PROBLEM 2-78 (page 55, 13th edition) Three forces act on the ring. The resultant force Fp has magnitude and direction as shown. · Determine the magnitude and coordinate direction angles of F3. 3 Z F2 = 110 N F3 FR= 120 N F1 = 80 N 3 5 45° y 4 1 30° x Figure: 02_P078-079 Copyright @2013 Pearson Education, publishing as Prentice Hall
2-78 (page 55, 13th edition) COMPLETION OF PROBLEM (page 1) · Forces (suppressed units) 0 3.00 F = 80 1 -i +-k 5 5 F2 = 0110 k F2 = Fi +Fj +Fk U Fp = 120(cos 45 sin 30 i + cos45[ cos 300 j + sin 45 k) U . Equations for the resultant force ? F = F i +F j+F k Rz Rx Ry F = IF : X 120 cos 45ª sin 300 = 80 4 + F Rx 5 # y F = IF : x FD = CF : Ry y 120 cos 45- cos 30! = F 110+ F Rz Z 120sin 45 = 80 3 5 Z
2-78 (page 55, 13th edition) COMPLETION OF PROBLEM (page 2): SOLVING THE EQUATIONS . Solve for F , F , F : N F = 120 cos 45 sin 30 1 80 4 0 = A X 5 F = 120 cos 45" cos 30! = B y F = 120sin 45 0 80 3 H 5 + 110= C `z · Magnitude and coordinate direction angles: F3 = VFX 2 + F2 + F2 = VA2+B2+C2 = F = 166N I > = cos (F / F3) 3 X = cos 1(A/F) = 97.54 13 = cos"1(F / F3) = cos11(B/F) = 63.70 3 cos11(F_/F3) = cos11(C/F) = 27.5