Resolução do Capítulo 10 - Método dos Trabalho Virtuais

Prévia do material em texto

PROBLEM 10.1 
Determine the vertical force P which must be applied at G to maintain the 
equilibrium of the linkage. 
 
SOLUTION 
 
Assuming 
 Ayδ 
it follows 
120 1.5
80C A A
y y yδ δ δ= = 
1.5E C Ay y yδ δ δ= = 
( )180 3 1.5 4.5
60D A A A
y y y yδ δ δ δ= = = 
( )100 100 1.5 2.5
60 60G A A A
y y y yδ δ δ δ= = = 
Then, by Virtual Work 
( ) ( )0: 300 N 100 N 0A D GU y y P yδ δ δ δ= − + = 
 ( ) ( )300 100 4.5 2.5 0A A Ay y P yδ δ δ− + = 
 300 450 2.5 0P− + = 
 60 NP = + 60 N=P W 
 
 
PROBLEM 10.2 
Determine the vertical force P which must be applied at G to maintain the 
equilibrium of the linkage. 
 
SOLUTION 
 Link ABC 
 
 Link DEFG 
 
 
 
 
 
Assume 
clockwiseδθ 
Then for point C 
( )5 in.Cxδ δθ= 
and for point D 
( )5 in.D Cx xδ δ δθ= = 
And for link DEFG 
15Dxδ δφ= 
 5 15δθ δφ∴ = 
or 1
3
δφ δθ= 
Then 44 2 2 in.
3G
δ δφ δθ = =    
Now cos 45G Gyδ δ= ° 
 4 2 cos 45
3
δθ = °   
 4 in.
3
δθ =    
Then, by Virtual Work. 
( ) ( ) ( ) ( )0: 80 lb in. 40 lb in. in. 0E GU x P yδ δθ δ δ= ⋅ − + = 
 10 480 40 0
3 3
Pδθ δθ δθ   − + =       
 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 
 
 Link ABC 
 
 Link DEFG 
 
 
 
 
Following the kinematic analysis of Problem 10.2, we have 0:U = 
( ) ( ) ( )0: 80 lb in. 40 lb in. 0EU x Mδ δθ δ δφ= ⋅ − + = 
 10 180 40 0
3 3
Mδθ δθ δθ   − + =       
 or 160 lb in.= ⋅M W 
 
 
 
 
 
PROBLEM 10.4 
Determine the couple M which must be applied to member DEFG to 
maintain the equilibrium of the linkage. 
 
SOLUTION 
 
Assuming 
 Ayδ 
it follows 
120 1.5
80C A A
y y yδ δ δ= = 
1.5E C Ay y yδ δ δ= = 
( )180 3 1.5 4.5
60D A A A
y y y yδ δ δ δ= = = 
1.5 1
60 60 40
E A
A
y y yδ δδφ δ= = = 
Then, by Virtual Work: 
( ) ( )0: 300 N 100 N 0A DU y y Mδ δ δ δφ= − + = 
 ( ) 1300 100 4.5 0
40A A A
y y M yδ δ δ − + =   
 1300 450 0
40
M− + = 
 6000 N mmM = + ⋅ 6.00 N m= ⋅M 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 xδ δ= ⇒ = 
4 4H D H Dx x x xδ δ= ⇒ = 
5 5I D I Dx x x xδ δ= ⇒ = 
(a) Virtual Work 0: 0G G SP IU F x F xδ δ δ= − = 
 or ( )( ) ( )20 lb 3 5 0D SP Dx F xδ δ− = 
 thus, 12.00 lb SPF T= W 
 Now SP IF k x= ∆ 
 or ( )12.00 lb 4 lb/in. Ix= ∆ 
 Thus, 3 in.Ix∆ = 
 and 
1 1
4 5D H I
x x xδ δ δ= = 
4 
5
∴ ∆ = ∆H Ix x 
 ( )4 3 in.
5
= or 2.40 in.Hx∆ = W 
(b) Virtual Work: 0: 0G G H H SP IU F x F x F xδ δ δ δ= + − = 
 or ( )( ) ( )( ) ( )20 lb 3 20 lb 4 5 0D D SP Dx x F xδ δ δ+ − = 
 thus, 28.0 lb SPF T= W 
 Now SP IF k x= ∆ 
 or ( )28.0 lb 4 lb/in. Ix= ∆ 
 Thus, 7 in.Ix∆ = 
 From part (a) 
4
5H I
x x∆ = ∆ 
 ( )4 7 in.
5
= or 5.60 in.Hx∆ = 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 xδ δ= ⇒ = 
4 4H D H Dx x x xδ δ= ⇒ = 
5 5I D I Dx x x xδ δ= ⇒ = 
(a) Virtual Work: 0: 0E E SP IU F x F xδ δ δ= − = 
 or ( )( ) ( )20 lb 2 5 0D SP Dx F xδ δ− = 
 thus, 8.00 lb SPF T= W 
 Now SP IF k x= ∆ 
 or ( )8.00 lb 4 lb/in. Ix= ∆ 
 Thus, 2 in.Ix∆ = 
 And 
1 1
4 5D H I
x x xδ δ δ= = 
4 
5
∴ ∆ = ∆H Ix x 
 ( )4 2 in.
5
= 
 or 1.600 in.Hx∆ = W 
(b) Virtual Work: 0: 0D D E E SP IU F x F x F xδ δ δ δ= + − = 
 or ( ) ( )( ) ( )20 lb 20 lb 2 5 0D D SP Dx x F xδ δ δ+ − = 
 thus, 12.00 lb SPF T= W 
 
PROBLEM 10.6 CONTINUED 
 Now SP IF k x= ∆ 
 or ( )12.00 lb 4 lb/in. Ix= ∆ 
 Thus, 3 in.Ix∆ = 
 From part (a) 
4
5H I
x x∆ = ∆ 
 ( )4 3 in.
5
= or 2.40 in.Hx∆ = 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 yδ δ= 
(a) Virtual Work: 
0: 0A CU P y F yδ δ δ= − = 
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 FΣ = + − = 
 75 N 300 N 0R + − = 
 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 Ayδ 
Have 160 1or
320 2C A C A
y y y yδ δ δ δ= = 
Since bar CD moves in translation 
E F Cy y yδ δ δ= = 
or 1
2E F A
y y yδ δ δ= = 
Virtual Work: 
 
( ) ( )0: 400 N 600 N 0A E FU P y y yδ δ δ δ= − + + = 
 ( ) ( )1 1400 N 600 N 0
2 2A A A
P y y yδ δ δ   − + + =       
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 θ= 
2 cosAx lδ θ δθ= 
and 3 cosFy l θ= 
3 sinFy lδ θ δθ= − 
Virtual Work: 0: 0A FU Q x P yδ δ δ= + = 
 ( ) ( )2 cos 3 sin 0Q l P lθ δθ θ δθ+ − = 
 3 tan
2
Q P θ= 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 lθ δ θ δθ= = − 
( )2 sin ; cos
2 2
CD l CD lθ θδ δθ= = 
Virtual Work: 
 ( )0: 0AU P y Q CDδ δ δ= − − = 
 ( )2 sin cos 0
2
P l Q l θθ δθ δθ − − − =   
 ( )
sin2
cos /2
Q P θθ= 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 lθ δ θδθ= = 
( )2 sin ; cos
2 2
CD l CD lθ θδ δθ= = 
Virtual Work: ( )0: 0AU P x Q CDδ δ δ= − = 
 ( )2 cos cos 0
2
P l Q l θθδθ δθ − =   
 ( )
cos2
cos /2
Q P θθ= 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′∆ 
tanA C a θ′ = 
( ) tanAy A C a θ′= − = − 
2cosA
ayδ δθθ= − 
For :CC B′∆ 
sinBC l A Cθ′ ′= − 
 sin tanl aθ θ= − 
sin tanBy BC l aθ θ′= = − 
2cos cosB
ay lδ θδθ δθθ= − 
Virtual Work: 
 0: 0A BU Q y P yδ δ δ= − = 
 2 2cos 0cos cos
a aQ P lδθ θ δθθ θ
   − − − − =       
 2 2coscos cos
a aQ P l θθ θ
   = −      3cos 1lQ P
a
θ = −  W 
 
 
 
PROBLEM 10.13 
A double scissor lift table is used to raise a 1000-lb machine component. 
The table consists of a platform and two identical linkages on which 
hydraulic cylinders exert equal forces. (Only one linkage and one 
cylinder are shown.) Each member of the linkage is of length 24 in., and 
pins C and G are at the midpoints of their respective members. The 
hydraulic cylinder is pinned at A to the base of the table and at F which is 
6 in. from E. If the component is placed on the table so that half of its 
weight is supported by the system shown, determine the force exerted by 
each cylinder when 30 .θ = ° 
 
SOLUTION 
 
 
First note 2 sin 24 in. (length of link)Hy L Lθ= = 
Then 2 cosHy Lδ θδθ= 
Now 
2 23 5cos sin
4 4AF
d L Lθ θ   = +       
 21 9 16sin
4
L θ= + 
Then ( )
2
2 16sin cos
4 2 9 16sin
AF
Ld
θ θδ δθθ= + 
 
2
sin cos4
9 16sin
L θ θ δθθ= + 
Virtual Work: 
 cyl
10: 0
2AF H
U F d W yδ δ δ = − =   
or ( )( )cyl 2sin cos4 500 lb 2 cos 09 16sinF L L
θ θ δθ θδθθ
  − =  + 
 
and cyl 2
sin 250 lb
9 16sin
F θ θ =+ 
Finally, cyl 2
sin 30 250 lb
9 16sin 30
F ° =
+ °
 
 or cyl 1803 lbF = W 
 
 
 
 
PROBLEM 10.14 
A double scissor lift table is used to raise a 1000-lb machine component. 
The table consists of a platform and two identical linkages on which 
hydraulic cylinders exert equal forces. (Only one linkage and one 
cylinder are shown.) Each member of the linkage is of length 24 in., and 
pins C and G are at the midpoints of their respective members. The 
hydraulic cylinder is pinned at A to the base of the table and at F which is 
6 in. from E. If the component is placed on the table so that half of its 
weight is supported by the system shown, determine the smallest 
allowable value of θ knowing that the maximum force each cylinder can 
exert is 8 kips. 
 
SOLUTION 
 
 
From the results of the Problem 10.13 
cyl 2
sin 250 lb
9 16sin
F θ θ =+ 
Then ( )
2
sin8000 lb 250 lb
9 16sin
θ
θ =+ 
or ( )2 232sin 9 16sinθ θ= + 
Thus, 2 9sin
1008
θ = 
 5.42θ = °W 
 
 
 
PROBLEM 10.15 
Derive an expression for the magnitude of the couple M required to 
maintain the equilibrium of the linkage shown. 
 
SOLUTION 
 
 
ABC: sin cosB By a y aθ δ θδθ= ⇒ = 
2 sin 2 cosC Cy a y aθ δ θδθ= ⇒ = 
CDE: Note that as ABC rotates counterclockwise, CDE rotates clockwise 
 while it moves to the left. 
Then Cy aδ δφ= 
or 2 cosa aθδθ δφ= 
or 2cosδφ θδθ= 
Virtual Work: 
 0: 0B CU P y P y Mδ δ δ δφ= − − + = 
 ( ) ( ) ( )cos 2 cos 2cos 0P a P a Mθδθ θδθ θδθ− − + = 
 or 3
2
M Pa= W 
 
 
 
 
PROBLEM 10.16 
Derive an expression for the magnitude of the couple M required to 
maintain the equilibrium of the linkage shown. 
 
SOLUTION 
 
 
 
 
Have 
cosBx l θ= 
 sinBx lδ θδθ= − (1) 
sinCy l θ= 
cosCy lδ θδθ= 
Now 1
2B
x lδ δθ= 
Substituting from Equation (1) 
1sin
2
l lθδθ δφ− = 
or 2sinδφ θδθ= − 
Virtual Work: 
 0: 0CU M P yδ δϕ δ= + = 
 ( ) ( )2sin cos 0M P lθδθ θδθ− + = 
or 1 cos
2 sin
M Pl θθ= 
 
2 tan
PlM θ= W 
 
 
 
PROBLEM 10.17 
Derive an expression for the magnitude of the couple M required to 
maintain the equilibrium of the linkage shown. 
 
SOLUTION 
 
 
Have 
sinBx l θ= 
cosBx lδ θδθ= 
cosAy l θ= 
sinAy lδ θδθ= − 
Virtual Work: 
 0: 0B AU M P x P yδ δθ δ δ= − + = 
 ( ) ( )cos sin 0M P l P lδθ θδθ θδθ− + − = 
 ( )sin cosM Pl θ θ= + W 
 
 
 
 
PROBLEM 10.18 
The pin at C is attached to member BCD and can slide along a slot cut in 
the fixed plate shown. Neglecting the effect of friction, derive an 
expression for the magnitude of the couple M required to maintain 
equilibrium when the force P which acts at D is directed (a) as shown, (b) 
vertically downward, (c) horizontally to the right. 
 
