Question

1) Draw a free body diagram of the figures and determine the reactions for each of these. Cables can still only hold axial loads 15 in B 10 TEOTTI 5 kips 30° 2 kips

Answer 1

Solution. - Free Body line Diagroom that for the first problem is 250 1 200 CO Ac sinzoo Acsim 300 T 7 To Co T 0 3000 300x - 500N 250 mm AC Cos30 Now considering ACB portion; 250 costo AD COSE A Ana V1300 250 singo Cx 130° FAD sino |200 sin Zoo 36 500N 250mm 200cosso 200 mm

Taking moment about "C" considering 26 - (500x 200 sin 30) - (FAD COSA x 250 sin 30) +(FAn sind x 250 cos 30) = 0 or, FAD = 500x200 simbo Sin 2 x 250 cosso - Cosex 250 sinzo From the initial picture, – – – (1) tano Ac sinzo +250 Accos 30° on, o=, - 11.250 sinzo +250) 250 cosso os, o = 60° putting this value in equation (1) and we get FAD = 400N Now applying {fx=0; + FAD Coso tx – 500=0 oh, (n = 500 - 400 cos 60 = 300N ) Applying {foso; - FAD sino + cy=0 Oh, Cy= 400 sim 60 = 346.410 (1)

Now considering joint Dong FAD & FAO singo 6 = 60 FAD COSSO Dy Applying Ifn=0; - FAD COS 60 + Da=0 og, Dr = 400 cos 60 = 200N ( Applying I Fo=0; + FAD sin 68 - Dy=0 of Dg= 400 simbo= 346.41N(+)

for the second probleon is Free Body Diagram line there 15m 10 in ) Acsin 20° 7516 PAC Cos20 700 Considering ACB portion; 15 in 15 sinzo 10 in 7516- 40 sinzo 20, No FAOLOSO Ico A FAD AD Sine. 15 cos 20 K20 COS2oo ?

Taling moment about "c" considering 26 - (75 x 15 cos 208) + (FAD Cose x 10 sinzo) & (FAN sina x 10 cos 20) = 0 on FAO = 75x15 cos 20" Cose x losinzo + sinoxio cos 20° Now from initial picture; forn o 12- Ac simzo ACCos20° or = tan /12- 10 sinzo 10 coszo on = 42.400 putting that value in equation Gg we get FAD = 119.29 lb Applying {fx=0; we get; AD COSO -(x=0 091, Coe = 119.29 cos 42:40 = 88.09.16 (6) Applying {fy=0; we get; - FAD sino tlg -75=0 or Cy=75+ 119.29 sin 42.40 on, Cy= 155.44 lb (A)

Considering FAD Joint "D" only: FAD Sin42.400 = 42.40 FAO COs 42.400 Do Applying {fx=0; - FAD COs 42.40 + x = 0 on, Da= 119.29 Cos42.48 oh, Da= 88.09 16 ) Appbing For O; FAD sin 42.40 - Dy=0 og Dy=119.29 sin 42.400 of, Dy=80.4406 (1)

Free Body Diagrion line that . for the third problem is 2001 T300 thuips A4 quips 16' 16' Considering AGB portion; Tsinto TAB 60+100 - 70° T Costoo (Guips zuips

from that figure; 164 16 = tem to hổ = b)-(216) - (54 Tro 7 32. on, ha 32 tan 30 Now taling moment about "A" considering 26 - (t on T = (2x16) + (532) (32 ton Boxsinto) – (32 costo) on, T= 29.923 hips. Applying {fx=0; we get An - Tsinto or, Axe = 29.923 sinto or An = 28.118 hips ) Applying {fy=0; we get Ao-2-5- Trosto og, A=2+5+ 29.923 cos 70 = 17.234 uips (A) Now considering Joint c onlng. Ž Fre=0 D & Tsinzo" x = + cos2o = 29.923 coszo l = 28.118 Lips (6) - 28-118 hips 200 Co= Tsim 20 = 29.923 sim 20 - 10.234 hips Tcos 200