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Unformatted text preview: PROBLEM 2.45 Two cables are tied together at C 'and are loaded as shown.
Knowing that P = 500 N and a = 60°, determine the tension in
(a) in cable AC, ([7) in cable BC. SOLUTION F recBody Diagram Force Triangle THC _ THC _500N Law of Sines: . — .  . sm35° sun 75° sm 70°
(3) TM; 2 3:07oNosin35" T“. =305 N 4
(13) TM. = sin 75° THC. : 514 N 4 PROBLEM 2.66 A lbO—kg load is supported by the ropeandpulley arrangement shown.
Knowing that a = 40°, determine (a) the angle ﬂ, ((3) the magnitude of
the force P that must be exerted on the free end of the rope to maintain
equilibrium. (See the hint for Problem 2.65.) SOLUTION
FreeBody Diagram: PulleyA (a) EFJ = 0: 2Psin sin ﬁ— Pcos 40": 0 sin ﬁ=éeos 40° 3:22.52” 5:225" 4 (b) EFy = 0: Psin 40°+2PC05 22.52°— 1569.60 N :0  3 P263051 4
=(lboxhC‘ialmiaz)
r I5c. .eow —
PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that a=25° and ﬁ=lﬁ° and that the tension in
cable CD is 80 N, determine (a) the combined weight of the
boatswain‘s chair and the sailor, (b) in tension in the support
cable ACB. SOLUTION —
FreeBody Diagram
Y
' i. ZFJ = 0: TAG; cos 15° — TAG; cos 25°  (80 N)cos 25“ = 0
Tm, =1216.E5 N
+1 Zﬁ, =0; (1216.15 N)sin15°+(]2l6.15 N)sin 25°
. +(so N)sin 25°ﬁW=0
W = 362.54 N
(a) W = 863 N 4
(b) TACB =1216N 4
PROBLEM 2.80 Determine the magnitude and direction of the force F : (240 N)i — (270 N)j + (680 N)k. l—SOLUTION F: J11} +Fy2 +Ff
F = ,/(240 N)2 + (—270 N)2 + (680 N) F = 770 N 4 cos Qx=£=—24—0§ 91:71.8” ‘
F 770N .
F _
cos 6,,=—y= WON 9 =110.5° 4
 F TION y
cosﬂz=£=ﬂ 6 =28.0°‘
F 770N z PROBLEM 2.114 A horizontal circular plate weighing 60 lb is suspended as shown from
three wires that are attached to a support at D and form 30° angles with
the vertical. Determine the tension in each wire. SOLUTION l'f zo=0= — JD (sin 30°)(sin 50°) 4» TBS (sin30°)(eos 40°) + Tm (sin 30°)(cos 60°) = 0 Dividing through by sin 30° and evaluating:
—0.76604TAD + 0.76604TBD + 0.5TCD = 0 (1) EF} = 0: —TMJ (cos 30°) — TED (cos 30°) — TCD (cos 30°) + 60 lb = 0 or Tw + 1'33 + Tm = 69.282 lb (2)
SF2 = 0: TAD sin 30° cos 50° + TED sin 30°sin 40° — Tm sin 30°sin 60° = 0 or 0.642797” + 0.64279TBD u 0.8660375 = o (3) Solving Equations (1), (2), and (3) simultaneously:
TAD = 29.5 lb 4 Tm, =10.251b 4
Tm =29.5 lb 4 ...
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This note was uploaded on 09/11/2009 for the course ENGRD 2020 taught by Professor Zehnder during the Fall '06 term at Cornell.
 Fall '06
 ZEHNDER

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