# HW3 - PROBLEM 2.46 Two cables are lied together at (f and...

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Unformatted text preview: PROBLEM 2.46 Two cables are lied together at (f and are loaded as shown Knowing that (1 7 30“. determine the tension (a) in cabh: AC”. (.5) in cable BC. SOLUTION Free-Body Diagram Force Triangle "f- , M AC TR M. 36 ° 0 If B o _Y__ . w: Book‘s (9.6! “73% = 2943 N Law ofSinesz T4:- , Tm; 2943 N \$11160c sinSﬁ” 511165” , 294? N , (a) I“ : _ Z smoO” = 2812.|9N T” = 2.81kN4 51116:)“ 9 3 Eb) m : 2 4mﬁsinss“ = 2659.98N I'm = 2.66 kN 4 sin 65" PROBLEM 2.51 I Two forces 1’ and Q arc applied as shown to an aircraft connection. i Knowing that the connection is: in equilibrium and the 1’ = 400 lb and Q : 520 lb‘ determine the magnitudes of the forces cxcrtcd on the rods A and B. SOLUTION Free-Body Diagram Resolving the forces into x and y directions: R~P+Q+FJ+FH —0 Substituting components: R = 4400 1b,“ + [[520 lblc0555°li [(520 lb]sin\$53]j + Full ‘ (Ficos559)i v (F.,siiisso)j .7 0 5.20119 [n they-direction [one unknown force) 400 [h — [5201b]sin55” + FrsinSSU 7 0 Thus, _ 400 lb + [5201b)5i1155° J = ——- —. 1008.3]b smSSJ [ﬂ 1008 lb 4 In 11'”; I—diroction: (5201b)cos§5° i F8 7 F‘ c0555” — 0 Thus. F}; : ‘L‘JCOSSSS — ib)C0555o -— (1008.31b)c0555” - (520 lb)cos55O : 280.08 lb FB .. 2801b 4 PROBLEM 2.63 For the structure and lnadlng of Problem 34 L delenninc (a) the value of a for which the tension in cable BC is as small 215 possible. (b) the P corresponding value of tho tension. ! C. ' a :1 \. . ‘2 , l! I ‘J i frat u-n a" . .1 ‘- -.“ % SOLUTION Tm must be perpendicular to F“ to be as small as possible. Free-Body Diagram: C Force Triangle is a right triangle T136 l 550 EC. 0L 25° 40:)“: (a) We observe: a = 55“ a 2 ST‘ (b) Tm. :(4001b]sinm° or TM .. 346.41‘0 Tm. 345 1b4 ...
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## This note was uploaded on 05/18/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW3 - PROBLEM 2.46 Two cables are lied together at (f and...

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