11–1.
The scissors jack supports a load P. Dete Determi rmine ne the axia axiall forc fo rce e in the the sc scre rew w ne nece cess ssar ary y fo forr eq equil uilib ibriu rium m wh when en th the e ja jack ck is in the position u. Each of the four links has a length 2 L and is pinpin-co conn nnec ected ted at at it itss ce cent nter er.. Poi oint ntss B and D can mo move ve horizontally.
P
SOLUTION x
=
2L cos u,
y
=
4L sin u,
dU -
F
=
dy
0;
P(4L cos u du) =
dx
2P cot u
-
-
=
=
Pdy
-
2L sin u du
Fdx
F( 2L sin u du) -
=
=
D
A
B u
4L cos u du -
C
0 0 Ans.
.
.
11–4.
The sprin The pring g has an un unst strretc etch hed lengt gth h of 0.3 0.3 m. D Dete etermin rmine e the an angle u f or e equilibrium quilibrium if th t he unif orm link linkss each each have a mass of of 5 5 kg.
C
0.6 m θ
θ
SOLUTION B
Free Body Diagram: Th The e syste stem m has only one deg egrree of fr free eed dom defin fine ed by the inde ind epend nde ent coordin coordinate ate u. Wh When u und unde ergoes a pos osiitiv ive e di disspl place acem ment du, only nly t th he sprin pring g f orce Fsp and nd t th he weights of of th the link linkss (49.05 N) N) do work rk.. Virtual Displacements: Th The e pos osiition o of f poin ints ts B, D and G are meas easur ure ed fr fro om t th he fixe fix ed poin intt A usin ing g pos osiition coo coordin rdinates ates xB , xD and and yG , res esp pect ectiv ive ely ly.. xB
=
0.1 0. 1 sin u
xD
=
210. 0.7 7 sin u2
-
yG
=
0.35 0.3 5 cos u
dyG = - 0.35
dxB =
0.1 0. 1 co coss udu
0.1 sin u
=
(1)
1.3 1. 3 sin u
dxD =
1.3 1. 3 co coss udu
(2)
sin udu sin
(3)
Virtual–Work Equation: When poin ints ts B, D and G und unde ergo pos osiitiv ive e vir virttual disspl di place acem ments dxB , dxD and dyG , the sprin pring g f orce Fsp that acts at poin intt B does pos osiitiv ive e work whil while e th t he sprin s pring g f orce Fsp that acts at poin intt D and nd t th he weight of o f link link N) do negat egativ ive e work rk.. AC A C and CE (49.05 N) dU =
21 - 49.05dyG2
0;
+
Fsp1dxB
- dxD2 =
0
(4)
Sub ubst stiitutin ing g Eq Eqs. s. (1), (2) a and nd (3) in into to (4) yi yie eld ldss
134 34.3 .335 35 sin u
-
1.2Fsp cos u2 du
=
0
(5)
However, fro How from the sprin pring g f ormul rmula, a, Fsp = kx = 4003210. 0.6 6 sin u2 = 480 sin sin u - 120. Sub ubst stiitutin ing g thi thiss valu lue e in into to Eq Eq.. (5) yi yie eld ldss
134. 34.335 335 sin u Sin ince ce du
Z
-
576 sin u cos u
+
144 co coss u2 du
=
144 14 4 co coss u
0
-
0.34
0
0, then 34.335 sin sin u u =
-
15.5°
and u
=
85.4°
576s 76sin in u cos u
+
=
Ans. Ans.
D
0.1 m k = k = 400 N/m A
E
. .
.
.
.
mass
50 kg
50 (9.81) N = 490.5 N
490.5
245.25 N
.
11–10.
The thin The t hin rod o of f weight W rest r est again against st th t he sm s moot ooth h wall ll a and nd P floo fl oorr. Dete etermin rmine e th the mag agni nittud ude e of of f f orce n nee eed ded to hold it in e in equilibrium quilibrium f or a giv give en a an ngle u.
B
SOLUTION
l
Free-Body Diagram: Th The e syste stem m has only one deg egrree of fr free eed dom defin fine ed by the inde ind epend nde ent coordin coordinat ate e u . When u und unde ergoes a pos osiitiv ive e di disspl place acem ment du, only nly t th he weight of of th the rod W and f orce P d do o work rk.. Virtual Displacements: Th The e weight of o f th t he rod W and f orce P are locate ocated d fr fro om t th he fixe fix ed poin ints ts A and B usin ing g pos osiition coo coordin rdinates ates yC and and xA, res esp pect ectiv ive ely yC
=
1 sin s in u 2
dyC
=
xA
=
l cos u
dxA
=
1 cos udu 2 -
l s sin in udu
(1)
(2)
Virtual-Work Equation: When poin Virtual-Work ints ts C and A und unde ergo pos osiitiv ive e vir virttual di disspl place acem ments dyC and dxA, the weight of of th the rod W and f orce F d do o negat egativ ive e work rk.. dU
=
0; - WdyC
-
PdyA
=
0
(3)
Sub ubst stiitutin ing g Eq Eqs. s. (1) a and nd (2) in into to (3) yi yie eld ldss
a Sin ince ce du
Z
Pl s sin in u
-
b
Wl cos u du 2
=
0
0, then Pl s sin in u P
=
-
Wl cos u 2
W cot u 2
=
0 Ans.
P
A
θ
11–11.
Dete etermin rmine e th t he f orce F act actin ing g on on t th he cord cord whi whicch is requir quire ed to main inta tain in equilibrium of the hori rizo zon nta tall 20 20-k -kg g bar AB. Hint: Expr Express ess th the total total constant vertical length l of th t he cord cord in te in term rmss of of p pos osiition coo coordin rdinates ates s1 and and s2. Th The e deriv rivat ativ ive e of of thi hiss equ equat atiion yi yie eld ldss a relat atiionship bet etw wee een n d1 and d2.
s
2
F s
1
SOLUTION
A
B
Free–Body Diagram: Only f orce F and nd t th he weight of of link link AB ( (98.1 98.1 N) N) do work rk.. Virtual Displacement: Fo Forrce F and nd t th he weight of of link link AB ( (98.1 98.1 N) N) a arre locate ocated d fr fro om the top to p o of f th t he fix fixe ed link usin ing g pos osiition coo coordin rdinates ates s2 and and s1 . Sin ince ce th t he cord co rd has a con co nsta stan nt lengt gth h, l , then
4s1
-
s2
=
4ds1
l
-
ds2
=
0
(1)
Virtual–Work Equation: When s1 and s2 und unde ergo pos osiitiv ive e vir virttual di disspl place acem ments ds1 and ds2 , the w we eight of link AB (98.1 N) N) a and nd f orce F d do o pos osiitiv ive e work rk a and nd negat egativ ive e work rk,, res esp pect ectiv ive ely ly.. dU
=
1 2 1 2
0;
196.2 ds1
-
F ds2
0
=
(2)
Sub ubst stiitutin ing g Eq Eq.. (1) in into to (2) yi yie eld ldss
1 196.2 Sin ince ce ds1
Z
-
2
4F d s1
=
0
0, then 196.2 F
=
-
4F
=
49..1 N 49
0 Ans.
20 (9.81) = 196.2 N
.
11–13.
Eac ach h memb mbe er o of f th t he pinpin-co conn nnecte ected d mec ech hani nissm has a mass of f 8 8 k kg. g. If th the sprin spring g is un unst strretc etch hed wh whe en u = 0°, dete etermin rmine e the angle u f or equilibrium quilibrium.. Set k = 300 3000 0 N m a nd M = 1 00 00 N # m.
