Hw 2 - E"1 I BC 2.1 m By Slmt at tnang es-=-I 61ON I...

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PROBLEM 2.46 75' Two eables are tied together at C and are loaded as shown. Detennine the tension (0) in cable AC, (b) in cable Be. .c7~"_-lC SOLUTION Free-Body Diagram Force Triangle ... W=mg = (200 kgX9.81 mI") =1962N T AC THe 1962 N Law ofsines: --= = sin 1.5" sin 105" sin 60" T _ (1962 N) sin 15" (a) = 586 N . .. T Ae AC - sin 60" 1: _ (1962 N),in 10~O (b) T Be =2190 N . .. Be - ",j. . ~o
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c lm i i 2 ,lm --t- 2 . PROBLEM 2.60 Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable that can be used to support the load shown if the tension in the cable is not to exceed 870 N. 120{) N SOLUTION Free-.Body Diagram: C (ForT = 725 N) 810 N 2T I elO'" +1 fE, =0: 2T, -1200N=0 --~-- " " " ~ =600N " " T . 1 +T , 2 =T 2 T; +(600 N)' = (870 N)' 1200N T,=630N I 1------"'1
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Unformatted text preview: E "1 . I BC 2.1 m By Slmt at tnang es: --=--I 61ON . ... / I 870N 630N r-7f I BC=2.90m T.... I I , , ' C rtJ30,.,┬Ě' - -t L = 2(BC) 2.lm"""'i =5.80m L=5.80m <II PROBLEM 2.67 A 600-lb crate is supported by seven.1 rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the him for Problem 2.65.) T (01 (bI (,,) (dl (., SOLUTION Free-Body Diagram of Pulley (oj -r-r +1 rF, ~o: 2T-(600 Ib)~O ┬ąbOOlb T ~+(600 lb) T~300lb ... (h) , T ┬ą 1,.00 II> +1 rF, =0: 2T-(600 Ib)=O I T=-(600 Ib) 2 T=3001b . .. (e) (d) T~~lb '~~OOlb +1 rF, =0: +1 rF.~O: 3T-(600Ib)=0 T=!(600 lb) 3 3T-(600 Ib)=O T =!(600 lb) 3 ,T = 200 Ib . .. T=2001b ... (e) +1 rF , =0: 4T-(600 Ib)=O l T=!(600 Ib) 4 T=150.0 lb . .. G,OOlb...
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Hw 2 - E"1 I BC 2.1 m By Slmt at tnang es-=-I 61ON I...

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