HW2_Solution

# HW2_Solution - PROBLEM 2.45 Two cables are tied together at...

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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 rec-Body 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 rope-and-pulley 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 Free-Body 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 —| Free-Body 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.

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HW2_Solution - PROBLEM 2.45 Two cables are tied together at...

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