>
A
M
k u
300
mm
B
SOLUTION 300
y1
=
0.15 0. 15 sin u
y2
=
0.3 sin u
dy1
=
0.15cos u du
dy2
=
0.3cos u du
dU
=
0;
300
2(78.48 78.48)) dy1
[2((78.48 [2 78.48)( )(0.15 0.15 cos u) 47.088 47. 088 cos u Fs
=
-
-
+
+
=
78.48dy2
-
78.48((0.3 cos u) 78.48
Fs(0.3 cos u)
3000(0.3 sin sin u)
47.088 47. 088 cos u
u
D
+
100
=
100du
=
0
Fs(0.3 cos u)
+
100] du
Fs dy2 -
+
=
0
0
900sin u
135 sin sin 2 2u
+
100
=
0
Solvin So lving, g,
u
=
42.7°
Ans.
or
u
=
54.8°° 54.8
Ans.
mm
C
mm
11–14.
Eac ach h memb mbe er o of f th t he pinpin-co conn nnecte ected d mec ech hani nissm has a mass of f 8 8 k kg. g. If th the sprin spring g is un unst strretc etch hed wh whe en u = 0°, dete etermin rmine e the requir quire ed st stiffn iffness ess k so that the mec ech hani nissm is in equilibrium wh whe en u = 25°. Set M = 0.
A
M
k u
300
mm
B
SOLUTION 300
y1
0.15 0. 15 sin u,
=
y2
dy1
=
0.15 0.1 5 cos u du ,
dU
=
0;
=
25°;
Fs
=
+
=
>
1.24 k 1.24 kN N m
+
=
78.48dy2
-
78.48((0.3 cos u) 78.48
+
=
u
D
0.3 0. 3 co coss u du
in 25° 25°)) k(0.3 sin
2(78.48 78.48)( )(0.15 0.15 cos 25°) 25°) k
0.3 sin u dy2
2(78.48 78.48))dy1
[2((78.48 [2 78.48)( )(0.15 0.15 cos u) u
=
300
Fs dy2 -
=
0
Fs(0.3 cos u)] du
=
0
0.1268k
78.48((0.3 cos 25°) 78.48 25°)
-
0.1268k(0.3 cos 25°) 25°)
=
0 Ans.
mm
C
mm
11–15.
The se The servi rvice ce wind windo ow at a f ast-f ast-f ood ood resta estaur ura ant co con nsistso stsof f g glass doo oorrs that open and closea oseau uto tom mat atiica callyu llyussin ing g a moto otorwhi rwhicch suppli upplie es a torqu rque e M to each doo oorr. Th The e f ar end nds, s, A and B, move along the hori rizo zon nta tall guid uide es. If a f ood ood tray beco ecom mes stu st uck bet etw wee een n the doo oorrs a s sh shown wn,, dete etermin rmine e the hori rizo zon nta tall f orce the d doo oorrs exert on the tray at the p pos osiition u .
a
a
C
M
SOLUTION x
=
2a cos u, dU
=
0;
dx
= - 2a
sin u du 0
-M
du
-
F dx
-M
du
+
F (2a sin u)du
F
=
M 2a sin u
=
=
a
u
0 Ans.
a u
A
B
M
D
11–16.
If the sprin pring g is un unst strretc etch hed wh whe en u = 30°, the mass of of the cylind ylinde er is 25 kg, and nd t th he mec ech hani nissm is in in e equilibrium quilibrium wh whe en u = 45°, d dete etermin rmine e the st stiffn iffness ess k of the sprin pring g. Rod AB slid lides es fr free eely ly t thr hro ough t th he coll colla ar at A. Ne Neg glect th the mass of of th the rods.
k A
Free-Body Free-Bod y Diagram: Diagram: When u und unde ergoes a pos osiitiv ive e vir virttual a an ngul ula ar di disspl place acem ment of of du, the das ash h lin line e confi config gur urat atiion s sh hown in in F Fiig. a i iss f orm rme ed. We ob obse serv rve e th that only only t th he sprin spring g f orce Fsp and the weight W of the cylind ylinde er do work wh whe en the vir virttual di disspl place acem ment tak ta kes pl place. ace. Virtual Displacement: Th The e pos osiition of the poin intt B at whi whicch WD and nd Fsp act is spec ecifi ifie ed by by t th he pos osiition coo coordin rdinates ates xB and and yB, meas easur ure ed fr fro om t th he fix fixe ed poin intt C . xB
=
0.45 sin sin u
dxB
=
0.45 cos udu
yB
=
0.45 cos u
dyB
= - 0.45
(1)
sin udu sin
(2)
Virtual–Work Equation: In thi Virtual–Work hiss case Fsp mu must st be reso esolv lve ed in into to its hori rizo zon nta tall and F F F F F vertica call co comp mpo onent, t, i i..e. ( sp)x = sp cos f and ( sp)y = sp sin f. Sin ince ce ( sp)x and W and nd act act tow toward rdss th the negat egativ ive e sen sense of of th their ir co corr rres esp pondin nding g vir virttual di disspl place acem ments ts,, their work is negat egativ ive.Ho e.How wever, (Fsp)y does pos osiitiv ive e work rk s sin ince ce it acts tow toward rdss th the pos osiitiv ive e sen se nse of of iits corr corres esp pondin nding g vir virttual di disspl place acem ment. Thu Thus, s, dU = 0;
- Fsp cos
fdxB
+
sin in fdyB Fsp s
+
Sub ubst stiitutin ing g W = 25 25((9.81 9.81))
=
- Fsp cos
Fsp s sin in f( - 0.45 sin sin udu)
f(0.45 cos udu)
du A 110.3625 sin u
-
+
-
f)
=
+
=
0
+
(3)
=
=
0
0
sin u sin f, the above equ quat atiion ca can n
0.45Fsp cos (u
-
245.25(( - 0.45 sin 245.25 sin udu)
-
sin u sin f) B
cos u cos f
du A 110.3625 sin u Z
( - WdyB)
245.25 N, Eq Eqs. s. (1) a and nd (2) in into to Eq Eq.. (3), we have
0.45Fsp( cos u cos f
Using Usin g the ind inde entity cos (u be rewri writte tten n as
Sin ince ce du
-
f) B
=
0
0, then 110.3625 sin u Fsp
=
-
0.45Fsp cos (u
-
f)
=
0
245.25 sin u cos (u - f)
(4)
Applyin pplying g th the law o of f cosin cosines es to th the geom geomet etry ry s sh hown in in F Fiig. b, we have AB
=
2 0.45 0.452
=
2 0.5625 0.5625
+
0.62
-
-
2(0.45 0.45)( )(0.6 0.6)) cos u
0.54 cos u
Thus, Thu s, the str stretc etch h o of f th the sprin spring g is giv give en by x
0.6 m
B
SOLUTION
=
2 0.5625 0.5625
=
0.1171 m
-
0.54 cos 45°
-
2 0.5625 0.5625
-
0.54 cos 30°
u
0.45 m C
11–16. (continued)
The magnitude of Fsp computed using the spring force formula is therefore Fsp
The angle f at u
Substituting Substi tuting u
=
=
=
kx
=
0.1171 k
45° can be obtained by referring to the geometry shown in Fig.c. tan f
=
f
=
0.6
0.45 cos 45° 0.45 sin 45° 41.53° -
45° and the result resultss for Fsp and f into Eq. (4), we have 0.1171 k k
=
=
245.25 sin 45° cos(45° - 41.53°)
>
1484 N m
=
>
1.48 kN m
Ans.
11–17.
