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assignment 8 solutions

# assignment 8 solutions - PROBLEM 4.5 A hand truck is used...

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Unformatted text preview: PROBLEM 4.5 A hand truck. is used to move .two barrels, each weighing 80 1b. Neglecting the weight of the hand truckgdetermine (a) the vertical force P which should be applied to the handle to maintain equilibrium when a = 35°, (1)) the corresponding reaction .at each of the two-wheels; SOLUTION Free-Body Diagram: ‘ a1 = (20 in.)sina — (8 in.)icosa a2 = (32 in.)cosa — (20 in.)sina (mesa: 4 A b = (64 in.)cosa (95») Costa ' I P t From free-body diagram of handtruck +‘) 2MB = 0: P(b)‘— W(a2)+ ”((01): 0‘ (1) “ML Amaze: P—2w+23=0 (2) a. 02 . For a = 35° l:— b a, = 205111350 — 8cos35° = 4.9183‘in. 02 = 3200535" — 20\$i1135° : 14.741311r. b = 64cos350 = 752.426 in. (a) From Equation (1) P(52.426 in.).— 80 lb(14.74-13 in.) + 301b(4.9183'in.) = 0": ,P =14.98961b or P“: 14.9911: [4'9 (b) From Equation (2) 14.9896 lb -— 2(801b) + 23 = o: B = 72.505115- or B = 72.5 1134‘ 4' t PROBLEM 4.9 A control rod is attached to a crank at A and cords are attached at B and C. For the given force in the rod, determine the range of values of the tension in the cord at C knowing that the cords must remain taut and that the maximum allowed tension in a cord is 36 lb. SOLUTION Free-Body Diagram: 53cm: 1' For (Tam, TB = 0 +) 2M0 = 0: (gram (4.8 in) — (801b)(2.4 in.) 0 [(13)max = 401b]>[Tmax = 36113] (TC)max = 36.01b 'For (TC) ,TB = Tm = 361b min +‘) 2M0 = 0: (Tam (4.8 in.) + (361b)(1.6 in.) — (801b)(2.4 in.) 0 (TC)min = 28.01b Therefore: 28.0 lb S TC 5 36.0 lb { .330 N I‘ZSO n1ni+1+250 mm" PROBLEM 4'17 riff; -11: 17- 300 mm F; i 1—0 2' Determine the reactions at A and B when (a) a = 0, (b) a = 90°, .1 ' '1 (c) a 2 30°. SOL UTI O N Free-Body Diagram: Equations of equilibrium: J) EMA : 0: —(330 N)(0.25 m) + Bsina(0.3 m) + Bcosa(0.5 m) = 0 (1) 22.25%: AX—Bsina20 (2) +i 2F) = 02 A,»- — (330 N) + Bcosoc = 0 (3) (a) Substitution a = Ointo (l), (2), and (3) and solving forA and B: B = 165.000 N, AX = 0, A}, = 165.0N or A=165.0N I! B:165.0N I 4 (1)) Substituting 0: = 90° into (1), (2), and (3) and solving for/1 and B: B = 275.00 N, Ax = 275.00 N, A), = 330.00 N A = .1213 + Af, = ./(275)2 + (330)2 = 429.56 N A. 9 = tan—117) = mfg—’37? = 50.1940 .'.A=430NA:50.2°, B=Z75N-4 PROBLEM 4.17 CONTINUED (c) Substituting 0: = 30° into (1), (2), and (3) and solving forA and B: B = 141.506 N, Ax = 70.753 N, A}, = 207.45 N, :> A = JAE + 43 = ./(70.753)2 + (207.45)2 = 219.18 N 9 = tan’1% : tan’l%77‘% = 71.168° A = 219 N A: 712°, B = 141.5N 3; 60° 4 7 PROBLEM 4.21 1 The required force to be exerted at A by lever ABC is 3 1b. Knowing that // a = 30° and that the spring has been stretched 1.2 in, determine (a) the {cg/j/ f a constant k of the spring ([7) the reaction at B. ifl‘ 1111+ C 0. S) m. 3- i<—-2 4 111 SOLUTION Free-Body Diagram: (a) 2 4 in. 2M = 0 e A — 0 9 in F = O ‘1") Br [ 00505 J ( ) 5p . 8 or Esp w 1b:kx:k(121n) c0530 Solving for k: k = 7.69800 lb/in. k = 7.70 lb/in. { (b) i, 211; 2 o: (31b)sin30° + B‘. +( 81" J: 0 ' ' 00530" or Bx = —10.7376 lb +1 >313, : 0: , (3 lb)cos30° + By = 0 or By 2 2.5981 lb (—10.7376)2 + (2.5981)2 = 11.047511), and ,1 2.5981 10.7376 Therefore: B = 11.051b b; 13.60” 4 0 = tan = 136020" ...
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