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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 FreeBody 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
FreeBody 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 theydirection [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
un a"
. .1 ‘
.“ % SOLUTION Tm must be perpendicular to F“ to be as small as possible. FreeBody 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.
 Spring '08
 Jenkins
 Statics

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