7 .324 Static 2 9 1 4 component Dallas Hallsten Pe=Pcos (46)-535(03 (469) Py = 409 & IN x-2 complement P+z = Psin (1/2") = 5 35 sin (46 ") Piz: 343.89N * comparant Pe=297.818U Px=Px= cos (30)=343.89609(300) 2 component P= = Pxz Sin (30°) = 343. 8 9 sin ( 30") P= = 171.945 N (b) Fy = P cos (40º) = 535cos(40°) 364) F =409.83 Q= cos ( 1 ) = COS (40 4. 83 ) = 40° Q 535 N - PxZ = 343. 89N Fv = Px= ces(30)=343,89cos(38,7065) Fr = P == sin (30) = 343.89 sin(30) 171.95 COS"(Ex)=Q 247.82 Cos' (p)= 0 535 - /171.95) cos/ 297.821 1 Fz- 38;) 605 (535 8W Q= = 71.25° 0% = 56.170 V
1 2) FAO = 303/45 Q4 =200 Fesol = FIABE Sin 20º)Los 40") = 30 3 sin (20º) cos (40") Fx= 79.391b FAQS = FIOREOs(20) = 303 COSE20"> Fy= 284,7316 Fee)= - FABSin (200) sin (40 ) = 363 sin (20") san (40") FZ= 66.611b 25) Q = cos-1 ( aBx) 11 =Los (By) Q2 =CD 5 /963 FAB LOS -21-79,39) 605- 1/ 30 3 303 Q =105.18° Q4 = 19.99° 20° 45 284. 73 6051/ - 66 61 363 102.69º 3) AB = 74. 216 ft AC=85.59Ft 5 985.59 = AB 541+18,364 ft 1-3.047;+1,016, +Zott AB - 67.36 3 X 4 d Z (3,8) (-59; +18; +36k) 67.35
457.08 247.06 4) A3 = (4 2) T-(5 W)X AB = 7 m Ac (24), -15. 6), (9.2) AL-1.am A1 /2 -(5-6)} - L3- 3)4 AD : 65- TABSTATS NAS FAIS 1.3 AC =/=329321-456766c.51757) Tmc " AD = (C.BL(54) -. 50769k ) TAB (-0.Li-0.5) SAD) EF =0; TAB +TAX + TROP; 2 0 LIGTATT.32932TAc)Ml-8TAB : 756765pc=86154IND+P)9+ 1.56757TAC -- 507691 AD YL =G -. 6TABT. 324TALSO -STAB -- 75676TAC-86TAD +P=0 15675TAC-0-5076TADO AC=457.08 AD-571 AB = 247.0L P= 983.79N 5) TAC = 72Pct TAx= 72569 To-72(,485)-1.457-364) Ircal 71 5.451-36ht.487) TAB=TA (48)-(-3277:36h) - Tax(.475-,5294+.7065) 9 (45)2+1.36) 3+14812 -(-43.21-34-56(+46680)N V1.3252+1.3654464853 FAD = TAD (,487)-(.251-362) = Two(-3851++5534 +, 738;) JE.25)2+1.36249,4992 TAB = 657 . 428 W 0,4706 TAB-43.2-0,385TAD=0 0,706TAB +46.08+, 738TAD =W=0 TAD=691,392N -0.529 TAB-34,56+0.553 FAD= 0 W=1020.472N