PROBLEM 4-53 (page 146, 13th edition) PROBLEM 4-55 (page 145, 12th edition) · Determine the moment of F about A. ! · 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 MOMENT OF THE FORCE ! · Moment M, of F about A (suppressing units): M =i ! F = ! ! ! ! i j k 4 . 3 !2 4 12 !3 =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 ! = 67.5° ! = 84.1º ! = 23.2° · Magnitude of È : F = 13.0lb Lever arm for È : d = M /F = 3.02ft
4-53 (page 146, 13th edition) 4-55 (page 145, 12th edition) COMPLETION OF PROBLEM MOMENT OF THE FORCE ABOUT AN AXIS . Magnitude MA AC Mac= uAciM AC of the moment of F about AC: AC A ! 42 + 32 ! ! ! ! ! IAC = 4i + 3j u = 'AC IAC = 0.8i+0.6j V MA=(15i + 4j + 36k)lb!ft MAC=14.4 lb !ft · Alternative method (without first calculating MA ): MAC= LAC? MA =LAC? (TAR! ?) ? = 0.8i + 0.6j 'AC ! ! ! ! r .= (4 i + 3j ! 2 k)ft AB ! ?= (4 i + 12j ! 3 k)lb M AC 0 0.8 0.6 4 3 !2 4 12 !3 lb "ft =14.4 lb"ft