# hw2 answers - PROBLEM 3.142 A worker tries to move a rock...

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Unformatted text preview: PROBLEM 3.142 A worker tries to move a rock by applying a 360—N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of pan a. SOLUTION (:1) Have Have where (b) Have where 2F: M: II 360 N(—sin40°i — cos40“j) = —(231.40 N)i — (275.78 N)j = F or F :360N 7 50°1 2M0: rB/D x R = M raw : {(0.65 m)cos30°]i + [(0.65 m)sin30°]j = —(0.56292 m)i + (0.32500 m) j i j k —0.56292 0.32500 0 N-m = [(155240 + 75.206)N-m]k —231.40 —275.78 0 (230.45 N~m)k or M = 230 N-m ) 4 2MB: M = MD x Fr1 . rm = {(1.05 m)cos30°]i + [(1.05 m)sin30°] j = —(0.90933 m)i + (0.52500 m) j PROBLEM 3.142 CONTINUED i ' k FA —0.90933 0.53500 0 N-m = [230.45 N-m]k 0 —1 0 or (0.90933FA)k = 230.45]: -. FA = 253.42N or FA = 253N 14 Have 2F: F = FA + FD —(231.40 N)i — (275.73 N)j = —(253.42 N)j + FD(—cosd9i — sin6j) From i: 231.4ON = F300519 (1) j: 22.36 N = FDsinB (2) Equation (2} divided by Equation ( 1) me = 0.096629 6 = 5.51930 or 6 = 5.52” Substitution into Equation (1) FD = ﬂ = 232.48 N cosS.5193° or F)D = 232 N 7 5.52°4 PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that 6 = 25°, Mx = —61 lb-ﬁ, and M: = —43 lb-ft, determine (9 and d. SOLUTION Have 2M0: 134/0 x F = M0 where rm = —(4 in.)i + (11 in.)j — (d)k F = F(c056005¢i — sinBj + costin¢k) For F = 70 lb, 9 : 25° F = (701b)J:(0.90631cos¢)i — 0.42262j + (0.90631sin¢)k] i j k M0 = (7011)) 4 11 —d in. —0.9063ICOS¢ —0.42262 0.9063lsin¢ : (70 1b)[(9.9694sin¢ — 0.42262d)i + (ﬁ0.90631dcos¢ + 3.62525in¢) j + (1.69048 __ 9.9694cos¢)k]in. and MA = (701b)(9.9694sin¢ — 0.422625!) in. = —(61 lb’ft)(12in./ft) (1) M). = (70 1b)( ~0.90631dcos¢v + 3.6252sin¢\$)in. (2) M: : (701b)(£.69048 — 9.9694cos¢) in. = —43lb-ﬁ(12in./ft) (3) PROBLEM 3.152 CONTINUED From Equation (3) ¢ = cos'[634'33 = 24.636° 697.86 or ((5 = 24.6° 4 From Equation (1) d = [102290 2 34.577 in. 29.583 or d = 34.6 in. 4 _[ 15 in. PROBLEM 4.34 Neglecting friction and the radius of the pulley. determine (a) the tension in cable ABD, (b) the reaction at C. SOLUTION A 7/;‘7jﬂ 30 lb 0' 00.4. 5 First note 15 = tan" — = 221520" a [36] ,8 = tan"[£] = 36.870" 20 (a) From f.h.d. of member ABC +‘) EMC = 0: {3010)(23 in.) — [Tsin 22.620°)(36 in.) —(Tsin36.870°](20 in.) = 0 T = 32.500 lb or T = 32.51bd (b) From f.b.d. of member ABC —+- 2F, = 0: C, + [32.500 lb}(00522.620° + cos 36.870°] = 0 -. c, = —56.000 lb or C, = 56.0001b -—~ +1 2F), = 0: C), — 30 lb + (32.500 Ib)(sia122.620° + sin36.8'.-'0°) = 0 c} = —2.0{)01lb or Cy = 2.0001 lb I Then c z W = (56.0); +(2.001)2 = 56.036 lb and 9 = tan-{3] = tan'] [in—J = 2.0454o 0; — 6.0 or C = 56.0 lb 7 2.DS°‘ PROBLEM 4.45 E_ ,_ *1 1m.— _,. 4.," A 20-11) weight can be supported in the three different ways shown L-z" #5 t—Séi: Encoaxglinfaiat the pulleys have a 4-' -1n. radius, determine the reaction at A SOLUTION (a) (a) From f.b.d. 0MB A:€AJ LEI-1:0: A_,=0 i: +323..=0: AJ.—2010=0 ’5 b“; or A). = 20.0 lb and A = 2001014 +‘j EMA = 0: M1 —(2010)(1.5 ﬁ) = 0 MA. = 30010-11 or M,‘ = 30.010034 2011: )Note: 4 mtgn'] = 0.33333 0 From £b.d. of AB 116?; 1.2550: A_.—201Ia=0 +7.b cl 20 lb or AI = 20.0 lb .—f 25‘. :0: AJ. —201b =0 or A}. = 20.0 lb Then A = 213+ Af. = {20.0)2 +(:10.0}2 = 28.284 lb A = 28.3 lb 4: 45°41 3: 2M,l = 0: MA, +(2010)(0.33333 a) ~(2010)(1.5 0 + 0.33333 11): 0 M_., = 30010-0 or M4 = 30.010034 ——————_—_—__—— PROBLEM 4.45 CONTINUED (5) From f.b.d. of A3 + _ . _ (C) -—-a- 2F; — 0. Ar - 0 qua. =0: AI‘,—2(}lb—201b=0 or A}. = 40.011: and A = 40011:]! +‘j 3M1 = 0: MA —[201b)(l.5 a — 0.33333 3) —{2otb)(1.5 n + 0.33333 ﬂ) = 0 . M,' = 60.0]b-ﬁ L or MA : 60.01b-ﬁ )4 PROBLEM 4.54 A slender rod A3. of weight W. is attached to blocks A and B. which move freely in the guides Show. The blocks are connected by an elasLic cord which passes over a pulley at C‘ (a) Express the tension in the cord in terms of W and 9. (2:) Determine the value of 8 for which the tension in the cord is equal to 3 W. SOLUTION 5 (a) From f.b.d. of mdAB +3 EMC = 0: Home) + W[[%]cosa] * meosa) = o : Woosﬂ 2(co59 — sine?) Dividing both numerator and denominator by cos 9, “it I ) l—tanﬂ {b} For T = 3W, awe—g]— (1 —tan0) I-tanﬂzl 6 9 = tan—{g} = 39.806“ or 9 = 39.3“‘ ...
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