Question

> Answer 3/83 The vertical plane containing the utility cable turns 30° at the vertical pole OC. The tensions T, and I are both 950 N. In order to prevent long-term leaning of the pole, guy wires AD and BE are utilized. If the two guy wires are adjusted so as to carry equal tensions I which together reduce the moment at O to zero, determine the net horizontal reaction at O. Determine the required value of 7. Neglect the weight of the pole. 10° 10° 1 NcL- - B T = 950N | T = 950 N ΛΙΑ 7- == Z300 x 15°E OA = 9 m OB = 11 m OC = 13 m OD = 8 m OE = 10 m Answer 0 = 144.9 , T = 471 N We've updated our read aloud feature! 3/84 Determine and plot the moment M required to rotate arm OA over the range o 38 < 180°. Find the maximum value of M and the angle 8 at which it occurs. The colla Give it a try here. darniandmation afthanh-ft in ito kasrina Tatarmina and nlat thamannituda aftha distributedtantinnl forna nunnantad lithin aallon artan thansmannaa Af Athananina AAA nTV ATIM NATHA Aa ho ©

Answer 1

Coordinates :- oco, o, o] m A :- [0, 0, 9 ] m. B :- 1o, o III m ..:-10, o 13 m D - [8 cos 15 , 8 sin 15, o m E :- I lo cos 15 10 sin15, o m Express all the forces in rector o = Oxi toy " + Of ki form. • E9.0) T = (-950 cos lo sin 3o) i +(950 cos 10 cos 3o) ; - (950 sin 10) k² T = -467.7837 i + 810.2251 i 164.9658 k N ... Eg.(2)

(950 co Tz - co0 - (950 cosio);" - (950 sin 10) ... Eq? (3) TAD = TAD WAD = TAD . Ił-A7 LIADI TAD = T 8 cosis i + 8 sin 15; - gk 145 Eq. (4) IBEL TBE = T lo cosi5 i + 10 sin 15. ... Eq. (5) of Moment o. all Now forces these Take @ Note Moco Myo pro =0 =0 o Giren

Moment of T, co = roc X T, = 13kx - 467.7837 l + 810.2251 i - 164.9658 F 7 = -10532.9263 - 6081.1881 tok ... Eq. A)] Moment of Tz @o = Voc x T2 = 13 th x 1 - 950 cosio ; - 950 sin 10 kg - 12162.3758 it on tok ...Eq. (B) Moment of TAD @o = TOA X TAD = 9 x T 8 cos is i + 8 sin 15 - 96 √145 ti 4+G) : +(63***59); -om (-1.5476 15.7755 XTAD XTAD Eq. [ C ] Moment of TBE @o = 8oB TBE = 11k X T locos 15 i + lo sinisi. √221 L - 11 k q = (-1.9151 +1.1473) tok = (x TBE (XTBE 1 . Egior

ZMx C o 0 ie 1 = 0 Eq. (A) 10 532.9263 + 12162 3758 - 1.5476 TAD - 1.9151 TBE = 0 Rearrange 1.5476 T + 1.9151 T = 1629.4495 ... JEq. (E) EMY eo =0 i.e. £ j = 0 for - Eq". (A) +5.7755 TAD To CD) - 6081. 1881 + 7. 1473 TBE = 0 Rearrange 5.7755 T + 7. 1473 T = 6081. 1881 Eq. (F) or Eq. (F), Solve Either Eq. (E) we get T = 470 5782 T ² 471 N N Ans

Now, use EFx = 0 force i.e. equilibrium Li=0 for Eq" (1) to (5) ox - 467.7837 + 8 costs T + ( 10 cosis) T = 0 221 Ox = -139.9578 N {fy = 0 i.e. £i" = 0 for Eq? . (1) to (5) Loy + 810.2251 - 950 coslo +( 8 sinus) T +( 10 sinis) T20 Oy = -37.5016 N Net = = Horizontal Reaction, √(x)2 + Coy) 2 144.895 N E 1449 r Ans