Question

Question Completion Status QUESTION 1 The metal pipe AB is supported by a thrust bearing at A, and a rope at B. The direction cosines for rope BC are as defined below. When the two forces are applied as shown, please determine the support reactions at A and the tension in rope BC. Please write all equilibrium equations and clearly indicate the direction and magnitudes of all forces and moments direction canine age = 55.5" dirertion corner - 112.09 03 meter 4 KN direction contrar 139.0 6 KN 1.5 meters rach File Browse My Computer Browse Content Collection

Answer 1

Foca cos x = Ž Fech -0.375 FBC Cosß Focy I Feel B A TRAX Amy 0.3m RAY Cosr= VFBC (-0.7567) MzTRAZ Fec z I Focl 0.566 FBC OKN 0.um {Mp=0. (Foly = Fee Cosa Foc Cos (55.5) (Foc ly = For Los B Cos (112) FBC = 0.5664 Foc = -0.3746 Foc { Fn=0 (Fec ) ₂ = Fee casi = Foc cos (139) -0.7547 FBC Elyo - 0.375 FBC +4 -Ray to Ray = 4-0.375 Foc 0.566 foc = RAR & Fzzo -0.7547 FGC 6't RAZ 2o. RAZ = 6+ 0.7547 FBC Eme=0. -0.375 Foc x0.3 + 4x0.3 = o. FBC 10. 667 KN CS Scanned with CamScanner

RAM 0.566 x Foc RAY = 4 - 0.375 EBC Ran = 6.0345 KN =4-01375 (10.667) | Ray = 0 KN RAZ = 6 +0.75 47 FOL z RAz = 14.05 kn Resultent 2 reaction at A= RA= V RAM + Ray +RAZ 15.29 in moment reaction at a -0.375 Foc a 1.5 + 4x 1.5 = oknum My -6* (15-0.4) - 0.7547x(10.667) - 0.566 FBC xo. 688 krom 2 My 20.688 kwam y. RAGE >RAX Tension in My e Fac = 10.6671N rope RA Thrust Bearing CS Scanned with CamScanner