SOLUTION 
 
 
 
 
Have cosDx l θ= 
sinDx lδ θδθ= − 
3 sinDy l θ= 
3 cosDy lδ θδθ= 
Virtual Work: ( ) ( )0: cos sin 0D DU M P x P yδ δθ β δ β δ= − − = 
( )( ) ( )( )cos sin sin 3 cos 0M P l P lδθ β θδθ β θδθ− − − = 
 ( )3sin cos cos sinM Pl β θ β θ= − (1) 
(a) For P directed along BCD, β θ= 
 Equation (1): ( )3sin cos cos sinM Pl θ θ θ θ= − 
 ( )2sin cosM Pl θ θ= sin 2M Pl θ= W 
(b) For P directed , 90β = ° 
 Equation (1): ( )3sin 90 cos cos90 sinM Pl θ θ= ° − ° 
 3 cosM Pl θ= W 
(c) For P directed , 180β = ° 
 Equation (1): ( )3sin180 cos cos180 sinM Pl θ θ= ° − ° 
 sinM Pl θ= W 
 
 
 
PROBLEM 10.19 
A 1-kip force P is applied as shown to the piston of the engine system. 
Knowing that 2.5 in.AB = and 10 in.BC = , determine the couple M 
required to maintain the equilibrium of the system when (a) 30 ,θ = ° 
(b) 150 .θ = ° 
 
SOLUTION 
Analysis of the geometry: 
 
 
 
 
 
 
Law of Sines 
sin sin
AB BC
φ θ= 
 sin sinAB
BC
φ θ= (1) 
Now 
cos cosCx AB BCθ φ= + 
 sin sinCx AB BCδ θδθ φδφ= − − (2) 
Now, from Equation (1) cos cosAB
BC
φδφ θδθ= 
or cos
cos
AB
BC
θδφ δθφ= (3) 
From Equation (2) 
cossin sin
cosC
ABx AB BC
BC
θδ θδθ φ δθφ
 = − −   
 
or ( )sin cos sin cos
cosC
ABxδ θ φ φ θ δθφ= − + 
Then ( )sin
cosC
AB
x
θ φδ δθφ
+= − 
 
PROBLEM 10.19 CONTINUED 
Virtual Work: 0: 0CU P x Mδ δ δθ= − − = 
 ( )sin 0
cos
AB
P M
θ φ δθ δθφ
 +− − − =  
 
Thus, ( )sin
cos
M AB P
θ φ
φ
+= (4) 
For the given conditions: 1.0 kip 1000 lb, 2.5 in., and 10 in.:P AB BC= = = = 
(a) When 2.530 : sin sin 30 , 7.181
10
θ φ φ= ° = ° = ° 
( ) ( ) ( )sin 30 7.1812.5 in. 1.0 kip 1.5228 kip in.
cos7.181
0.1269 kip ft
M
° + °= = ⋅°
= ⋅
 
 or 126.9 lb ft= ⋅M W 
(b) When 2.5150 : sin sin150 , 7.181
10
θ φ φ= ° = ° = ° 
( ) ( ) ( )sin 150 7.1812.5 in. 1.0 kip 0.97722 kip in.
cos7.181
M
° + °= = ⋅° 
 or 81.4 lb ft= ⋅M W 
 
 
 
PROBLEM 10.20 
A couple M of magnitude 75 lb ft⋅ is applied as shown to the crank of 
the engine system. Knowing that 2.5 in.AB = and 10 in.BC = , 
determine the force P required to maintain the equilibrium of the system 
when (a) 60 ,θ = ° (b) 120 .θ = ° 
 
SOLUTION 
From the analysis of Problem 10.19, ( )sin
cos
M AB P
θ φ
φ
+= 
Now, with 75 lb ft 900 lb in.M = ⋅ = ⋅ 
(a) For 60θ = ° 
2.5sin sin 60 , 12.504
10
φ φ= ° = ° 
( ) ( ) ( ) ( )sin 60 12.504900 lb in. 2.5 in.
cos12.504
P
° + °⋅ = ° 
 or 368.5 lbP = 
 369 lb=P W 
(b) For 120θ = ° 
2.5sin sin120 , 12.504
10
φ φ= ° = ° 
( ) ( ) ( ) ( )sin 120 12.504900 lb in. 2.5 in.
cos12.504
P
° + °⋅ = ° 
 or 476.7 lbP = 
 477 lb=P W 
 
 
 
 
 
PROBLEM 10.21 
For the linkage shown, determine the force P required for equilibrium 
when 18 in.,a = 240 lb in.,M = ⋅ and 30 .θ = ° 
 
SOLUTION 
Consider a virtual counterclockwise rotation δφ of link AB. 
 Then B aδ δφ= 
 Note that 
 cosB Byδ δ θ= 
 cosa θ δφ= 
 
If the incline were removed, point C would move down Cyδ as a result of the virtual rotation, where 
cosC By y aδ δ θ δφ= = 
For the roller to remain on the incline, the vertical link BC would then have to rotate counterclockwise. Thus, 
to first order: 
( )totalC Cy yδ δ≈ 
Then 
( )total
sin
C
C
y
S
δδ θ= 
 cos
sin
a θ δφ
θ= 
 
tan
a δφθ= 
Now, by Virtual Work: 0: 0CU M P Sδ δφ δ= − = 
or 0
tan
aM Pδφ δφθ
 − =   
or tanM Paθ = 
With 240 lb in., 18 in., and 30M a θ= ⋅ = = ° 
 ( ) ( )240 lb in. tan 30 18 in.P⋅ ° = or 7.70 lb=P 30.0°WPROBLEM 10.22 
For the linkage shown, determine the couple M required for equilibrium 
when 2 ft,a = 30 lb,P = and 40 .θ = ° 
 
SOLUTION 
From the analysis of Problem 10.21, tanM Paθ = 
Now, with 30 lb, 2 ft, and 40 ,P a θ= = = ° we have 
( )( )tan 40 30 lb 2 ftM ° = 
 or 71.5 lb ft= ⋅M W 
 
 
 
PROBLEM 10.23 
Determine the value of θ corresponding to the equilibrium position of 
the mechanism of Problem 10.10 when 60 lbP = and 75 lb.Q = 
 
SOLUTION 
 
 
From geometry 
2 cos , 2 sinA Ay l y lθ δ θ δθ= = − 
( )2 sin , cos
2 2
CD l CD lθ θδ δθ= = 
Virtual Work: 
 ( )0: 0AU P y Q CDδ δ δ= − − = 
 ( )2 sin cos 0
2
P l Q l θθ δθ δθ − − − =   
or ( )
sin2
cos /2
Q P θθ= 
With 60 lb, 75 lbP Q= = 
( ) ( ) ( )
sin75 lb 2 60 lb
cos /2
θ
θ= 
( )
sin 0.625
cos /2
θ
θ = 
or ( ) ( )( )
2sin /2 cos /2
0.625
cos /2
θ θ
θ = 
36.42θ = ° 
 36.4θ = °W 
(Additional solutions discarded as not applicable are 180 )θ = ± ° 
 
 
 
PROBLEM 10.24 
Determine the value of θ corresponding to the equilibrium position of 
the mechanism of Problem 10.11 when 20 lbP = and 25 lb.Q = 
 
SOLUTION 
 
 
2 sinAx l θ= 
2 cosAx lδ θ δθ= 
2 sin
2
CD l θ= 
( ) cos
2
CD l θδ δθ= 
Virtual Work: 
 ( )0: 0AU P x Q CDδ δ δ= − = 
 ( )2 cos cos 0
2
P l Q l θθ δθ δθ − =   
or ( )
cos2
cos /2
Q P θθ= 
With 20 lb and 25 lbP Q= = 
( ) ( ) ( )
cos25 lb 2 20 lb
cos /2
θ
θ= 
or ( )
cos 0.625
cos /2
θ
θ = 
Solving numerically, 56.615θ = ° 56.6θ = °W 
 
 
 
 
 
PROBLEM 10.25 
A slender rod of length l is attached to a collar at B and rests on a portion 
of a circular cylinder of radius r. Neglecting the effect of friction, 
determine the value of θ corresponding to the equilibrium position of the 
mechanism when 300 mm,l = 90 mm,r = 60 N,P = and 120 N.Q = 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
Geometry OC r= 
cos
B
OC r
OB x
θ = = 
cosB
rx θ= 
2
sin
cos
θδ δθθ=B
rx 
cos ; sinA Ay l y lθ δ θδθ= = − 
Virtual Work: 
 ( )0: 0A BU P y Q xδ δ δ= − − = 
 2
sinsin 0
cos
rPl Q θθ δθ δθθ− = 
 2cos Qr
Pl
θ = (1) 
Then, with 300 mm, 90 mm, 60 N, and 120 Nl r P Q= = = = 
( )( )
( )( )2
120 N 90 mm
cos 0.6
60 N 300 mm
θ = = 
or 39.231θ = ° 39.2θ = °W 
 
 
 
PROBLEM 10.26 
A slender rod of length l is attached to a collar at B and rests on a portion 
of a circular cylinder of radius r. Neglecting the effect of friction, 
determine the value of θ corresponding to the equilibrium position of 
the mechanism when 280 mm,l = 100 mm,r = 300 N,P = and 
600 N.Q = 
 
SOLUTION 
From the analysis of Problem 10.25 
2cos Qr
Pl
θ = 
Then with 280 mm, 100 mm, 300 N, and 600 Nl r P Q= = = = 
( )( )
( )( )2
600 N 100 mm
cos 0.71429
300 N 280 mm
θ = = 
or 32.311θ = ° 32.3θ = °W 
 
 
 
 
 
PROBLEM 10.27 
Determine the value of θ corresponding to the equilibrium position 
of the mechanism of Problem 10.12 when 600 mm,l = 
100 mm,a = 100 N,P = and 160 N.Q = 
 
SOLUTION 
 
 
For :AA C′∆ tanA C a θ′ = 
( ) tanAy A C a θ′= − = − 
2cosA
ayδ δθθ= − 
For :CC B′∆ 
sinBC l A Cθ′ ′= − 
 sin tanl aθ θ= − 
sin tanBy BC l aθ θ′= = − 
2cos cosB
ay lδ θ δθ δθθ= − 
Virtual Work: 
 0: 0A BU Q y P yδ δ δ= − − = 
 2 2cos 0cos cos
a aQ P lδθ θ δθθ θ
   − − − − =       
 2 2coscos cos
a aQ P l θθ θ
   = −       
or 3cos 1lQ P
a
θ = −   
With 600 mm, 100 mm, 100 N, and 160 Nl a P Q= = = = 
( ) ( ) 3600 mm160 N 100 N cos 1
100 mm
θ = −   
or 3cos 0.4333θ = 
 40.82θ = ° 40.8θ = °W 
 
 
 
PROBLEM 10.28 
Determine the value of θ corresponding to the equilibrium position of 
the mechanism of Problem 10.13 when 900 mm,l = 150 mma = 
75 N,P = and 135 N.Q = 
 
SOLUTION 
 
 
For :AA C′∆ tanA C a θ′ = 
( ) tanAy A C a θ′= − = − 
2cosA
ayδ δθθ= − 
For :BB C′∆ 
sinB C l A Cθ′ ′= − 
 sin tanl aθ θ= − 
sin tan
tan tan
B C l aBB θ θθ θ
′ −′ = = 
cosBx BB l aθ′= = − 
sinBx lδ θ δθ= − 
Virtual Work: 
 0: 0B AU P x Q yδ δ δ= − = 
 ( ) 2sin 0cos
aP l Qθ δθ δθθ
 − − − =   
 2sin cosPl Qaθ θ = 
or 2sin coslQ P
a
θ θ= 
With 900 mm, 150 mm, 75 N, and 135 Nl a P Q= = = = 
( ) 2900 mm135 N 75 N sin cos
150 mm
θ θ= 
or 2sin cos 0.300θ θ = 
Solving numerically, 19.81θ = ° and 51.9°W 
 
 
 
 
 
 
PROBLEM 10.29 
Two rods AC and CE are connected by a pin at C and by a spring AE. The 
constant of the spring is k, and the spring is unstretched when 30 .θ = ° 
For the loading shown, derive an equation in P, ,θ l, and k that must be 
satisfied when the system is in equilibrium. 
 
