Question

Q. 1 Determine the x and z components of reaction at the journal bearing A and the tension in cords BC and BD necessary for equilibrium of the rod. (0.5 point) 3 m 3 m 2 m 6 m 4 m F - (3503) N F-1-800k) N

Answer 1

• The FBD for given Problem, 2 Fi= - 800 FN Ax F2=350 JN • FBC Az NOW x Foc - FBC. UBC - 3] + uh UBL= BC B IF 18BCI 9 SBC = 8 - 0B Point B as a reference. SRC= (-3) + uk)-(oltojitok -3jtu FBC = FBC (-23 + 4 FBD = FBD. UBD FBD FAD Ca Stufe

& NOW apply equation of equilibrium EFx=0 {Fy=0 350-0.6FBc +0.6FBD = 0 cilj EFz=0 Ą - 800+ 0.8 Foc + 0.8F8D = 0 clii) { Mac=0 MAX + 0.6FBD (6) + 0.8 FBC (6) = 800 (6) -- civ) { My=0 (-0.8FBcX2) - (0.8 FBD x 2) +(800X2) 20 EM2=0 MAZ+ 0.6FBD (2) - 0.6FBc (22 = 0 (vi) sel FBC+ FBD from ean (ii) a.st FBC + FBD) -350 & put in ean cuis MAZ + 2 (-350) = 0 MAZ = TOO Nom (vilo) FBD + FBC = 10 OON. used this ean in (iii) Az 800 + 0.8(1000)0 Az=0 in ean qui 4800 MAX + 4-8(1000) MAX=0 olve ean ciis & ean (vii) FBC = 793 N FBD = 208N
i hope this will help you.???