{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

assignment 6

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 < ...
View Full Document

{[ snackBarMessage ]}