SOLUTION 
 
 
cosEy l θ= 
sinEy lδ θ δθ= − 
Spring: 
Unstretched length 2l= 
( )2 2 sin 4 sinx l lθ θ= = 
4 cosx lδ θ δθ= 
 ( )2F k x l= − 
( )4 sin 2F k l lθ= − 
Virtual Work: 
 0: 0EU P y F xδ δ δ= − = 
 ( ) ( )( )sin 4 sin 2 4 cos 0P l k l l lθ δθ θ θ δθ− − − = 
 ( )sin 8 2sin 1 cos 0P klθ θ θ− − − = 
or ( ) cos1 2sin
8 sin
P
kl
θθ θ= − 
1 2sin
8 tan
P
kl
θ
θ
−= W 
 
 
 
 
 
PROBLEM 10.30 
Two rods AC and CE are connected by a pin at C and by a spring AE. The 
constant of the spring is 300 N/m, and the spring is unstretched when 
30 .θ = ° Knowing that 200 mml = and neglecting the mass of the rods, 
determine the value of θ corresponding to equilibrium when 160 N.P = 
 
SOLUTION 
From the analysis of Problem 10.29, 1 2sin
8 tan
P
kl
θ
θ
−= 
Then with 160 N, 0.2 m, and 300 N/mP l k= = = 
( )( )
160 N 1 2sin
8 300 N/m 0.2 m tan
θ
θ
−= 
or 1 2sin 1 0.3333
tan 3
θ
θ
− = = 
Solving numerically, 
 24.98θ = ° 25.0θ = °W 
 
 
 
 PROBLEM 10.31 
Solve Problem 10.30 assuming that force P is moved to C and acts 
vertically downward. 
 
SOLUTION 
 cos , sinC Cy l y lθ δ θδθ= = − 
 Spring: 
 Unstretched length 2= l 
 ( )2 2 sin 4 sinθ θ= =x l l 
 4 cosδ θδθ=x l 
 ( )2= −F k x l 
 ( )4 sin 2θ= −F k l l 
Virtual Work: 
 0: CU P y F xδ δ δ= − − 
 ( ) ( )( )sin 4 sin 2 4 cos 0θδθ θ θδθ− − − − =P l k l l l 
 ( )sin 8 2sin 1 cos 0θ θ θ− − =P kl 
or ( ) cos2sin 1
8 sin
θθ θ= −
P
kl
 
With 200 mm, 300 N/m, and 160 Nl k P= = = 
( )
( )( ) ( )
160 N cos2sin 1
8 300 N/m 0.2 sin
θθ θ= − 
or ( ) cos 12sin 1
sin 3
θθ θ− = 
Solving numerically, 39.65θ = ° 
and 68.96θ = ° 
 39.7θ = °W 
 and 69.0θ = °W 
 
 
PROBLEM 10.32 
For the mechanism shown, block A can move freely in its guide and rests 
against a spring of constant 15 lb/in. that is undeformed when 45 .θ = ° 
For the loading shown, determine the value of θ corresponding to 
equilibrium. 
 
SOLUTION 
First note ( )10sin in.Dy θ= 
Then ( )10cos in.δ θδθ=Dy 
Also ( )2 12cos in.θ=Ax 
Then ( ) ( )0 24 in. cos 45Ax = ° 
and ( )24sin in.δ θδθ= −Ax 
With 0:δθ < Virtual Work: ( )0: 60 lb 0D SP AU y F xδ δ δ= − = 
where ( )0 = − SP A AF k x x 
 ( )( )( )15 lb/in. 24cos 24cos 45 in.θ= − ° 
 ( )( )360 lb cos cos 45θ= − ° 
Then ( )( ) ( ) ( )60 10cos 360 cos cos 45 24sin 0θδθ θ θδθ − − ° =  
or ( )5 72 tan cos cos 45 0θ θ− − ° = 
Solving numerically, 15.03 and 36.9θ θ= ° = °W 
 
 
 
 
 
 
 
 
PROBLEM 10.33 AND 10.34 
10.33: A force P of magnitude 150 lb is applied to the linkage at B. The 
constant of the spring is 12.5 lb/in., and the spring is unstretched when 
AB and BC are horizontal. Neglecting theweight of the linkage and 
knowing that 15 in.,l = determine the value of θ corresponding to 
equilibrium. 
10.34: A vertical force P is applied to the linkage at B. The constant of 
the spring is k, and the spring is unstretched when AB and BC are 
horizontal. Neglecting the weight of the linkage, derive an equation in ,θ 
P, l, and k that must be satisfied when the linkage is in equilibrium. 
 
SOLUTION 
 
 
 
 
 
 
2 cos 2 sinθ δ θδθ= = −C Cx l x l 
sin cosθ δ θδθ= =B By l y l 
( ) ( )2 2 1 cosθ= = − = −CF ks k l x kl 
Virtual Work: 0: 0C BU F x W yδ δ δ= + = 
 ( )( ) ( )2 1 cos 2 sin cos 0kl l W lθ θδθ θδθ− − + = 
 ( )24 1 cos sin cosθ θ θ− =kl Wl 
or ( )1 cos tan
4
θ θ− = W
kl
 
Problem 10.33: Given: 0.3 m, 600 N, 2500 N/m= = =l W k 
Then ( ) ( )( )
600 N1 cos tan
4 2500 N/m 0.3 m
θ θ− = 
or ( )1 cos tan 0.2θ θ− = 
Solving numerically, 40.22θ = ° 
 40.2θ = °W 
Problem 10.34: From above ( )1 cos tan
4
W
kl
θ θ− = W 
 
 
 
 
PROBLEM 10.35 
Knowing that the constant of spring CD is k and that the spring is 
unstretched when rod ABC is horizontal, determine the value of θ 
corresponding to equilibrium for the data indicated. 
150 lb, 30 in., 40 lb/in.P l k= = = 
 
SOLUTION 
 sinθ=Ay l 
 cosδ θδθ=Ay l 
 Spring: v CD= 
 Unstretched when 0θ = 
 so that 0 2v l= 
 For :θ 
 902 sin
2
v l θ° + =    
 cos 45
2
v l θδ δθ = ° +   
Stretched length: 
0 2 sin 45 22
s v v l lθ = − = ° + −   
Then 2sin 45 2
2
θ  = = ° + −    F ks kl 
Virtual Work: 
 0: 0AU P y F vδ δ δ= − = 
 cos 2sin 45 2 cos 45 0
2 2
θ θθδθ δθ    − ° + − ° + =        Pl kl l 
or 1 2sin 45 cos 45 2 cos 45
cos 2 2 2
θ θ θ
θ
      = ° + ° + − ° +            
P
kl
 
 1 2sin 45 cos 45 cos 2 cos 45
cos 2 2 2
θ θ θθθ
      = ° + ° + − ° +             
 
cos 45
21 2
cos
θ
θ
 ° +  = − 
 
PROBLEM 10.35 CONTINUED 
Now, with 150 lb, 30 in., and 40 lb/in.P l k= = = 
( )
( )( )
cos 45150 lb 21 2
40 lb/in. 30 in. cos
θ
θ
 ° +  = − 
or 
cos 45
2 0.61872
cos
θ
θ
 ° +   = 
Solving numerically, 17.825θ = ° 17.83θ = °W 
 
 
 
 
PROBLEM 10.36 
Knowing that the constant of spring CD is k and that the spring is 
unstretched when rod ABC is horizontal, determine the value of θ 
corresponding to equilibrium for the data indicated. 
600 N, 800 mm, 4 kN/m.P l k= = = 
 
SOLUTION 
From the analysis of Problem 10.35, we have 
cos 45
21 2
cos
P
kl
θ
θ
 ° +  = − 
With 600 N, 800 mm, and 4 kN/mP l k= = = 
( )
( )( )
cos 45600 N 21 2
4000 N/m 0.8 m cos
θ
θ
 ° +  = − 
or 
cos 45
2 0.57452
cos
θ
θ
 ° +   = 
Solving numerically, 30.98θ = ° 31.0θ = °W 
 
 
 
 
PROBLEM 10.37 
A horizontal force P of magnitude 160 N is applied to the mechanism at 
C. The constant of the spring is 1.8 kN/m,k = and the spring is 
unstretched when 0.θ = Neglecting the mass of the mechanism, 
determine the value of θ corresponding to equilibrium. 
 
SOLUTION 
 
 
 
 
 
 
 
Have s r s rθ δ δθ= = 
 F ks krθ= = 
and sinθ=Cx l 
cosδ θδθ=Cx l 
Virtual Work: 
 0: 0CU P x F sδ δ δ= − = 
 ( )cos 0θδθ θ δθ− =Pl kr r 
or 2 cos
θ
θ=
Pl
kr
 
( )( )
( )( )2
160 N 0.24 m
cos1800 N/m 0.1 m
θ
θ= 
2.1333
cos
θ
θ= 
Solving numerically, 1.054 rad 60.39θ = = ° 60.4θ = °W 
 
 
 
 
PROBLEM 10.38 
A cord is wrapped around drum A which is attached to member AB. 
Block D can move freely in its guide and is fastened to link CD. 
Neglecting the weight of AB and knowing that the spring is of constant 
4 lb/in. and is undeformed when 0,θ = determine the value of θ 
corresponding to equilibrium when a downward force P of magnitude 
96 lb is applied to the end of the cord. 
 
SOLUTION 
 
 
 
Have ( )15 tan in.Cy θ= 
Then ( )215sec in.Cyδ θδθ= 
Virtual Work: 0: 0P SP CU P s F yδ δ δ= − = 
where ( )3 in.Psδ δθ= 
and SP CF ky= 
 ( )( )4 lb/in. 15 in. tanθ= 
 ( )60 tan lbθ= 
Then ( )( ) ( ) ( )296 lb 3 in. 60 tan lb 15sec in. 0δθ θ θδθ  − =    
or 23.125 tan sec 1θ θ = 
Solving numerically, 16.41θ = °W 
 
 
 
 
 
PROBLEM 10.39 
The lever AB is attached to the horizontal shaft BC which passes through 
a bearing and is welded to a fixed support at C. The torsional spring 
constant of the shaft BC is K; that is, a couple of magnitude K is required 
to rotate end B through 1 rad. Knowing that the shaft is untwisted when 
AB is horizontal, determine the value of θ corresponding to the position 
of equilibrium when 400 lb,P = 10 in.,l = and 150 lb ft/rad.K = ⋅ 
 
SOLUTION 
 
 
 
 
 
Have sinθ=Ay l 
cosδ θδθ=Ay l 
Virtual Work: 
0: 0AU P y Mδ δ δθ= − = 
 cos 0θδθ θδθ− =Pl K 
or 
cos
θ
θ =
Pl
K
 (1) 
With 400 lb, 10 in., and 150 lb ft/radP l K= = = ⋅ 
( ) 10 in.400 lb
12 in./ft
cos 150 lb ft/rad
θ
θ
   = ⋅ 
or 2.2222
cos
θ
θ = 
Solving numerically, 61.25θ = ° 61.2θ = °W 
 
 
 
PROBLEM 10.40 
Solve Problem 10.39 assuming that 1.26 kips,P = 10 in.,l = and 
150 lb ft/radK = ⋅ . Obtain answers in each of the following quadrants: 
0 90 ,θ< < ° 270 360 ,θ° < < ° and 360 450 .θ° < < ° 
 
SOLUTION 
Using Equation (1) of Problem 10.39 and 
1.26 kip, 10 in., and 150 lb ft/radP l K= = = ⋅ 
we have 
( ) 10 in.1260 lb
12 in./ft
cos 150 lb ft/rad
θ
θ
   = ⋅ 
or 7 or 7cos
cos
θ θ θθ = = (1) 
The solutions to this equation can be shown graphically using any appropriate graphing tool, such as Maple, 
with the command: ( ){ }( )plot theta, 7 * cos theta , 0..5 * Pi/2 ;t = 
Thus, we plot and 7cos in the rangey yθ θ= = 
50
2
πθ≤ ≤ 
 
We observe that there are three points of intersection, which implies that Equation (1) has three roots in the 
specified range of .θ 
 0 90 ; 1.37333 rad, 78.69
2
πθ θ θ ≤ ≤ ° = = °   78.7θ = °W 
 3270 360 2 ; 5.65222 rad, 323.85
2
πθ θ π θ θ ≤ ≤ ° ≤ ≤ = = °   324θ = °W 
 5360 450 2 ; 6.61597 rad, 379.07
2
πθ π θ θ θ ≤ ≤ ° ≤ ≤ = =   379θ = °W 
 
 
 
 
 
 
PROBLEM 10.41 
The position of crank BCD is controlled by the hydraulic cylinder AB. 
For the loading shown, determine the force exerted by the hydraulic 
cylinder on pin B knowing that 60 .θ = ° 
 
SOLUTION 
 
 
 
 
Have 
2 2315 108 333 mmACd = + = 
108tan
315
φ = 
or 18.9246φ = ° 
Now, let α θ φ= + 
Then, by the Law of Cosines 
( )( )2 2333 150 2 333 150 cosABd α= + − 
or ( ) ( )213.3389 9.990cos 10 mmABd α= − × 
and 
( ) ( )
499.5sin mm
13.3389 9.990cos
ABd
αδ δαα= − 
With 0δα > 
Virtual Work: cyl0: 0D ABU P y F dδ δ δ= − = 
where 480 N,P = and 
D CDy dδ δα= 
Then 
( )( ) ( )cyl
499.5sin480 N 120 mm mm 0
13.3389 9.990cos
F αδα δαα
    − =  −   
 
or ( ) ( )3cyl499.5sin 57.6 10 13.3389 9.990cosFα α= × − 
With 60 : 60 18.9246θ α= ° = ° + ° 
 
 
 
 
 
PROBLEM 10.41 CONTINUED 
have 
 ( ) cyl499.5sin 60 18.9246 F ° + °  
 ( ) ( )357.6 10 13.3389 9.990cos 60 18.9246= × − ° + ° 
or cyl 397.08 NF = 
and 
100 13.3389 9.990cos78.9246 337.93 mmABd = − ° = 
Then, by the Law of Sines 
150 337.93
sin sin 78.9246β = ° 
or 25.824β = ° 
 cyl 397 N=F 44.7°W 
 
 
 
 
 
PROBLEM 10.42 
The position of crank BCD is controlled by the hydraulic cylinder AB. 
Determinethe angle θ knowing that the hydraulic cylinder exerts a 
420-N force on pin B when the crank is in the position shown. 
 
