MAE 130A HW2 prob 3.87_3.95_3.118_3.121_3.147_3.153 soln

# MAE 130A HW2 prob 3.87_3.95_3.118_3.121_3.147_3.153 soln -...

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Unformatted text preview: PROBLEM‘W 5‘“ Three cables attached to a disk exert on it the forces shown. (0) Replace the three forces with an equivalent force-couple system at A. (6) Determine the single force which is equivalent to the force-couple system obtained in part a, and specify its point of application on a line drawn through points A and D. \omr/ (0‘1 . (a) Hate 2F: F3 + FC + FD = FA Since FB = -130 FA =1?f =110N 420° or FA = 110.0151 .1: 20.0°< Have EMA: uFBT (r)— Far [r)+ Fm (r) 2 MA {(140 N)sm15°](0.2 m) — [(110 N)sin 25°](0.2 m) + [[140 N)sin45°](0.2 m) = MA MA = 3.2545 N-m or 01LI = 3.25 N-m‘} 4 (1») Have 2F: F,r = FE or FE = 110.0 N A: 20.0% EM: MA = [FgcosZODHa] 3.2545 NAm = [(110 N)cosZO°](a) . a = 0.031435 m or a = 31.5 mm below A 4 l_._____ _- __I ' PROBLEM 3.95 A 315-N force F and 70-N - m Couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner D. 60 mm 36“ mm \“ /£3<1§0 mm 1 \ero mm = 7‘11}? (0.360 m)i — (0.120 m)j + (0.130 m)k 0.420 m (315 'N) = (750 N)(O.360i — 0.1203 + 0.180k) 0r FD = (270 N): — (90.0 N)j + (135.0 N)k < 2MB: 1v1+rm3><laz1wD M I lAcM _ (0.240 m)i — (0.180 m)k - [70.0N-m) = (70.0 N-m)(0.800i — 0.600k) rm = (0.360 m)k i j k MD=(70.0N-m)(0.8i-0.6k)+ 0 0 0.36(750N.m) 0.36 —o.12 0.18 = (56.0 N-m)i — (42.0 N-m)k + [(32.4 N-m)i + (97.2 N-m)j] 0r MD = (88.4 N‘m)i + (97.2N-m]j — (42.0 N-m)k 4 PROBLEM 3.118 While USing a pencil sharpener, a student applies the forces and couple shoWn. ((1) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A con5isting of the force R = (3.9 lb)i + Ryj -— (1.1 lh)k and the couple Mi: =M.i+(1.5lb-ﬁ)j—(1.116-e)k.. (0} Find the corresponding values of Ry 311de. SOLUTION Have 2F: B + C = R 2sz Bx+cx=19lb or B, =3.9]b—Cx (I) my: Cy a Ry (2) 212: C2 == —1.1 lb (3) Have EMA: rmxB+rme+M3=M§ l i j k I 1 j k E x 0 45 +135 0 2.0 +(21b-ﬁ)i =Mxi+(1.5lb-ﬁ)j—(1.11b-ﬁ)k Bx 0 0 C, C_,, —1.1 (2 ~ 0.16666?Cy)i + (0.3753: + 0.16666'1qr + 0.36667) j + (0.33333CJ,)k = Mxi + (1.5)j —(l.1)k From i-coefﬁcient 2 — 0.1666670}, : M; {4} j — coefﬁcient 0.3753x + 0.1666676; + 0.3666? = 1.5 (5) k -coeﬁioient 0.33333C). = —1.1 or Cy = —3.3 lb (6) (a) F mm Equations {1) and (5): 0.375(39 ~ C1) + 0.1666671; = 1.13333 Cx = 0329” = 1.580001b 0.20833 From Equation (1): .1r = 3.9 —1.53000 2 2.32 lb B = (2.321b)i4 c = (1.530 lb)i — (3.30 16)] —' (1.1 ib)k 4 (b) From Equation {2): R}. = q. = —3.30 lb or Ry = —(3.3016)j4 From Equation (4): M. = —0.166667(—3.30) + 2.0 = 2.55001b-ﬁ or Mx = (2.5516-t1)i4 PROBLEMm= 'M‘P—l As an adjustable brace BC is used to bring a wall into plumb, the force- coupie system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A knowing that R = 21.2 lb and M = 13.251134}, 2F: R = RA = R139 ‘ i ' __ - I u _ . I k Where MC = 106 m. = 2"26‘b(42i ~96] — 16k) R .4 or Rh, = (8.40lb}i —(19.20 lb]j —{3.20 lb)k ( Have EMA: ranR+M=MA where rm = (42 in): + (43 in.)k = %(42i + 48k) a = (3.5 &)i + [4.0 ﬂ)k R = (8.40 [bk-(19.20 lb]j—(3.201b)k M Z —lBC‘M : "Hi + M03 25 Ib.ﬁ) 106 I z —(5.25 [b-ﬁ)i + (12 lb»ﬁ]j + (2 lb-fl)k ' . 7 PROBLEMW 5 H The triangular plate ABC is Supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 220 lb, determine the moment of that force about the line joining points D and B. [(36 in.)i — (24 in.)j + (3 i“-)k]( TAB Z XAETAE = 44in 2201b) = (1801b)i —(1201b)j +(401b)k 0.960 —0.2so 0 -. M03 = 0 —4 s lb-in. 180 —120 40 = (0.960)[(—4)(40) _ (ax-120)] + (—0.2so)[8(180) — 0] : 364.81b-in. 01' M08 = lb'in.‘ PROBLEMM 3453 Three children are standing on a 15 x lS-ﬁ raft. The Weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a fourth child of weight 95 lb climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four Weights is to pass through the center of the raﬁ. mw H:m+n+m+%=n —(85 lb)j-(601b)j—(90lb)j —(95 1b)j = R R = ~(3301b)j Have EMI: FA(ZA) + FB[23) + Fc(zc) + 170(20): R(2H) (3511:.)(9 ﬁ)+[601b)(1.5 ft)+(901b)(14.25 ft) +(95 lb)(zD) = (3301b)(?.5 it) -. 29 = 3.5523 ft or 25 = 3.55 ft 4 I3w: 2M3: FA(xA)+FB(xB)+FC(xC)+FD(xD)=R(xH) [851b)[3 ft)+(601b)(4.5 ﬁ)+ [901b)(14.25 ft)+(951b)(xD) = (3301b)[7.5 ﬂ) '. x0 = 7.0263 ft or xD = 103 ﬂ ‘ —'--'——._._.___—‘_—‘—_—‘____—____———____ it ...
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## This note was uploaded on 03/11/2010 for the course MAE 130A 614117 taught by Professor Lubarda during the Spring '08 term at UCSD.

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MAE 130A HW2 prob 3.87_3.95_3.118_3.121_3.147_3.153 soln -...

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