>
If the spring has a torsional stiffness of k = 300 N # m rad and it is unstretched when u = 90°, determine the angle u when the frame is in equilibrium.
k
300 N m/ rad rad
B 0.5 m 0.5 m M
SOLUTION
600 N m
u
Free-Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configurati configuration on shown in Fig. Fig. a is formed. We observe that only only couple moment M and the torque M sp developed in the torsional spring do work when the virtual displacement takes place. The magnitude of M sp can be computed using the spring force formula, Msp
=
k(2a)
c a2 bd
300 2
=
Virtuall Displacement: Virtua Displace ment: Since a
p
=
u
-
=
300(p
-
2u)
u, then
-
2
p
da
(1)
= - du
Virtual–Work Equation: Since M and M sp act towards the negative sense of their Virtual–Work corresponding angular virtual displacements, displacements, their work is negative. negative. Thus Thus,, dU
=
0;
- Mdu +
Substituting M
=
600 N # m , Msp
du[ - 600
Since du
Z
+
=
0
(2)
300(p
-
2u) , an and Eq. (1 (1) into Eq. (2 (2), we we have
2[300(p
-
2u)]( - du)
=
- 600du -
2( - Mspda)
600(p
-
2u)]
=
=
0
0
0, then - 600 +
u
=
600(p
1.071 rad
=
2u) 61.4°
=
0 Ans.
A
C
11–18.
A 5-kg uniform serving serving table is supported supported on each side by pair pa irss of tw two o id iden enti tica call li link nkss, AB an and d CD, an and d sp spri ring ngss CE. If the bowl has a mass of 1 kg, determine the angle u where the table is in equilibrium. The springs springs each have a stiffness of k = 400 N m and are unstretc unstretched hed when u = 90°. Neglect the mass of the links.
250 25 0 mm
>
150 15 0 mm
E A
k
C
250 mm u
u
SOLUTION
D B
Free - Body Free Body Diagram: When u undergoes a positive virtual angular displacement of configuration shown in Fig. Fig. a is formed formed.. We observe observe that only only the du, the dash line configuration spring force F sp, the weight weight Wt of the table, table, and the the weight weight Wb of the bowl bowl do work work when the virtual displacement takes place. The magnitude of F sp can be computed using the spring force formula, Fsp = kx = 400A 0. 0.25 25 co coss u B = 100 co coss u N. Virtual Displacement: The position of points of application of Wb, W t , an and d F sp are specified by the position coordinates yGb, yGt, and xC , resp respectiv ectively ely.. Here Here,, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb
=
0.25 0.2 5 sin u
+
b
dyGb =
0.25 0.2 5 cos udu
(1)
yGt
=
0.25 sin u
+
a
dyGt =
0.25 0.2 5 cos udu
(2)
xC
=
0.25 0.2 5 cos u
dxC = - 0.2 0.25 5
sin udu
(3)
Virtual Work Equation: Since Wb, Wt , an and d F sp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus Thus,, dU =
0;
Substituting Wb
=
Fsp
=
- WbdyG
a 12 b (9.81)
b
=
A - WtdyG B
+
t
4.905 4. 905 N , Wt
A - FspdxC B
+
=
a 52 b (9.81)
=
=
0
(4)
24.525 N ,
100 co coss u N, Eqs Eqs.. (1) (1),, (2) (2),, and (3) int into o Eq. Eq. (4) (4),, we have have
- 4.905(0.25
cos udu) - 24.525(0.25 cos udu)
A
du - 7.3 7.3575 575 cos u +
Since du
Z
25 si sin n u cos u
B=
-
100 co coss u( - 0.2 0.25 5 sin udu)
=
0
0
0, then - 7.3575
cos u
+
cos u( - 7.3575
+
25 si sin n u cos u 25 sin u)
=
=
0
0
Solving the above equation, cos u
=
0
- 7.3575 +
u =
17.1°
u =
90°
25 si sin nu
Ans. =
0 Ans.
150 mm
11–19.
A 5-kg uniform serving serving table is supported supported on each side by two tw o pa pair irss of id iden enti tica call li link nkss, AB an and d CD , an and d sp spri ring ngss CE. If the bowl has a mass of 1 kg and is in equilibrium when u = 60°, determine the stiffness k of each spring.The springs are unstretc unstretched hed when u = 90°. Neglect the mass of the links links..
250 25 0 mm
150 15 0 mm
E A
k
C
250 mm u
u
SOLUTION
D B
Free - Body Free Body Diagram: When u undergoes a positive virtual angular displacement of configuration shown in Fig. Fig. a is formed formed.. We observe observe that only the the du, the dash line configuration spring force F sp, the weight weight Wt of the table, table, and the the weight weight Wb of the bowl bowl do work work when the virtual displacement takes place. The magnitude of F sp can be computed using the spring force formula, Fsp = kx = k A 0.2 0.25 5 cos u B = 0.25 k cos u. Virtual Displacement: The position of points of application of Wb, W t , an and d F sp are specified by the position coordinates yGb, yGt, and xC , respe respectiv ctively ely.. Here Here,, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb
=
0.25 sin u
+
b
dyGb =
0.25 cos udu
(1)
yGt
=
0.25 sin u
+
a
dyGt =
0.25 0.2 5 cos udu
(2)
xC
=
0.25 cos u
dxC = - 0.2 0.25 5
sin udu
(3)
Virtual Work Equation: Since Wb, W t , an and d F sp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus Thus,, dU =
0;
Substituting Wb
=
Fsp
=
- WbdyG
a 12 b (9.81)
+
b
=
A - WtdyG B t
4.90 4. 905 5 N, Wt
+
=
A - FspdxC B
a 52 b (9.81)
=
=
0
(4)
24.5 24 .525 25 N,
0.25k cos u N, Eqs Eqs.. (1) (1),, (2) (2),, and (3) (3) into into Eq. Eq. (4) (4),, we have have
- 4.905(0.25
cos udu) - 24.525(0.25 cos udu) - 0.25k cos u( - 0.2 0.25 5 sin udu)
A
du - 7.3 7.3575 575 cos u +
Since du
Z
0.0625k sin u cos u B
k =
0
0
0, then - 7.3575
When u
=
=
=
cos u 117.72 sin u
+
0.0625k sin u cos u
60°, then k
=
117.72 sin 60°
=
>
13 6 N m
=
0
Ans.
150 mm
The e “ “Nurem Nuremberg berg scissors” scissors” is subjected subjected to a 11–20. Th horizontal horizo ntal force of P = 600 N. Determ Determine ine the angle u for equilibrium. The spring has a stiffness of k = 15 kN m and is unstretched when u = 15°.
>
E u
200 mm
A B k 200 mm
D
C
P
The “Nurem “Nuremberg berg scissor scissors” s” is subjec subjected ted to a horizontal force of P = 600 N. Determine the stiffness k of the spring for equilib equilibrium rium when u = 60° . The spring is unstretched when u = 15°. 11–21.
E u
200 mm
A B k 200 mm
D
C
P
11–22.
The dumpster has a weight W and a center of gravity at G. Determine the force in the hydraulic cylinder needed to hold it in the the general position u .
b a
SOLUTION s
(a2
ds =
(a
=
(a
dU =
0;
F(a2 F
=
+
a
+
+
+
Fds
W(a
90°)
2a c sin u
+
d cos u
-
d sin u
-
Wdy
=
du
0
1
2a c sin u)- 2 ac cos u du +
c
1
b) co coss u du
+
+
2a c sin u)- 2 ac cos u du
b) si sin nu
+
c2
2a c cos (u
-
c2
c2
+
+
dy =
c2
+
2 a2
=
y
θ
2 a2
=
b - d tan u ) ac
b 2
a2
+
-
c2
W(a
+
+
b) cos u du
2a c sin u
+
Wd sin u
du =
0
Ans.
