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Unformatted text preview: TBQ A pineenneeted system
of bars supports a EDD
lb lead as shown in Fig.
PTEQ. Determine the
reactions at supports s
and E and the force
exerted by the pin at C on member ACE. EGLUTIDN From a freebody diagram
for the complete system: + C'EMA = B!{2ﬂ} 3Dﬂ{30} = D B = 450 1b = 45H 1h * I + + EF = s + EJlr : A  45D = ﬂ 1'. H K s = 45D it = 450 lb + II Frem a freebody diagram
for pin F: + —s EF . = T  30s sin 45° = ﬂ
1r EF TEF : 212.1 1b 3 212 1b [T] From a freehsdy diagram
fer bar ACE: + c EH: = —Ayilﬂi + dﬁﬂilﬂl  212.1[1Dfees 45”] = n s3 = lﬁﬂ.ﬂ lb = 15D.ﬂ ]h T + T SE = n + e  212.1 sin 45“
‘11" ‘1'" 1" = 150 u + CH  212.1 sin is“ = D c = ﬂ ?*BQ {Continued} Fran the freehudy diagram
far the canplete system: +TEF =13. +A3ﬂﬂ
3 3 3
= E? + 150.ﬂ ~ Bﬂﬂ = D E = iﬁﬂ.ﬂ lb = 150.0 lb T ".Ir r—’——‘
1 = {13312 + {Aria = {1133.313 + 1133.1113 = 131.3 13 a 171 lb _ 1 1513.0 _ I: _ a
3A  tam 451M _ 13.131 I as 131 13 .3 13.13 Ana.
2 2 E 2
3 = 1’13“: +131“: = 1111333} + 1153.1}: = 131.313 3: 131111
_ ~1 133.3 _ a a
BB  tan 4513.0 . 131.33 E a: 131 11:. 3 13.13 .1113.
C = ﬁﬂﬂ 1b +— Ans. E a 212 13 3 15.3" 1n1. T11ﬂ* A cylinder with a mass
of 150 ks; is supported
by a two—bar frame as
shown in Fig. PT—llﬂ. Determine all forces acting 0n member ACE. rig. P7410 SGLUTI 0N '5 From a freebody diagram I x for the cylinder: 54 = mg = 150(93071 = 14:1.1m
+—EF =nsin45"Esin45°=u
+ T BF? =21: cos 45° 1471.1 = o D = E = 10¢0.2 1b E a 1040 1b *5 45.0“ (On menher ACE] From a freebody diagram for the campiete frame: FIWJH
1040.2’1L
+ 4: ENE = M2! 14?1.1{1}= B
A = 735.6 N = $35.6 N T D E C:
I. i T36 [i T Ans. Q
c A
From a free—body diagram 1*
for bar ACE:
T.
+ (,7 EMC = Tm — 735.5(1:  1040.2{031 = o 7353;,”
1‘ = 156T.E N z 1553 N T as 1563 w —» Ans.
+ > 22F?r = c“ + T + 1040.2 cos 45°
= :3x + 155?.3 + 1040.5 cos 45" = a c“ = —2303 N 2 2300 N «—
+ ’1‘ BF? = c? + T35.6 a 1040.2 sin 45° = 0
c :0 Eatzaoumt— Ana. 3" ?BT* A pinconnected system
of levers and hers is
used as a teggle fer a
press as shown in Fig.
P?—3T. Determine the
force 3 exerted on the can at A when a force e : 100 lb is applied
to the lever at G. Fm.Plﬂ? SOLUTION From a Free—body diagram
for the lever: + G EMF = lUﬂiﬂﬂ}  FD£:31 = a FDE = 3T5 1h From a freebody diagram
for the pin at D: + *4 EFx = F :03 6T0  F Gee TED  3?5 = U ED CD
_ o . H c _
+ 1‘ EFT — FBD am ST + FCD em PB  G ______E
FED = 639.5 lb
FED = 6{}1.3 lb U
‘E'Z
From a. freehunch.r diagram nu
for the piston at B:
T :  .r ' D :
+ EF? A? 639 J Bin 6? u H: A? = 533.? 1b = 533.? 1b T Force on the can: F 3 533 1b ¢ Ans. T95* Forces of 25 1b are
applied to the handles
of the pipe pliers
shown in Fig. p795.
Determine the force
exerted on the pipe
at D and the farce
exerted on handle
DAB by the pin at A. SDLUTIDN Fran a free—body diagram
for nember DAB: + c Ema = 5:1.25}  25(51 = D
D = 130.ﬂ lb Ans.
t .4 22 = D sin 35“ + A = U
x x
A = —D sin 330 X I! 150.0 sin 22° llﬂ.32 lb = 110.32 1b *— LI + T EF = 5 — D 555 32“  25 = u
if .‘I’ 5 cos 32° + 25 .1:
II 150.5 cos 33° + 25 ll 166.34 lb = 156.34 lb T 2 = {53:2 + {5?}: = 21415.22}2 + {155.54}2 = 255.25 15 a 255 15 _ 2 155.34 _ a
an _ tan :TTETEE _ 122.55 K = 200 lb 5 55.4“ Ans. TBE* Fereee pf 5 N are applied
he the handlee ef the paper
punch ahown in Fig. PTBﬁ.
Determine the farce exerted on the paper at D and the feree exerted on the pin at E by handle ABC. SDLUTIUN Fran a freehedy diagram
for handle EBB: +GEMB=DidﬂI 'EITU} =U
D = 3.?5 N T +——}EF :B :12]
x x
E = U I + T EF = B + 5 + E.T5 = U
r x El:Ir = “IET5 1h = 13.?5 1h i Farce exerted on the
paper by handle EED: ﬁ = E.T5 N i Ana. Force exerted en the pin
at a by handle nae: E = 1am lb l ens. ...
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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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