HW25 - PROBLEM 8.36 I=30in.is...

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Unformatted text preview: PROBLEM 8.36 ThuslundarrudABoflanglh I=30in.is attachedtnacollaratBand reﬁsunasmallwheelbcswdatahoﬁzontaldistamea-4hﬁunthe verlicalrodonwhichﬂlccoﬂnrslidchEnowingﬂmthcoocfﬁcimtof shﬁcﬁ-icﬁonbetweenﬂnecollarandtheverﬁcalmdisﬂlsm neglectingmeradimofﬂaewhaeLdmﬁlﬁneﬂnmngcofvaluesofPfor whichaquilibriumismaimainedwhen Q=251b andﬁ=30°. SOLUTION a 4 in. FBDrml+eullan ' ' Human—- ‘ sine £1130” Neglect weights ofmdandoollar. (_ m, = o: [30 in.)(einso°}(2s tb} - [a in.)c = o c = 46.375 It: ' ' h—I-EF; = o: N—Cuos30" = u ' ' N - (46.815 Ib)eoe30= = 40.595 lb Impending mntion up: ’F; = M = 025(40595 1b) = mid-Mb 121-} = o: —25 lb + (46.375 lb)einso°-P-m.149 lb =6 mﬁinﬁaoAC=22in or _ 'P'='-1.563 Ila-10.149 lb = —11.71 lb .mdhgmaiondéwmnieaﬂmofFismwupmwsﬁllhwe '= IFI = em = 10.14911: 1215; a o: —25 [bi-{46.875 Ib)ein30°—P+m.149 lb =0 P = -l.563 lb + 10.14911: = 8.59 1b 3. quﬁlibriumfor 41.7111: 5 P 5 8.59m ‘ PROBLEM 3.47 __| “L |_1,,I+"#___1 Solve Problem 3.46mmm1ingﬂlatfwcal'isdj1mdtuﬂlelsﬁ. n - c -' n F-h=€mw[ (_ :10}; = 0: (12 511.][00 11:) — [16 in.)(401b] —(211n.)R,, 130524.035“ — (2 111.391.124.035" = 0 of RA = 16.005 Th [15) Lv IF; = 0: c, — [16.005 Ib)ain24.036° - 0 1:;r = 6.52 DJ Hi: 1 2F, = 0: c’ w 0010 — 40 lb + (16.005 Ib)cus[24.036“] = 0 C, = 105.4 b [4 FED wedge: 5'“ =- \$90-\$20 {on 45, at 29.035 " L. 4:, = Max " I : 11,1403? r (16.005 lb)oos24.036= = 0 x“, =15.0671b : (16.0051b)si1124.036“ +(15.067 lb}sinl4.036° —p = 0 ' nil-10.17115“ SOLUTION FBI) pipe: PROBLEM 8.80 A15” wedgois fumedlmdaralﬂﬂ-Ibpipe as shomT‘hceocﬂicimtof static friction a! all surfaces is 0.20. Determine (a) at which surface slipping of the pipe will ﬁrst occur, (b) the force 1' for which motion of thawedge is impmding. . (a) (maﬁa: rFA—rF3=0 or . .FA=F}I Bmitisappsrsntlhst NH > Nd,sosinoe (5)4 = (yak. ' motionmustﬁrstimpendatdi and FSEFA=FENA=02N‘ (b) (_ EH3 = a: (rsin15°]W + r(1+ 31.1150}?M —[rm15°)1v1 = 0_ 0.2533000 lb] + 1.2588(03NA) - 0.9559115, = 0 or NA = 352m and _ FA = 7.25 1h \ :13. -=_ o: N, u NAsinIS" +15 cos15° — Wmslﬁ" = o N, = (36.241b]sin15° +(7251h +100 1b)cos15= = 112.97 lb (noteNa > NA assisted, andFa < paNB) f2}; = 0: NW +[7.251b)sjn15° — (112.971b)cos15° - a 111W = 1012411: Impending slip: 1'} = 14an = o.2(1m.24] .- 21.45 lb —-r as; . o: 21.451]:+(7251b}cos15°+(112.971b)si1115"—P= o P = 57.7]b-—" ...
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HW25 - PROBLEM 8.36 I=30in.is...

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