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 Document
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 ...
View
Full
Document
This note was uploaded on 09/22/2009 for the course MGMT 209 taught by Professor Swim during the Spring '08 term at Texas A&M.
 Spring '08
 SWIM

Click to edit the document details