11–1.
SOLUTION
of lamp (10 lb force) do work.
Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B
using position coordinates yAand xA.
(1)
(2)
Virtual–Work Equation: When yAand xAundergo positive virtual displacements
and ,the 10 lb force and horizontal component of do
positive work while the vertical component of does negative work.
(3)
Substituting Eqs. (1) and (2) into (3) yields
Since ,then
At the equilibrium position ,
Ans.F
AC =10 cos 45°
0.5 cos 45° +0.8660 sin 45° =7.32 lb
u=45°
F
AC =10 cos u
0.5 cos u+0.8660 sin u
ldu Z0
(10 cos u–0.5F
AC cos u–0.8660F
AC sin u)ldu =0
dU=0; 10dyA–F
AC sin 30°dyA+F
AC cos 30°,
dxA=0
F
AC,F
AC sin 30°,
FAC,FAC cos 30°,dxA
dyA
yA=lsin udyA=lcos udu
xA=lcos udxA=-lsin udu
Use the method of virtual work to determine the tensions in
cable AC.The lamp weighs 10 lb.
B
A
C
45°30°
11–2.
The scissors jack supports a load P. Determine the axial
force in the screw necessary for equilibrium when the jack
is in the position Each of the four links has a length Land
is pin-connected at its center.Points Band Dcan move
horizontally.
u.
CD
P
11–3.
Ifaforce of is applied to the handle of the
mechanism, determine the force the screw exerts on the cork
of the bottle.The screw is attached to the pin at Aand passes
through the collar that is attached to the bottle neck at B.
P=5lb
SOLUTION
Free – Body Diagram: When undergoes a positive virtual angular displacement of
, the dash line configuration shown in Fig. ais formed. We observe that only the
force in the screw Fsand force Pdo work when the virtual displacements take place.
Virtual Displacement: The position of the points of application for Fsand Pare
specified by the position coordinates yAand yD, measured from the fixed point B,
respectively.
(1)
(2)
Virtual Work Equation: Since Pacts towards the positive sense of its corresponding
virtual displacement, it does positive work. However, the work of Fsis negative
since it acts towards the negative sense of its corresponding virtual displacement.
Thus,
(3)
Substituting , Eqs. (1) and (2) into Eq. (3),
Since , then
Ans.F
S=15 lb
cos udu Z0
cos udu(90 –6F
S)=0
5(18 cos udu)F
S(6 cos udu)=0
P=5lb
dU=0; PdyD+
A
–F
SdyA
B
=0
yD=6(3 sin u)dyD=18 cos udu
yA=2(3 sin u)dyA=6 cos udu
du
u
3 in.
D
B
A
u30°
P5lb
*11–4.
T
h
e
di
s
k
h
as a we
i
g
h
t of 10
lb
an
d
i
s su
bj
ecte
d
to a vert
i
ca
l
force and a couple moment
Determine the disk’s rotation if the end of the spring
w
raps around the periphery of the disk as the disk turns.
The spring is originally unstretched.
u
M=8lb#ft.P=8lb
SOLUTION
Since
W
here
Ans.u=0.7407 rad =42.4°
13.33 =12(1.5u)
x=1.5u
F=kx
20–1.5F=0F=13.33 lb
du Z0
u(20 –1.5F)=0
8(1.5 du)+8du –F(1.5 du)=0
dU=0;
PdyP+Mdu –FdyF=0
dyF=dyP=1.5du
P8lb
k12 lb/ft
M8lbfift
1.5 ft
11–5.
The punch press consists of the ram R, connecting rod AB,
and a flywheel. If a torque of
M=75 N #m
is applied to the
flywheel, determine the force F applied at the ram to hold
the rod in the position
u=60°.
SOLUTION
F
200 mm
M
B
R
A
600 mm
u
1128
11–6.
The flywheel is subjected to a torque of
M=75 N #m
.
Determine the horizontal compressive force F and plot the
result of F (ordinate) versus the equilibrium position
u
(abscissa) for
0°…u…180°.
SOLUTION
u
u
M=75 N #m.
u
F
200 mm
M
B
R
A
600 mm
u
11–7.
SOLUTION
spring forces Fsp,the weight of the block (50 lb),the weights of the links (10 lb) and
the couple moment Mdo work.
Virtual Displacements: Thespring forces Fsp,the weight of the block (50 lb) and the
weight of the links (10 lb) are located from the fixed point Cusing position
coordinates y3,y2and y1respectively.