SOLUTION 
From Problem 10.41, we have 
( ) ( )3cyl499.5sin 57.6 10 13.3389 9.990cosFα α= × − 
Then, with cyl 420 NF = 
We have 
( ) ( )3499.5sin 420 57.6 10 13.3389 9.990cosα α= × − 
or 
( )23.64219sin 13.3389 9.990cosα α= − 
or 
( )213.2655 1 cos 13.3389 9.990cosα α− = − 
or 
213.2655cos 9.990cos 0.0734 0α α− + = 
Then 
( ) ( )( )
( )
29.990 9.990 4 13.2655 0.0734
cos
2 13.2655
α ± − −= 
or 
41.7841 and 89.5748α α= ° = ° 
Now and 18.9246θ α φ φ= − = ° 
so that 22.9 and 70.7θ θ= ° = °W 
 
 
 
 
 
PROBLEM 10.43 
For the linkage shown, determine the force P required for equilibrium 
when 40 N m.M = ⋅ 
 
SOLUTION 
 
 
 
 
For bar ABC, we have 
 where 375 mm
2
cy a
a
δδα = = 
and for bar CD, using the Law of Cosines 
2 2 2 2 cos55C D C Da L L L L= + − ° 
Then, noting that constant,a = we have 
( ) ( )0 2 2 2 cos55 2 cos55C C D D C D C DL L L L L L L Lδ δ δ δ= + − ° − ° 
Then, because :C CL yδ δ= − 
( ) ( )cos55 cos55C D C D C DL L y L L Lδ δ− ° = − ° 
For the given position of member CD, CDE∆ is isosceles. 
 and 2 cos55D CL a L a∴ = = ° 
Then 
( ) ( )22 cos55 cos55 2 cos 55C Da a y a a Lδ δ° − ° = − ° 
or 
2
cos55
1 2cos 55D C
L yδ δ°= − ° 
Now, Virtual Work: 0: 0DU M a P Lδ δ δ= − = 
or 
2
cos55 0
2 1 2cos 55
C
C
yM P y
a
δ δ°   − =   − °    
which gives 
21 2cos 55
2 cos55
MP
a
− °= ° 
Then ( )
240 N m 1 2cos 55
2 0.375 m cos55
P ⋅ − °= ° 
 or 31.8 N=P 35.0°W 
 
 
 
 
 
PROBLEM 10.44 
A cord is wrapped around a drum of radius a that is pinned at A. The 
constant of the spring is 3 kN/m, and the spring is unstretched when 
0.θ = Knowing that 150 mma = and neglecting the mass of the drum, 
determine the value of θ corresponding to equilibrium when a downward 
force P of magnitude 48 N is applied to the end of the cord. 
 
SOLUTION 
 
 
 
First note 
90θ β+ = ° 
90 α β α θ+ = ° ⇒ = 
 Length of cord unwound for rotation s aθ θ∴ = 
Now ( )0 1 cos , the distance moves down for rotation y a Oθ θ= − 
 P Oy y s= + 
( ) 1 cos is the distance moves down for rotation Py a a Pθ θ θ∴ = + − 
Then ( )sinPy a aδ θ δθ= + 
Now, by the Law of Cosines 
( ) ( ) ( )( )2 22 4 2 2 4 2 cosSPL a a a a θ= + − 
or 
2 5 4cosSPL a θ= − 
Then 
4sin2
2 5 4cosSP
L a θδ δθθ= − 
 4 sin
5 4cos
a θ δθθ= − 
Finally 
 ( )0SP SP SPF k L L = −  
 ( )2 5 4cos 2k a aθ= − − 
 ( )2 5 4cos 1ka θ= − − 
Thus, by Virtual Work: 0: 0P SP SPU P y F Lδ δ δ= − = 
 
 
 
PROBLEM 10.44 CONTINUED 
or 
( ) ( ) 4 sin1 sin 2 5 4cos 1 05 4cosaPa ka θθ δθ θ δθθ + − − − = −  
or 
( )1 sin sin 5 4cos sin 0
8
P
ka
θ θ θ θ + − − + =   
Substituting given values: 
( )( ) ( )
48 N 1 sin sin 5 4cos sin 0
8 3000 N/m 0.15 m
θ θ θ θ + − − + =   
 
or 
( )1 1 sin sin 5 4cos sin 0
75
θ θ θ θ + − − + =   
Solving numerically, 
 15.27θ = °W 
 
 
 
PROBLEM 10.45 
The telescoping arm ABC is used to provide an elevated platform for 
construction workers. The workers and the platform together have a mass 
of 204 kg, and their combined center of gravity is located directly 
above C. For the position when o20 ,θ = determine the force exerted on 
pin B by the single hydraulic cylinder BD. 
 
SOLUTION 
 
 
 
 
 
 
 
In :ADE∆ 
0.9 mtan
0.5 m
AE
DE
α = = 
 60.945α = ° 
0.9 m 1.0296 m
sin 60.945
AD = =° 
From the geometry: 
( ) ( )5 m sin , 5 m cosC Cy yθ δ θδθ= = 
Then, in triangle BAD: Angle BAD α θ= + 
Law of Cosines: 
( )( ) ( )2 2 2 2 cosBD AB AD AB AD α θ= + − + 
or ( ) ( ) ( )( ) ( )2 22 2.4 m 1.0296 m 2 2.4 m 1.0296 m cosBD α θ= + − + 
 ( )( )2 2 26.82 m 4.942cos mBD α θ= − + (1) 
 
 
 
PROBLEM 10.45 CONTINUED 
And then 
 ( )( ) ( )( )2 4.942sinBD BDδ α θ δθ= + 
 ( )( )
4.942sin
2
BD
BD
α θδ δθ+= 
Virtual work: 0: 0C BDU P y F BDδ δ δ= − + = Substituting ( )( )2000 N 5 m cos BFθδθ− +
or ( )
cos4047 N/m
sinBD
F BDθα θ
 =  +  
 (2) 
Now, with 20θ = ° and 60.945α = ° 
Equation (1): 
( )2 6.82 4.942cos 60.945 20BD = − ° + ° 
 2 6.042BD = 
 2.46 mBD = 
Equation (2) 
( ) ( )
cos 204047 2.46 m N/m
sin 60.945 20BD
F
 °=  ° + °  
 
or 9473 NBDF = 9.47 kNBD =F W 
 
 
 
PROBLEM 10.46 
Solve Problem 10.45 assuming that the workers are lowered to a point 
near the ground so that o20 .θ = − 
 
SOLUTION 
Using the figure and analysis of Problem 10.45, including Equations (1) and (2), and with 20 ,θ = − ° we have 
Equation (1): ( )2 6.82 4.942cos 60.945 20BD = − ° − ° 
 2 3.087BD = 
 1.757 mBD = 
Equation (2): ( )( ) ( )
cos 20
4047 1.757
sin 60.945 20BD
F
− °= ° − ° 
 10196 NBDF = 
or 10.20 kNBD =F W 
 
 
 
 PROBLEM 10.47 
A block of weight W is pulled up a plane forming an angle α with the 
horizontal by a force P directed along the plane. If µ is the coefficient of 
friction between the block and the plane, derive an expression for the 
mechanical efficiency of the system. Show that the mechanical efficiency 
cannot exceed 12 if the block is to remain in place when the force P is 
removed. 
 
SOLUTION 
 
 
 
 
 Input work P xδ= 
( )Output work sinW xα δ= 
Efficiency: 
 sin sinorW x W
P x P
αδ αη ηδ= = (1) 
 0: sin 0 or sinxF P F W P W Fα αΣ = − − = = + (2) 
 0: cos 0 or cosyF N W N Wα αΣ = − = = 
cosF N Wµ µ α= = 
Equation (2): ( )sin cos sin cosP W W Wα µ α α µ α= + = + 
Equation (1): ( )
sin 1or
sin cos 1 cot
W
W
αη ηα µ α µ α= =+ + W 
If block is to remain in place when 0,P = we know (see page 416) that 
sφ α≥ or, since 
tan , tansµ φ µ α= ≥ 
Multiply by cot :α cot tan cot 1µ α α α≥ = 
Add 1 to each side: 1 cot 2µ α+ ≥ 
Recalling the expression for ,η we find 1
2
η ≤ W 
 
 
 
 
 
PROBLEM 10.48 
Denoting by sµ the coefficient of static friction between collar C and the 
vertical rod, derive an expression for the magnitude of the largest couple 
M for which equilibrium is maintained in the position shown. Explain 
what happens if tansµ θ≥ . 
 
SOLUTION 
 
 
 
 
Member BC: Have cosBx l θ= 
 sinBx lδ θδθ= − (1) 
 and sinCy l θ= 
 cosCy lδ θδθ= (2) 
Member AB: Have 1
2B
x lδ δφ= 
Substituting from Equation (1), 
1sin
2
l lθδθ δφ− = 
or 2sinδφ θδθ= − (3) 
Free body of rod BC 
For max,M motion of collar C impends upward 
 ( )( )0: sin cos 0B sM Nl P N lθ µ θΣ = − + = 
tan sN N Pθ µ− = 
tan s
PN θ µ= − 
Virtual Work 
 ( )0: 0s CU M P N yδ δφ µ δ= + + = 
 ( ) ( )2sin cos 0sM P N lθδθ µ θδθ− + + = 
( )
max
tan
2 tan 2 tan
s
s s
PP
P N
M l l
µµ θ µ
θ θ
++ −= = 
or ( )max 2 tan s
PlM θ µ= − W 
If tan , ,s Mµ θ= = ∞ system becomes self-locking 
 
 
 
PROBLEM 10.49 
Knowing that the coefficient of static friction between collar C and the 
vertical rod is 0.40, determine the magnitude of the largest and smallest 
couple M for which equilibrium is maintained in the position shown 
when o35 ,θ = 30 in.,l = and 1.2 kips.P = 
 
SOLUTION 
From the analysis of Problem 10.48, we have ( )max 2 tan s
PlM θ µ= + 
With 35 , 30 in., 1.25 kipsl Pθ = ° = = 
( )( )
( )max
1200 lb 30 in.
59,958.5 lb in.
2 tan 35 0.4
M = = ⋅° − 
 4996.5 lb ft= ⋅ 
 4.9965 kip ft= ⋅ 
 max 5.00 kip ftM = ⋅ W 
For min,M motion of C impends downward and F acts upward. The equationsof Problem 10.48 can still be 
used if we replace sµ by .sµ− Then 
( )min 2 tan s
PlM θ µ= + 
Substituting, ( )( )( )min
1200 lb 30 in.
16,360.5 lb in.
2 tan 35 0.4
M = = ⋅° + 
 1363.4 lb ft= ⋅ 
 1.3634 kip ft= ⋅ 
 min 1.363 kip ftM = ⋅ W 
 
 
 
 PROBLEM 10.50 
Derive an expression for the mechanical efficiency of the jack discussed 
in Section 8.6. Show that if the jack is to be self-locking, the mechanical 
efficiency cannot exceed 12 . 
 
