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# Resolução do Capítulo 10 - Método dos Trabalho Virtuais

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```PROBLEM 10.1
Determine the vertical force P which must be applied at G to maintain the

SOLUTION

Assuming
Ay\u3b4
it follows
120 1.5
80C A A
y y y\u3b4 \u3b4 \u3b4= =
1.5E C Ay y y\u3b4 \u3b4 \u3b4= =
( )180 3 1.5 4.5
60D A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = =
( )100 100 1.5 2.5
60 60G A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = =
Then, by Virtual Work
( ) ( )0: 300 N 100 N 0A D GU y y P y\u3b4 \u3b4 \u3b4 \u3b4= \u2212 + =
( ) ( )300 100 4.5 2.5 0A A Ay y P y\u3b4 \u3b4 \u3b4\u2212 + =
300 450 2.5 0P\u2212 + =
60 NP = + 60 N=P W

PROBLEM 10.2
Determine the vertical force P which must be applied at G to maintain the

SOLUTION

Assume
clockwise\u3b4\u3b8
Then for point C
( )5 in.Cx\u3b4 \u3b4\u3b8=
and for point D
( )5 in.D Cx x\u3b4 \u3b4 \u3b4\u3b8= =
15Dx\u3b4 \u3b4\u3c6=
5 15\u3b4\u3b8 \u3b4\u3c6\u2234 =
or 1
3
\u3b4\u3c6 \u3b4\u3b8=
Then 44 2 2 in.
3G
\u3b4 \u3b4\u3c6 \u3b4\u3b8\uf8eb \uf8f6= = \uf8ec \uf8f7\uf8ed \uf8f8
Now cos 45G Gy\u3b4 \u3b4= °
4 2 cos 45
3
\u3b4\u3b8\uf8eb \uf8f6= °\uf8ec \uf8f7\uf8ed \uf8f8
4 in.
3
\u3b4\u3b8\uf8eb \uf8f6= \uf8ec \uf8f7\uf8ed \uf8f8
Then, by Virtual Work.
( ) ( ) ( ) ( )0: 80 lb in. 40 lb in. in. 0E GU x P y\u3b4 \u3b4\u3b8 \u3b4 \u3b4= \u22c5 \u2212 + =
10 480 40 0
3 3
P\u3b4\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
or 40 lb=P W

PROBLEM 10.3
Determine the couple M which must be applied to member DEFG to
maintain the equilibrium of the linkage.

SOLUTION

Following the kinematic analysis of Problem 10.2, we have 0:U =
( ) ( ) ( )0: 80 lb in. 40 lb in. 0EU x M\u3b4 \u3b4\u3b8 \u3b4 \u3b4\u3c6= \u22c5 \u2212 + =
10 180 40 0
3 3
M\u3b4\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
or 160 lb in.= \u22c5M W

PROBLEM 10.4
Determine the couple M which must be applied to member DEFG to
maintain the equilibrium of the linkage.

SOLUTION

Assuming
Ay\u3b4
it follows
120 1.5
80C A A
y y y\u3b4 \u3b4 \u3b4= =
1.5E C Ay y y\u3b4 \u3b4 \u3b4= =
( )180 3 1.5 4.5
60D A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = =
1.5 1
60 60 40
E A
A
y y y\u3b4 \u3b4\u3b4\u3c6 \u3b4= = =
Then, by Virtual Work:
( ) ( )0: 300 N 100 N 0A DU y y M\u3b4 \u3b4 \u3b4 \u3b4\u3c6= \u2212 + =
( ) 1300 100 4.5 0
40A A A
y y M y\u3b4 \u3b4 \u3b4\uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7\uf8ed \uf8f8
1300 450 0
40
M\u2212 + =
6000 N mmM = + \u22c5 6.00 N m= \u22c5M W

PROBLEM 10.5
An unstretched spring of constant 4 lb/in. is attached to pins at points C
and I as shown. The pin at B is attached to member BDE and can slide
freely along the slot in the fixed plate. Determine the force in the spring
and the horizontal displacement of point H when a 20-lb horizontal force
directed to the right is applied (a) at point G, (b) at points G and H.