(1)
(2)
(3)
Virtual–Work Equation: When y1,y2and y3undergo positive virtual displacements
,and ,the spring forces Fsp,the weight of the block (50 lb) and the weights
of the links (10 lb) do negative work. The couple moment Mdoes negative work
when the links undergo a positive virtual rotation .
(4)
Substituting Eqs.(1),(2) and (3) into (4) yields
Since ,then
At the equilibrium position , .
Ans.M=sin 20°[4(8)+120] =52.0lb #ft
F
sp =kx =2(4)=8lbu=20°
M=sin u(4F
sp +120)
8F
sp sin u+240 sin u–2M=0
du Z0
(8F
sp sin u+240 sin u–2M)du =0
dU=0;
–2F
spdy3–50dy2–20dy1–2Mdu =0
du
dy3
dy2
dy1
y1=2cos udy1=-2sin udu
y2=0.5+4 cos udy2=-4sin udu
y3=1+4cos udy3=-4sin udu
Whenthe50-lb uniform block compresses the
twoverticalsprings 4in. Iftheuniform linksAB and CD
eachweigh10 lb,determinethe magnitudeoftheapplied
couplemoments Mneededto maintain equilibrium when
u=20°.
u
=20°,
BD
MM
1ft
1ft
k2lb/in.k2lb/in.
1ft
2ft
uu
*11–8.
The bar is supported by the spring and smooth
collar that allows the spring to be always perpendicular to
the bar for any angle If the unstretched length of the
spring is determine the force Pneeded to hold the bar in
the equilibrium position Neglect the weight of the bar.u.
l0,
u.
C
A
B
k
l
a
u
11–9.
Since
Ans.
and
Ans.u=35.8°
u=16.6°
sin u–0.707 tan u–0.075 =0
60 cos u=8001cos u–cos 45°2sin u
F
s=k14 cos u–4 cos 45°2=20014 cos u–4 cos 45°2
F
s=60acos u
sin ub
3–F
s1–4sin u2–6014 cos u24du =0
–F
sdx–30dy–30dy=0dU=0;
dx=-4sin ududy=4 cos udu,
The 4-ft members of the mechanism are pin-connected at
their centers.If vertical forces act at Cand
Eas shown,determine the angle for equilibrium. The
spring is unstretched when Neglect the weight of
the members.
u=45°.
u
P
1=P
2=30 lb
P1P2
k= 200 lb/ft
CE
DA
θ
2ft
2ft
1132
11–10.
The thin rod of weight Wrest against the smooth wall and
floor.Determine the magnitude of force Pneeded to hold it
in equilibrium for a given angle
SOLUTION
(1)
(2)
Virtual-Work Equation:When points Cand Aundergo positive virtual displacements
and ,the weight of the rod Wand force Fdo negative work.
;(3)
Substituting Eqs.(1) and (2) into (3) yields
Since ,then
Ans.P=W
2 cot u
Pl sin u–Wl
2 cos u=0
du Z0
aPl sin u–Wl
2 cos ub du =0
–WdyC–PdyA=0dU=0
dxA
dyC
dxA=-l sin uduxA=l cos u
dyC=1
2 cos uduyC=1
2 sin u
l
B
u.
2
11–11.
If each of the three links of the mechanism have a mass of
4 kg, determine the angle
u
for equilibrium. The spring,
which always remains vertical, is unstretched when
u=0°.
SOLUTION
u
u
200 mm
200 mm
200 mm
C
D
A
M 30 N fi m
k 3 kN/m
B
u
u
11–11. Continued
*11–12.
The disk is subjected to a couple moment M. Determine the
disk’s rotation
u
required for equilibrium. The end of the
spring wraps around the periphery of the disk as the disk
turns. The spring is originally unstretched.
u
u
k 4 kN/m
M 300 N fi m
0.5 m
11–13.
A
5
-kg uniform serving table is supported on each side by
pairs of two identical links, and ,and springs .If
the bowl has a mass of , determine the angle where the
table is in equilibrium. The springs each have a stiffness of
and are unstretched when .Neglect
the mass of the links.
u=90°k=200 N>m
u1kg
CECDAB
SOLUTION
,the dash line configuration shown in Fig.aisformed. We observe that only the
spring force Fsp,the weight Wtofthe table, and the weight Wbofthe bowl do work
when the virtual displacement takes place.The magnitude of Fsp can be computed
using the spring force formula,
Virtual Displacement: The position of points of application of Wb,Wt,and Fsp are
specified by the position coordinates and xC,respectively.Here,
are measured from the fixed point Bwhile xCis measured from the fixed point D.
