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Unformatted text preview: 610 Draw a free—body diagram for the diving board shown in Fig. P610. SOLUTION The action of the pin at support A is represented by force components —. A and Ky. Support B exerts a X normal force 3 on the board. The diver exerts a normal force 3 on ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 619 Draw a free—body diagram for
I {a} bar BE shown in Fig. P6—19.
(b1 bar DF shown in Fig. P6—19. SOLUTION (a) The link at the left end of bar
BE exerts a force g on the bar
that is directed along the axis
of the link. The action of the
pin support at point C of bar BE is represented by forces ﬁx and Cy. The pin at E exerts a force E on bar BE that is
normal to the surface of the slot in bar DE. The action of the pin support
at the right end of bar DF is
represented by force components
Fx and F3. The pin at E exerts
a force E on bar DF that is
normal to the surface of the slot in the bar. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—34 A beam is loaded and supported
as shown in Fig. P634. The
beam has a uniform cross section
and a mass of 180 kg. Determine the reaction at support A. SOLUTION A free—body diagram of the beam is
shown at the right. The reaction
at support A is represented by force components Xx and Ky and a , moment H . The weight 9 = mg of
the beam acts through the center
of gravity G of the beam and is
directed toward the center of
the earth. mg = 180(9.807)
= 1765.3 N a 1.765 kN A  W — 2
y Ay  1.7653 — 2 = 0 3.7653 kN E 3.77 kN = 3.77 3 kN = 3.77 kN T MA — 3  1.7653(2) — 2(4) = 0 14.531 kNm E 14.53 kN'm 14.53 E kN'm = 14.53 kNm C ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 637 A structural member is loaded and supported as awn shown in Fig. P6—37. The member has a uniform
cross section and weighs
208 lb. Determine the reactions at supports A
and B. SOLUTION A free—body diagram of the beam is
shown at the right. The reaction
at support A is represented by
force components Xx and Ky. The reaction at B is a vertical force 3. The weight w of the beam acts
through the center of gravity G of the beam and is directed toward
the center of the earth. /( ,\
. ‘ /. ls
L1d1 + Lad2 + Lads 6(3) + 4(6) + 3(7.5) G = 4.962 ft A + 200  300 = 0 X 100 lb = ~Ay(2)  208(2.962) — 200(2)  100(7)  300(2) = 0 1158.0 lb = 1158 lb A: + 2 /(100)2 + (1158)2 = 1162.3 1b a 1162 lb y
tan1  35.03° z 85.1° X = 100 T  1158 A 1b 1162 lb s 85.10 + C EN = 8(2) — 208(4.962)  200(2)  100(9)  300(2) = 0 1466.0 lb = 1466 lb A
B:
B = 1466 3 lb = 1466 lb T ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—38 A beam is loaded and
supported as shown in
Fig. P638. The beam
has a uniform cross
section and a mass of
20 kg. Determine the
reaction at support A
and the tension T in
the cable. SOLUTION The action of the pin at support A
is represented by force components Ax and Ky. 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
the center of gravity G of the
beam and is directed toward the
center of the earth. W = mg = 20(9.807) = 196.14 N + C EMA = T sin 24°11 000) — 196.14(1.440)
— 750(1.750) + T sin 70° (2880) =
512.3 N z 512 N A + T cos 24°  T cos 700 X A + 512.3 cos 24°  512.3 cos 70° = o X 292.8 N 3 293 N = Ay + T sin 24°  w  750 + T sin 70° = Ay + 512.3 sin 24° 196.14 — 750 + 512.3 sin 700 = 0 256.4 N a 256 N
/ 2 / 2 2
= Ax + A = (—292.8) + (256.4) = 389.2 N 3 389 N = ' 138.79° a 138.80 K = 293 i + 256 3 1b = 389 lb 5 41.2° ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 640* An angle bracket is loaded
and supported as shown in
Fig. P640. Determine the reactions at supports A and B. 350 N
Fig. P640 SOLUTION ,1 From a freebody diagram
for the angle bracket: + C EMA = B(0.200) — 500(0.100)
— 350(0.220) = 0 = 635 N B
§ = 635 j N = 636 N «— =A +500—B X A + 500  635 = X 135 N = A  350 = 0
y 350 N y = /(135)2 + (350)2 = 375.1 lb a 375 lb '1 350 0 tan —§§ = 68.9 3 lb = 375 lb a 68.9° ENGINEERING MECHANICS  STATICS, 2nd. Ed. 645 A rope and pulley system is
used to support a body as
shown in Fig. P6—45. Each
pulley is free to rotate and
the ropes are continuous over
the pulleys. Determine the
force p required to hold the
body in equilibrium if the
weight w of the body is 400 lb. SOLUTION From a free—body diagram
for pulley A: + T 2F = 2T  400 = 0
y ‘1 T1 = 200 lb From a free—body diagram
for pulley B: + T 2F = 2T  200 = 0
y 2 T2 = 100 lb From a free—body diagram
for pulley C: +T2Fy=2P100=O P = 50 lb Ans. W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES The wrecker truck of Fig. P6—63 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a, tangential component Bx while the force exerted on the front wheel
consists of a normal force Ay only. Determine the maximum pull P
the wrecker can exert when 6 : 300 if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backwards (the front wheels remain in contact with the ground).
SOLUTION From a freebody diagram
for the wrecker truck: For impending tipping: + C 2MB = W(8) — p sin 30° (10) — P cos 30° (5)
= 15,000(8)  p sin 30° (10)  P cos 30° (5) = 0 12,862 lb 8 12.86 kip = By — W  P cos 300 By  15,000  12,862 cos 30° = 0 26,139 lb 2 26.1 kip —3x + p sin 30°
Bx + 12,862 sin 30° 6431 lb 3 6.43 kip
Bx(max) = 0.88V = 0.8(26,139) = 20,911 lb > 6431 lb Therefore: P 12.86 kip 45'} ...
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 Spring '08
 McVay

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