (continued) Apply Newton’s 2 nd Law to block m 2 :  F y = m 2 a y = m 2 a (no accel. in x–direction)  m 2 g – T = m 2 a (2) Combining equations (1) and (2):  m 2 g – m 1 a = m 2 a  a = [m 2 / (m 1 + m 2 )]g and, plugging in for a:  T = [m 1 m 2 / (m 1 + m 2 )]g

Example Problem #5.97 Free–body diagram of car: x y R W N f (a)  F x = ma x = m(v 2 /R)  f = m(v 2 /R)  R = mv 2 /f  f =  s N =  s W =  s mg  R = mv 2 /  s mg = v 2 /  s g = (35.8 m/s) 2 /(0.76)(9.8 m/s 2 ) = m (about 563 ft.) (b) From above, v max 2 =  s gR  v max = (  s gR) 1/2 = [(0.20)(9.8 m/s 2 )(171.7 m)] 1/2 = m/s = 41.0 mph (c) v max = (  s gR) 1/2 = [(0.37)(9.8 m/s 2 )(171.7 m)] 1/2 = m/s = 55.8 mph  The posted speed limit is evidently designed for wet road conditions.