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assignment 6 - PROBLEM 3.50 A force P is applied to the...

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Unformatted text preview: PROBLEM 3.50 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that My = —20 N m and M2 = —3. 5 N m, determine the moment M ofP about the x axis when 9: 60°. Based on M,r = (Pcos¢)[(0.225 m)sin9] — (Psin¢)[(0.225 m)cos0] My =—(Pcos¢)(0 125 m) M: —(Psin¢r)(0. 125 m) ByEqua’tion(3) —.(Psinq>)(0 125) ByEquation(2) M: =—)((Pcos¢ 0125) —3. 5 — = tan 9. 9262° or _20 4:; (I): From Equation (3): -3.5 N-m = —(Psin9.9262°)(0.125 m) P = 162.432 N From Equation (1): Mx : (162.432 N)(0.225 m)(cos9.9262°sin60° — sin9.9262°cos60°) = 28.027 N-m or M,c = 28.0 N-m 4 PROBLEM 3.60 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EG at E is 24.6 Eb, determine the moment of that force about the line joining points D and I. MDI_ — 7‘01 (Tc/1 X TEG) where 7L0] = fl: (4-8 ft)l-(1.2 fl), DI (4.8 £02 + (—1.2 1-1)2 = 0.97014 i — 0.24254j r0), = —(35.1 11)]: [(36fi) i)— (.108fi)j—(35.1fi)k] T =T —.(2461b) EG MEG (6 11)+ “(10311) +—2(35.1ft) = (2.4 1b)i — (7.2 lb) j — (23.4 lb)k Then 0.97014 —0.24254 0 M0, = 0 0 —35.1 lb-fi 2.4 —7.2 —23.4 = 20.432 — 245.17 = —224.74 lb-fi or MD] = —2251b~ft 4 PROBLEM 3.72 The two shafls a speed-reducer unit are subjected to couples of magnitude MI 2 18 N m and M2 = 7.5 N-m , respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis SOLUTION Based on M = M1 + M; M=USNmk M2 = (7.5 N.m)i aMnUfiNmfifimNmm M = and (7.5 N-m)2 + (18 Nm)2 = 19.5 N-m mM=mwNm< wm A=M=flflflflifififlk M 1%Nm m B “m Cm®=% gg=m3m c036}! = 3. 9y = 90° m%=%' ne=nam 01' 9x = 67.40, 9y : 90.00, 62 = 22.60 < —_——‘——w-——--—— r PROBLEM 3.95 A 63-lb force F and 560-1b«in. couple M are applied to corner A of the block shown. Replace the given force~couple system with an equivalent force-couple system at corner D. SOLUTION Have 2F: F = FD AI where F = F— A1 1441— 4.8j + 7.2k = (63 lb)—~——~— (14.4)2 + (—4.8)2 + (7.2)2 So that F = (54.01b)i—(18.001b)j+ (27.01b)k 1 Have 2MB: M + rm x 1—? = MD Where M = M E AC 9.6i — 7.2k (9.6)2 + (— 7.2)2 = (5601b-in.) = (4481b-in.)i—(3361b-m.)k i j k Then MD 2 (44816111.)i—(336 1b-in.)k+ 0 0 14.4 113m 54 —18 27 = (448 lb-in.)i — (336 1b-m)k + [(259.2 lb-in.)i + (7776 16m) j] or MD = (7071b-in.)i+ (7781b-in.)j—(336 lb-én)k < ...
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