SOLUTION
First note:
3 3G D G Dx x x x\u3b4 \u3b4= \u21d2 =
4 4H D H Dx x x x\u3b4 \u3b4= \u21d2 =
5 5I D I Dx x x x\u3b4 \u3b4= \u21d2 =
(a) Virtual Work 0: 0G G SP IU F x F x\u3b4 \u3b4 \u3b4= \u2212 =
or ( )( ) ( )20 lb 3 5 0D SP Dx F x\u3b4 \u3b4\u2212 =
thus, 12.00 lb SPF T= W
Now SP IF k x= \u2206
or ( )12.00 lb 4 lb/in. Ix= \u2206
Thus, 3 in.Ix\u2206 =
and
1 1
4 5D H I
x x x\u3b4 \u3b4 \u3b4= =
4
5
\u2234 \u2206 = \u2206H Ix x
( )4 3 in.
5
= or 2.40 in.Hx\u2206 = W
(b) Virtual Work: 0: 0G G H H SP IU F x F x F x\u3b4 \u3b4 \u3b4 \u3b4= + \u2212 =
or ( )( ) ( )( ) ( )20 lb 3 20 lb 4 5 0D D SP Dx x F x\u3b4 \u3b4 \u3b4+ \u2212 =
thus, 28.0 lb SPF T= W
Now SP IF k x= \u2206
or ( )28.0 lb 4 lb/in. Ix= \u2206
Thus, 7 in.Ix\u2206 =
From part (a)
4
5H I
x x\u2206 = \u2206
( )4 7 in.
5
= or 5.60 in.Hx\u2206 = W

PROBLEM 10.6
An unstretched spring of constant 4 lb/in. is attached to pins at points C
and I as shown. The pin at B is attached to member BDE and can slide
freely along the slot in the fixed plate. Determine the force in the spring
and the horizontal displacement of point H when a 20-lb horizontal force
directed to the right is applied (a) at point E, (b) at points D and E.

SOLUTION
First note:
3 3G D G Dx x x x\u3b4 \u3b4= \u21d2 =
4 4H D H Dx x x x\u3b4 \u3b4= \u21d2 =
5 5I D I Dx x x x\u3b4 \u3b4= \u21d2 =
(a) Virtual Work: 0: 0E E SP IU F x F x\u3b4 \u3b4 \u3b4= \u2212 =
or ( )( ) ( )20 lb 2 5 0D SP Dx F x\u3b4 \u3b4\u2212 =
thus, 8.00 lb SPF T= W
Now SP IF k x= \u2206
or ( )8.00 lb 4 lb/in. Ix= \u2206
Thus, 2 in.Ix\u2206 =
And
1 1
4 5D H I
x x x\u3b4 \u3b4 \u3b4= =
4
5
\u2234 \u2206 = \u2206H Ix x
( )4 2 in.
5
=
or 1.600 in.Hx\u2206 = W
(b) Virtual Work: 0: 0D D E E SP IU F x F x F x\u3b4 \u3b4 \u3b4 \u3b4= + \u2212 =
or ( ) ( )( ) ( )20 lb 20 lb 2 5 0D D SP Dx x F x\u3b4 \u3b4 \u3b4+ \u2212 =
thus, 12.00 lb SPF T= W

PROBLEM 10.6 CONTINUED
Now SP IF k x= \u2206
or ( )12.00 lb 4 lb/in. Ix= \u2206
Thus, 3 in.Ix\u2206 =
From part (a)
4
5H I
x x\u2206 = \u2206
( )4 3 in.
5
= or 2.40 in.Hx\u2206 = W

PROBLEM 10.7
Knowing that the maximum friction force exerted by the bottle on the
cork is 300 N, determine (a) the force P which must be applied to the
corkscrew to open the bottle, (b) the maximum force exerted by the base
of the corkscrew on the top of the bottle.

SOLUTION

From sketch
4A Cy y=
Thus, 4A Cy y\u3b4 \u3b4=
(a) Virtual Work:
0: 0A CU P y F y\u3b4 \u3b4 \u3b4= \u2212 =
1
4
P F=
( )1300 N: 300 N 75 N
4
F P= = =
75.0 N=P W
(b) Free body: Corkscrew
0: 0yF R P F\u3a3 = + \u2212 =
75 N 300 N 0R + \u2212 =
225 N=R W

PROBLEM 10.8
The two-bar linkage shown is supported by a pin and bracket at B and a
collar at D that slides freely on a vertical rod. Determine the force P
required to maintain the equilibrium of the linkage.

