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Extra_1 - PROBLEM 4.69 For the flame and loading shown...

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Unformatted text preview: PROBLEM 4.69 For the flame and loading shown, determine the reactions at C and D. 150 N SOLUTION Since member ED is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and be opposite in direction for ED to be in equilibrium. The force B acting at B of member ABC will be equal in magnitude but opposite in direction to force B acting on member BD. Member ABC is a three-force body with member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate'the above conditions. The 1 force triangle for member ABC is also shown. The angles a and B are found from the member dimensions: a = tan’1 Oi?) = 26.565° 1.0 m ,6 = tan—1(1'5—m) = 56.310° 1.0m Applying the law of Sines to the force triangle for member ABC, 150 N C B sin(fl — a) _ sin(90° + a) ‘ sin(90° — ,8) or 150 N _ C _ B sin29.745° sin116.565° sin33.690° 150 N ' 1 . ° C = —( _)5m 16 565 = 270.42N sm29.745° or c = 270N I; 56.3°< 150N ‘ 33.690° and D=B=w~—=lé7.704N sm29.745° or D = 167.7 N V: 26.6°{ SOLUTION E PROBLEM 4.77 A small hoist is mounted on the back of a pickup truck and is used to lift a 120-kg ‘ crate. Determine (a) the force exerted on the hoist by the hydraulic cylinder BC, (b) the reaction at A. First note W= mg: (120 kg)(9.81m/s2) = 1177.2 N From the geometry of the three forces acting on the Small hoist xAD = (1.2 m)cos30° = 1.03923 m yAD = (1.2 m)Sin30° = 0.6m and yBE = xADtan75° = (1.03923 m)tan75° = 3.8785 In Then a = tan—1 M = tan_l[ 347“ J = 73.366° xAD 1.03923 ,8 = 75° — a = 75° — 73.366° =1.63412° 0 = 180° - 15° -,B = 165° - 1.63412° =163.366° Applying the law of Sines to the force triangle, W B A sinfl = sinH = sin15° or 1177.2N _ B _ A sinl.63412° sinl63.366° sin15°. (a) B =11816.9N or B =11.82kN >2 75.0°< (b) A = 10 684.2 N or A =10.68kN V 73.4°4 PROBLEM 4.1061 y 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. a P4.105 Two steel pipes AB and BC, each having a weight per unit length D of 5 lb/fi, are welded together at B and are supported by three wires. A Knowing that a = 1.25 ft, determine the tension in each wire. a SOLUTION First note WAR = (51b/fi)(2 ft) = 10 lb WBC = (51b/fi)(4 a) = 20 lb From f.b.d. of pipe assembly EFy :0: TA +TC+TD—101b—201b =0 TA +TC +TD =301b zMx = o: (101b)(1fi)— TA(2 it) = 0 or TA = 5.0011: From Equations (1) and (2) TC + TD = 25 1b 2M, = 0: TC(4 a) + TD(amax) — 201b(2 fl) = 0 (4 fi)TC + TDamax = 40113-11 PROBLEM 4.106 CONTINUED Using Equation (3) to eliminate TC 4(25 — TD) + TDamax = 40 By observation, a is maximum when TD is maximum. From Equation (3), (T D )max occurs when TC = 0. Therefore, (T D) = 25 lb and max Results: (a) (1mm 2 1.600 R { (b) TA = 5.00 lb { TC = 0 4 TD = 25.01b 4 ...
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