G d
11–23.
The cranks The ksh haf t is subjecte bjected d to a tor to rque of M = 50 N # m. Dete etermin rmine e the hori rizo zon nta tall co compr mpress essiv ive e f orce F appli pplie ed to the pi pisto ston n f or eq equili uilib brium wh whe en u = 60 60°°.
400 mm
100 mm
u F M
SOLUTION (0.4 0.4))2 0
=
0
= +
(0. 0.1) 1)2 2x dx
dU =
Forr Fo
u =
60°°, 60
x =
+ x +
0;
2
-
2(0. 0.1)( 1)(x)( )(cos cos u)
0.2x sin u du -
50du
-
0.2 0. 2 co coss u dx
- Fdx =
0
0.4405 m 0.09 976 769 9 du dx = - 0.0 ( - 50 F =
+
0.09 0.0 976 769 9F) du
512 N
Ans. =
0 Ans.
11–24.
The cr cran anks ksha haft ft is su subj bjec ecte ted d to a to torq rque ue of M = 50 N # m. Dete De term rmin ine e th the e ho hori rizo zont ntal al co comp mpre ress ssiv ive e fo forc rce e F an and d pl plot ot the resul t of F (ordinate) versus u (abscissa) for 0° … u … 90°.
400 mm 100 mm
u F M
SO LUTI O N (0.4)2
=
(0.1)2
+
x2
0
+
2x dx
+
0.2x sin u du
=
0
dx dU - 50du -
F
=
F
50(2x
=
=
2(0.1)(x)(cos u)
-
0.2 0. 2 co coss u d x
a 0.0.220.2cocoss sin 2 b u
x
u
0;
x
-
- 50du -
a 0.0.220.2cocoss sin 2 b u
x
u
-
-
x
-
du
Fdx
du
(1)
=
=
0,
0 du
Z
0
0.2cos u)
0.2x sin u
From Fro m Eq. (1) x2
-
x
=
x
=
F
=
0.2x cos u
-
0.15
=
0
0.2cos u ; 2 0.04 0 .04 co coss2 u 2 0.2cos u
+
0.6
+
2 0.04 0 .04 co coss2 u
,
+
0.6
+
0.6
since 2 0.04 0 .04 co coss2 u
+
0.6
7
0.2 cos u
2 500 2 0.04 0 .04 co coss2 u (0.2 cos u
+
2 0.04 0 .04 co coss2 u
+
0.6)sin u
Ans.
11–25.
Rods AB and BC have centers of mass located at their midpoints. If all contacting surfaces are smooth and BC has a mass of of 100 kg, kg, determ determine ine the appro appropriate priate mass of AB required for equilibrium.
C
A
1.5 m
0.75 m
SO LUTI O N
B
x
=
1.25 1.2 5 cos f;
3
-
1.25 1.2 5 cos f
1.25 sin f df 1.25
a 0.75 b 1.25
0.75df
3 =
=
1m
2.5 2. 5 co coss u
sin u du
= - 2.5
a 1.5 b 2.5
du
= - 1.5du
= - 2du
df
y1
=
a1.252 b sin
y2
=
1.25 1.2 5 sin u
f
dy1
=
0.625 cos f df
dy2
=
1.25 cos u du
dU
=
0;
- m(9.81)dy1 -
- m(9.81)(0.625
cos f df)
- m(9.81)(0.625)
[m(9.81) m
x
2.5 2. 5 co coss u - 2. 2.5 5
=
df
-
=
-
a 1.251 b (
981]du
100 10 0 kg
=
-
981dy2
=
0
981(( 1. 981 1.25 25 co coss u du)
- 2du) -
981(1.25)
=
a 2.52 b
0
du
=
0
0 Ans.
2m
If the potential energy for a conservative onedegree-of-freedom system is expressed by the relation V = (3y 3 + 2y 2 - 4y + 50) J, wh where ere y is given in meters, determine the equilibrium positions and investigate the stability at each position. 11–26.
11–27.
If the potential function for a conservative one-degree-offreedom system is V = 8x3 - 2x2 - 10 J, where x is given in meters,dete rmine the positions for equilibrium and investigate the stability at each of these positions.
1
2
SOLUTION V =
8x3
dV =
dx
124
24x2
x -
x =
0
2x2
-
-
2
4
x =
10
-
4x
=
0
0
and x
=
0.16 0. 167 7m
Ans.
2
d V dx
2
=
48x
-
4
2
d V
x =
0,
x =
0.167 m,
dx
2 = -4 6
0
Unstable
Ans.
2
d V dx
2 =
4
7
0
Stable
Ans.
11–28.
If the potential function for a conservative one-degree-offreedom system is V = (1 (12 2 sin 2u + 15 co coss u) J,where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.
SOLUTION 12 si sin n 2u
V = dV du
=
1
24 1
0;
-
15 co coss u
24 co coss 2u
2
2 sin2 u
48 si sin n2 u Choosing the angle 0°
+
15 si sin nu
15 si sin nu
-
15 si sin nu
+
-
-
24
=
=
=
0
0
0
180°
6 u 6
u =
34.6°
Ans.
u =
145°
Ans.
and
2
d V 2
du
= -
48 si sin n 2u
-
15 co coss u
2
u =
34.6°,
u =
145°,
d V 2
du
= -
57.2
6
0
Unstable
Ans.
Stable
Ans.
2
d V 2 =
du
57.2
7
0
If the potential energy for a conservative onedegree-of-freedom system is expressed by the relation V = (24 sin u + 10 cos 2u) J, 0° … u … 90° , det determ ermine ine the the equilibrium positions and investigate the stability at each position. 11–29.
. .
.
.
11–30.
If the potential function for a conservative one-degree-offreedom system is V = 10 co coss 2u + 25 si sin n u J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.
1
2
SOLUTION 10 co coss 2u
V =
25 sin u
+
For equilibrium: dV du
1
-
= -
20 si sin n 2u
40 sin u
u =
sin-1
+
25 co coss u
2
25 cos u
+
a 2540 b
=
=
=
0
0
38.7° 38. 7° and 141° 141°
Ans.
and u =
cos-1 0
=
90°
Ans.
Stability: 2
d V 2
du
= -
40 cos 2u
-
25 sin u
2
u =
38.7°,
u =
141°,
u =
90°,
d V 2
du
= -
24.4
6
0,
Unstable
Ans.
0,
Unstable
Ans.
Stable
Ans.
2
d V
2 = - 24.4 6
du 2
d V 2 =
du
15
7
0,
11–31.
Determine the angle u for equilibrium and investigate the stability of the mechanism in this position. The spring has a stiffness of k = 1.5 kN m and is unstretched when u = 90°. The block A has a mass mass of 40 kg. kg. Neglect the mass of the link linkss.
600 mm
>
A k D
B F
450 mm
SOLUTION C
Potential Function: With reference to the datum, Fig. a, the gravitational gravitational potential potential energy of block A is positive since its center of gravity is located above the datum. Here, y = (0.45 sin u + b) m, where b is a constant. Thus Thus,, Vg
mgy
=
=
40(9.81)(0.45 sin u
b)
+
=
176.58 sin u
+
392.4 b
The eleastic potential energy of spring BF can be computed using Ve s
=
=
0.45 cos u m.Thus, Ve
1 (1500)(0.45 cos u)2 2
=
=
1 2 ks , where 2
151.875 cos2 u
The total potential energy of the system is V
=
Vg
Ve
+
=
176.58 sin u
+
151.875 cos2 u
+
392.4 b
Equilibrium Configuration: Taking the first derivative of V , we have have dV du
Equilibrium requires
dV du
176.58 cos u
=
cos u
=
176.58
0
dV du
u
-
303.75 sin u)
=
=
=
=
0
0
=
90°
303.75 sin u
-
35.54°
Stability: We ca can n wr write ite
303.75 cos u sin u
-
cos u(176.58
=
303.75 cos u sin u
0 . Thus Thus,,
=
176.58 cos u
u
-
Ans. =
0
35.5°
Ans.
