Dallas Hallston statics 1 T1=830 75 +40 º,Q=1800 sino = 12 5775 T2= 884.5916 830 sino TOR Since sinuo sin 65 5.440 R=588-6616 Sin650 - 1 2 830 5.7.75 830 = R TAL = TAB TAB=1.9 TAC Sin 30 Sin 25 5:136 1.9 AL=2.25KN Sin 25 TAB R 1.9 R Sin25 sinr25 R= 3.68 KM sin250 - sin/25° sinto 50N= Sin 25 A = 23.80 53N Q+B +25° = 18GB = 131.5- R P R =53 R=93,9N Simp sin 25 Sin1315 5-425 Fx =- 615Ncos60 =- 307.5N Fy=615 Nsin60 =532.605N Fx=815Ncos70=278.746NFC=815NS070 = 765.8498N FX=365NL0525=330.8024N FY=365 Nsin25°= 154.2556N 100 Ax=100N cosQ Fn = - 100 N Sin. Rx = - 21.254N 150 Fx = 150NLCS30 +Q Fx = - 150 s. h3O+Q Ry = - 318,62N 200 Fx =- ZOON cosa Fy =- 200 sino Ry 2 = 319N tank= Rx =86.20 Rx 1 sinQ
90Ncost+ 175N sind 115+90since-175ca3D=C 115NT90 Nsina - 175 Nc03# 23+18 sbQ-35casff = 0 199, Q+ 23 = 35 VT- sin2Q 324 sin 210+828sinQ+529=1225(1-sin2(1) 1549sin=A+8283inQ-696= 6 SinQ= = 4544 A = 27.024° R = 90cas (27.024)+175 sin(27-024) R=159.68N T ca E FCB P 50(30" 390 Singo = Tc 9 singo Siglo p tension on ALS34315 5.760 TCB = P TCB sin40 Sin 50 - 390 sin 80 Sin40 tension on BL = 2551b 0 150 0 W=mg=120kgx9.8/m/s2=1177,2N Jin130 Tac= Toc = 1177.2 Tar = 1.1772KN TBC Sinzo sinzo 5 .4 20 11 1,1772K (sin 30) TALE2.6KM Sinzo (5.4/30) FBC=1.721KN
y r FB FA Fc = 6,96KN Fo = 5.22kN Fo FD=PFOTOFA ??× ZFx=O?3 FB·Fc-IFA = 6 FC= FB-?FA { Fy=O ?EDIEFB FAO ? D