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Unformatted text preview: PROBLEM 4.69
For the ﬂame and loading shown, determine the reactions at C and D. 150 N SOLUTION Since member ED is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and
be opposite in direction for ED to be in equilibrium. The force B acting at B of member ABC will be equal in
magnitude but opposite in direction to force B acting on member BD. Member ABC is a threeforce body with
member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate'the above conditions. The 1 force triangle for member ABC is also shown. The angles a and B are found from the member dimensions: a = tan’1 Oi?) = 26.565°
1.0 m ,6 = tan—1(1'5—m) = 56.310° 1.0m Applying the law of Sines to the force triangle for member ABC,
150 N C B sin(ﬂ — a) _ sin(90° + a) ‘ sin(90° — ,8) or 150 N _ C _ B
sin29.745° sin116.565° sin33.690°
150 N ' 1 . °
C = —( _)5m 16 565 = 270.42N
sm29.745°
or c = 270N I; 56.3°<
150N ‘ 33.690°
and D=B=w~—=lé7.704N
sm29.745°
or D = 167.7 N V: 26.6°{ SOLUTION
E PROBLEM 4.77 A small hoist is mounted on the back of a pickup truck and is used to lift
a 120kg ‘ crate. Determine (a) the force exerted on the hoist by the
hydraulic cylinder BC, (b) the reaction at A. First note W= mg: (120 kg)(9.81m/s2) = 1177.2 N From the geometry of the three forces acting on the Small hoist xAD = (1.2 m)cos30° = 1.03923 m yAD = (1.2 m)Sin30° = 0.6m and yBE = xADtan75° = (1.03923 m)tan75° = 3.8785 In
Then a = tan—1 M = tan_l[ 347“ J = 73.366°
xAD 1.03923 ,8 = 75° — a = 75° — 73.366° =1.63412°
0 = 180°  15° ,B = 165°  1.63412° =163.366° Applying the law of Sines to the force triangle, W B A sinﬂ = sinH = sin15° or 1177.2N _ B _ A
sinl.63412° sinl63.366° sin15°.
(a) B =11816.9N
or B =11.82kN >2 75.0°<
(b) A = 10 684.2 N or A =10.68kN V 73.4°4 PROBLEM 4.1061 y 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. a P4.105 Two steel pipes AB and BC, each having a weight per unit length
D of 5 lb/ﬁ, are welded together at B and are supported by three wires. A Knowing that a = 1.25 ft, determine the tension in each wire.
a SOLUTION First note WAR = (51b/ﬁ)(2 ft) = 10 lb WBC = (51b/ﬁ)(4 a) = 20 lb From f.b.d. of pipe assembly EFy :0: TA +TC+TD—101b—201b =0
TA +TC +TD =301b
zMx = o: (101b)(1ﬁ)— TA(2 it) = 0
or TA = 5.0011:
From Equations (1) and (2) TC + TD = 25 1b
2M, = 0: TC(4 a) + TD(amax) — 201b(2 ﬂ) = 0 (4 ﬁ)TC + TDamax = 4011311 PROBLEM 4.106 CONTINUED
Using Equation (3) to eliminate TC 4(25 — TD) + TDamax = 40 By observation, a is maximum when TD is maximum. From Equation (3), (T D )max occurs when TC = 0. Therefore, (T D) = 25 lb and max Results: (a) (1mm 2 1.600 R {
(b) TA = 5.00 lb {
TC = 0 4 TD = 25.01b 4 ...
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 Spring '08
 Jenkins

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