PROBLEM 4-48 (page 144, 14th edition) Force F acts perpendicular to the inclined plane. · Determine F as a Cartesian vector. · Determine the moment of F about A as a Cartesian vector. U . Determine the magnitude of the moment of F about A. . Determine the coordinate direction angles of the moment of F about A. Z A 3 m F = 400 N 3 m B C x 4 m y 04_PROB_048-049 Copyright @2016 Pearson Education, All Rights Reserved
4-48 (page 144, 14th edition) COMPLETION OF PROBLEM (page 1) FORCE AS A CARTESIAN VECTOR · F is perpendicular to the inclined plane. That is, F points in the direction of II ICB CA · Suppressing units: That is, where X = 400/ 42 + 32 + 42 CA CB = L r U i j k = 3(4i + 3j + 4k) U 0 4 3 3 4 0 F= (41 + 3j + 4k) X · Re-inserting units: F = (250 i + 187 j + 250k)N
4-48 (page 144, 14th edition) COMPLETION OF PROBLEM (page 2) FORCE AS A CARTESIAN VECTOR ALTERNATIVE DERIVATION · F is perpendicular to the inclined plane. That is, F is perpendicular to ? CA and to I CB u . U Fil =0 CA and CB Fif =0 · Writing ? ? ? F =ai + bj + ck it follows that and SO b = 3a/4 and Fif A =14bb+ 3c = 0 Filep =3al 4b = 0 c = a L . It follows that F can be written in the form given on the previous page: where ? ? F=(4i + 3j + 4k) X ? X = 400/ 42 + 32 + 42