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Unformatted text preview: PROBLEM 2.17 Solve Problem 2.2 using trigonometry Problem 2.2: The cable stays AB and AD help support pole AC. Knowing
that the tension is 500 N in AB and 160 N in AD, determine graphically
the magnitude and direction of the resultant of the forces exerted by the
stays atA using (a) the parallelogram law, (b) the triangle rule. SOLUTION From the geometry of the problem: tan"i = 38.66“ a :
2 5
ﬂ = tan—1% = 30.96°
Now: 6 = 180° — (38.66 + 3036") = 110.38 And, using the Law of Cosines: R2 = (500 N)2 + (160 N)2 — 2(500 N)(160 N)cosllO.38” = 331319 N2
R = 575.6N Using the Law of Sines: 160N _ 575.6N
siny sinllO.38° siny =[ 160 N Jsin11038° 575.6N
: 0.2606
y =15.1°
¢ = (90” 7 a) + y : 66.44° R : 576N 7 66.4% _ PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.23. Problem 2.23: Determine the x and y components of each of the forces shown.
IR in. 56 ill lv—Fﬂlin — SOLUTION The components of the forces were
determined in Problem 2.23. F204 = —(48.01b)i + (90.0 lb) j
Fm = (112.01b)1 + (180.01b)j Fm = —(3201b)i —(2401b)j 0( Rzgo“: Thus P  __
'—' R = R:r + R).
=  an,
Ex 37' R : (2561b)1+(30.01b)j
Now:
1... 2 £12
a = tan’lgﬂ : 6.68”
256 and R = (—256 111)2 +(30.0113)2 = 257.75 1b R : 2581b In 6.68°4 J ...
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 Fall '08
 C.Trujilliano
 Statics

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