176.58 cos u
-
151.875 sin 2u. Th Thus us,, th the e se seco cond nd
derivative of V is d2V d 2u
At u
At u
= - 176.58
=
90°,
sin u
d 2V d 2u
=
127.17
=
35.54°,
7
2
-
303.75 cos 2u
= - 176.58
sin 90°
-
303.75 cos 180°
=
127.17
7
0
u = 0°
0
Stable
d 2V d2u
= - 201.10 6
2
= - 176.58
sin 35.54°
-
Ans.
303.75 cos 71.09°
u = 35.54°
0 Unstable
Ans.
u
E
u
11–32.
The spring of the scale has an unstretched length of a. Determine the angle u for equilibriu equilibrium m when when a weight W is supported supported on the platform. platform. Neglec Neglectt the weight of the member mem berss. Wh What at value value W would be required to keep the scale in neutral equilibrium when u = 0°?
W
L
L
k
SO LUTI O N Potential Function: The datum is established at point A. Sin Since ce the weigh weightt W is above the the datum, its potential potential energy energy is positive. positive. Fro From m the geometry geometry,, the spring spring stretches x = 2L sin u and y = 2L cos u. V
=
Ve
Vg
=
1 2 kx 2
=
1 (k)(2 L sin u)2 2
=
2kL2 sin 2 u
+
+
Wy
+
W(2L cos u)
+
2WL cos u
dV du
=
4kL2 sin u cos u
dV du
=
2kL2 sin 2u - 2WLsin u
dV du
-
2WL sin u
=
=
=
0.
0
0
Solving, 0°
or
u =
cos-1
Stability: To have neutral equilibrium at u d2V du2 d 2V du2 W
2
=
u -
=
4kL2 cos 2 u
=
4kL2 cos 0°
-
-
=
a2 b
0°,
W kL
d 2V du2
2
Ans.
= u -
0.
0°
2WL cos u
2WLcos 0°
=
0
0°
2kL
L u
u
a
Equilibrium Position: The system is in equilibrium if
u =
L
Ans.
.
.
.
–
.
The unif unifo orm beam eam has has mas mass M . If the contacting surfaces are smooth, determine determine the angle for equilibrium and investigate the stabil stability ity of the beam beam when when it is in this position. The spring has an unstretched length of . Units Used: 3
kN 10 N Given: M 200 kg
0.5 m
kN
k 1.2
m
l 2 m
Solution:
l sin 2
V M g d
1 2
k l cos l cos
2
l cos k l cos l cos l sin l sin 0 2
V V' M g
d 2
d
2
l sin k l 2 sin 2 k l cos l cos l cos l cos 2
V V'' M g
d
There are 2 equilibrium points Guess
30 deg Find
1
Guess
60 deg
2
Find
l cos k l cos l cos l sin l sin 0 2
M g
Given 1
36.4 36.4 deg
l cos k l cos l cos l sin l sin 0 2
M g
Given 2
Ans.
62.3 62.3 deg
Ans.
Check Stability
l sin k l 2 sin 2 k l cos l cos 1 l cos l cos 1 1 1 2
V'' 1 M g
l sin k l 2 sin 2 k l cos l cos 2 l cos l cos 2 2 2 2
V'' 2 M g
V'' 1 1.624 kN m
Unstable
V'' 2 1.55 kN m
Stable
Ans.
11–36.
Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 3 kg and the suspended block D has a mass of 7 kg. Cord DC has a total length length of of 1 m.
500 mm
500 mm
u
u
A
C
500 mm
SOLUTION l
=
500 50 0 mm
y1
=
l sin u 2
y2
=
l
V
=
2Wy1
=
Wl sin u
dV du
=
l(W cos u
tan u
=
W 2WD
u =
12.1°
d2V du2 u =
12.1°,
=
d 2V du2
+
D
2l(1 -
=
-
cos u)
-
-
WD l(3
-
-
2cos u)
2WD sin u)
3(9.81) 14(9.81)
=
0
=
0.2143 Ans.
-
2WD cos u)
0.5[ - 3(9.81) sin 12.1° = -
2 co coss u)
WDy 2
l( - W sin u
=
l(3
=
70.2
6
0
-
14(9.81) cos 12.1°]
Unstable
Ans.
11–37
.
Each of the two springs has an unstretched length . Determine the mass M of the cylinder when it is held in the equilibrium position shown, i.e., y = y = a. Given: a 1m b 500 mm 500 mm k 200
N m
Solution: V 2
dV d y
k 2
y b
y b
2k
2
2
2
2
2
Mg y
Set
y a
Guess
M 1 kg
y 2
y b
M g
2
Given
2k
2
2
y b
M Find ( M )
y 2
M g 0
2
y b
M 22.5 22.5 kg
Ans.
11–38.
A homoge homogeneous neous block block rests rests on top top of the the cylindrical cylindrical surface. Derive the relationship between the radius of the cylinder, r , and the dimensio dimension n of the block, block, b, fo forr stab stable le equilibrium. Hint : Establ Establish ish the potential potential energy function function forr a sm fo smal alll an angl gle e u, i. i.e e., ap appr prox oxim imat ate e sin u L 0, an and d cos u L 1 - u2 2.
b
b
>
r
SOLUTION Potential Function: The datum is established at point O. Since the center center of gravity for the blo block ck is is abov above e the the datum datum,, its pot potent ential ial ene energy rgy is posi positiv tive. e. Her Here, e, b y = r + cos u + ru sin u. 2
a
b
V
=
Wy
For small angle u, sin u
V
W
=
= u
=
W
=
W
ca
r
and u
b 2
+
2 u
1
=
b cos
-
2
B a 2 b ¢1 a2 4 r
b
+
ru2
-
bu2
-
r
2
≤ b 2
+
+
ru2
=
a
W r
-
b 2
b
Stability: To have stable equilibrium, d 2V du2
2
=
d 2V
a b
r
-
6
du2
a
W r
u = 0°
b 2
2r
0
u =
b
2
R
u =
7
dV du
=
0.
0°
0.
u = 0°
-
7
[1]
b
Equilibrium Position: The system is in equilibrium if dV du
d
. Th Then Eq. [1] becomes 2 u
+
ru sin u
u +
b 2
b
7
0
0 Ans.
11–39
.
The uniform rod AB rod AB has a mass M . If = spring DC spring DC is unstretched at = 90 deg, determine the angle for for equilibrium and investigate the stability at the equilibrium position. The spring always acts in the horizontal position due to the roller guide at D. at D. Units Used: 3
kN 10 N Given: M 80 kg k 2
kN
a 1m b 2m
m
Solution:
a b sin 2
V M g
V'
V''
1 2
k a cos
2
a b cos k a2 sin 2 V M g 2 2 d d
d
2 2
a b sin k a2 cos 2 2
V M g
d Equilibrium Guess
30 deg
Given
Guess
70 deg
Given
a b cos 2 a b cos M g 2 M g
k 2 a sin 2 0 2 k 2 a sin 2 0 2
Check Staibility
a b sin k a2 cos 2 1 1 2
V'' 1 M g
a b sin k a2 cos 2 2 2 2
V'' 2 M g
1 36.1 36.1 deg
V'' 1 1.3 kN m
Unstable
Ans.
