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2.21 Determine the tensions developed in wires CD, CB, and BA and the angle o required for equilibrium of the 15-kg cyli

Posted: Wed Apr 27, 2022 7:36 am
by answerhappygod
2 21 Determine The Tensions Developed In Wires Cd Cb And Ba And The Angle O Required For Equilibrium Of The 15 Kg Cyli 1
2 21 Determine The Tensions Developed In Wires Cd Cb And Ba And The Angle O Required For Equilibrium Of The 15 Kg Cyli 1 (56.15 KiB) Viewed 25 times
2.21 Determine the tensions developed in wires CD, CB, and BA and the angle o required for equilibrium of the 15-kg cylinder E and the 30-kg cylinder F. D 30° E Equations of Equilibrium: Applying the equation of equilibrium along the x and y axes to the free-body diagram of joint C shown in Fig. (a). ΣF = 0; Fac cos 6-Fcocos 30º = 0 (1) +ΤΣF = 0; - Fac sin 8+ Fco sin 30° - 15(9.81) = 0 (2) By referring to the free-body diagram of joint B shown in Fig. (b). ΕΣF = 0; (3) Fu cos 45º - Fec cos 0 = 0 FRA sin 45° + Focsin - 30(9.81) = 0 +TEF, = 0; Solving Eqs. (1) through (4), yields Ans Ans Fru = 395.8 N Fco = 323.2 N FR = 280.2 N 8= 2.95° Ans Ans FED • FEA Foc 30°C C tot 45° х B o Fec 15(9.81) N 30(9.81) N la) (6)

2.42 The springs on the rope assembly are originally unstretched when 6 = 0º. Determine the tension in each rope when F = 450 N. Neglect the size of the pulleys at B and D. -0.6 m 0.6 m D A A YA k = 500 N/m k = 500 N/m 0.6 1 = cos 0.6 0.6 m T= kx = k(1-1) = 500 -0.6 = 300 1 cos - 1 (1) cos ser +TEF, = 0; 2 T sin 0-450=0 (2) Substituting Eq. (1) into (2) yields: 600 (tan 8-sin o) - 450 = 0 T tan 8-sin = 0.75 By trial and error: -X = 57.957° From Eq. (1), F=450 N T = 300 1 cos 57.957° -1 = 265.4 N Ans

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