Question

3. 3D RIGID BODY Determine the support reactions at O (ball-and-socket) and the tensions in cables AB and AC. (Helpful Hints: Point Clies on the xy-plane, which means the z-coordinate for Cis 0. The rod is vertical.) Dimensions in meters 20 KN 22

Answer 1

Z A ath TAB F= 20KN 02 Solution:- Stept:- Draw FBD of rod OA:- point o(0,0,0), A(0,0, 8) BC3, -5,2), ((-7,-4,0) • unit vectors along cables! - ij AciA = (-7-0) 1 + (4-035 - B (7)2+642+(8)2 Î clA==0.61 -0.355-0.7 3î - 59+(2-8) (3)+(5) HOP hop = 0-36-0-69-0-72m Yox y so x i Anin fig: FBD BIA in TARE TO ABA All forces in vector form: - / ² = 2o Ĵ KN ii) ∞ = Oxâ+ oyộtozik - (Reaction at o) iii) JAC = Tc AclA position Veetors of point A and point o from origin o : - ij g = 4 Step2!? Now using equilibrium equations (in 30) and solve for an unknowns: - so (Net moment about 0 =0) ij JA= 8k A e may » 8 x8+ TAXTAC +8AX TAR TO

→ 4R X (209) + 8 $*(-0.62€-0-359-0.7k). To + BÅ x(0.36 ê - 0.69 -0.728) TB=0 + - 80 i + (-4.969+2.8 JTc+[2-889+4.Bî]To=o => [-80 +2. 8TC+4.8T6]î + (-4.967c+2-88T8] 9 =0 solving this yi j - 4.9656 = -2.88TB => Tc= 0-581TB) ii) -80+ 2.8 Tc+ 4.8TR= 0 -> - 80 +2.8X0-581TB+ 4.BTB=0 To= 12.45 kN) - Tension in AB and Fc= 7-23 KN Tension in AC y Any B) E Fau=o (Vector sum of all forces =o) > Řtot TACT TAB > 209+ Oxît oyš točk +7.23(6-622 –0-359-0-78) + 12:45X(0-36î -0.69 -0-72k)=0 solving this jj Ox= 7.23x0.62 -12.45X0-36 10x ? OKNI Ang -20 +7.23X0-35 + 12.45 X0:6 loy= -10KN Ary ) 02= 7.23X0.7+12:45X0.72 Oz = 14.025 KN) Any ii) Oy
Ox, Oy and Oz are the support reactions at ball and socket joint at O.

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