SOLUTION 
 
 
 
Recall Figure 8.9a. Draw force triangle 
 
( )tan sQ W θ φ= + 
tan so that tany x y xθ δ δ θ= = 
( )Input work tan sQ x W xδ θ φ δ= = + 
( )Output work tanW y W xδ δ θ= = 
Efficiency: ( )
tan ;
tan s
W x
W x
θδη θ φ δ= + ( )
tan
tan s
θη θ φ= + W 
From page 432, we know the jack is self-locking if 
sφ θ≥ 
Then 2sθ φ θ+ ≥ 
so that ( )tan tan 2sθ φ θ+ ≥ 
From above ( )
tan
tan s
θη θ φ= + 
It then follows that tan
tan 2
θη θ≤ 
But 2
2 tantan 2
1 tan
θθ θ= − 
Then 
( )2 2tan 1 tan 1 tan
2 tan 2
θ θ θη θ
− −≤ = 1 
2
η∴ ≤ W 
 
 
 
 
PROBLEM 10.51 
Denoting by sµ the coefficient of static friction between the block 
attached to rod ACE and the horizontal surface, derive expressions in 
terms of P, ,sµ and θ for the largest and smallest magnitudes of the 
force Q for which equilibrium is maintained. 
 
SOLUTION 
 
 
 
 
 
 
For the linkage: 
 0: 0 or2 2
A
B A
x PM x PΣ = − + = =A 
Then: 1
2 2s s s
PF A Pµ µ µ= = = 
Now 2 sinAx l θ= 
2 cosAx lδ θ δθ= 
and 3 cosFy l θ= 
3 sinFy lδ θ δθ= − 
Virtual Work: 
 ( )max0: 0A FU Q F x P yδ δ δ= − + = 
 ( ) ( )max 1 2 cos 3 sin 02 sQ P l P lµ θ δθ θ δθ − + − =   
or max
3 1tan
2 2 s
Q P Pθ µ= + 
 ( )max 3tan2 s
PQ θ µ= + W 
For min,Q motion of A impends to the right and F acts to the left. We 
change sµ to sµ− and find 
 ( )min 3tan2 s
PQ θ µ= − W 
 
 
 
 
 
 
PROBLEM 10.52 
Knowing that the coefficient of static friction between the block attached 
to rod ACE and the horizontal surface is 0.15, determine the magnitudes 
of the largest and smallest force Q for which equilibrium is maintained 
when o30θ = , 8 in.,l = and 160 lb.P = 
 
SOLUTION 
Using the results of Problem 10.52 with 
30 , 8 in., 160 lb, and 0.15sl Pθ µ= ° = = = 
We have ( ) ( )max 160 lb 3tan 30 0.15 150.56 lb2Q = ° + = 
 max 150.6 lbQ = W 
and ( ) ( )min 160 lb 3tan 30 0.15 126.56 lb2Q = ° − = 
 min 126.6 lbQ = W 
 
 
 
 
 
PROBLEM 10.53 
Using the method of virtual work, determine separately the force and the 
couple representing the reaction at A. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
:yA Consider an upward displacement Ayδ of ABC 
ABC: A B Cy y yδ δ δ= = 
CDE: 
1 ft 2.5 ft
C Ey yδ δ= 
or 2.5E Ay yδ δ= 
EFG: 
0.8 ft 1.2 ft
E Gy yδ δ= 
or ( )1.2 ft 2.5
0.8 ftG A
y yδ δ= 
 3.75 Ayδ= 
Virtual Work: 
( ) ( )0: 240 lb 60 lb 0y A B GU A y y yδ δ δ δ= + − = 
or ( ) ( )240 lb 60 lb 3.75 0y A A AA y y yδ δ δ+ − = 
or 15 lby =A 
:xA Consider a horizontal displacement Axδ : 
Virtual Work: 0: 0x AU A xδ δ= = 
or 0xA = 15.00 lb∴ =A W 
:AM Consider a counterclockwise rotation about A: 
ABC: 2 , 3B A C Ay yδ δθ δ δθ= = 
CDE: 
1 ft 2.5 ft
C Ey yδ δ= 
or ( )2.5 3E Ayδ δθ= 
 7.5 Aδθ= 
EFG: 
0.8 ft 1.2 ft
E Gy yδ δ= 
 
 
 
 
PROBLEM 10.53 CONTINUED 
or ( )( ) ( )
1.2 ft
7.5
0.8 ftG A
yδ δθ= 
 11.25 Aδθ= 
Virtual Work: ( ) ( )0: 240 lb 60 lb 0A A B GU M y yδ δθ δ δ= + − = 
or ( )( ) ( )( )240 lb 2 60 lb 11.25 0A A A AM δθ δθ δθ+ − = 
 or 195.0 lb ftA = ⋅M W
 
 
 
 
 
PROBLEM 10.54 
Using the method of virtual work, determine the reaction at D. 
 
SOLUTION 
 
 
 
 
 
 
 
Consider an upward displacement Eyδ of pin E. 
CDE: 
1 ft 3.5 ft
D Ey yδ δ= 
or 1
3.5D
y Eδ δ= 
EFG: 
0.8 ft 1.2 ft
E Gy yδ δ= 
or 1.5G Ey yδ δ= 
Virtual Work: 
 0: 60 0D GU D y yδ δ δ= + = 
or ( )( )1 60 lb 1.5 0
3.5 E E
D y yδ δ  + =   
or 315 lb=D W 
 
 
 
 PROBLEM 10.55 
Referring to Problem 10.41 and using the value found for the force 
exerted by the hydraulic cylinder AB, determine the change in the length 
of AB required to raise the 480-N load 18 mm. 
 
SOLUTION 
From the solution to Problem 10.41 
cyl 397.08 NF = 
And, Virtual Work: cyl0: 0AB DU F S P yδ δ δ= − = 
where 0ABSδ < for 0Dyδ > 
Then ( ) ( )( )397.08 N 480 N 18 mm 0ABSδ − = 
 ( )or 21.8 mm shortenedABSδ = W 
 
 
 PROBLEM 10.56 
Referring to Problem 10.45 and using the value found for the force 
exerted by the hydraulic cylinder BD, determine the change in the length 
of BD required to raise the platform attached at C by 50 mm. 
 
SOLUTION 
 
 
 
 
 
Virtual Work: Assume that both Cyδ and BDδ increase 
 ( )0: 2000 N 0C BD BDU y Fδ δ δ= − + = 
( )0.05 m and 9473 N from Problem 10.45C BDy Fδ = = 
( )2000 0.05 m 9473 0BDδ− + = 
0.010556 mBDδ = 
 10.556 mm= 
The positive sign indicates that BD gets longer. 
 10.56 mmBDδ = W 
 
 
 
 
 
PROBLEM 10.57 
Determine the vertical movement of joint D if the length of member BF is 
increased by 75 mm. (Hint: Apply a vertical load at joint D, and, using 
the methods of Chapter 6, compute the force exerted by member BF on 
joints B and F. Then apply the method of virtual work for a virtual 
displacement resulting in the specified increase in length of member BF. 
This method should be used only for small changes in the lengths of 
members.) 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
Apply vertical load P at D. 
 ( ) ( )0: 12 m 36 m 0HM P EΣ = − + = 
3
P=E 
 
30: 0
5 3y BF
PF FΣ = − = 
 5
9BF
F P= 
Virtual Work: 
We remove member BF and replace it with forces BFF and BF−F at pins 
F and B, respectively. Denoting the virtual displacements of points B and 
F as Bδ r and ,Fδ r respectively, and noting that P and Dδ
JJJG
 have the 
same direction, we have 
Virtual Work: ( )0: 0BF F BF BU P Dδ δ δ δ= + ⋅ + − ⋅ =F r F r 
 cos cos 0BF F F BF B BP D F r F rδ δ θ δ θ+ − = 
 ( )cos cos 0BF B B F FP D F r rδ δ θ δ θ− − = 
where ( )cos cos ,B B F F BFr rδ θ δ θ δ− = which is the change in length of 
member BF. Thus, 
0BF BFP D Fδ δ− = 
( )5 75 mm 0
9
P D Pδ  − =   
41.67 mmDδ = + 
 41.7 mmDδ = W 
 
 
 
 
PROBLEM 10.58 
Determine the horizontal movement of joint D if the length of member 
BF is increased by 75 mm. (See the hint for Problem 10.57.) 
 
SOLUTION 
 
 
 
 
 
 
 
 
Apply horizontal load P at D. 
 ( ) ( )0: 9 m 36 m 0H yM P EΣ = − = 
4y
PE = 
 
30: 0
5 4y BF
PF FΣ = − = 
 5
12BF
F P= 
We remove member BF and replace it with forces BFF and BF−F at pins 
F and B, respectively. Denoting the virtual displacements of points B and 
F as Bδ r and ,Fδ r respectively, and noting that P and Dδ
JJJG
 have the 
same direction, we have 
Virtual Work: ( )0: 0BF F BF BU P Dδ δ δ δ= + ⋅ + − ⋅ =F r F r 
 cos cos 0BF F F BF B BP D F r F rδ δ θ δ θ+ − = 
 ( )cos cos 0BF B B F FP D F r rδ δ θ δ θ− − = 
where ( )cos cos ,B B F F BFr rδ θ δ θ δ− = which is the change in length of 
member BF. Thus, 
0BF BFP D Fδ δ− = 
( )5 75 mm 0
9
P D Pδ  − =   
 31.25 mmDδ = 31.3 mmDδ = W 
 
 
 
 
 PROBLEM 10.59 
Using the method of Section 10.8, solve Problem 10.29. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
Spring: 
( )2 2 sin 4 sinAE x llθ θ= = = 
Unstretched length: 
0 4 sin 30 2x l l= ° = 
Deflection of spring 
 0s x x= − 
( )2 2sin 1s l θ= − 
21
2 E
V ks Py= + 
( ) ( )21 2 2sin 1 cos
2
V k l P lθ θ = − + −  
( )24 2sin 1 2cos sin 0dV kl Pl
d
θ θ θθ = − + = 
( ) cos2sin 1 0
sin 8
P
kl
θθ θ− + = 
 1 2sin
8 tan
P
kl
θ
θ
−= W 
 
 
 PROBLEM 10.60 
Using the method of Section 10.8, solve Problem 10.30. 
 
SOLUTION 
Using the result of Problem 10.59, with 
160 N, 200 mm, and 300 N/mP l k= = = 
1 2sin
8 tan
P
kl
θ
θ
−= 
or ( )( )
1 2sin 160 N
tan 8 300 N/m 0.2 m
θ
θ
− = 
 1
3
= 
Solving numerically, 25.0θ = °W 
 
 
 
 
 
 
PROBLEM 10.61 
Using the method of Section 10.8, solve Problem 10.31. 
 
SOLUTION 
 
Spring: ( )2 2 sin 4 sinθ θ= = =AE x l l 
Unstretched length: 0 4 sin 30 2= ° =x l l 
Deflection of spring 0s x x= − 
 ( )2 2sin 1s l θ= − 
 21
2 C
V ks Py= + 
 ( ) ( )21 2 2sin 1 cos
2
θ θ = − + k l P l 
 ( )222 2sin 1 cosθ θ= − +V kl Pl 
( )24 2sin 1 2cos sin 0θ θ θθ = − − =
dV kl Pl
d
 
( ) cos1 2sin 0
sin 8
θθ θ− + =
P
kl
 
2sin 1
8 tan
θ
θ
−=P
kl
 
 
 
PROBLEM 10.61 CONTINUED 
With 160 N, 200 mm, and 300 N/mP l k= = = 
Have ( )( )( )
160 N 2sin 1
8 300 N/m 0.2 m tan
θ
θ
−= 
or 2sin 1 1
tan 3
θ
θ
− = 
Solving numerically, 39.65 and 68.96θ = ° ° 
 
39.7
69.0
θ
θ
= °
= °W 
 
 
 
 
 
PROBLEMS 10.62 AND 10.63 
10.62: Using the method of Section 10.8, solve Problem 10.33. 
10.63: Using the method of Section 10.8, solve Problem 10.34. 
 
SOLUTION 
 
Problem 10.62 
Have 150 lb, 15 in., and 12.5 lb/in.P l k= = = 
Then ( ) ( )( )
150 lb1 cos tan
4 12.5 lb/in. 15 in.
θ θ− = 
 0.2= 
Solving numerically, 40.2θ = °W 
Problem 10.63 21
2 B
V ks Py= + 
( )21 2
2
= − +C BV k l x Py 
2 cos and sinθ θ= = −C Bx l y l 
Thus, ( )21 2 2 cos sin
2
θ θ= − −V k l l Pl 
 ( )222 1 cos sinθ θ= − −kl Pl 
( )22 2 1 cos sin cos 0θ θ θθ = − − =
dV kl Pl
d
 
or ( )1 cos tan
4
θ θ− = P
kl
W 
 
 
 
 
PROBLEM 10.64 
Using the method of Section 10.8, solve Problem 10.35. 
 
