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 =FI F = A TAB i k j 4 3 2 4 12 3 =1 [3(13)112(12)][ j[4(03)[4(02)]+ k[4(12)14(3)] MA = (15 i + 4 j + 36 k)lb ft Magnitude and coordinate direction angles of M : A: MA= 39.2lbIft I = 67.5° I = 84.1º I = 23.2° · Magnitude of F : F =13.01tb Lever arm for F : 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 M =u iM `AC of the moment of F about AC: 'AC A 1 42 + 32 =0.8i+0.6j 'AC IAC V 4i+ 3j = TAC = MA = (15 i + 4 j + 36 k)lb ft MAC =14.4 lb ft · Alternative method (without first calculating M ): AC = uAciMA =u i(FARI F) ?= 0.8i +0.6j TAR= (4 i + 3j 12 k)ft F = (4 i + 12 j 3 k) lb MAC= AC 0.8 006 0 4 3 2 4 12 03 lb ft =14.4 lb ft