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2. Determine the unstretched length (length with no force developed) in the spring assuming the box has a mass of 40kg, is in


1. A force of 800N is used to pull point A. Determine the tensions in each cable (DA, BA and CA) required to hold point A in


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Equilibrium of Joint D (40X9.81) X = tan (2) 3 33.69 ° = tool 2 = 45 I 2 .. ToB sina + Toc sinB = 40x9.81. ... Eq. (1) {fx zNow, Tension in length of DB I stretched ] DB = 283 N. = (3) 2 + (2) = 3.6056 m F = k. 283 = 180 x 8 . d= 1.5722 m StretchedI points A, B, C and D. Write coordinates of o [ o, o, o ] m. Alo, o, s]m B[ 4, 4, olm C[-2, 3, o m D[-4, -5, o] m. find UnitFind unit vector, UAB. .. 2 AB = B- A TABU - = 4² +4 0 - 5² √(4)²+(4)²+(-572 - st 142 +4 - 5x] TAB TA ( 4 + 47 - sr] Eq. (2)Also Force, F = 800k N. 1 Eq. (4) NOW for Equilibrium of Joint A, we - have , £Fx = 0 Fy zo Ź FZ . for ZFx = 0; Equate i compSolve Eq s. (A), (B) and TAD = 565.1505 TAB = 577.7265 TAC = 85.7658 c N N N

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