HW27 - PROBLEM 8.132 p,=035 Consideringorfly values offl...

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Unformatted text preview: PROBLEM 8.132 p,=035 Consideringorfly values offl lessthan 90°, dctemfincmesmallestvalu: A=°30 of 0 requiredto startflme blockmovingtoflmrightwhcn(a) W = 751b, (b) H" =1001b. SDLU'HON FBI) block: (motion Impending) 30m“ W . Sin“ — smog—é!) . W14.036° “(9 ' ’3) “ T W or in a —14.036° - — a ( ) 123mm (a) FF = 75 lb: 0 =14.ose° + sin-141.5113... 123.6951b 100“) W =1ooIb: 9 =14.036° + sin“-—--—— 123.695 1h SOLUTION FBD'B Block A: 3::- lb PROBLEM 8.136 The m-lbmockA smith: 30-lbbluck3arc mppnrlndby mincline wlfichishcldintheposifionshomenowingthatfllecoefficiemfistafiu friction is 0.15 between all surfaces ofconract, determine the vaiue of 6 forwhichmotionis impending. ' A: /ZF,',=0:INA—(201b)m9=0 or NA=[201b)cosfl B: / 2E, =0: N, -NJ-[3le)cusfl=0 I _ N5 = Nll + (an Ib)ms0 = (501b]cosa Impendingmutionatallsmfaccs;3impends\: FA = ANA = (o.15)(zolb)oos9 = (3 @0089 F; a 15.1% = (o.15)(501b)cos9 = (7.5 lb]cos9 A: \ 2F, =0: (201i!)sin9+F1-T=O fig B: \zfié'o: (SDIb)sinB—FJ-F3—T=D So ' _ (lOIb)sin9—2FA—FB=0 (10 lb)sim9 = 2(3 1b)ch + [7.5 [mouse = 13.51b m“? lOlb I135: 0 = 53.5” { PROBLEM 8.141 Two lfl-lb blocks A and}! are connected by a slender rod of negligible wagiin The coefficimt of static firictiou is 0.30 between all surfaces ofcomact. andtherodforms anangla 6=30”' withtheverfical. (a) Show that the system is in equilibrium whén P .-= o. (b) Detcm'line 1h: largesth ofPforwlfichcquilihriumismaintamcd. SOLUTION b F P ,motionimpcndsatbofisurfaons FBDbluckB: U m m B: 1203, = 0: 0;.ll -Ile—Ffloo530° - 0 N, grungy}; ' fiction: I"; = EN, = 0.3M, --—- H; ;= 0: F, :Fflsinw“ =0 Slim ' . a, = 2173 = 0.6M; Solving (1) and (2) N3 '= 10 10 + 7(0005) ' FED block A: = 20.8166 113 P _ Then F” =, 0.005, =12.49{}01b H l I so _-_IA:_v3_I:-f-—*-_ 2F, =..__.0: Fflsin3ou—NA all] NA = g1?” - $114900 lb) = 0.2450 lb Impandingmotion: FA = “N; - 0.3(6.24501b)=¥1.8735 lb _ W» .= .F4_.+Fu°°:s3"‘_‘ .-. P —.1°Ib.-"° 45 P=FJ+TFA3—101b =1.s735 10 +'73(12.4900 lb] — 10 lb = 2.59 lb P=2.69lb‘ Since P = 2.691th initiate motitm. equilibrimnaximwifll P = 0" ...
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This note was uploaded on 02/12/2010 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW27 - PROBLEM 8.132 p,=035 Consideringorfly values offl...

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