2 90.0 90.0 deg
V'' 2 0.82 kN m
Stable
Ans.
1 Find 2 Find
11–40.
A spring with a torsional stiffness k is attached attached tothe pin at B. It is unstretched when the rod assembly is in the vertical position. Determine the weight W of the block that results in neutral equilibrium. Hint: Establish the potential energy function functi on for a small angle u , i.e. i.e.,, appro approximate ximate sin u L 0 , and cos u L 1 - u2 2.
L 2
>
A
L 2
SO LUTI O N k
Potential Function: With referen reference ce to the datum, Fig Fig.. a, the gravitat gravitational ional potentia potentiall
B
energy of the block is positive since its center of gravity is located above the datum. Here, the rods are tilted with a small angle u. Th Thus, y However, for a small angle u , cos u Vg
=
Wy
=
1
W
u2
-
2
=
L cos u 2
+
L cos u
=
3 L cos u. 2
. Thus,
a b a1 3 L 2
-
u2
2
b
=
a
3WL 1 2
-
u2
2
b
The elastic potential energy of the torsional spring can be computed using 1 Thus, s, Ve = k b 2, where b = 2u. Thu 2 Vg
=
1 k(2u)2 2
2ku2
=
The total potential energy of the system is V
=
Vg
+
Ve
a
3WL 1 2
=
-
u2
2
b
+
2ku2
Equilibrium Configuration: Taking the first derivative of V , we have dV du
= -
3WL u 2
+
4ku
u
=
a
-
3WL 2
+
4k
b
dV = 0 . Thus Thus,, du 3WL u + 4k = 0 2
Equilibrium requires
a
u
b
=
0°
Stability: The second derivative of V is d 2V 2
du
= -
3WL 2
To have neutral equilibrium at u 3WL + 4k 2 8k W = 3L -
+
=
=
4k
0° ,
d2V du2
2
=
0 . Th Thus,
u = 0°
0 Ans.
The equilibrium configuration of the system at u 8k d 2V 8k d2V W < W > 0 and is unstable if > < 0 . 3L du2 3L du2 Note:
a
b
a
b
=
0° is stable if
L 2 C
11– 41
.
Each bar has a mass per length of m0. Determine the angles at which they are suspended in equilibrium. The contact and at at A at A is smooth, and both are pin con-nected at B. Solution: atan
V
3l 2
m0
1 2
3l cos l m l cos 0 4 2 2
d
V d Guess
Given
9m0 l 8
l 2
m0 l cos l cos
l 4
sin
2
2
sin m0 l sin
10 deg atan
Find
m0 l
cos 0
8
10 deg
1 2
9 8
sin sin
9.18 deg 17.38
Ans.
1 8
cos 0
11–42.
The small postal scale consists of a counterweight W1, connected to the members having negligible weight. Determine the weight W2 that is on the pan in terms of the angles u and f and the dimensions shown. All members are pin connected.
W 2
a
u
SOLUTION
b
y1
=
y2
=
b cos u
W 1
a sin f
where g is a constant and f V
f
=
= - W1y1 + = - W1 b
(90°
=
W1 b sin u
W2
=
W1
+
-
sin a b cos b a
u
-
g)
W2 a sin (90°
-
u
-
u
-
-
g)
W2y2
cos u
dV du
a sin (90°
=
W2 a cos (90°
u f
-
u
-
-
g)
g)
Ans.
a
f
11–43.
>
If the spring has a torsional stiffness k = 300 N # m rad and is unwound when u = 90°, determine the angle for equilibrium if the sphere has a mass of 20 kg. Investigate the stability at this position. Collar C can slide freely along the vertical guide. Neglect the weight of the rods and collar C .
300 mm
D
C
300 mm
SOLUTION u
Potential Function: With reference to the datum, Fig. a, the gravita gravitational tional potenti potential al energy of the sphere is positive its center of gravity is located above the datum. Here, y = (0.3 sin (u 2) + 0.6 sin (u 2))m = (0.9 sin (u/2)). Thus Thus,,
>
>
Vg
mgy
=
>
20(9.81)(0.9 sin u 2)
=
>
The elastic potential energy of the torsional spring can be computed using Ve where b
=
p
2
-
A
176.58 sin (u 2)
=
=
1 kb2 , 2
u .Thus,
Ve
=
¢
p 1 300 2 2
-
u
≤
2 =
150u2
150pu
-
+
37.5p2
The total potential energy of the system is V
=
Vg
+
Ve
176.58 sin (u/2)
=
+
150u2
-
150pu
+
37.5p2
Equilibrium Configuration: Taking the first deri vative of V , we have have dV du
Equilibrium requires
dV du
88.29 cos (u/2)
=
=
300u
+
-
150p
0, Thus,
88.29 cos (u/2)
+
300u
-
150p
=
0
Solving by trial and error, we obtain u
=
1.340 rad
=
76.78°
=
76.8°
Ans.
Stability: The second derivative of V is d 2V du2
= - 44.145
At the equilibrium configuration of u d 2V du2
2
= - 44.14
u = 76.78°
=
sin (u/2)
+
300
=
272.58
76.78°,
sin 38.39°
+
300
7
0 Stable
Ans.
k B
300 mm
11–44.
The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade u along which it can park without overturning and investigate the stability in this position.
G
3.5 3. 5m
SO LUTI O N Potential Function: The datum is established at point A. Since the center center of gravity for the truck is above above the datum, datum, its potenti potential al energy energy is positive positive.. Here Here,, y = (1.5 sin u + 3. 3.5 5 co coss u) m. V
=
Vg
=
Wy
=
(1.5 5 sin u W(1.
+
3.5 3. 5 co coss u)
Equilibrium Position: The system is in equilibrium if dV du
Since W
Z
=
W(1.5 cos u
-
3.5 3. 5 si sin n u)
=
dV du
=
0
0
0, 1.5 1. 5 co coss u u =
-
23.20°
3.5 3. 5 si sin nu =
=
0
23.2°
Ans.
Stability: d2V
=
du2 d2V du2
2
W( - 1. 1.5 5 si sin nu
=
-
3.5 3. 5 co coss u)
W( - 1.5 sin 23.20°
-
3.5 cos 23.20° 23.20°))
= - 3.81W 6
0
u = 23.20°
equilibrium at u Thus,, the truck is in unstable equilibrium Thus
=
23.2°
Ans.
1.5 m 1.5 1. 5m u
11–45.
The cylinder is made of two materials such that it has a mass of m and a center of gravity at point G. Show that that when when G lies above the centroid C of the cylinder, cylinder, the equilibrium equilibrium is unstable.
G C r
SOLUTION Potential Function: The datum is established at point A. Since the center center of gravity gravity of the cy cylind linder er is is above above the dat datum, um, its pot potent ential ial ener energy gy is pos positiv itive. e. Her Here, e, y = r + d cos u. V
=
Vg
Wy
=
=
mg(r
+ d
cos u)
Equilibrium Position: The system is in equilibrium if dV du
sin u
= - mgd
sin u
=
0
u =
=
dV du
=
0.
0
0°
Stability: d 2V
cos u
= - mgd
du2 d 2V du2
2
cos 0°
= - mgd
= - mgd 6
0
u = 0°
Thus,, the cylinder is in unstable equilibrium at u Thus
=
0°
(Q.E.D.)
a
11–46.
