HW7_Solutions

HW7_Solutions - P\o L M 8.14 The 20-lb block A and the 40~lb block.8 are at rest on an incline as shown Knowing that the coefﬁcient of static

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Unformatted text preview: P \o L M 8.14 The 20-lb 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 square-threaded 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 = 144.51b-in. ’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 self-locking, 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.

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HW7_Solutions - P\o L M 8.14 The 20-lb block A and the 40~lb block.8 are at rest on an incline as shown Knowing that the coefﬁcient of static

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