Question

Cables CE and BF along with the ball and socket at A support the rigid bar ABCD. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) z 6 m 4 m 2 ms 2 m F = -(175 NK Determine the tension in each cable and the reaction components the ball and socket joint exerts on ABCD at A in newtons. (Indicate the direction with the sign of your answer.) z z zz z

Answer 1

rectors, Stept coordinates of various points ACHA, 7A, 2A) = A(0,0,0); Cohe, ye, Ze) 22(0,6,0) B[948, 48, 78) = B(0,4,0); 0620,40,20)=0(0,8.6) E CHE , Yes ?)-E (8,0,9) ;F (MF, 47,2)=f4-6,0,4) Step I Dusplacement vectors, position vectors & wito let î j k be unit vectors along x, y, z. "vector AB, RAB= (26k tlys-YAJJ+CZE ZAJTE AB=(0-0) it (4-0) it (0-0) Å = AB= oitaj toka Pectorac, rac = (x,-HA) itly.-YA)j +(Ze-ZAJE rac = oitoj toim Similarly vector AB, RAD= oi +8jtorm. *Unit vectors BF & C Ê Cû or &ûce) Unit vector û BF = (26-28), it CYF-Y8) jt (2,2k 1 Vera-26]²+ (ys-Y3)²+(2,26 ů Br=-62-45+41 û Br=-32-23 +21 V66)+(-49€ +1432 L V •Similarly unit vector ūCE = OLEK) û + Cye Yc)j +(2220? - VODE 26²+(YEYJ²+(225 2 ae - (-0)+(^-8)3+(2-0)* 0 - 4A-38 ++ s(8)+(-63*76437 que )

Step In urriting forces in vector form • Reaction at A, PA = An ît. Ay it AzÊN • Force at D, f = -175 PN van fijn RF •Tension in cable BF, 7 of = magnitude Xunit vector alag Per = TBF Û Bf Por=TBf 1-3 2-21 +2F) N • Also tension in cable CE, PLE = TCE û ce - Per=Tie (A2-38 +26)N Stef te Equititorium Conditions moment of force, m = ² x 7 cross product Using figl ci Net moment of all forces about A should be zero for equilibrium of A B C D bar, EM=0 h ABX Port RacX PETRAX Po

All the displacement vectors &force vectors are found in step #fI so EñA=0 > 10 4 oltlob oltlo 8ol -1751 17 13 FAXTBF FoxPce za Box coa toj top IBE oa olta o 6olt lo8 ol TBE 7- 3221 4 -321 lo o last zoitoj to solving the determinants we get: » Top (8i+oſ+2x) + Tee (zitoſ-247) +41400 i toj tof)=0 i fojtor? comparing x, y, z components of left hand side (Les) with Hy, 2 components of fight hand side (RUS) des Stereo +12 TEE -140020 - y- o=0 2 7 12TBE -24 TCE =0 -6 क 129 V29

solving eg 0 &0 TE=412.31060] TCE=269.25820 (11) Another equilibrium condition is that force on berascD =0 > EF 20 & Port TCE +P+PA=0 3 412.3106/-7-22 722) +269.2582/42 - 3jt2K) VIA Ir29 , +(oitoj=175E) + (A. U + Aggt Az R) zoitoj toile comparing xy, z of this with 2,4, 2 of RHS =- X=-3X412-3106 +269-258284 1+0+ Age=0 1A = loon] = fax yo2X412-3106 -3_x269.2582 to +Ay=0 - Ay = 350 N = Fay 2+2x412-3106 +2 x269.8582-175+Az=0 A2= -125 N = FAZ TBF=412.31060 TCE = 269.25820 | Ax=loon , Ay= 350N ,A₂=-1250 UTA Jag ^ Answers