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Using summation of force components, ∑Fx = 0: Ax = TCx Ax = 126.380 cos 30° Ax = 109.448 lb ∑Fy = 0: Ay = NB Ay = 164.172 lb +↑ ∑Fz = 0: Az + TCz − 400 = 0 Az + 126.380 sin 30° − 400 = 0 Az = 336.810 lb For the total reaction at A: RA = √Ax 2 + Ay 2 + Az 2 RA = √109.4482 + 164.1722 + 336.8102 RA = 390. 349 lb Situation 2 – Refer to FIG. MECH0409 A 5 x 8 ft sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. 4. Determine the tension (lb) in cable BD. a. 101.3 c. 405.6 b. 315.0 d. 353.1 5. Determine the tension (lb) in cable CE. a. 101.3 c. 405.6 b. 315.0 d. 353.1 6. Determine the reaction (lb) at A. a. 101.3 c. 405.6 b. 315.0 d. 353.1 Solution: Drawing the free body diagram of the bar,

Using summation of force components, + →∑Fx = 0: Ax − TCEx − TBDx = 0 Ax − 6 7 TCE − 2 3 TBD = 0 Ax − 6 7 (315) − 2 3 (101.240) = 0 Ax = 337.483 lb +↑ ∑Fy = 0: Ay + TCEy + TBDy − 270 = 0 Ay + 3 7 TCE + 1 3 TBD − 270 = 0 Ay + 3 7 (315) + 1 3 (101.24) − 270 = 0 Ay = 101.248 lb + →∑Fz = 0: Az + TBDz − TCEz = 0 Az + 2 3 TBD − 2 7 TCE = 0 Az + 2 3 (101.24) − 2 7 (315) = 0 Az = 22.499 lb For the total reaction at A: RA = √Ax 2 + Ay 2 + Az 2 RA = √337.4832 + 101.248 2 + 22.4992 RA = 353. 061 lb Situation 3 – Refer to MECH42.501 A transmission tower is held by three guy wires attached at a pin at A and anchored by bolts B, C, and D. The tower exerts an upward vertical force of 1800 lb at A. 7. Determine the tension (lb) in wire AB. a. 974 c. 533 b. 531 d. 840 8. Determine the tension (lb) in wire AC. a. 974 c. 533 b. 531 d. 840 9. Determine the tension (lb) in wire AD. a. 974 c. 533 b. 531 d. 840

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