PROBLEM 4-53 (page 146, 13th edition) PROBLEM 4-55 (page 145, 12th edition) • Determine the moment of F about A. • Determine the lever arm for F . ! • Determine the magnitude of the moment of F about an axis extending from A to C. Z, A 4 ft y x 3 ft C 2 ft B F = {4i + 12j - 3k} lb PROB04_054-055.jpg Copyright @ 2010 Pearson Prentice Hall, Inc.
4-53 (page 146, 13th edition) 4-55 (page 145, 12th edition) COMPLETION OF PROBLEM (page 1) MOMENT OF THE FORCE AND LEVER ARM • Moment M of F about A (suppressing units) ! ! = LAB' ! ! i ! ! ! j k 1 3 !2 !3 4 12 =i[3(!3)! 12(!2)]! j[4(!3)! 4(!2)]+ k[4(12)! 4(3)] MA= (15i + 4j + 36k)lb!ft · Magnitude and coordinate direction angles of M MA= 39.2lb !ft A ! = 67.5° ! = 84.1º ! = 23.2° • Magnitude of F F = 13.0lb • Lever arm for F d = M /F = 3.02ft
4-53 (page 146, 13th edition) 4-55 (page 145, 12th edition) COMPLETION OF PROBLEM (page 2) MOMENT OF THE FORCE ABOUT AN AXIS ! of the moment of F about AC . Magnitude M AC ! ! = ‘AC' A AC ! ! ! u = LAC = 4i + 3j ! -= 0.8i+0.6j AC IAC V 42 + 32 ! ! MA=(15i + 4j + 36k)lb!ft M =14.4 lb !ft · Alternative method (without first calculating MA) IAC= UACIMA =LAci(? ! F) ! ! ! = 0.8i+0.6j 'AC ! r B= (4i + 3j !2 k)ft AB ! F=(4i + 12j ! 3 k)lb ! ! ! = 4 AC M 0.8 0.6 0 !3 !2 4 12 !3 lb "ft =14.4 lb"ft