Question

A cable is attached at B to the right-angle pipe OAB in the figure. The tension in the cable is 730 lb. (a) Determine the angle between the cable and member BA. Determine the components of FBC (the cable force acting at B) (b) parallel to member BA, (c) perpendicular to member BA. 1.2 ft 1.7 ft. 0.7 ft 30 0.8 ft (a) degrees (b) Ib (c) Ib

Answer 1

co-ordinates of points The C = A = (1.7ft,0,0) (0, 0.7 ft, 1-2 ft) that oriented as follows, 0.8 (-cos 38 Ã - sin 30 ;) can See the vector AB Now, we yz plane -) lies parallel to У A B = А z 38 к لم co-ordinates of चि CB = B-C -0.4 i - 0.692 o.fi BA=0.49+0.692 = B = 1.7 à 0.47 +0.692 Å point B is ( 107 ft, -0.4 ft , -0.692 ft) 107 ê - lol 9 - 10892 - 1.7 ê + lol + 101 g +1.892 BÁ = 0.4 ŷ + 0.692 Å Angle between cable and angle b/w vector BC and_BA 1.749 0.788 - 0-37.9 IBC IBAI 2.77108 B C = K BA is BC BA cos o = ñ

-447.85ê +289.733 Face 1 730 [-1.7 h + 1.1; + 1.892 ² 7301-*-* 2-771 +498.43 to B FBC parallel FA - B will be, Unit vector along along Ba (FA ( ВА = 576-03 lbs IBAI (BA) component of = 57603 ỳ + √3 FBC along BA Dott 2 288.01 j +498.85 ВА will be, (c) Foc to Foc - Fochi parallel component of FB to B BA (-447.85 î + 2 89.78ị +498.439)-(288.01ſ +498.5657 - 447.85ê + 1,77 ŷ o - 0.42 Å o