This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 783* Determine all forces acting
on member BCD of the linkage
shown in Fig. P783. SOLUTION Member AC is a twoforce member;.
therefore, the line of action of
force C is known as shown on the
free—body diagram for member BCD: + C 2MB = c cos 45° (2.0) — 40 cos 30° (3.5) = o c = 85.73 1b a 85.7 lb
C a 85.7 lb 7 45° Ans.
= Bx + 40 cos 30° — 85.73 cos 45° = 0
25.98 1b a 26.0 lb —+
= B  85.73 sin 45° + 40 sin 30° = 0 V
40.62 lb 3 40.6 lb T AB”)2 + (By)2 = /(25.98)2 + (40.82)2 = 48.22 lb ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 788* A block with a mass of 150
kg is supported by a cable
which passes over a 150mm
diameter pulley that is
attached to a frame as shown
in Fig. P7—88. Determine all forces acting on member BCD
_of the frame. SOLUTION .r
From a freebody diagram
for the block: 147).}
W = mg = 150(9.807) = 1471.1 N
T 2 : — , =
+ Fy T 1471 1 0 ”7).,” T = 1471.1 N 5 From a freebody diagram i_____)( for the pulley at C: + —+ 223 c  1471.1 cos 20° = 0 X C = 1382.4 N = 1382.4 N —# ' X + T 2Fy = cy  1471.1 — 1471.1 sin 20° = 0 Cy = 1974.2 N = 1974.2 N T From a freebody diagram
for the bar BCD: + g 2M5 = B(0.450) + 1332.4(0.300)  1974.2(0.400) = 0 V/ B = 833.2 N a 333 N y/// + —+ 2F = D  1382.4  833.2 = 0 X X D  1974.2 = 0 D = 1974.2 N T L/
y y /(D )2 + (D )2 = /(2215.6)2 + (1974.2)2 = 2967.5 N z 2970 N
x v 1 1974.2 _ o  _ 0
tan '2215.6 — 41.71 D  2970 N E 41.7 Ans. /(c )2 + (c )2: /(—1332.4)? + (1974.2)2 = 2410.1 N z 2410 N
x y 1 1974.2 _ _ o _ o
 tan 31555::  125.01 C _ 2410 N r 55.0 Ans. JJZEB ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P7101. Determine the force exerted on the bolt at E and all forces acting on the handle ABC. FmP7401 SOLUTION From a freebody diagram
for member CDE: +—>EFX=C =0 X C = 0 X + C ZMD = Cy13)  E(2) = o  2
cy  3 E From a free—body diagram
for handle ABC: + C 2MB = cy(1) — 50(20) cy = 1000 1b = 1000 C = 1000 lb T By + 50 + 1000 = 0 —1050 1b 3 = 1050 lb ¢ 1.50y = 1.5(1000) = 1500 lb Force on the bolt: E = 1500 lb T (‘18 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7114 Figure P7114 is a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN
and have a center of gravity at H. Arm ABCD has a weight of 2 kN
and a center of gravity at B; arm DEFG has a
weight of 1 kN and a center of gravity at E. The weight of the hydraulic cylinders can‘
be ignored. Calculate the force in the horizontal cylinders CJ
and EI and all forces
acting on arm DEFG for the position shown. SOLUTION From a freebody diagram for
the complete mechanism: + C 2MA = P(1.8 cos 30°) — 2(0.9 sin 30°)  1(3.3 sin 30°) — 10(0.3 + 4.5 sin 30°) = 0 F = 18.00 kN <— Ans. "2"
30° From a free—body diagram for 0.3!".
the bucket and its contents: G: )OKN + C ZMG = E(1.2 cos 30°)  10(0.3) = 0 E = 2.887 kN = 2.887 kN —+ (on DEFG) E e 2.89 kN —+ Ans. + 9 2Fx Gx  2.887 = 0 G = 2.887 kN = 2.887 kN + (on DEFG) X + 1‘ my = G — 10 = 0 G = 10.00 kN = 10.00 kN ¢ (on DEFG) ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7114 (Continued) 0 = /(Gx)2 + (0y)2 = /(2.887)2 + (—10.00)'2 = 10.408 kN _ 1 —10.00 _ _ o o
66 _ tan _2.887 _ 106.10 G a 10.41 kN P 73.9 Ans. From a freebody diagram
for arm DEFG: . o
0 = tan'1 9L§—§lE—§9— = 19.107° 3.0 cos 60o + C EMD = F[1.2 cos (30° — 19.1o7°)]  2.887(1.2 cos 30°)  1(0.6 sin 30° — 10(1.8 sin 30° = 0 F = 10.438 kN F a 10.44 RN 3 19.11° Ans.
+ —+ 23x =‘Dx + 2.887 + 10.438 cos 19.1o7°  2.887 = o
0 = 9.863 kN
X
+ T EFy = 0y + 10.438 sin 19.107°  1  10 = 0
D = 7.583 kN
Y
D = (0x)2 + (my)2 = /(9.863)2 + (7.583)2 = 12.441 kN
_ 1 7.583 _ o — _ . ~ 0 .
eD _ tan _9.863  142.44 D _ 12.44 1N a 31.6 Ans. ¢4££L ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7121 A block weighing 1000
lb is supported by a
fivebar truss as
shown in Fig. P7121.
Determine the force
in each member of the truss. SOLUTION From a free—body diagram
for the pulley at C: +—>2Fx=c —1000=0 X 1000 lb = 1000 lb + C  1000 = 0
y = 1000 lb = 1000 lb T From a free—body diagram
for joint C: + ‘\ ZFy, = 1000 sin 30°  TCD sin 30°  1000 sin 60° = o + /” 2F , 1000 cos 30°  T — T cos 30° — 1000 cos 60°
x BC CD 1000 cos 30°  TBC — (732.05) cos 30° — 1000 cos 60° = o TBC = 732.05 N E 732 N (C) Ans. From a free—body diagram
for joint B: + /” EFX, = —TAB  732.05 = 0 TAB = 732.05 N E 732 N (C) + ‘\ EFy, = TED = 0 Tan = 0 From a freebody diagram
for joint D: + —» 2px  732.05 cos 60° = o TAD = 366.0 N E 366 N (C) ...
View
Full Document
 Spring '08
 McVay

Click to edit the document details