SOLUTION 
 
Spring 902 sin
2
v l θ° + =    
2 sin 45
2
v l θ = ° +   
Unstretched ( )0θ = 
0 2 sin 45 2v l l= ° = 
Deflection of spring 0 2 sin 45 22
s v v l lθ = − = ° + −   
( )
2
2 21 1 2sin 45 2 sin
2 2 2A
V ks Py kl P lθ θ  = + = ° + − + −     
2 2sin 45 2 cos 45 cos 0
2 2
θ θ θθ
    = ° + − ° + − =        
dV kl Pl
d
 
2sin 45 cos 45 2 cos 45 cos
2 2 2
θ θ θ θ      ° + ° + − ° + =            
P
kl
 
cos 2 cos 45 cos
2
P
kl
θθ θ − ° + =   
Divide each member by cosθ 
cos 45
21 2
cos
P
kl
θ
θ
 ° +  − = 
 
 
PROBLEM 10.64 CONTINUED 
Then with 150 lb, 30 in. and 40 lb/in.P l k= = = 
( )( )
cos 45
2 150 lb1 2
cos 40 lb/in. 30 in.
θ
θ
 ° +  − = 
 0.125= 
or 
cos 45
2 0.618718
cos
θ
θ
 ° +   = 
Solving numerically, 17.83θ = °W 
 
 
 
 
 
PROBLEM 10.65 
Using the method of Section 10.8, solve Problem 10.36. 
 
SOLUTION 
Using the results of Problem 10.64 with 600 N, 800 mm, and 4 kN/m= = =P l k , have 
 
cos 45
21 2
cos
P
kl
θ
θ
 ° +  − = 
 ( )( )
600 N
4000 N/m 0.8 m
= 
 0.1875= 
or 
cos 45
2 0.57452
cos
θ
θ
 ° +   = 
Solving numerically, 30.985θ = ° 31.0θ = °W 
 
 
 
 
 
PROBLEM 10.66 
Using the method of Section 10.8, solve Problem 10.38. 
 
SOLUTION 
 
 
 
Spring 21
2SP C
V ky= 
where tan 15 in.C AC ACy d dθ= = 
2 21 tan
2
θ∴ =SP ACV kd 
Force :P = −P PV Py 
where 3 in.Py r rθ= = 
 θ∴ = −PV Pr 
Then = +SP PV V V 
 2 21 tan
2
θ θ= −ACkd Pr 
Equilibrium 2 20: tan sec 0θ θθ = − =AC
dV kd Pr
d
 
or ( )( ) ( )( )2 24 lb/in. 15 in. tan sec 96 lb 3 in. 0θ θ − = 
or 23.125 tan sec 1 0θ θ − = 
Solving numerically, 16.4079θ = ° 16.41θ = °W 
 
 
 
 
PROBLEM 10.67 
Show that the equilibrium is neutral in Problem 10.1. 
 
SOLUTION 
 
We have =Ay u 
 4.5= −Dy u 
 2.5=Gy u 
Have ( ) ( ) ( )300 N 100 N 0= + + =A D EV y y P y 
( ) ( )300 100 4.5 2.5 0= + − + =V u u P u 
 ( )150 2.5= − +V P u 
150 2.5 0 so that 60 NdV P P
du
= − + = = 
Substitute 60 N=P in expression for V: ( )150 2.5 60V u = − +  
 0= 
 ∴ V is constant and equilibrium is neutral W 
 
 
 
 
 
PROBLEM 10.68 
Show that the equilibrium is neutral in Problem 10.2. 
 
SOLUTION 
 
Consider a small disturbance of the system so that 1θ � 
Have , 5 15θ φ= �C Dx x 
or 
3
θφ = 
Potential energy θ= − +E GV M Qx Py 
where ( )10 in. φ=Ex 
 10 in.
3
θ =    
and ( )4 2 in. cos 45φ = ° Gy 
 
 
PROBLEM 10.68 CONTINUED 
Then 10 4
3 3
θ θ θ= − +V M Q P 
 10 4
3 3
θ = + +  M Q P 
and 10 4
3 3θ = − +
dV M Q P
d
 
For equilibrium 10 40: 0
3 3
dV M Q P
dθ = − + = 
∴ At equilibrium, 0,=V a constant, for all values of .θ 
Hence, equilibrium is neutral 
 Q.E.D.W 
 
 
 
 
 
PROBLEM 10.69 
Two identical uniform rods, each of weight W and length L, are attached 
to pulleys that are connected by a belt as shown. Assuming that no 
slipping occurs between the belt and the pulleys, determine the positions 
of equilibrium of the system and state in each case whether the 
equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
Let each rod be of length L and weight W. Then the potential energy V is 
sin cos 2
2 2
θ θ   = +      
L LV W W 
Then 
cos sin 2
2
θ θθ = −
dV W L WL
d
 
For equilibrium 
0: cos sin 2 0
2
θ θθ = − =
dV W L WL
d
 
or cos 2sin 2 0θ θ− = 
Solving numerically or using a computer algebra system, such as Maple, gives four solutions: 
 1.570796327 rad 90.0θ = = ° 
 1.570796327 rad 270θ = − = ° 
 0.2526802551 rad 14.4775θ = = ° 
 2.888912399 rad 165.522θ = = ° 
Now 
2
2
1 sin 2 cos 2
2
θ θθ = − −
d V WL WL
d
 
 1 sin 2cos 2
2
θ θ = − +  WL 
 
PROBLEM 10.69 CONTINUED 
At 14.4775θ = ° 
( )2 2 1 sin14.4775 2cos 2 14.47752θ
  = − ° + °   
d V WL
d
 
 ( )1.875 0WL= − < 14.48 , Unstableθ∴ = ° W 
At 90θ = ° 
2
2
1 sin 90 2cos180
2θ
 = − ° + °  
d V WL
d
 
 ( )1.5 0= >WL 90 , Stableθ∴ = ° W 
At 165.522θ = ° 
( )2 2 1 sin165.522 2cos 2 165.5222θ
 = − ° + × °  
d V WL
d
 
 ( )1.875 0= − <WL 165.5 , Unstableθ∴ = ° W 
At 270θ = ° 
2
2
1 sin 270 2cos540
2θ
 = − ° + °  
d V WL
d
 
 ( )2.5 0= >WL 270 , Stableθ∴ = ° W 
 
 
 
 
 
PROBLEM 10.70 
Two uniform rods, each of mass m and length l, are attached to gears as 
shown. For the range o0 180θ≤ ≤ , determine the positions of 
equilibrium of the system and state in each case whether the equilibrium 
is stable, unstable, or neutral. 
 
SOLUTION 
 
Potential energy cos1.5 cos
2 2
l lV W W W mgθ θ   = + =       
 ( ) ( )1.5sin1.5 sin
2 2
θ θθ = − + −
dV Wl Wl
d
 
 ( )1.5sin1.5 sin
2
θ θ= − +Wl 
( )2 2 2.25cos1.5 cos2 θ θθ = − +
d V Wl
d
 
For equilibrium 0: 1.5sin1.5 sin 0θ θθ = + =
dV
d
 
Solutions: One solution, by inspection, is 0,θ = and a second angle less than 180° can be found numerically: 
 2.4042 rad 137.8θ = = ° 
Now ( )2 2 2.25cos1.5 cos2
d V Wld
θ θθ = − + 
 
 
PROBLEM 10.70 CONTINUED 
At 0:θ = ( )2 2 2.25cos0 cos02
d V Wl
dθ = − ° + ° 
 ( ) ( )3.25 0
2
Wl= − < 0,θ∴ = Unstable W 
At 137.8 :θ = ° ( )2 2 2.25cos 1.5 137.8 cos137.82
d V Wl
dθ  = − × ° + °  
 ( ) ( )2.75 0
2
Wl= > 137.8 ,θ∴ = ° StableW 
 
 
 
PROBLEM 10.71 
Two uniform rods, each of mass m, are attached to gears of equal radii as 
shown. Determine the positions of equilibrium of the system and state in 
each case whether the equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
 Potential Energy 
 sin cos
2 2
l lV W W W mgθ θ   = − + =       
 ( )cos sin
2
lW θ θ= − 
 ( )sin cos
2
dV Wl
d
θ θθ = − − 
 ( )2 2 sin cos2
d V Wl
d
θ θθ = − 
For Equilibrium: 0: sin cosdV
d
θ θθ = = − 
or tan 1θ = − 
Thus 45.0 and 135.0θ θ= − ° = ° 
Stability: 
At 45.0 :θ = − ° ( )2 2 sin 45 cos 452
d V Wl
dθ  = − ° − °  
 2 2 0
2 2 2
Wl  = − − <   
 
 45.0 ,θ∴ = − ° UnstableW 
At 135.0 :θ = ° ( )2 2 sin135 cos1352
d V Wl
dθ = ° − ° 
 2 2 0
2 2 2
Wl  = + >   
 
 135.0 ,θ∴ = ° StableW 
 
 
PROBLEM 10.72 
Two uniform rods, AB and CD, are attached to gears of equal radii as 
shown. Knowing that 3.5 kgABm = and 1.75 kg,CDm = determine the 
positions of equilibrium of the system and state in each case whether the 
equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
 Potential Energy 
 ( ) ( )2 23.5 kg 9.81 m/s sin 1.75 kg 9.81 m/s cos2 2l lV θ θ   = × − + ×       
 ( ) ( )8.5838 N 2sin cosl θ θ= − + 
 ( ) ( )8.5838 N 2cos sindV l
d
θ θθ = − − 
 ( ) ( )2 2 8.5838 N 2sin cosd V ld θ θθ = − 
Equilibrium: 0: 2cos sin 0dV
d
θ θθ = − − = 
or tan 2θ = − 
Thus 63.4 and 116.6θ = − ° ° 
Stability 
At 63.4 :θ = − ° ( ) ( ) ( )2 2 8.5838 N 2sin 63.4 cos 63.4d V ldθ  = − ° − − °  
 ( ) ( )8.5838 N 1.788 0.448 0l= − − < 
 63.4 ,θ∴ = − ° UnstableW 
At 116.6 :θ = ° ( ) ( ) ( )2 2 8.5838 N 2sin 116.6 cos 116.6d V ldθ  = ° − °  
 ( ) ( )8.5838 N 1.788 0.447 0l= + > 
 116.6 ,θ∴ = ° StableW 
 
 
 
PROBLEM 10.73 
Using the method of Section 10.8, solve Problem 10.39. Determine 
whether the equilibrium is stable, unstable or neutral. (Hint: The 
potential energy corresponding to the couple exerted by a torsional 
spring is 21 ,
2
Kθ where K is the torsional spring constant and θ is the 
angle of twist.) 
 
SOLUTION 
 Potential Energy 
 21 sin
2
V K Plθ θ= − 
 cosdV K Pl
d
θ θθ = − 
 
2
2 sin
d V K Pl
d
θθ = + 
Equilibrium: 0: cosdV K
d Pl
θ θθ = = 
For 400 lb, 10 in., 150 lb ft/radP l K= = = ⋅ 
( )
150 lb ft/radcos
10400 lb ft
12
θ θ⋅=    
 
 0.450θ= 
Solving numerically, we obtain 
1.06896 rad 61.247θ = = ° 
61.2θ = °W 
Stability 
( ) ( )2 2 10150 lb ft/rad 400 lb ft sin 61.2 012
d V
dθ
 = ⋅ + ° >   
 Stable∴ W 
 
 
 
PROBLEM 10.74 
In Problem 10.40, determine whether each of the positions of equilibrium 
is stable, unstable, of neutral. (See the hint for Problem 10.73.) 
 
