PROBLEM 3-43 (page 109, 13th edition) · Determine the magnitude and coordinate direction angles of the force P required to keep the concurrent force system in equilibrium. Z (-1.5 m, 3 m, 3 m) P F2 = 0.75 kN F3 = 0.5 kN 45° 120° y 60° 1 F1= 2 kN x Figure: 03_P043 Copyright @2013 Pearson Education, publishing as Prentice Hall
3-43 (page 109, 13th edition) COMPLETION OF PROBLEM (page 1): FORCES AND EQUATIONS OF EQUILIBRIUM · Forces (suppressing units) x y z k U U F = 2(cos45 i + cos60 j + cos120 k) ? F2 = (01.5i + 3j + 3k) A L A = 0.75/ v1.52 +32 + 32 ? L F2 = 0.5j · Equations of equilibrium U U F = F i +F +F k =0 R Rx Ry Rz F .= IF = 0: × P + 2cos 45 01.5A = 0 Rx X F. = IF =0: P + 2 cos60 + 3A 10.5= 0 Ry y y F =IF =0: Rz Z P +2cos120 +3A = 0 Z
3-43 (page 109, 13th edition) COMPLETION OF PROBLEM (page 2): SOLVING THE EQUATIONS OF EQUILIBRIUM . Solve for P , P , P : P = 12cos45 +1.5A = B X y P = 02 cos 60 13A + 0.5= C P= 2cos120 13A = D N · Magnitude and coordinate direction angles: P= P2+P2+P2 = VB2+C2 +D2 = 1.61kN [ p = cos"1(P /P) = cos11(B/P) = 1369 Ip = cos"1(Py /P) = cos11(C/P) =128! Ip = cos11(P /P) = cos11(D/P) = 72.00