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Unformatted text preview: Problem 1:
(a) see ﬁgure. L g ) (b) Apply the cosine law for the triangle ABC: BC = \/a2 + b2 — 2abcos(a + 90°] with a = 1 m, b = 0.75 m, and a = 30°. Thus, BC = 1.52 m. Next, apply the sine law to ﬁnd the angle 3: sin [3 _ sin(a + 90°)
a BC from which we obtain that sin/3 = 0.57 or [3 = 34.7°. ( (( > ___________________. To ﬁnd the tension T, consider the moment about A. By equilibrium, ZMAszsinﬂ—gWasina=0 (4.)
f_/__________,/ Thus, T: 1.17 kN. ( 2 _> (c) The length of the spring at the equilibrium is L = BC = 1.52 m. The unstretched length of the
spring is ._.——;—'—""’ C 1’)
L0 =\/az+b2 =1.25m ( ._.—"__""' Thus, the spring constant is T
L—L0 k: 24.33kN/m ( (L .__._.——__#__ (d) From the free body diagram: ZFX = AX — Tees/3 = 0 , which gives that AX = 0.96 kN; (' f) ZFy=Ay—Tsin/3—W=0,whichgivesthatAy=2.67kN. (I) ‘7’
A: See ﬁgure below. Ignore the radius ofthe pulleys. Find the forces in members CB, CG, and Ell/16:0: 3FCB+3T +m5(3+3JZ~c.>s/r°)= 0
r9 3) PCB :—fé.2§ lc/V (C) U) 1, /2'. See ﬁgure below for a bolt cutter. ( O
(a) Draw free body diagrams for the four parts (separately). (45*) (b) Find the force in member CD (10)
(c) Find the force exerted on the bolt. (15) C OVdCWvLQ/l Uga ._ a (r)
ZMB : 0 ’9 éoho Cost9 “(‘20 Fcpsfnl9+42§aa : 0 —~ 42:00 (£9 3) F = W : 424 n/ ‘
C9 60 C0“? 1" Zosinﬁ (COW/(Preston) (3) _ 63‘ ‘
{'uwev’r‘; 3) gone3(9‘763 014i 605(930‘64’6 2(0), U92 FBO —/.. :25' (1°)
ZMA=O => —?0P.AEDCo;(9 a 407—309;“(9:o
: \23‘u59 + 403614 9
> P : " TED I 375‘ N 0’) ...
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This note was uploaded on 04/18/2011 for the course E M 306 taught by Professor Rodin during the Spring '08 term at University of Texas.
 Spring '08
 Rodin

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