If each each o of f th the thr t hree ee lin links ks of of th the mec ech hani nissm has a weight W, dete etermin rmine e the angle u f or equilibrium. The The sprin pring, g, whi whicch alw lwa ays remain inss vertica call, is un unst strretc etch hed wh whe en u = 0°. k u
a
a
SOLUTION y1
=
a sin u
y2
=
2a
+
a sin u
y3
=
2a
+
2a sin u
Fs
=
ka s sin in u
dy1 =
a cos u du dy2 =
dy3 =
0; ( 0; (W
(W
ka s sin in u)a cos u du
Assum Ass ume eu
-
6
2 a
2a cos u du 2 a
dU =
-
a cos u du
Fs)dy1
90°° , so cos u 90
+
Wdy2
+
Wdy3
=
+
Wa cos u du
4W
-
sin in u ka s
u =
sin-1
Z
+
0 W(2a)cos u du
=
0
0. =
0
a4 b
Ans.
90°° 90
Ans.
W ka
or u =
11–47.
If the uniform rod OA has a mass of 12 kg, kg, determine the mass m that will hold the rod in equilibrium when u = 30°. Point C is coincident with B when OA is horizontal. horizontal. Neglect the size of the pulley at B.
B
C m
3m
SO LUTI O N Geometry: Using the law of cosines, =
2 1
lAB
=
2 1
l
=
lAB
lA¿B
2
2 1 3 cos 90°
2
=
2 10 10 m
lA¿B
=
2 10 10
+
-
3
1 21 2 1
-
+
2
3
2
A
2 10 10
-
2
- u
10 2 10
=
-
6 si sin nu
6 si sin nu
-
Potential Function: The datum is established at point O. Since the center center of gravity of the rod and the block are above the datum, datum, their potential energy energy is positive. positive. Here, y1 V
=
3 12
=
3
-
Vg
=
W1y1
=
9.81 m 3
=
29.43 m
l
3
=
10 10
-
-
2 10 10
-
24
6 si sin n u m and y2
=
0.5 sin u m.
W2y2
+
3 12
10 10
-
-
-
1
2 10 10
9.81 m 2 10 10
-
24
6 si sin nu
2 10 10
-
-
1
2
117.72 0.5 sin u
+
2
6 si sin nu
+
58.86 58 .86 sin u
Equilibrium Position: The system is in equilibrium if
`
dV du dV du
=
c 1
=
1 10 2
2 10 10
-
21
-
29.43 m cos u
= -
At u
0.
u = 30°
= - 9.81 m -
6 si sin nu
6 si sin nu
+
1 -2
-6
2d
coss u co
+
58.86 cos u
58.86 58. 86 cos u
30°,
`
dV du m
= -
u = 30°
=
5.29 5. 29 kg
29.43 m cos 30° 10 2 10
-
6 sin 30 30°°
+
58.86 58.8 6 cos 30° 30°
=
1m
0 Ans.
O
11–48.
The triangular block of weight W rests on the smooth corners which are a distance a apart. If the block has three three equal sides of length d, det determi ermine ne the angl angle e u for equilibrium.
d 60
G u
SOLUTION
a
AF
=
AD sin f
AD sin a
=
a sin 60°
AD
=
a (sin (60° sin 60°
+
u))
AF
=
a (sin (60° sin 60°
+
u)) sin (60°
=
a (0.75 cos2 u sin 60°
y=
d
2 3
V dV du
=
-
0.5774 d
-
-
-
u)
-
u)
0.25 sin2 u)
Wy
c
=
AD sin (60°
cos u - AF
W ( - 0.5774 d) si sin nu
Require, sin u or
=
=
-
0
a ( - 1. 1.5 5 si sin n u cos u sin 60° u
a ( - 2 cos u) sin 60° u
=
=
=
0°
-
d
0.5 sin u cos u)
=
0
Ans.
0
cos-1
a4 b d a
Ans.
60
11–49.
Two uniform uniform bars, bars, each having having a weight W , are pin pin-connected at their ends. ends. If they are placed over a smooth cylindrical cylind rical surfac surface, e, show that the angle u for equilib equilibrium rium must satisfy the equation cos u sin3 u = a 2r .
>
>
a
a
r
SOLUTION V =
dV du
r
a
a
r
csc u
2W
a
-r
b
=
2W
=
cos u 2
sin
cos u sin3 u
u
a
a =
2r
2
a -
2
cos u
csc u cot u
b a
+
2
sin u
b
=
0
sin u
(Q.E.D)
.
The uniform bar AB weighs weighs 50 N ( ≈ 5 kg). If the attached spring is upstretched when θ = 90°, use the method of virtual work and determine the angle θ for equilibrium. Note that the spring always remains in the vertical position due to the roller guide 11-51.
B
= K =
100 N/m K = 100 N/m
11m m
1m
1m
A
B
F1 1m
1
1m
1m
W
1
A
1m
1 (1 1 – 1sinθ ) 100
–50
[–50 + 100(1 – sinθ )](1cos )](1cosθ δθ ) = 0 50 – 100 sinθ = 0
.
y
11-51
y = 1sinθ V = 50( 50(1sin 1sinθ ) + dV d θ
K = 100 N/m
=
50cos 50cosθ
100 N/m 1m
d θ
1m
50 co co sθ cosθ θ
=
1m
100( 100(1 1 − 1sin 1sinθ )(− 1co 1coss θ )
dV
Require, = K =
2
=
−
=
W
0 A
1m
10 0( 0(1 − 1s 1si nθ ) c os os θ = 0
0
90 ,
or
50 − 100(1 − sin θ ) = 0
or
θ = 30
Ans.
1m
d 2V d θ 2 A
B
F1
(100) 100)((1 − 1sin 1sinθ )2
1m
B
1m
+
1
d 2V d θ 2
50sin 50sinθ
+
100( 100(1 1 − 1sin 1sinθ )(1s )(1sin in θ) + 100( 100(− 1co 1coss θ)(− 1cos 1cosθ )
50sin 50sinθ
+
100( 100(1 1 − 1si 1sin n θ )sin )sinθ + 100co 100coss 2 θ
=−
=−
θ = 90
,
θ = 30
,
d 2V d θ 2 d 2V d θ 2
=−
=
50 < 0
75 < 0
Unstable
S table
Ans.
Ans.
y
1m
11–53.
The uniform right circular cone having a mass m is suspended from the cord as shown. shown. Determine the angle u at which it hangs from the wall for equilibrium. Is the cone in stable equilibrium? a
u
SO LUTI O N a
V dV du
= -
= -
a
3a cos u 2
a
-
=
0.5 0. 5 co coss u
tan u
=
0.1667
u =
d2V du2
+
u
b
a cos u mg 4
9.46°
= -
u =
+
3a sin u 2
3 si sin nu
b
a sin u mg 4
a
-
9.46°,
a
=
0
Ans.
3a cos u 2 d2V du2
=
-
1.52 a mg
Stable
b
a sin u mg 4 7
0
Ans.
11–54. Th Thee asse mblyyshsohwn asse owncon cosinsts cyelind er as sembl stssists of aof seamise cymi 11–46. lind r and a rnecta N atnd r block. heock block recta blaock. If thIf weigw hesig8hslb40and he and gulanrgul e tbl inv thmi e se cylind s 1, inv 0 N,est est statybili tyewh en lind igate wh n th se cymi er weerigwhesig2h lb ga teigate the stthaebili e he mbl i r in in h quilibrium p i i n S tasse asseymbl y s est g t e e os t o . et. is resting in the equilibrium p osition. Set 4 in h = 100 mm.
h
250 mm 100 mm
W b = 40 N
4(100) d = = 42.44 1 mm 3π yb = (100 + 50 cos ) mm
V = = V g = 10(100 – 42.441 cos ) + 40(100 + 50 cos ) dV d
W s = 10 N
d = 42.44 1 mm
= 424.4 1 sin – – 2000 sin = = 0 100 mm
sin = = 0 = 0° = d 2V 2
d
50 mm
(equilibrium position) y s = (100 – 42.44 1 cos ) mm
= 424.4 1 cos – – 2000 cos
At = = 0°,
d 2V d 2
= –1575.6 < 0
Unstable
Ans.