SOLUTION 
 Potential Energy 
 21 sin
2
V K Plθ θ= − 
 cosdV K Pl
d
θ θθ = − 
 
2
2 sin
d V K Pl
d
θθ = + 
Equilibrium 0: cosdV K
d Pl
θ θθ = = 
For 1260 lb, 10 in., and 150 lb ft/radP l K= = = ⋅ 
( )
150 lb ft/radcos
101260 lb ft
12
θ θ⋅=    
 
or cos
7
θθ = 
Solving numerically, 1.37333 rad, 5.652 rad, and 6.616 radθ = 
or 78.7 , 323.8 , 379.1θ = ° ° ° 
Stability At 78.7 :θ = ° ( ) ( )2 2 10 150 lb ft /rad 1260 lb ft sin 78.712
d V
dθ
 = ⋅ + °   
 1179.6 ft lb 0= ⋅ > 78.7 ,θ∴ = ° StableW 
 At 323.8 :θ = ° ( ) ( )2 2 10150 lb ft/rad 1260 lb ft sin 323.812
d V
dθ
 = ⋅ + °   
 470 ft lb 0= − ⋅ < 324 ,θ∴ = ° UnstableW 
 At 379.1 :θ = ° ( ) ( )2 2 10 150 lb ft/rad 1260 lb ft sin 379.112
d V
dθ
 = ⋅ + °   
 493.5 ft lb 0= ⋅ > 379 ,θ∴ = ° StableW 
 
 
PROBLEM 10.75 
Angle θ is equal to 45° after a block of mass m is hung from member AB 
as shown. Neglecting the mass of AB and knowing that the spring is 
unstretched when 20 ,θ = ° determine the value of m and state whether 
the equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
 Potential Energy 
 Have 21
2 SP B
V kx mgy= + 
 where ( )0 0, 100 mm, 20 rad9SPx r r
πθ θ θ= − = = ° = 
 cos , 450 mmθ= =B AB ABy L L 
Then ( )22 01 cos2 θ θ θ= − + ABV kr mgL 
and ( )2 0 sinθ θ θθ = − − AB
dV kr mgL
d
 
2
2
2 cosθθ = − AB
d V kr mgL
d
 
With 800 N/m, 45θ= = °k 
Equilibrium: ( )( ) ( )( )2 20: 800 N/m 0.1 m 9.81 m/s 0.45 m sin 04 9 4dV md π π πθ  = − − =   
Then 1.11825 kgm = 1.118 kgm = W 
Stability 
Now ( )( ) ( )( )( )2 2 22 800 N/m 0.1 m 1.118 kg 9.81 m/s 0.45 m cos 4d Vd πθ = − 
 4.51 J 0= > 
 Stable∴ W 
 
 
 
 
PROBLEM 10.76 
A block of mass m is hung from member AB as shown. Neglecting the 
mass of AB and knowing that the spring is unstretched when 20 ,θ = ° 
determine the value of θ corresponding to equilibrium when 3 kg.m = 
State whether the equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
Using the general results of Problem 10.76 and noting that now 
 03 kg, and 20m θ= = ° 
we have 
Equilibrium ( )2 00: sin 0ABdV kr mgLd θ θ θθ = − − = 
 ( )( ) ( )( )( )2 2800 N/m 0.1 m 3 kg 9.81 m/s 0.45 m sin 09πθ θ − − =   
or 1.65544sin 0
9
πθ θ − − =   
Solving numerically, 1.91011 radθ = 
 109.441= ° 
 or 109.4θ = °W 
Stability 
 
2
2
2 cosAB
d v kr mgL
d
θθ = − 
 ( )( ) ( )( )( ) ( )2800 N/m 0.1 m 3 kg 9.81 m/s 0.45 m cos 109.4= − ° 
 12.41 J 0= > 
 Stable∴ W 
 
 
 
 
 
PROBLEM 10.77 
A slender rod AB, of mass m, is attached to two blocks A and B which can 
move freely in the guides shown. Knowing that the spring is unstretched 
when 0y = , determine the value of y corresponding to equilibrium 
when 12 kg, 750 mm, and 900 N/m.m l k= = = 
 
SOLUTION 
 
Deflection of spring = s, where 2 2s l y l= + − 
2 2
ds y
dy l y
=
−
 
Potential Energy: 21
2 2
yV ks W= − 
1
2
dV dsks W
dy dy
= − 
( )2 2 2 2 12dV yk l y l Wdy l y= + − −+ 
 
2 2
11
2
lk y W
l y
  = − − + 
 
Equilibrium 
2 2
10: 1
2
dV l Wy
dy kl y
  = − = + 
 
Now ( )( )212 kg 9.81 m/s 117.72 N, 0.75 m, and 900 N/mW mg l k= = = = = 
Then ( )
( )
( )2 2
117.72 N0.75 m 11
2 900 N/m0.75 m
y
y
  − =  + 
 
or 
2
0.751 0.6540
0.5625
y
y
  − = + 
 
Solving numerically, 0.45342 my = 
 453 mmy = W 
 
 
 
PROBLEM 10.78 
The slender rod AB of negligible mass is attached to two 4-kg blocks A 
and B that can move freely in the guides shown. Knowing that the 
constant of the springs is 160 N/m and that the unstretched length of each 
spring is 150 mm, determine the value of x corresponding to equilibrium. 
 
SOLUTION 
 
First note ( ) ( )2 20.4 0.22 0.4 my x = − − −   
 ( )20.4 0.8 0.1116 mx x= − − + − 
Now, the Potential Energy is 
 ( ) ( )2 21 10.15 0.15 0.4
2 2 A B
V k x k y m g m gy= − + − + + 
 ( ) ( )22 21 10.15 0.25 0.8 0.11162 2k x k x x= − + − − + − 
 ( )20.4 0.4 0.8 0.1116A Bm g m g x x+ + − − + − 
For Equilibrium 
( ) ( )2 2 0.8 20: 0.15 0.25 0.8 0.1116 2 0.8 0.1116dV xk x k x xd x xθ  −= − + − − + − −  − + −  
 
2
0.8 2 0
2 0.8 0.1116
B
xm g
x x
−− =
− + −
 
Simplifying, 
( ) ( )20.4 0.8 0.1116 4 0.4 0Bk x x x m g x− + − + − + − = 
Substituting the masses, 0.4 kg,A Bm m= = and the spring constant160 N/m:k = 
( )( ) ( )( )( )2 2 2160 N/m 0.4 0.8 0.1116 m 4 4 kg 9.81 m/s 0.4 m 0x x x x− + − + − + − = 
 
 
 
PROBLEM 10.78 CONTINUED 
or ( ) ( )20.4 0.8 0.1116 0.981 0.4 0x x x x− + − + − + − = 
Simplifying, ( ) ( )2 220.8 0.1116 0.7924 1.981x x x− − = − 
or 24.92436 3.93949 0.739498 0x− + = 
Then 
( ) ( )( )
( )
23.93949 3.93949 4 4.92436 0.739498
2 4.92436
x
± − −= 
or 0.49914 mx = and 0.30086 mx = 
Now 0.4 m 301 mmx x≤ ⇒ = W 
 
 
 
PROBLEM 10.79 
A slender rod AB, of mass m, is attached to two blocks A and B that can 
move freely in the guides shown. The constant of the spring is k, and the 
spring is unstretched when AB is horizontal. Neglecting the weight of the 
blocks, derive an equation in ,θ m, l, and k that must be satisfied when 
the rod is in equilibrium. 
 
SOLUTION 
 
Elongation of Spring: 
sin coss l l lθ θ= + − 
( )sin cos 1s l θ θ= + − 
Potential Energy: 21 sin
2 2
lV ks W W mgθ= − = 
 ( )221 sin cos 1 sin
2 2
lkl mgθ θ θ= + − − 
 ( )( )2 1sin cos 1 cos sin cos
2
dV kl mgl
d
θ θ θ θ θθ = + − − − (1) 
Equilibrium: 
( )( )0: sin cos 1 cos sin cos 0
2
dV mg
d kl
θ θ θ θ θθ = + − − − = 
or ( )( )cos sin cos 1 1 tan 0
2
mg
kl
θ θ θ θ + − − − =   W 
 
 
 
 
 
PROBLEM 10.80 
A slender rod AB, of mass m, is attached to two blocks A and B that can 
move freely in the guides shown. Knowing that the spring is unstretched 
when AB is horizontal, determine three values of θ corresponding to 
equilibrium when 125 kg,m = 320 mm,l = and 15 kN/mm.k = State 
in each case whether the equilibrium is stable, unstable, or neutral. 
 
SOLUTION 
Using the results of Problem 10.79, particularly the condition of equilibrium 
( )( )cos sin cos 1 1 tan 0
2
mg
kl
θ θ θ θ + − − − =   
Now, with ( )( )2125 kg 9.81 m/s 1226.25 N, 320 mm,W mg l= = = = and 15 kN/m,k = 
Now ( )( )
1226.25 N 1.2773
2 2 15000 N/m 0.32 m
W
kl
= = 
so that ( )( )cos sin cos 1 1 tan 1.2773 0θ θ θ θ + − − − =  
By inspection, one solution is cos 0 or 90.0θ θ= = ° 
Solving numerically: 0.38338 rad 9.6883 and 0.59053 rad 33.8351θ θ= = ° = = ° 
Stability 
 ( ) ( ) ( ) ( )2 22 1cos sin cos sin sin cos 1 sin cos sin2
d V kl mgl
d
θ θ θ θ θ θ θ θ θθ  = − − + + − − − +  
 2 2 2 2 2cos sin 2sin cos sin cos 2sin cos sin cos sin
2
mgkl
kl
θ θ θ θ θ θ θ θ θ θ θ = + − − − − + + +   
 2 1 sin cos 2sin 2
2
mgkl
kl
θ θ θ  = + + −     
 ( ) ( ) ( )215 N/m 0.32 m 1 127.73 sin cos 2sin 2θ θ θ = − + −  
Thus, at 
At 90 :θ = ° 
2
2 89.7 0
d V
dθ = > 90.0 ,θ∴ = ° StableW 
At 9.6883 :θ = ° 
2
2 0.512 0
d V
dθ = > 9.69 ,θ∴ = ° StableW 
At 33.8351 :θ = ° 
2
2 0.391 0
d V
dθ = − < 33.8 ,θ∴ = ° UnstableW 
 
 
 
PROBLEM 10.81 
Spring AB of constant 10 lb/in. is attached to two identical drums as 
shown. Knowing that the spring is unstretched when 0,θ = determine 
(a) the range of values of the weight W of the block for which a position 
of equilibrium exists, (b) the range of values of θ for which the 
equilibrium is stable. 
 
SOLUTION 
 
 
 
 
 
Have 
block
21
2 SP y
V kx W= − 
where 2 sin , 6 in.SP A Ax r rθ= = 
and block , 8 in.y r rθ= = 
Then ( )21 2 sin
2 A
V k r Wrθ θ= − 
 2 22 sinAkr Wrθ θ= − 
and ( )22 2sin cosAdV kr Wrd θ θθ = − 
 22 sin 2Akr Wrθ= − 
 
2
2
2 4 cos 2A
d V kr
d
θθ = (1) 
For equilibrium 20: 2 sin 2 0A
dV kr Wr
d
θθ = − = 
Substituting, ( )( ) ( )22 10 lb/in. 6 in. sin 2 8 in. 0Wθ − = 
or 90sin 2 (lb)W θ=
(a) From Equation (2), with 0:W ≥ 0 90 lbW≤ ≤ W 
(b) From Stable equilibrium 
2
2 0
d V
dθ > 
 Then from Equation (1), cos 2 0θ > 
 or 0 45θ≤ ≤ °W 
 
 
 
 
 
 
 
PROBLEM 10.82 
Spring AB of constant 10 lb/in. is attached to two identical drums as 
shown. Knowing that the spring is unstretched when 0θ = and that 
40 lb,W = determine the values of θ less than 180° corresponding to 
equilibrium. State in each case whether the equilibrium is stable, 
unstable, or neutral. 
 
SOLUTION 
See sketch, Problem 10.81. 
Using Equation (2) of Problem 10.81, with 40 lbW = 
40 90sin 2θ= (for equilibrium) 
Solving 13.1939 and 76.806θ θ= ° = ° 
Using Equation (1) of Problem 10.81, we have 
At 13.1939 :θ = ° ( )2 22 4 cos 2 13.1939 0Ad V krdθ = × ° > 13.19 , Stableθ∴ = ° W 
At 76.806 :θ = ° ( )2 22 4 cos 2 76.806 0Ad V krdθ = × ° < 76.8 , Unstableθ∴ = ° W 
 
 
 
PROBLEM 10.83 
A slender rod AB of negligible weight is attached to two collars A and B that 
can move freely along the guide rods shown. Knowing that o30β = and 
100 lb,P Q= = determine the value of the angle θ corresponding to 
equilibrium. 
 
SOLUTION 
 
Law of Sines ( ) ( )sin 90 sin 90A
y L
β θ β=° + − − 
( )cos cosA
y L
θ β β=− 
or ( )cos
cosA
y L
θ β
β
−= 
From the figure: ( )cos cos
cosB
y L L
θ β θβ
−= − 
Potential Energy: ( ) ( )cos coscos
cos cosB A
V Py Qy P L L QL
θ β θ βθβ β
 − −= − − = − − −  
 
( ) ( )sin sinsin
cos cos
dV PL QL
d
θ β θ βθθ β β
 − −= − − + +  
 
 ( ) ( )sin sin
cos
L P Q PL
θ β θβ
−= + − 
Equilibrium ( ) ( )sin0: sin 0
cos
dV L P Q PL
d
θ β θθ β
−= + − = 
or ( ) ( )sin sin cosP Q Pθ β θ β+ − = 
( )( )sin cos cos sin sin cosP Q Pθ β θ β θ β+ − = 
 
 
 
PROBLEM 10.83 CONTINUED 
or ( )cos sin sin cos 0P Q Qθ β θ β− + + = 
sin sin 0
cos cos
P Q
Q
β θ
β θ
+− + = 
 tan tanP Q
Q
θ β+= (2) 
With 100 lb, 30P Q β= = = ° 
200 lbtan tan 30 1.1547
100 lb
θ = ° = 
 49.1θ = °W 
 
 
 
PROBLEM 10.84 
A slender rod AB of negligible weight is attached to two collars A and B 
that can move freely along the guide rods shown. Knowing that 
o30 ,β = 40 lb,P = and 10 lb,Q = determine the value of the angle θ 
corresponding to equilibrium. 
 
