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Unformatted text preview: P \o L M 8.14 The 20lb block A and the 40~lb block .8 are at rest on an incline as
shown. Knowing that the coefﬁcient of static friction is 0.25 between all surfaces of contact, determine the value of 9 for which motion is
impending. SULU FBD’S: ION 20/!) Note: Slip must impend at both surfaces simultaneously. i2ﬁ}=0: N1~201b=0, N12201b
Impending slip: H 2 MN] : (0.25j(20 lb) z 5 lb
«23:0: —T+51b=0, T2511)
f = 0: N2 — (20 1b + 40 ib)cos9 —(51b)sin9 = 0
N2 2 (6O lb)cose «(51b)sin8
Impending slip: 172 : ,uSN2 : (0.25)(600036 — 53in6) lb
\ 211,, z 0: F2 » 5 lb — (5 lb)c059 + (20 lb + 40 1b)sin9 = 0 *ZOCOSQ + 58.755in6 — 5 = 0 Solving numerically, 9 = 23.40 4 ’ PROBL M 8.67 The squarethreaded worm gear shown has a mean radius of 1.5 in. and a
lead of 0.375 in. The larger gear is subjected to a constant clockwise couple of 7.2 kipin. Knowing that the coefﬁcient of static friction
between the two gears is 0.12, determine the couple that must be applied
to shaft AB to rotate the large gear counterclockwise. Neglect friction in
the bearings at A, B, and C. SOLUTION
FBD large gear: C = 0: (12 in.)I/V a 7.2 kipin. = 0, W = 0.600 kips : 600 lb _ #1 0.375 in. 9  tan ‘ = 2.27850
2750.5 in.) (p = tarfl H. = tan—1 0.12 = 6.84280
S is Q = Wtan(9 + (is) =(6001b)tan9.12130 = 96.333 lb 94 m 15 z ‘
1:: r 1.5 in.
Q. a ’D ” 2M = 0: (1.5 in.)(96.333 lb)w M = 0 l
\K 4? y M = 144.51bin. ’ROBLEM 8.114 A differentiai hand brake is used to controi the speed of a drum. P Determine the minimum value of the coefﬁcient of static friction for
which the brake is self~locking when the drum rotates counterclockwise. SOLUTION
'FBD Lever:
7 If brake is selflocking, no force P is required
73 A' r r , .
N £5} 5 2w“ D 21MB 5: 0: In.) TC " In.) 77/; z 0
ml? A. ——~D
130° 
x A” C“ TC 2 3.75?) For impending slip on drum: TC 2 TAeWG em? = 3.75, or #5 = $1113.75 With E z HS 2 0.361 In i PROBLEM 10.37 Knowing that the constant of spring CD is k‘ and that the spring is unstretched
when 9 = 0, determine the value of 9, Where 0 S 9 S 900, corresponding
to equilibrium for the given data. P=600N,Z=800mm, k=4l<N/m From geometry: yA = 151119
5yA = I 005959 xC 210059 + [5in9
= l(cos9+ sin9)
yC = 15in9— 10059 21(5in9 — 0059) [CD = Z\/(cos9 + 5in9)2 + [(sin9 — 0059) — (4)]2 = H3 + 25in9 — 20059 005 9 + sin9 43 + 25in9 — 20059 and FZSP = — = kit/3 + 25in9 — 20059 — 1) £11: 51GB 2 Work: 0: '" FEP5ZCD : 0 or P(lcos969) — kl(1 /3 + 2st — 20059 —1)[z3—C%SQ—%§%9—959 = 0
+ SH] — COS or l————.—1———<l+tan9)=£
1/3 + 251119 — 20059 kl 600 N
(4000 N/m)(0.8m) = 0.1875 Solving numerically PROBLEM 10.50 Denoting by its the coefﬁcient of static friction between the block
attached to rod ACE and the horizontal surface, derive expressions in
terms of P, ,us, and 6 for the largest and smallest magnitudes of the force Q for which equilibrium is maintained. SOLUTION For the linkage: +5 2MB :0: Then: Now and 01” For Qmm, motion of A impends to the right and F acts to the left. We
change ,uS to #15 and ﬁnd Virtual Work:
5U = O: (Qmax — F)5XA + P5yF = 0 l
Qmax "— “Ausp —XA+“XAP:0 01‘
2
P 1
F = ‘USA : [ls—2” : xA =ZZsin6 SxA = 21003956
yF = 3lcose
5ny = ~3lsin969 2 3 l
Qmax : EPtanQ + Qmax Qmin : 2 )(21c0s950)+ P(—3lsin659) = 0 2 5< 33(3th + as) 4 3tan9 — as) 4 ...
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This note was uploaded on 01/08/2011 for the course ME 35a taught by Professor Bhattacharya during the Fall '09 term at Caltech.
 Fall '09
 Bhattacharya
 Statics

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