PROBLEM 2-79 (page 54, 12th edition) · Specify the magnitude of F3 and its coordinate direction angles &3, B3, Y3 so that the resultant force is 9 j kN. Z F2 = 10 kN F3 13/15 Y3 B3 12 Q3 30° y F1 = 12 kN x PROB02_079.jpg Copyright 2010 Pearson Prentice Hall, Inc.
2-79 (page 54, 12th edition) COMPLETION OF PROBLEM (page 1) · Forces (suppressed units) F1 = 12(cos30° ] - sin 30°k) F2 =10|- 13 12 - 5 it-k 13 1 F3=F}[+F,j+F,k R FR=9] · Equations for the resultant force Fp = Fp i+Fp j+FR k FRY = EF .: 0 = 13 x 120 + F 9 = 12cos 30° + F. y FRY = >F: Ry FRy = EF .: 0 =- 12sin 30° + 50 + F Rz 13 Z
2-79 (page 54, 12th edition) COMPLETION OF PROBLEM (page 2): SOLVING THE EQUATIONS . Solve for F., F., F .: 120 F = 120 = A 13 F =9-12cos30° = B y 50 F =12sin 30° = C 2 13 · Magnitude and coordinate direction angles: F3= VF?+F2+F2 = VA2 +B2 +C2 = F = 9.58 kN "3 = cos-1(F}/F3) = cos-1(A/F)=15.5° B3 = cos-1(F, /F3) = cos-1(B/F)=98.4° Y3= cos-1(F_/F3) = cos-1(C/F)=77.0°