SOLUTION

Assume Ay\u3b4
Have 160 1or
320 2C A C A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= =
Since bar CD moves in translation
E F Cy y y\u3b4 \u3b4 \u3b4= =
or 1
2E F A
y y y\u3b4 \u3b4 \u3b4= =
Virtual Work:

( ) ( )0: 400 N 600 N 0A E FU P y y y\u3b4 \u3b4 \u3b4 \u3b4= \u2212 + + =
( ) ( )1 1400 N 600 N 0
2 2A A A
P y y y\u3b4 \u3b4 \u3b4\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
or 500 NP =
500 N=P W

PROBLEM 10.9
The mechanism shown is acted upon by the force P; derive an expression
for the magnitude of the force Q required for equilibrium.

SOLUTION

Virtual Work:
Have 2 sinAx l \u3b8=
2 cosAx l\u3b4 \u3b8 \u3b4\u3b8=
and 3 cosFy l \u3b8=
3 sinFy l\u3b4 \u3b8 \u3b4\u3b8= \u2212
Virtual Work: 0: 0A FU Q x P y\u3b4 \u3b4 \u3b4= + =
( ) ( )2 cos 3 sin 0Q l P l\u3b8 \u3b4\u3b8 \u3b8 \u3b4\u3b8+ \u2212 =
3 tan
2
Q P \u3b8= W

PROBLEM 10.10
Knowing that the line of action of the force Q passes through point C,
derive an expression for the magnitude of Q required to maintain
equilibrium

SOLUTION

Have 2 cos ; 2 sinA Ay l y l\u3b8 \u3b4 \u3b8 \u3b4\u3b8= = \u2212
( )2 sin ; cos
2 2
CD l CD l\u3b8 \u3b8\u3b4 \u3b4\u3b8= =
Virtual Work:
( )0: 0AU P y Q CD\u3b4 \u3b4 \u3b4= \u2212 \u2212 =
( )2 sin cos 0
2
P l Q l \u3b8\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6\u2212 \u2212 \u2212 =\uf8ec \uf8f7\uf8ed \uf8f8
( )
sin2
cos /2
Q P \u3b8\u3b8= W

PROBLEM 10.11
Solve Problem 10.10 assuming that the force P applied at point A acts
horizontally to the left.

SOLUTION

Have 2 sin ; 2 cosA Ax l x l\u3b8 \u3b4 \u3b8\u3b4\u3b8= =
( )2 sin ; cos
2 2
CD l CD l\u3b8 \u3b8\u3b4 \u3b4\u3b8= =
Virtual Work: ( )0: 0AU P x Q CD\u3b4 \u3b4 \u3b4= \u2212 =
( )2 cos cos 0
2
P l Q l \u3b8\u3b8\u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6\u2212 =\uf8ec \uf8f7\uf8ed \uf8f8
( )
cos2
cos /2
Q P \u3b8\u3b8= W

PROBLEM 10.12
The slender rod AB is attached to a collar A and rests on a small wheel at
C. Neglecting the radius of the wheel and the effect of friction, derive an
expression for the magnitude of the force Q required to maintain the
equilibrium of the rod.

SOLUTION

For :AA C\u2032\u2206
tanA C a \u3b8\u2032 =
( ) tanAy A C a \u3b8\u2032= \u2212 = \u2212
2cosA
ay\u3b4 \u3b4\u3b8\u3b8= \u2212
For :CC B\u2032\u2206
sinBC l A C\u3b8\u2032 \u2032= \u2212
sin tanl a\u3b8 \u3b8= \u2212
sin tanBy BC l a\u3b8 \u3b8\u2032= = \u2212
2cos cosB
ay l\u3b4 \u3b8\u3b4\u3b8 \u3b4\u3b8\u3b8= \u2212
Virtual Work:
0: 0A BU Q y P y\u3b4 \u3b4 \u3b4= \u2212 =
2 2cos 0cos cos
a aQ P l\u3b4\u3b8 \u3b8 \u3b4\u3b8\u3b8 \u3b8
\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 \u2212 \u2212 \u2212 =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
2 2coscos cos
a aQ P l \u3b8\u3b8 \u3b8
\uf8eb \uf8f6 \uf8eb \uf8f6= \u2212\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8```