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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 Nm = —(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 Nm or M,c = 28.0 Nm 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: (48 ft)l(1.2 ﬂ), DI (4.8 £02 + (—1.2 11)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 lbﬁ
2.4 —7.2 —23.4 = 20.432 — 245.17
= —224.74 lbﬁ or MD] = —2251b~ft 4 PROBLEM 3.72 The two shaﬂs a speedreducer unit are subjected to couples of magnitude MI 2 18 N m and M2 = 7.5 Nm , 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 Nm)2 + (18 Nm)2
= 19.5 Nm
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 63lb force F and 5601b«in. couple M are applied to corner A of the
block shown. Replace the given force~couple system with an equivalent
forcecouple 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 = (5601bin.) = (4481bin.)i—(3361bm.)k i j k
Then MD 2 (44816111.)i—(336 1bin.)k+ 0 0 14.4 113m
54 —18 27 = (448 lbin.)i — (336 1bm)k + [(259.2 lbin.)i + (7776 16m) j] or MD = (7071bin.)i+ (7781bin.)j—(336 lbén)k < ...
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 Fall '07
 Delph

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