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Unformatted text preview: PROBLEM 4.70 For the frame and loading shown, deteimine the reactions at A and C. n 7 7' D
+40 llllil+l<i (7‘0 mm SOLUTION 130 M IBON Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibritun. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three—force body with
member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle ,8 is found from the member dimensions: 60 m
100 m ﬂ : tan’l[ ) = 30964" Applying of the law of sines to the force triangle for member BCD, # 130N _ B _ C r
sin(45°—ﬂ) sinﬂ sinl35° or 130 N # i B _ C
sin14.036° si1130.964° sinl35O
(130 N)sin30.964°
= 2 f, r = 275.78N
sm14.036°
(130 N)sinl35°
and C : .— = 379.02 N sinl47036° or A = 276N V: 4500‘ or C : 379N 5. 59.004 PROBLEM 4.73 A ZOOlb crate is attached to the trolleybeam system shown. Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (17) the reaction
at B. SOLUTION From geometry of forces yBE = 20 ‘ J’DE = 2.0 ~ l.5tan35°
{194; 11:75:: : 0.94969 ft ﬂ = ta11"£0‘94969) = 32.339°
and a = 90° ~ 6 : 90° — 32.339o = 57.6610 49 = [3 + 35" 2 32.3390 + 35° : 67.339o Applying the law of sines to the force triangle, 200 lb _ T _ 713777
sine sina sin 55°
(200 lb) T B
01‘ ¥ = *W :
sin 67.339” si1157.661° sin55°
200 lb v' 57.661“
(a) T:£¢n_)=183.1161b
smG7.339°
01' T = 183.11b4
2001b ' 55°
(b) B : w = 177.536 lb sin 67339" 01‘ B = 177.5 lb ‘5 323° { PROBLEM 4.106 H For the pile assembly of Problem 4.105, determine (a) the largest
permissible value of a if the assembly is not to tip, (b) the corresponding
tension in each wire. P4.105 Two steel pipes AB and BC, each having a weight per unit length
of 5 lb/ft, are welded together at B and are supported by three wiresi
Knowing that a = 1.25 ft, determine the tension in each wire. SOLUTION First note WAR 2 (5 lb/ﬁ)(2 ﬁ) = 10 lb
W3C : (5 lb/ft)(4 ﬁ) = 201b
From f.b.d. of pipe assembly
217), :0: T4+TC+TD~101b72Olb=O
TA+TC+TD=301b (1)
2114‘. = o: (10 ]b)(1 ft) — TAO ft): 0
or T4 2 5.00 lb (2)
From Equations (1) and (2) TC + T D = 25 lb (3) 2M: = 0: TC(4 It) + T[,(anm) — 201b(2 ﬂ) : 0 or (4 ﬁ)TC + TDamax : 401M: (4) PROBLEM 4.106 CONTINUED
Using Equation (3) to eliminate Tc 4(25 — TD) + TDa = 40 max 60
or amax : 4 # F
D
By observatic '1 is maximum when TD is maximum. From Equation (3), (71))“Ix occurs when TC. : 0, Therefore, ’ X : 251b and a : _@
"“X 25
= 1.600 ft
R< anm : 1.600 ft 4
7",, : 5.00 lb 4
TC : 0 4 TD : 25.01b 4 ...
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This note was uploaded on 09/06/2011 for the course EGM 2511 taught by Professor Jenkins during the Fall '08 term at University of Florida.
 Fall '08
 Jenkins

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