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The sign has a mass of 90 kg with center of mass at G (Figure 1) Part A Determine the z, y, z components of reaction at the ball-and- socket joint A Enter the z, y, and z components of the reaction using three significant figures separated by commas. A, A A: 0,550,433 Figure 1 of 1 > swers Incorrect; Try Again: One attempt remaining Part B Detertmine the tension in wires BC and BD Express your answer using three significant figures separated by a comma. Трс, TBD: 530,265 N I m 3:11 PM

Answer 1

(-2,0,1) (1) D, 2) TG WB 6,2,0) ya W = 90x9.81 2882.90 free body diagram At À , (Ax , Ay, A₂) , K = Ani + hy jt Ask Ho -882.9K N UBD & UBC are be unit vector along 130 &BC FSO = FBO TURO FBc = FBC Tube JBP = (-2,0,1) - 20, 20). Į (-21 -2j+k) 12² +2²+12

ов и С,9, 2)-(, 2,5 - у ?+ 2*+21 г/; -2 -1 2 * бар Као (-a-aj• %) би га (i-j++) 2 га го іс Erl, - Pag x + * * * * * да го лв е j, бл. * і те, јx (-si-+) бар + 2x (і-зја) + jх (-882 г: ( 4к +25) Кво + (-x + 4 ) тек – 582 vi

Equating ; jk components to zero 1 2 3 F30 +41 FBC - 882.970 0 1. / F3P - 2 - FBC =0 solving the se equations € 0 & 2 co FBD = 264.87 N FBC = 529.74 N force equilibrium EF=O » FA & Foot For +W zo it; (Axi + Ayj + A₃1) + 2 (~1.j + 2 ) F30 * 3 (1-23 +21) EBC + -882.4 K =0 Ax = 2 3 F30 -13 FBC = 23 x 264.84 -1 23x524.34 - Ax=0 Ay = 2/3 EBD + 2 FBC 2 2 x (529.74+264.87 ₂529.74N Az = - 3 FRD - 2 3 3 +63 2:4 = 441.45N

PART A Components of Reactions at A Ax= 0 Ay - 529.I4N A2 = 441.450 PART B Tension in wire Bo, FBO = 264.87% Tension is wire BC, FRC = 529.7

Answers

PART A

Components of reaction at A

Ar = 0

Ay = 529.74N

A = 441.45N

PART B

Tension in wire BC = 529.74N

Tension in wire BD = 264.87N