(1)
(2)
(3)
Virtual Work Equation: Since Wb,Wt,and Fsp act towards the negative sense of
their corresponding virtual displacement, their work is negative.Thus,
(4)
Substituting ,,
,Eqs.(1),(2),and (3) into Eq. (4),we have
Since ,then
Solving the above equation,
Ans.
Ans.u=36.1°
–7.3575 +12.5sin u=0
cos u=0u=90°
cos u(–7.3575 +12.5 sin u)=0
–7.3575 cos u+12.5sin ucos u=0
du Z0
du
A
–7.3575 cos u+12.5sin ucos u
B
=0
–4.905(0.25 cos udu)–24.525(0.25 cos udu)–50 cos u(–0.25 sin udu)=0
F
sp =50 cos uN
W
t=a5
2b(9.81)=24.525 NW
b=a1
2b(9.81)=4.905 N
dU=0;–W
bdyGb+
A
–W
tdyGt
B
+
A
–F
spdxC
B
=0
xC=0.25 cos udxC=-0.25 sin udu
yGt=0.25 sin u+adyGt=0.25 cos udu
yGb=0.25 sin u+bdyGb=0.25 cos udu
yGband yGt
yGb,yGt,
F
sp =kx =200
A
0.25 cos u
B
=50 cos u N.
du
ACk
250 mm
250 mm 150 mm
150 mm
B
D
E
uu
11–14.
SOLUTION
,the dash line configuration shown in Fig.aisformed. We observe that only the
spring force Fsp,the weight Wtofthe table, and the weight Wbofthe bowl do work
when the virtual displacement takes place.The magnitude of Fsp can be computed
using the spring force formula, .
Virtual Displacement: The position of points of application of Wb,Wt,and Fsp are
specified by the position coordinates and xC,respectively.Here,
are measured from the fixed point Bwhile xCis measured from the fixed point D.
(1)
(2)
(3)
Virtual Work Equation: Since Wb,Wt,and Fsp act towards the negative sense of
their corresponding virtual displacement, their work is negative.Thus,
(4)
Substituting ,,
,Eqs.(1),(2),and (3) into Eq. (4),we have
Since ,then
When ,then
Ans.k=117.72
sin 45° =166 N>m
u=45°
k=117.72
sin u
–7.3575 cos u+0.0625ksin ucos u=0
du Z0
du
A
–7.3575 cos u+0.0625ksin ucos u
B
=0
–4.905(0.25 cos udu)–24.525(0.25 cos udu)–0.25kcos u(–0.25 sin udu)=0
F
sp =0.25kcos uN
W
t=a5
2b(9.81)=24.525 NW
b=a1
2b(9.81)=4.905 N
dU=0;–W
bdyGb+
A
–W
tdyGt
B
+
A
–F
spdxC
B
=0
xC=0.25 cos udxC=-0.25 sin udu
yGt=0.25 sin u+adyGt=0.25 cos udu
yGb=0.25 sin u+bdyGb=0.25 cos udu
yGband yGt
yGb,yGt,
F
sp =kx =k
A
0.25 cos u
B
=0.25 kcos u
du
A
5
-kg uniform serving table is supported on each side by
two pairs of identical links, and ,and springs .If
the bowl has a mass of and is in equilibrium when
,determine the stiffness of each spring.The springs
are unstretched when .Neglect the mass of the links.u=90°
ku=45°
1kg
CECDAB
ACk
250 mm
250 mm 150 mm
150 mm
B
D
E
uu
11–15.
SOLUTION
Ans.F=
M
2asin u
–Mdu +F(2asin u)du =0
dU=0;–Mdu –Fdx=0
x=2acos u,dx=-2asin udu
Theservice windowat a fast-foodrestaurantconsists ofglass
doorsthat openand close automatically usingamotor which
supplies atorqueMto eachdoor. Thefarends, Aand B,
movealongthehorizontalguides.Ifafoodtraybecomes
stuckbetweenthedoorsasshown,determinethehorizontal
force thedoorsexertonthetrayat thepositionu.
MA
aa
aa
CB
D
M
uu
*11–16.
