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Unformatted text preview: 638 A beam is loaded and supported as shown in
Fig. P638. The beam ‘,
has a uniform cross A H . section and a mass of s, I r
‘ L— 1000 mm —>L— 750 mm 750 mm 380 20 kg. Determine the W
reaction at support A mm
and the tension T in 750 N the cable. FmP638 SOLUTION The action of the pin at support A
is represented by force components —. Ax and Ay. The cable is continuous over the pulley; therefore, the
force in the cable is constant. At
points B and C the cable exerts
tensile forces T on the beam that
are tangent to the cable. The
weight w of the beam acts through i the center of gravity G of the } beam and is directed toward the 1 \ center of the earth.
W = mg = 20(9.807) = 196.14 N + C EMA = T sin 24°(1.000) — 196.14(1.440)
 75011.750) + T sin 70° (2880) = o
T = 512.3 N e 512 N Ans. + ~+ 2F = A + T cos 24o — T cos 700 X X A + 512.3 cos 24°  512.3 cos 70° = o X 292.8 N s 293 N 23'
II + T XFY = Ay + T sin 24° ~ w — 750 + T sin 70° = Ay + 512.3 sin 240 196.14  750 + 512.3 sin 700 = 0 Ay = 256.4 N z 256 N
A = /A: + A: = /(292.8)2 + (256.4)2 = 389.2 N z 389 N Ans.
_ 1 256.4 _ o 0
ex  tan _292.8  138.79 2 138.8 K = —293 i + 256 3 lb = 389 lb 5 41.2° Ans. Bar AB of Fig. P664 has a uniform cross section, a mass of 25 kg, and a length of 1 m long. Determine the angle 9 for equilibrium. SOLUTION From a freebody diagram
for the bar: W = mg = 25(9.807) = 245.2 N + —+ ZFX = A cos 60o  B cos 450 = 0 + T ZFY = A sin 60° + B sin 45° — w = A sin 60° + B sin 45° — 245.2 2 0 Solving yields: A = 179.50 N B = 126.92 N + C EMA = B sin 450 «L cos 6) + B cos 45° (L sin 6) — W(0.5L cos 9)
= 126.92 sin 45° (1 cos 6) + 126.92 cos 45° (1 sin e)  245.2(0.5 cos 6) = 0 H 32.85 cos 6 '1 _
89.75 ~ 89.75 sin e O 9 20.1 Ans. tan ...
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This note was uploaded on 07/05/2010 for the course ENGR 120 taught by Professor Jones during the Spring '10 term at Florida State College.
 Spring '10
 jones
 Statics

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