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

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