This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTION re, PROBLEM 2.55 The cabin of an aerial tramway is suspended from a set of wheels that can
roll freely on the support cable ACE and is being pulled at a constant speed by cable DE. Knowing that a = 40° and ,6 = 35°, that the
combined weight of the cabin, its support system, and its passengers is
24.8 kN, and assuming the tension in cable DF to be negligible,
determine the tension (51) in the support cable ACE, (b) in the traction
cable DE. Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If
considered as a rigid body (Chapter 4) it would be found that its center of
gravity should be located to the left of the centerline for the line CD to be vertical. Now i. 2F, = 0: TACB[cos3S° — cos40°) — TDE cos40° = 0 and +1213, 2 o: TACB(sin4O° — sin35°) + TDEsin40° — 24.8kN = 0 0.0692110? + 0.6143be = 24.8kN From (1)
TACB = 14.426TDE
Then, from (2)
0.0692(14.426TDE) + 0.643TDE = 24.8 kN (b) TDE =1s.1kN4 (a) TACB = 4 ‘ t ﬂ—ﬂ
PROBLEM 2.77 y A horizontal circular plate is suspended as shown from three wires which
are attached to a support at D and form 30° angles with the vertical.
Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (13) the angles 1%,
By, and 6; that the force exerted at A forms with the coordinate axes. 3 I
SOLUTION (0) Fr = Fsin30°sin50° = 220.6 N (Given)
F = ——~—. 220'6‘N = 575.95 N
sm30° $111150O
F = 576 N 4
(b) 0056’Y = = 220'6 : 0.3830
' F 575.95
8x = 67.5°4
Fy = Fcos 30° = 498.79 N
F
cost = "3 : 49??? = 0.86605
' F 573.93
49}. 2 300° 4
F: : ——Fsin30°cos§0°
= —(575.95 N)sin30°cos§0°
= —185.107 N
c056, = TF4 = = —0.32139
' F 575.95
t9_,~ = 108.7°4 f w, PROBLEM 2.101 The support assemny shovm is bolted in place at B, C, and D and
supports a downward force P at A. Knowing that the forces in members
AB, AC, and AD are directed along the respective members and that the
force in member AB is 146 N, determine the magnitude of P. SOLUTION
Note that AB, AC, and AD are in compression.
Have
2134 = (—220 mm)2 + (192 mm)2 + (0)2 = 292 mm
dm =1/(192 mm)2 + (192 mm)2 + (96 mm)2 = 288mm
dCA = (0)2 + (192 mm)2 + (—144 mm)2 = 240 mm
146 N . .
and PM = FBAABA = 292 mm[(—220 mm)1 + (192 mmh]
= —(110N)i + (96 N)j
_ _ FCA  _
Fm _ 1?“ch _ 240 mm[(192 my (144 mm)k]
: FCA(0.80j — 0.60k)
F . .
FDA = Fmim = 2880;m[(192 mm)1 + (192 mm); + (96 mm)k:
= FDA [0.66667i + 0.66667j + 0.33333k]
With P = —Pj
AtA: 2F=0z FBA+FCA+FDA+P20
i—Component: —(110N) + 0.66667FDA = 0 or FDA =165 N
jcomponent: 96 N + 0.80% + 0.66667(i65 N) — P = 0 (1)
kcomponent: —0.6OFCA + 0.33333(165 N) = 0 (2) Solving (2) for FCA and then using that result in (1), gives P = 279 N 4 mt—w—"mﬁv _.__ # PROBLEM 2.133 Two cables tied together at C’ are loaded as shown. Determine the range
of values of Q for which the tension will not exceed 60 1b in either cable. I M—i SOLUTION Have
or / ’X Then for 01' From 01" For or
Thus, Therefore, EFI = 0: TCA ~Qoos30° = 0
TCA = 0.8660 Q
TCA : 601b
0.8660Q < 601b
stam
25 = 0: ICE = P — Qsin30° TCB = 7511» — 0.50Q
TUB 3 601b 7511: — 0.50Q 3 601b
0.50Q 2151b Q2301b 30.0 s Q s 69.31134 \ l. PROBLEM 2.139
frame ABC is supported in part by cable DBE which paSSes through a A frictionless ring at B. Determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension / in the cable is 385 N. p r F :
/ \ fat E i
/ :5 v l ﬂ 100 lllltl ASK} "1W E SOLUTION The force in each cable can be written as the product of the magnitude of the force and the unit vector along
the cable. That is, with B = —(0.48 m)i + (0.51m)j i (0.32 m)k BD : (ﬁ—OAS mf + (0.51 m)3 + (—0.32 m)3 = 0.77 m 373’ T .  . .
TED = MED = 8055 : D 711:]0m(_(0.zts m)1+(0.31m)J—(0.32 m)k] TED = TBD(—0.6234i + 0.6623j — 0.4156k) and BE = —(0.27 m)i + (0.40 m)j  (0.6 m)k 7 1‘ " BE = (ii—0.27 m) + (0.40 m) + (—0.6 m) = 0.770 In 8D 31"
i = : _ — 2 35
85 BE 35 BD 0.770 m [—(026 m)i + (0.40 mli  (06 mlkl T35 : TBE(_—0.3506i + 0.5 i95j — 0.7792k) Now, because of the frictionless ring at B, TEE = TED : 385 N and the force on the support due to the two cables is F = 385 N(—0.6234i + 0.6623j — 0.4156k — 0.3506i + 0.5195j — 0.7792k) —(375 N): + (455 NH — (46G N)k 1 PROBLEM 2.139 CONTINUED The magnitude of the resultant is 'v 1 F; + F}? + F; = {—375 Nf + (455 N)‘ + (—460 N) = 747.83N or F = 748 N 4
The direction of this force is:
8x = (:0?1 7:378: or 6x =120.1D 4
611‘ = cos—1 7:73:33 or Q, = 525° 4
= 05’1 72:16:; or 92 = 128.0°4
. _i ...
View
Full
Document
This note was uploaded on 10/09/2008 for the course COE 2001 taught by Professor Valle during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 VALLE

Click to edit the document details