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Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—84* Determine all forces acting
on member ABC of the linkage
shown in Fig. P784. SOLUTION From a freebody diagram
for member ABC: .z + C EMA = By(200) — 500(400) o B = 1000 N = 1000 N T Ay + 1000  500 = 0 500 N = 500 N ¢ +
_;
M
’13 < II A + B = 0 H + —e 2F
X X X From a freebody diagram
for member BD: + C 2MB = Bx(300) + 1000(400)  250(200) = 0 B = —1166.7 N 3 1167 N 9— (on ABC) 1166.7 N 3 1167 N —4 /(Ax)2 + (Ay)2 = /(1166.712 + (500.0)2 = 1269.3 N A =
e _ t “1 f1 _ ta 1 —500.0 _ _23 200
A ‘ a Ax ' " 1166.7 ' '
K a 1269 N % 23.2° Ans.
2 2 2 2
B = (Bx)> (By) = /(1166.71 + (1000) = 1536.6 N
e _ t 1 E1 _ t 1 1000.0 _ 139 400
a ‘ a" B ' 3“ —1166.7 ' ° B a 1537 N z 40.6° (£213 ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. 787* A pinconnected system ﬁﬁﬁﬁﬂﬂﬂﬁéﬁﬁt of levers and bars is
used as a toggle for a
press as shown in Fig. force F exerted on the Determine the can at A when a force
ﬂ = 100 lb is applied
to the lever at G. SOLUTION From a freebody diagram
for the lever: + C EMF = 100(30)  FDE(8) = o F : DE 375 lb From a freebody diagram
for the pin at D: cos 67°  F cos 73° — 375 = 0 CD , O
Sln 67 + FCD + —+ 2F = —F 80 F . 0 __
BD Sln 78 — O — 639.5 lb 601.8 lb From a freebody diagram
for the piston at B: + T EFy = A  639.5 sin 67° = 0 Y Ay = 588.7 lb = 588.7 lb T Force on the can: F E 589 lb ¢ D. STURGES ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 795* Forces of 25 lb are
applied to the handles
of the pipe pliers
shown in Fig. P795.
Determine the force
exerted on the pipe
at D and the force
exerted on handle F. . P795
DAB by the pin at A. % SOLUTION / From a freebody diagram
for member DAB: + C EMA = D(1.25) — 25(9) = o D 180.0 lb Ans. D sin 38° + Ax = o
_ D sin 38°
180.0 sin 38° 110.82 lb = 110.82 lb +— Ay  D cos 38° — 25 = o
D cos 38° + 25
180.0 cos 38° + 25 166.84 lb = 166.84 lb T /(Ax)2 + (Ay)2 = /(—110.82)2 + (155.84)2 = 200.29 1b a 200 lb 1 166.84 0
m  13359 tan ‘ = 200 lb 8 56.4° Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7103* The folddown chair of Fig.
P7103 weighs 25 lb and has
its center of gravity at G.
Determine all forces acting on member ABC. SOLUTION From a freebody diagram
for the chair: + C EMA = 25(3) — E(Zl) = o
, E = 3 571 lb = 3.571 lb +—
= A  3 571 = o
X 3.571 lb = 3.571 lb 4 = 25 lb = 25 lb T 
‘X _ 2 2
 (Ax) + (Av) I /(3.571)2 + (25)? = 25.25 lb —1 25
ex ta“ 3. 71 K = 25.3 lb 8 81.9° = 81.87° Member BD is a two—force member;
therefore, the line of action of
force 3 is known as shown on the
freebody diagram for member ABC: + C inc = B cos 39.81° (9) — 3.571(21) = o
B 10.846 lb B = 10.85 lb 8 e 39.8° c + 3.571  10.846 cos 39.81° = 0 . c = 4.761 lb X X = Cy + 25  10.846 sin 39.810 = 0 CV = 18.056 lb /(cx)2 + (Cy)? = /(4.761)2 + (18.056)2 = 18.673 lb 1 —18.056 .___~ 0 = o
4.761  75.23 C 18.67 lb 2 75.2 tan ...
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