Question

2) The system shown is in equilibrium. It is known that the breaking strength of cables AC and CB are 1000lbs each. Also know is that the total deformation of springs 1 and 2 is 1.5 ft. Find; A) largest value of P (or cable CD) for which neither AC or CB will break, b) Using the value of CD obtained in part a and knowing that the moment at E is zero, find the value of GH and Wi, c) normal force acting on W2, d) value of W2, e) spring constant k2, f) deformation in each spring. Ki=1000lb/ft 500lb 1500lb Box in your answers 76 400 G S 8ft 350 1800lb A450 300 B

Answer 1

Gueslion Solutont Given Te syleu shoon 's in equilibriuus EPCB he bredeing stresuq obes Ae e blep to d e, sisq parls Nouse hau e CaSolution have to daexuine Re laxa ue opp. s.beak Coe CB hiek neithex AC PaiAow gise e A,B,C ate we TeD mabap FAC FBC qut Coble teusion Dm To Rau diogow

sin 300 Bc Ac FA Gos5° Fec above SubitikoDe Fc FAc J을 sin 300 TD ooo Max given aboue expicaio Subed: hDe t, voe sin 5 loo0 TCD o% n's Sin 3 + 5H "'S TeD [ th TcD loo0 Ten: 10p6 o.70106 +o.H08248 Tep Jo0o TCD = ,115. 35 2b (b) soluTion have to ind D vofue of G H Now aoe

Hdiagou ou e give diagrou we Parl to H Soo b Tep b d'ogiow daa above EMe sin p0°x10 Cos Hpo t FaH S00 x7 kuow TeD- \\15 .35 )b. we 15. 35 X 8- 500 x 9 = So Nouo 8 FGH Cas + - (s.128.555 8,922.8 5,422.8 12-556231 FGH 5, 422. 8 12.55623 ST 2olb

Now take past G do Aou giveu 2. 350 Bou abbove diogroto, dalo. feu Co Hoo . + T os 350 20 88 IEdf Ts Cas sS° I431. 88 6s 4o So Nlous 20 EP36 xo 366 +T (0.819152) 13 330. 82 oo 13 (o. 819152) = 13 ip,300 66 l00 13 Ts Co.ej9|52) = L2 0o. 66 3 Ts Co SI9 51) p200 66 = L 9t68H9 01 4200 66 Ts = 10.648 926 Ts 39 4.466 b

Fy IT Sih 350 = 20 3 afhrody ind FGH 3.8. Lb loe 394. eb6 Į6. Subrtike Re bsvave above epeio 2 + LP31-88 Sn 40 O 39 4. 466 Sin 350 3 lo 394 466 x0.5735I 6 240 7h9.Ox88.1EHT 13 240 2 I2.60 38 22 6. 256 360 8 8 49 4 2HO 226. 256 = WI 1 3 3848.84q4 226. 256 1 29 328 38H8.8u94 3 90. 521H 13 69.809 3 solurliou e) - reave to nd R Nouo OYe e oding On

N23 Ts afveo dy Pind = 394. 66 lb twe So Now 39 Lp. 466 w, (o 5) 39 Lp. 66 89 4. 466 789 오b. CD) soltiou Poce olutiouNorn on Now 89 YO.SG 6 NE Ce) solulon fowalion ofsping 1 de Nos given 5 Cquismespring Ts -

15 looo k2 39 4. H66 loookz S94, H66 -+39 4. 466k 1500 k2-4466 k2 39H H66 394,466 = l05.534 394, 466 2. Do5.53 ]$b)P 356 8/0 Ts pring 39Hp. 466 JDoo O 39 HH66 Ts DeforaDio In Spring 39 t. 466 556. 8)0