Thee 210-N lb se(mi 1-kg) lindse heupp bl ock h ock h as ermi supp cylind ortsert s orts whi thecbl 11–55. 11–47. Th cy 3 a spcec c w a sepigec t ocf weigh80 t of lb =f t13 . D .5 ete kN/rmin e ete thermin whi h ifi has hifi m3. D h eigehtth he = h f t h will whipr ducepr noedu utce ralnequilibrium utral equilibrium in the h o oef itghhet bl ock th whi e blcock chowill in pos itoiown n s. hown. thiteiopnos sh
h
250 mm 100 mm
W b = [13.5 (10–6) (h) (200) (250)] N
4(100)
d =
3π
= 42.44 1 mm
yb = 100 +
h
2
cos mm W s = 10 N
V = = V g = 10(100 – 42.441 cos )
1 h + 13.5 6 h(200)(250) 100 + cos 2 10 dV d
100 mm
= 424.4 1 sin – – 0.3375 h2 sin = = 0 y s = (100 – 42.44 1 cos ) mm
sin = = 0 = 0° = d 2V d 2
d = 42.44 1 mm
(equilibrium position)
= 424.4 1 cos – – 0.3375 h2 cos = = 0
h =
424.41 0.3375
= 35.46 mm
Ans.
11–56
.
If the potential energy for a conservative one-degree-ofone-degree-of-freedom freedom system is expressed by the relation V = = (ax (ax3 + bx2 + cx + cx + d ), ), determine the equilibrium positions and investigate the stability at each position.
Given:
a
N
4
m
b
2
1
N
c
m
3N
d
10N m
Solution: V
3
a x
=
2
bx c x d
Required Position : dV =
dx
3
a x
2
2 b x c
2 b
x1
x1
0.590 m
4
=
0
2
4 ( 3 a) c 2 3 a b
Ans.
2 b
4
x2
x2
0.424 m
b
2
4 3 a c
2 3
a
Ans.
Stability : 2
d V dx
At x
At x
2
=
V''
=
6
a x
2 b N
x1
V''1
6 a x1 2 b
V''1
12.2
x2
V''2
6 a x2 2 b
V''2
12.2
m N m
V''1
0
Stable
V''2
0
Unstable
Ans.
Ans.
11–57. 11–22. D Determin rmine thhe weighht of f block G required to balance tthhee diff 10 F kg) diff eerreennttiiaall lelevveerr wh wheennthtehe100 lb (l is 20-N oad loa d F r nisce in wh baelanntce he the lpeavne.rTh e lebvaela pl d iosnpltace he pda onn. Th i s in hewh loaedn t ace and lbl d aa nd rehenot haekle ver. T1ake oa rebl nock levoenr t ock ot ona t .T 2 in x. = 300 mm.
100 mm 100 mm
x C
A
G
B
E
D
50 mm
F
Free – Body Diagram: When the lever undergoes a virtual angular displacement of a about point B, the dash line configuration f block G and the weight WF of l f load F do work when the virtual shown in Fig. a is f ormed. We observe that only the weight WG of bl displacements take place. Virtual Displacement: Since yG is very small, the vertical virtual displacement of bl f block G and load F ca can be approximated as yG = (300 + 100) = 400
(1)
yF = 50
(2)
Virtual Work Equation: Since WG acts to wards the positive sense of i f its corresponding virtual displacement, its work is positive. f its corresponding virtual displacement. Thus, However, f orce WF does negative work since it acts towards the negative sense of i U = = 0; U
W G yG + (–W F yF ) = 0
(3)
Substituting W G = 100 N a nd Eqs. (1) and (2) into Eq. (3), W G(400 ) – 100(50 ) = 0 (400 (400 W G – 5000) = 0
Since ≠ 0, then
400 W G – 5000 = 0 W G = 12.5 N
Ans.
11–58. If t nd tthhe bl block G F w thhee lload 00 lb N aand 11–23. If weeiigghhss 120 weighs 120 lbN, rminee its its ppos quilibrium of the etermin osiittiioonn x , ddete f or equilibrium diff erential l ever. The llever i s in in balance wh when tthe lload aand block are not on the lever.
100 mm 100 mm
x C
A
G
B E
D
50 mm
F
about point B, the dash line configuration Free – Body Diagram: When the lever undergoes a virtual angular displacement of a shown in Fig. a is f ormed. We o bserve that only the weight WG of bl f block G and the weight WF of l f load F do work when the virtual displacements take place. f block G and load F ca can be approximated as Virtual Displacement: Since yG is very small, the vertical virtual displacement of bl yG = (100 + x)
(1 )
yF = 50
(2)
Virtual Work Equation: Since WG acts towards the positive sense of i f its corresponding virtual displacement, its work is positive. f its corresponding virtual displacement. Thus, However, f orce WF does negative work since it acts to wards the negative sense of i U U = = 0;
W G yG + (–W F yF ) = 0
(3)
Substituting W F = 100 N, W G = 10 N, E qs. (1) and (2) into Eq. (3), 10(100 + x) – – 100(50 ) = 0 [10(100 + x) – 5000] = 0
Since ≠ 0, then 10(100 + x) – 5000 = 0 x = 400 mm
Ans.
11–59
.
A force P is applied to the end of the lever. Determine the horizontal force on the piston for equilibrium. Solution: s 2 l x 2 l cos x 2 l sin U P s F x 0
P 2l F 2l sin l sin 0 F P csc P csc
Ans.
F
11–60. A 5-kg unif orm serving table is supported o n each side by pairs of t two identical links, AB and CD, and s prings CE. If the bowl has a mass of 1 kg, d etermine the angle u where the table is in equilibrium. The springs each have a stiffness of k = 200 N m and are unstretched when u = 90°. Neglect the mass of t the links ks..
>
250 mm
150 mm
E A
k
C
250 mm u
u
D B
150 mm
11–61. A 5-kg uniform serving table is supported on each side si de by tw two o pa pair irss of id iden enti tica call li link nkss, AB an and d CD , an and d sp spri ring ngss equilibrium m when CE. If the bowl has a mass of 1 kg and is in equilibriu u = 45°, determine the stiffness k of each spring.The springs are unstretch unstretched ed when u = 90° . Neglect the mass of the links links..
250 mm
150 mm
E A
k
C
250 mm u
u
D B
150 mm
11– 62
.
Rods AB Rods AB and and BC BC have have centers of mass located at their midpoints. If all contacting surfaces are smooth and BC has has mass m BC determine the appropriate mass m AB of AB of AB required required for equilibrium. Given: m BC 15 0 kg a 0.8 m b 1m c 2m d 1.6 m
Solution: Use as the independent independent variable
Define
2
L1
2
a b
L2
Then
L1 cos L2 cos b c
Thus
Also
2
2
c d
d c
atan
a b
atan
L1 sin L2 sin 0
L2 sin L1 sin
y1
y2
L1 2 L2 2
sin
y1
sin
y2
L1 2 L2 2
cos
L2 sin cot 2
cos
L2 sin cot L2 m BC cos 0 U m AB g y 1 m BC g y 2 gm AB 2 2 m AB m BC tan cot
m AB 15 0 kg
Ans.
