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Unformatted text preview: PROBLEM 4.70 For the frame and loading shown, deteimine the reactions at A and C. n 7 7' D +40 llllil+l<i (7‘0 mm SOLUTION 130 M IBON Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibritun. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three—force body with member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle ,8 is found from the member dimensions: 60 m 100 m ﬂ : tan’l[ ) = 30964" Applying of the law of sines to the force triangle for member BCD, # 130N _ B _ C r sin(45°—ﬂ) sinﬂ sinl35° or 130 N # i B _ C sin14.036° si1130.964° sinl35O (130 N)sin30.964° = 2 f, r = 275.78N sm14.036° (130 N)sinl35° and C : .— = 379.02 N sinl47036° or A = 276N V: 4500‘ or C : 379N 5. 59.004 PROBLEM 4.73 A ZOO-lb crate is attached to the trolley-beam system shown. Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (17) the reaction at B. SOLUTION From geometry of forces yBE = 2-0 ‘ J’DE = 2.0 ~ l.5tan35° {194; 11:75:: : 0.94969 ft ﬂ = ta11"£0‘94969) = 32.339° and a = 90° ~ 6 : 90° — 32.339o = 57.6610 49 = [3 + 35" 2 32.3390 + 35° : 67.339o Applying the law of sines to the force triangle, 200 lb _ T _ 713777 sine sina sin 55° (200 lb) T B 01‘ ¥ = *W : sin 67.339” si1157.661° sin55° 200 lb v' 57.661“ (a) T:£¢n_)=183.1161b smG7.339° 01' T = 183.11b4 2001b ' 55° (b) B : w = 177.536 lb sin 67339" 01‘ B = 177.5 lb ‘5 323° { PROBLEM 4.106 H For the pile assembly of Problem 4.105, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. P4.105 Two steel pipes AB and BC, each having a weight per unit length of 5 lb/ft, are welded together at B and are supported by three wiresi Knowing that a = 1.25 ft, determine the tension in each wire. SOLUTION First note WAR 2 (5 lb/ﬁ)(2 ﬁ) = 10 lb W3C : (5 lb/ft)(4 ﬁ) = 201b From f.b.d. of pipe assembly 217), :0: T4+TC+TD~101b72Olb=O TA+TC+TD=301b (1) 2114‘. = o: (10 ]b)(1 ft) — TAO ft): 0 or T4 2 5.00 lb (2) From Equations (1) and (2) TC + T D = 25 lb (3) 2M: = 0: TC(4 It) + T[,(anm) — 201b(2 ﬂ) : 0 or (4 ﬁ)TC + TDamax : 401M: (4) PROBLEM 4.106 CONTINUED Using Equation (3) to eliminate Tc 4(25 — TD) + TDa = 40 max 60 or amax : 4 # F D By observatic '1 is maximum when TD is maximum. From Equation (3), (71))“Ix occurs when TC. : 0, Therefore, ’ X : 251b and a : [email protected] "“X 25 = 1.600 ft R< anm : 1.600 ft 4 7",, : 5.00 lb 4 TC : 0 4 TD : 25.01b 4 ...
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