SOLUTION 
Using Equation (2) of Problem 10.83, with 40 lb, 10 lb, and 30 ,P Q β= = = ° we have 
( )( )
( )
40 lb 10 lb
tan tan 30 2.88675
10 lb
θ = ° = 
 70.89θ = ° 70.9θ = °W 
 
 
 
 
PROBLEM 10.85 
Collar A can slide freely on the semicircular rod shown. Knowing that the 
constant of the spring is k and that the unstretched length of the spring is 
equal to the radius r, determine the value of θ corresponding to 
equilibrium when 20 kg,m = 180 mm,r = and 3 N/mm.k = 
 
SOLUTION 
 
Stretch of Spring 
 s AB r= − 
 ( )2 coss r rθ= − 
 ( )2cos 1s r θ= − 
Potential Energy: 21 sin 2
2
V ks Wr W mgθ= − = 
 ( )221 2 cos 1 sin 2
2
V kr Wrθ θ= − − 
( )2 2cos 1 2sin 2 cos 2dV kr Wr
d
θ θ θθ = − − − 
Equilibrium 
( )20: 2cos 1 sin cos 2 0dV kr Wr
d
θ θ θθ = − − − = 
( )2cos 1 sin
cos 2
W
kr
θ θ
θ
− = − 
Now 
( )( )
( )( )
220 kg 9.81 m/s
0.36333
3000 N/m 0.180 m
W
kr
= = 
Then ( )2cos 1 sin 0.36333
cos 2
θ θ
θ
− = − 
Solving numerically, 0.9580 rad 54.9θ = = ° 54.9θ = °W 
 
 
 
PROBLEM 10.86 
Collar A can slide freely on the semicircular rod shown. Knowing that the 
constant of the spring is k and that the unstretched length of the spring is 
equal to the radius r, determine the value of θ corresponding to equilibrium 
when 20 kg,m = 180 mm,r = and 3 N/mm.k = 
 
SOLUTION 
 
Stretch of spring 
( )2 coss AB r r rθ= − = − 
( )2cos 1s r θ= − 
 21 cos 2
2
V ks Wr θ= − 
( )221 2cos 1 cos 2
2
kr Wrθ θ= − − 
( )2 2cos 1 2sin 2 sin 2dV kr Wr
d
θ θ θθ = − − + 
Equilibrium 
( )20: 2cos 1 sin sin 2 0dV kr Wr
d
θ θθθ = − − + = 
 ( ) ( )2 2cos 1 sin 2sin cos 0kr Wrθ θ θ θ− − + = 
or ( )2cos 1 sin
2cos
W
kr
θ θ
θ
− = 
Now 
( )( )
( )( )
220 kg 9.81 m/s
0.36333
3000 N/m 0.180 m
W
kr
= = 
Then 2cos 1 0.36333
2cos
θ
θ
− = 
Solving 38.2482θ = ° 38.2θ = °W 
 
 
 
 
PROBLEM 10.87 
The 12-kg block D can slide freely on the inclined surface. Knowing that 
the constant of the spring is 480 N/m and that the spring is unstretched 
when 0,θ = determine the value of θ corresponding to equilibrium. 
 
SOLUTION 
 
 
 
First note, by Law of Cosines 
( ) ( )222 0.4 0.4sin 2 0.4 0.4sin cos
2 2 2
d θ θ θ   = + −       
or 20.4 1 sin sin m
2
d θ θ= + − 
Now 
 21
2 SP D D
V kx m gy= − 
 ( ) ( ) ( )2 01 0.4 sin 602 A D Dk r m g y dθ  = − + − °  
( )2 2 201 0.4 0.4 1 sin sin sin 602 2A D Dkr m g y
θθ θ  = − + − + − °      
 
For equilibrium 0:dV
dθ = 
2
2
12 sin cos cos
2 2 20.4 sin 60 0
2 1 sin sin
2
A Dkr m g
θ θ θ
θ θ θ
  −    + ° =
+ −
 
or 2
2
sin 2cos0.1 sin 60 0
1 sin sin
2
A Dkr m g
θ θθ θ θ
−+ ° =
+ −
 
 
 
 
 
 
PROBLEM 10.87 CONTINUED 
Substituting, 
 ( )( )2 2480 N/m 0.050 m 1 sin sin
2
θθ θ+ − 
 ( )( )( ) ( )2 30.1 m 12 kg 9.81 m/s sin 2cos 02 θ θ+ − = 
or ( )21 sin sin 8.4957 sin 2cos 0
2
θθ θ θ θ+ − + − = 
Solving numerically, 1.07223 radθ = 
 or 61.4θ = °W 
 
 
PROBLEM 10.88 
Cable AB is attached to two springs and passes through a ring at C. 
Knowing that the springs are unstretched when 0,y = determine the 
distance y corresponding to equilibrium. 
 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
First note that the tension in the cable is the same throughout. 
1 2 F F∴ = 
or 1 1 2 2k x k x= 
or 12 1
2
kx x
k
= 
 1
960 N/m
480 N/m
x= 
 12x= 
Now, point C is midway between the pulleys. 
∴ ( ) ( )2 22 1 210.2 0.22y x x = + + −   
 ( ) ( )21 2 1 210.2 4x x x x= + + + 
 ( ) ( )21 1 1 110.2 2 24x x x x= + + + 
 ( )2 21 190.6 m4x x= + 
 
PROBLEM 10.88 CONTINUED 
Now 
 2 21 1 2 2
1 1
2 2
V k x k x mgy= + − 
( )22 21 1 2 1 1 11 1 12 2.4 92 2 4k x k x mg x x = + − +   
 ( ) 2 21 2 1 1 11 14 2.4 92 4k k x mg x x = + − +   
For equilibrium 
( ) 11 2 1 21 1 1
2.4 180: 4 0
2 2.4 9
dV xk k x mg
dx x x
 + = + − = + 
 
or ( ) ( )( )( )( ) ( )( )( )( )2 21 1 1 11980 4 490 N/m m 2.4 9 m 10 kg 9.81 m/s 1.2 9 m 02x x x x+ × × + − + = 
or ( )21 1 1 1288 2.4 9 5.886 1 7.5 0x x x x+ − + = 
Solving, 1 0.068151 mx = 
Then ( ) ( )22 90.6 0.068151 0.068151
4
y = + or 227 mmy = W 
 
 
 
 
PROBLEM 10.89 
Rod AB is attached to a hinge at A and to two springs, each of constant k. 
If 50 in.,h = 24 in.,d = and 160 lb,W = determine the range of 
values of k for which the equilibrium of the rod is stable in the position 
shown. Each spring can act in either tension or compression. 
 
SOLUTION 
 Have sin cosC Bx d y hθ θ= = 
 Potential Energy: 212
2 C B
V kx Wy = +   
 2 2sin coskd Whθ θ= + 
 Then 22 sin cos sindV kd Wh
d
θ θ θθ = − 
 2 sin 2 sinkd Whθ θ= − 
 and 
2
2
2 2 cos 2 cos
d V kd Wh
d
θ θθ = − (1) 
For equilibrium position 0θ = to be stable, we must have 
2
2
2 2 0
d V kd Wh
dθ = − > 
or 2 1
2
kd Wh> (2) 
Note: For 2 1 ,
2
kd Wh= we have 
2
2 0,
d V
dθ = so that we must determine which is the first derivative that is not 
equal to zero. Differentiating Equation (1), we write 
3
2
3 4 sin 2 sin 0 for 0
d V kd Wh
d
θ θ θθ = − + = = 
4
2
2 8 cos 2 cos
d V kd Wh
d
θ θθ = − + 
 
 
PROBLEM 10.89 CONTINUED 
For 0:θ = 
4
2
4 8
d V kd Wh
dθ = − + 
Since 
4
2
4
1 , 4 0,
2
d Vkd Wh Wh Wh
dθ= = − + < we conclude that the equilibrium is unstable for 
2 1
2
kd Wh= 
and the > sign in Equation (2) is correct. 
With 160 lb, 50 in., and 24 in.W h d= = = 
Equation (2) gives ( ) ( )( )2 124 in. 160 lb 50 in.
2
k > 
or 6.944 lb/in.k > 
 6.94 lb/in.k > W 
 
 
 
PROBLEM 10.90 
Rod AB is attached to a hinge at A and to two springs, each of constant k. 
If 30 in.,h = 4 lb/in.,k = and 40 lb,W = determine the smallest 
distance d for which the equilibrium of the rod is stable in the position 
shown. Each spring can act in either tension or compression. 
 
SOLUTION 
Using Equation (2) of Problem 10.89 with 
30 in., 4 lb/in., and 40 lbh k W= = = 
( ) ( )( )2 14 lb/in. 40 lb 30 in.
2
d > 
or 2 2150 ind > 
12.247 in.d > 
 smallest 12.25 in.d = W 
 
 
 
PROBLEM 10.91 
The uniform plate ABCD of negligible mass is attached to four springs of 
constant k and is in equilibrium in the position shown. Knowing that the 
springs can act in either tension or compression and are undeformed in 
the given position, determine the range of values of the magnitude P of 
two equal and opposite horizontal forces P and −P for which the 
equilibrium position is stable. 
 
SOLUTION 
 
 
 
 
 
Consider a small clockwise rotation θ of the plate about its center. 
Then 2 4P SPV V V= + 
where cos
2P
aV P θ =    
 ( )1 cos
2
Pa θ= 
and 21
2SP SP
V ky= 
Now 
2
2
2
ad a = +   
 5
2
a= 
and 180 90
2
θα φ  = ° − + ° −     
 90
2
θφ = ° − −   
Then 5 sin
2SP
ay θ α  =      
 5 sin 90
2 2
a θθ φ  = ° − −     
 5 cos
2 2
a θθ φ = −   
 
 
 
 
 
PROBLEM 10.91 CONTINUED 
and 
2
1 5 cos
2 2 2SP
aV k θθ φ  = −     
 2 2 25 cos
8 2
ka θθ φ = −   
2 2 25 cos cos
2 2
V Pa ka θθ θ φ ∴ = + −   
Then 
 2 25sin 2 cos
8 2
dV Pa ka
d
θθ θ φθ
  = − + −    
 2 1 cos sin
2 2 2
θ θθ φ φ      + − − −           
 ( )2 2 25 1sin 2 cos sin 2
2 2 2
Pa ka θθ θ φ θ φ θ  = − + − + −     
2
2 2
2
5cos 2cos
2 2
d V Pa ka
d
θθ φθ
  = − + −    
 ( )12 cos sin sin 2
2 2 2
θ θθ φ φ θ φ θ     − − − − + −           
 ( )21 cos 2
2
θ φ θ − −  
 ( )2 25 3cos 2cos sin 2
2 2 2
Pa ka θθ φ θ φ θ  = − + − + −    
 ( )21 cos 2
2
θ φ θ − −  
( )2 23 5 1 3sin 4 cos sin sin 22 2 2 2 2
d V Pa ka
d
θ θθ φ φ φ θθ
      = + − − − + −           
 
 ( ) ( ) ( )23 1cos 2 cos 2 sin 2
2 2
θ φ θ θ φ θ θ φ θ − − − − + −  
 ( ) ( )25 1 5sin sin 2 cos 2
2 2 2
Pa kaθ φ θ θ φ θ= + − − − 
 ( )21 sin 2
2
θ φ θ + −  
 
 
 
 
PROBLEM 10.91 CONTINUED 
When 0, 0dV
d
θ θ= = for all values of P. 
For stable equilibrium when 0,θ = require 
( )2 2 22 50: 2cos 02d V Pa kad φθ > − + > 
Now, when 0,θ = 12cos
55
2
a
aφ = = 
2 15 0
5
Pa ka  ∴ − + >   
or P ka< 
When ( )for 0 :P ka θ= = 
0dV
dθ = 
2
2 0
d V
dθ = 
3
2
3
5 sin 2 0 unstable
4
d V ka
d
φθ = > ⇒ 
 ∴ Stable equilibrium for 0 P ka≤ < W 
 
 
 
 
 
PROBLEM 10.92 
Two bars are attached to a single spring of constant k that is unstretched 
when the bars are vertical. Determine the range of values of P for which 
the equilibrium of the system is stable in the position shown. 
 
SOLUTION 
 
 
 
Spring: 2sin sin
3 3
L Ls φ θ= = 
For small values of and :φ θ 2φ θ= 
 22 1cos cos
3 3 2
L LV P ksφ θ = + +   
 ( ) 21 2cos 2 2cos sin
3 2 3
PL Lkθ θ θ = + +    
( ) 222sin 2 2sin sin cos
3 9
dV PL kL
d
θ θ θ θθ = − − + 
 ( ) 222sin 2 2sin sin 2
3 9
PL kLθ θ θ= − + + 
(