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Friction

# Friction - 17 Friction I I Friction I W FBD of Block at...

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Unformatted text preview: 17 Friction I I Friction I W FBD of Block at rest on a rough horizontal surface If at rest, O=ZF As P increases so does F so long as F S p S -N F: “k' N Value of P for Impending Motion No Motion Friction I -W-sin(Q) +, F=o . —-W-cos(6) + N=O At impending motion F: 118- N tan (0):};S / Friction I Frictionless pulley ,» ' ‘7 us= .2 t:ermpending motion @ W (lb—N Weight of G=? Problem 1 Friction I Problem 1 —W-sin(20.deg) + Q- F=O V 1) —W-cgs(20-deg) t N=0 \ 2) N=W- cos ( 20- deg ) N=46.985_\1b ' For Impendlﬂg m0t10n F ' “sf N Substitute F into 1) .=H -N=.2-(46.9§5) . - . S” G=F t W-sm(20-deg) F=9.39'Z 1b G=9.397+ 50- sin(20- deg ) G=26.4981b \ Friction I ...................................................................................... .Q .................................. F rictionless pulley ' A) u s- ‘2 for impending motion up the slope Recall that answer Weight of G= 26.498 lb 6.3 FrlCtlon I Problem7 in Class III» Frictionless ulle s _ P Y “S. .2 W= 50' lb For impending motion up the slope 7:. H'éﬁ‘? T \$315.0 ’ gal-O HT... F+-S‘a, @4513: @ HT:- AC 50— (7.! == 4 59kmch E-‘GM i—GQ.7_ N = 940. 3gb. h i ! —433—Frs “V‘s-DONio :ﬁuomw / 334C575qN3 Fan-x =' 13:0” - E N9 [email protected] B‘tkaq FBAF’ / , ME“ NOT "mg CASE \oao QHP Waxy, eczema; pr- B hwo Fkovﬂ. QNQTEQUILIED / K a; Friction I WA: QOOON WE: H 000 M BETE «Mm/5:3 (N EQUkLlBYUOh-u Put ya: lsoow / \\ TH-e s‘rsrcpm ls l 692%}: o NH—D‘oooNﬁ-s—ooNfQwa'ﬁcfzo hiNAZ \/ 40ft No \$ks? Ai-B El‘ 2 I‘ISONCVSB ' a "54—— =S'RS'N bf .200 €wCL- FA< PM 091V THE: SPSTFW w cake {:5 /' l_v, / A K £3 THE Mﬁx FO'GCF'? 5‘02 WHICH rv E‘GLU {Laﬁwum “h w /! Na \$k|P® Ai—G,‘ The weights of blocks A and B are WA = 1000 lb and WB = 100 lb. If the coefﬁcient of friction is 0.30 at all surfaces, determine the force P required to produce impending motion of block A. N, —/oaoLL—F;,= a' N‘ﬁ—Iaoo kb—QBX N )5: o :0 , g2 — 0 1 Q75: 13‘— N393 50+.‘3N3C-309): c 74 ‘ .0 I _ F! Ms 1M / G: 7- 9 ® QIBXND ' N L: ° " \/ 3‘ Hag/71kt: s ‘amwss CTN-d) SeMuArawe‘musk? L \ @F:o N=‘9\7,9‘§N=°11,3' ‘3 .1 a 5 345% r 75‘ Lb _p +4314: W71+ Nam Iv°=é -100 .. p +.3(17s.3}_+ 361910.954 425nm 3P= 336.0HLBSJ / Friction I Problem 4 in Class Determine value of 2 required to start the blocks moving 7—, SOLUTION Friction I ‘A free-body diagram for bloek A is shown at the right. A w = mAg'3R50(9.807) = 490.4 N For impending motion: F — 490.4 cos 30° = 0 n T — Ff — W sin 300 r - 0.10(424.7) - 490.4 sin 30° 0 ll Problem in m Determine value of 2 required to start the blocks 287.7 N moving (.11- A free-body diagram for block B is shown at the right. WE = mBg + 100(9.807) = 980.7 N For impending motion: 1 + 2F = 9 sin 25° + Fn — 980.7 = o l + M '11 I] P cos 25o - Ff - T H P cos 25° — 0.109n — 287.7 = 0 Solving yields: P = 406.7 N E 407 N Ans. Impending Motion By Sliding P _>_ F_=j(/Ls :gg-‘u-wyg- .-r-,-. m u“ 1 q ‘ ‘ "M -.~v~"’ ' .-'-': -.- if“ if 1‘ i 1 n. "unit *6 IglglgﬂII-IIIIF-I (J? ' ‘ ‘I M s n - r “a IIIII l5,- '1 d — L.“ . ‘ lirfa'w I “i ' -. . 4 : ., - / I‘ ‘ -.-. up: I at ‘ r . -' «’74- P ' : a"! :' « 4 - ‘ : I . ' ‘ . i . - a” .' I " - I I. . “a, -. L“: x My" '/ 1 7 : Ext-17" “T- ------------------------------------- -—-i—Iii3-iLiii-l-iv¢'--’ vl-H‘-—-—-—-—---—-—-—————---—-—-—-—-—--- 2 ~11 Impending Motion By Tipping N <WE._MW M4 MW ( The ZOO-lb crate is being moved by a rope that passes over a smooth pulley. The C06fﬁC1uu[ of friction between the crate and the ﬂoor is 0.30. a. Assume that h = 4 ft and determine the force P necessary to produce impending motion. b. Determine the value of h for which impending motion by slipping and by tipping would occur simultaneously. CL) (1 CASE” :1: NoTlPPm/Q eNk‘f \$kl? @BQTTQVA‘ \$M? 0N {501‘wa .3 F=ﬂs N Fr—L. BX'MJOLQ . F: 60 L3. @553" 53 ?-F‘0 ME. SMW‘W . 09 4f” CLASSES.“ TH? ~\ ~ +7.00 ‘ {-T) 1‘0 ?= RBSVFOQ’ZE N g ' o o 'I'IF 'P \S \$\$~0Wktf {Meme-93¢} IT MIN—L. rte-mu €01.53 b ‘0 J MIG) .“ x ( @in = o N—‘Loo =- a N: 7.00ka Fm; 6 kW ?( WC; 380 \ Friction I Problem 3 ml in Class For both surfaces 1.1 S = .25 Determine P required to cause impending motion Friction I WA: 50(9.807)=4<90.35 N ﬁ* 0=ZFX ///A?7 T-cos(45-deg )=O Impending Motion Af=.%5-An .r’ 1) f—kos(45-deg)=0 V f 0=ZF An +_ T-sin(45-deg) ~ WA=0 . 2) 11:),31n(45-deg) —‘.4‘9Q.35‘=0 i/ 3’ Solving Eq 1)& 2) Gives: An=392.27N T=138.692N ~ \ F BD of block A Problem 3 in Class X m B n B' _ 735.53_ 392.27=0 1'1 Bn — 735.53! 392.27=0 Bn=1127.80N V Problem 3 in Class WB:_75-(9.8)=735.53N Impending Motion P i .25-(39227) i .25-(1127.80b=0 P=380-N ‘L/ ...
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