The members of the mechanism are pin connected. If a
vertical force of 800 N acts at A, determine the angle
u
for
equilibrium. The spring is unstretched when
u=0°.
Neglect
the mass of the links.
1 m
1 m
1 m
k 6 kN/m
D
B
u
SOLUTION
u
u
u=23.58°=23.6°
11–17.
When
u=30°
, the 25-kg uniform block compresses the two
horizontal springs 100 mm. Determine the magnitude of the
applied couple moments M needed to maintain equilibrium.
Take
k=3 kN>m
and neglect the mass of the links.
A
B
D
C
M
M
300 mm
200 mm
100 mm
100 mm
50 mm
k
k
u
u
SOLUTION
1141
11–18.
SOLUTION
, the dash line configuration shown in Fig. ais a formed. We observe that only the
spring force Fsp acting at points Aand Band the force Pdo work when the virtual
displacements take place.The magnitude of Fsp can be computed using the spring
force formula,
Virtual Displacement: The position of points Aand Bat which Fsp acts and point C
at which force Pacts are specified by the position coordinates yA,yB, and yC,
measured from the fixed point E, respectively.
(1)
(2)
(3)
Virtual Work Equation: Since Fsp at point Aand force Pacts towards the positive
sense of its corresponding virtual displacement, their work is positive.The work of
Fsp at point Bis negative since it acts towards the negative sense of its
corresponding virtual displacement. Thus,
(4)
Substituting ,,Eqs.(1), (2), and (3) into Eq.(4),
Since , then
Ans.u=41.2°
–2400(sin u–0.2588) +960 =0
cos udu Z0
cos udu
C
–2400(sin u–0.2588) +960
D
=0
6000(sin u–0.2588)(0.2 cos udu –0.6 cos udu)+600(1.6 cos udu)=0
P=600 NF
sp =6000(sin u–0.2588)
dU=0; F
sp dyA+
A
–F
spdyB
B
+PdyC=0
yC=8(0.2 sin u)dyB=1.6 cos udu
yB=3(0.2 sin u)dyB=0.6 cos udu
yA=0.2 sin udyA=0.2 cos udu
F
sp =kx =15(103)
C
2(0.2 sin u)–2(0.2 sin 15°)
D
=6000(sin u–0.2588) N
du
The “Nuremberg scissors” is subjected to a horizontal force
of . Determine the angle for equilibrium. The
spring has a stiffness of and is unstretched
when .u=15°
k=15 kN>m
uP=600 N
P
200 mm
200mm
A
C
D
E
B
k
u
u=41.2°
11–19.
The “Nuremberg scissors” is subjected to a horizontal force
of . Determine the stiffness kof the spring for
equilibrium when .The spring is unstretched when
.u=15°
u=60°
P=600 N
SOLUTION
, the dash line configuration shown in Fig. ais formed. We observe that only the
spring force Fsp acting at points Aand Band the force Pdo work when the virtual
displacements take place.The magnitude of Fsp can be computed using the spring
force formula.
Virtual Displacement: The position of points Aand Bat which Fsp acts and point C
at which force Pacts are specified by the position coordinates yA,yB, and yC,
measured from the fixed point E, respectively.
(2)
(3)
Virtual Work Equation: Since Fsp at point Aand force Pacts towards the positive
sense of its corresponding virtual displacement, their work is positive.The work of
Fsp at point Bis negative since it acts towards the negative sense of its
corresponding virtual displacement. Thus,
(4)
Substituting ,,Eqs. (1), (2), and (3) into Eq. (4),
Since , then
When ,
Ans.k=6000
sin 60° –0.2588
=9881N>m=9.88 kN>m
u=60°
k=6000
sin u–0.2588
–0.16k(sin u–0.2588) +960 =0
cos udu Z0
cos udu
C
–0.16k(sin u–0.2588) +960
D
=0
(0.4)k(sin u–0.2588)(0.2 cos udu –0.6 cos udu)+600(1.6 cos udu)=0
P=600 NF
sp =k(sin u–0.2588)
dU=0; F
sp dyA+
A
–F
spdyB
B
+PdyC=0
yC=8(0.2 sin u)dyB=0.6 cos udu
yB=3(0.2 sin u)dyB=0.6 cos udu
F
sp =kx =k
C
2(0.2 sin u)–2(0.2 sin 15°)
D
=(0.4)k(sin u–0.2588) N
du
P
200 mm
200mm
A
C
D
E
B
k
u