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ENGINEERING MECHANICS ‘ STATICS. 2nd. Ed. H. F. RILEY AND L; D. STURGES 6—73‘ Determine the reaction 51) Hi
at support A of the Pipe system shown in
Fig. P6—73 when the
force applied to the
pipe wrench is 50 lb. SOLUTION From a freebody diagran
for the pipe system: For force equilibrium: 2? = R + P
A 7 i + 23 3 + 10 E
R = F AP BIA x P = (7 i 4 23 3 + 10 i) x (—50 E)
= —1150 i — 350 3 in.'lb
For moment equilibrium 23 = CA M1” :01“ — 1150) i + (11Ay  350) 3 + MAI E z 0 f!
u A MA, 1 + MAY 3 + HA! 2 = 1150 1 + 350 3 1n.lb Ans. (M )2 + 1n 12 = /(1150)2 + 1350)2 = 1202 in.lb
Ax AY Bar AB is used to support
an BSD—lb load as shown in Fig. P6—79. End A of
the bar is supported with
a ball and socket joint.
End B of the bar is
supported with two cables.
Deteruine the components
of the reaction at support
A and the tensions in the
two cables. 21. In mui fig. 1'647‘) Si—aA—i‘cs SumI~79 677‘ The bent bar shown in Fig.
P677 is supported with
three brackets that exert
only force reactions on
the bar. Determine the
reactions at supports A,
B, and C. Fig. P677 SOLUTION )aoih 1' From a free—body diagram
for the bar: moment equilibrium:  (EB/A x E) + (EC/A x C) + (FD/A x D) + (FE/A x E)
[(—3 1 + e E) x (Bx i + By 3)} [(8 i + 9 3 + 14 E) x (cx i + C1 E1] [(8 i) x 175 3)] + [(8 i + 14 E) x (50 i — 100 211 (GB + 9c i i + (GB + 14c + as — 100) 3
y z x x z
+ (—813y  94:x — 600) E = O For force equilibrium: 2F=K+B+C+D+E = 4 3 + A E + B i + B 3 + c i + c E + 75 3 + so Y  100 E
Y z x y x z = (B + c 1 so; i + (A + a + 75) 3 + (A + c — 100) i = O
x x y y 1 1
Solving yields A = 450 lb
Y A = 450 lb 2 450 3 + 450 E lb 400 i < 350 2 lb  —450 i — 525 3 lb ’Zﬁ SOLUTION From a free—body diagram
for the bar: For moment equilibrium: u (16 1 + 26 i) x (—TBC 3) x (32 3 + 26 E) x (—180 i)
+ (16 i + 32 31 x 1~350 E) A = (26 TBC — 27.200) 1 + (~26 TED + 13,600) 3
+ (—16 TBC + 32 TED) E = O From which: 18c 1046.2 lb a 1045 lb T —1046 3 lb TED 523.1 lb 3 523 lb —523 1 lb For force equilibrium:
2F = X + TBC + Tan + 9
= Axl + 4y3 + Ali  TBC 3 — T201  850 E (Ax ' 523.!) i + (A — 1046.2) 3 + (AZ v 850) R : 0 Y
which: 523.1 lb 2 523 lb
1046.2 lb 3 1046 lb 850 lb 523 i + 1046 3 + 950 2 lb GBl‘ A shaft is loaded through a pulley and a lever (see Fig. PG—Bll
that are fixed to the shaft. Friction between the belt and pulley
prevents slipping of
the belt. Determine
the force F required
for equilibrium and
the reactions at
supports A and B.
The support at A is
a ball bearing and
the support at B is
a thrust hearing.
The bearings exert
only force reactions
on the shafL 501! “1 IS“ lh Fig. P6411 SOLUTION From s freebody disgran
for the system: {My = P(14) 4 150(6)  500(6) = 0 P = 150 lb i : —150 E lb Ans. For moment equilibrium: {HE = [(—10 3) x (Ax i + A2 E11 + [(30 3 v 6 E) x (500 3)]
+ [(30 3 — 6 E) x (—150 3)} + ((14 i + 14 3) x 1200 3  150 E1] = (—10.1z  2100) i + (131x  16,700) E = 0 927.8 1 — 116.67 E lb a 923 3  116.7 E lb A = H9218)2 + (—116.67)2 = 935.1 lb 3 935 lb For force equilibrium: Ans. 2? = 927.3 1  116.67 E + Bx i + By 3 + BZ E  500 i — 150 3 + 200 3 — 150 E = (Bx 1 277.0) 3 + (By + 200) 3 + (a2 — 266.67) E = O 3 + 61 E = 277.3 3  200 3 + 266.67 E lb
a —278 3 — 200 3 + 267 E lb B = /(—277.8)2 + (—200)2 + (256.67)2 = 1 Y
Ans. 433.9 lb 3 434 lb 6~87* The plate shown in Fig.
P6—B7 weighs 150 lb and
is supported in a
horizontal position by
two hinges and a cable.
The hinges have been
properly aligned;
therefore, they exert
only force reactions on
the plate. Assume that
the hinge at B resists
any force along the axis
of the hinge pins.
Determine the reactions
at supports A and B and
the tension in the cable. SOLUTION TC=Te =C (20)2 + (—22)2 + 113)2 = 0.5754TC 1 — 0.6330Tc J + 0.5179TC E Pro. a freebody diagram
for the plate: For moment equilibrium: ER = (F B C/B x X) + (F x 9) + _
XTC) (rA/B 0/9 [(8 i + 22 3) x (0.5754TC 1 ~ 0.6330Tc J n 1 0.51791c E11 + [(24 I) x (Ay 3 + AZ E)! + {112 i + 11 3) x (v150 E11 (11.3936TC  1650) i + (4.1432TC — 24Az + 1800) 3 + (—17.7228T + 24A ) E = U
C v Solving yields: a
n 144.82 1b I [44.8 lb TC 144.82(0.5754 i — 0.6330 3 + 0.5179 R) M 33.33 3  91.67 3 + 75.00 E lb A1 = 106.98 lb
A1 = 50.00 lb
K = 106.98 3 + 50.00 E lb 2 107.0 3 + 50.0 E lb A = /(106.96)2 + (50.00)2 = For force equilibrium: 116.09 lb I l18.l lb 2F = K + B + TC + D = 106.98 3 + 50.00 E + B 3 + B 3 + B y z + 83.33 i  91.67 3 + 75 00 E — 150 E = (Bx + 03.33) i + (By + 106.98 — 91.67) 3
+ (01 + 50.00 + 75.00 — 150.00) E = O B = B i + B 3 + 0 E
x y z = —03.33 3 ~ 15 31 3 + 25 00 E lb 2 03.3 i  15.31 3 + 25.0 E lb B = /(83.33)2 + (15.31)2 + 125.0012 = 88.34 lb x 99.3 lb Ans. Ans. E Ans. ...
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This note was uploaded on 10/13/2009 for the course ME 274 taught by Professor Valry during the Spring '09 term at Iowa State.
 Spring '09
 Valry
 Statics

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