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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 Wsin(Q) +, F=o . —Wcos(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 —Wsin(20.deg) + Q F=O V 1) —Wcgs(20deg) 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 Wsm(20deg)
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
’ galO HT... F+S‘a, @4513:
@ HT: AC 50— (7.! == 4 59kmch E‘GM i—GQ.7_ N = 940. 3gb. h i
! —433—Frs “V‘sDONio
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B hwo Fkovﬂ. QNQTEQUILIED / K a; Friction I WA: QOOON
WE: H 000 M BETE «Mm/5:3 (N EQUkLlBYUOhu Put
ya: lsoow / \\ THe s‘rsrcpm ls l 692%}: o NH—D‘oooNﬁs—ooNfQwa'ﬁcfzo
hiNAZ \/ 40ft No $ks? AiB 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 $kP® 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+.‘3N3C309): c
74 ‘ .0 I _
F! Ms 1M /
G: 7 9
® QIBXND ' N L: ° " \/ 3‘ Hag/71kt: s ‘amwss CTNd) 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 freebody 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 freebody 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‘uwyg .r,. m u“ 1 q ‘ ‘ "M
.~v~"’ ' .'': . if“
if 1‘ i 1
n. "unit
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‘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 : Ext17" “T
 —i—Iii3iLiiiliv¢'’ vlH‘———————————————— 2
~11 Impending Motion By Tipping N <WE._MW M4 MW ( The ZOOlb 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 {Meme93¢} IT MIN—L. rtemu €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 Tcos(45deg )=O
Impending Motion Af=.%5An .r’ 1) f—kos(45deg)=0 V f 0=ZF An +_ Tsin(45deg) ~ WA=0 .
2) 11:),31n(45deg) —‘.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=380N ‘L/ ...
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 Spring '08
 McVay

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