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 cos2PaV P θ =

( )1 cos2

Pa θ=

and 212SP SPV ky=

Now 2

2

2ad a = +

52a

=

and 180 902θα φ

= ° − + ° −

902θφ = ° − −

Then 5 sin2SPay θ α

=

5 sin 902 2a θθ φ

= ° − −

5 cos2 2a θθ φ = −

PROBLEM 10.91 CONTINUED

and 2

1 5 cos2 2 2SP

aV k θθ φ = −

2 2 25 cos8 2

ka θθ φ = −

2 2 25 cos cos2 2

V Pa ka θθ θ φ ∴ = + −

Then

2 25sin 2 cos8 2

dV Pa kad

θθ θ φθ

= − + −

2 1 cos sin2 2 2

θ θθ φ φ + − − −

( )2 2 25 1sin 2 cos sin 22 2 2

Pa ka θθ θ φ θ φ θ = − + − + −

22 2

25cos 2cos2 2

d V Pa kad

θθ φθ

= − + −

( )12 cos sin sin 22 2 2

θ θθ φ φ θ φ θ − − − − + −

( )21 cos 22θ φ θ − −

( )2 25 3cos 2cos sin 22 2 2

Pa ka θθ φ θ φ θ = − + − + −

( )21 cos 22θ φ θ − −

( )2

23

5 1 3sin 4 cos sin sin 22 2 2 2 2

d V Pa kad

θ θθ φ φ φ θθ

= + − − − + −

( ) ( ) ( )23 1cos 2 cos 2 sin 22 2θ φ θ θ φ θ θ φ θ − − − − + −

( ) ( )25 1 5sin sin 2 cos 22 2 2

Pa kaθ φ θ θ φ θ= + − − −

( )21 sin 22θ φ θ + −

PROBLEM 10.91 CONTINUED

When 0, 0dVd

θθ

= = for all values of P.

For stable equilibrium when 0,θ = require

( )2

2 22

50: 2cos 02

d V Pa kad

φθ

> − + >

Now, when 0,θ = 12cos55

2

a

aφ = =

2 15 05

Pa ka ∴ − + >

or P ka<

When ( )for 0 :P ka θ= =

0dVdθ

=

2

2 0d Vdθ

=

32

35 sin 2 0 unstable4

d V kad

φθ

= > ⇒

∴ Stable equilibrium for 0 P ka≤ <

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 sin3 3L Ls φ θ= =

For small values of and :φ θ 2φ θ=

22 1cos cos3 3 2L LV P ksφ θ = + +

( )21 2cos 2 2cos sin

3 2 3PL Lkθ θ θ = + +

( ) 222sin 2 2sin sin cos3 9

dV PL kLd

θ θ θ θθ

= − − +

( ) 222sin 2 2sin sin 23 9

PL kLθ θ θ= − + +

( )2

22

44cos 2 2cos cos 23 9

d V PL kLd

θ θ θθ

= − + +

When 2

22

6 40:3 9

d V PL kLd

θθ

= = − +

For stability: 2

22

40: 2 09

d V PL kLdθ

> − + >

209

P kL≤ <

PROBLEM 10.93 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

First note 2 sin sin3 3L La θ φ= =

and sin3Ls θ=

For small values of and :φ θ 2φ θ=

22 1cos cos3 3 2L LV P ksθ φ = + +

( )212cos cos 2 sin

3 2 3PL Lkθ θ θ = + +

( )2

2sin 2sin 2 sin cos3 9

dV PL kLd

θ θ θ θθ

= − − +

( )22 sin sin 2 sin 2

3 18PL kLθ θ θ= − + +

( )2 2

22 cos 2cos 2 cos 2

3 9d V PL kLd

θ θ θθ

= − + +

When 2 2

20: 29

d V kLPLd

θθ

= = − +

For stability: 2 2

2 0: 2 09

d V kLPLdθ

> − + >

1018

P kL≤ <

PROBLEM 10.94 Bar AC is attached to a hinge at A and to a spring of constant k that is undeformed when the bar is vertical. Knowing that the spring can act in either tension or compression, determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION

Consider a small disturbance of the system defined by the angle .θ

Have

2 sin sinCx a aθ φ= =

For small :θ 2θ φ=

Now, the Potential Energy is

212 B EV kx Py= +

where sinBx a θ=

and /E C E Cy y y= +

2 cos 2 cosa aθ φ= +

( )2 cos cos 2a θ θ= +

Then ( )2 21 sin 2 cos cos 22

V ka Paθ θ θ= + +

and ( ) ( )21 2sin cos 2 sin 2sin 22

dV ka Pad

θ θ θ θθ

= − +

( )21 sin 2 2 sin 2sin 22

ka Paθ θ θ= − +

( )2

22 cos 2 2 cos 4cos 2d V ka Pa

dθ θ θ

θ= − +

For 0θ = and for stable equilibrium:

2

2 0d Vdθ

>

or ( )2 2 1 4 0ka Pa− + >

PROBLEM 10.94 CONTINUED

or 110

P ka<

1 010

P ka∴ ≤ <

Check stability for 10kaP =

( )3

23 2 sin 2 2 sin 8sin 2d V ka Pa

dθ θ θ

θ= − + +

( )4

24 4 cos 2 2 cos 16cos 2d V ka Pa

dθ θ θ

θ= − + +

Then, with 0 and 10kaPθ = =

0dVdθ

=

2

2 0d Vdθ

=

3

3 0d Vdθ

=

( )( )4

24 4 2 1 16

10d V ka ka adθ

1 = − + +

20.6 0 Unstableka= − < ⇒

PROBLEM 10.95 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of P for which the equilibrium of the system is stable in the position shown when

300 mm,a = 400 mm,b = and 90 N.Q =

SOLUTION

First note sin sinA a bθ φ= =

For small values of and :θ φ a bθ φ=

or ab

φ θ=

( ) ( )cos 2 cosV P a b Q a bφ θ= + − +

( ) cos 2 cosaa b P Qbθ θ

= + −

( ) sin 2 sindV a aa b P Qd b b

θ θθ

= + − +

( )2 2

2 2 cos 2 cosd V a aa b P Qbd bθ θ

θ = + − +

When 0:θ = ( )2 2

2 2 2d V aa b P Qd bθ

= + − +

Stability: 2 2

2 20: 2 0d V a P Qd bθ

> − + >

2

22 bP Qa

< (1)

or 2

22aQ Pb

> (2)

With 90 N, 300 mm, and 400 mmQ a b= = =

Equation (1): ( )( )

( )2

2400 mm

2 90 N 320 N300 mm

P < =

For stability 0 320 NP≤ <

PROBLEM 10.96 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of Q for which the equilibrium of the system is stable in the position shown when

480 mm,a = 400 mm,b = and 600 N.P =

SOLUTION

Using Equation (2) of Problem 10.95 with 600 N, 480 mm, and 400 mmP a b= = =

Equation (2) ( )( )

( )2

2480 mm1 600 N 432 N

2 400 mm> =Q

For stability 432 NQ >

PROBLEM 10.97 Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed and the system is in equilibrium when 1 2 0.θ θ= = Determine the range of values of P for which the equilibrium position is stable.

SOLUTION

Have sinBx l θ=

1 2sin sinθ θ= +Cx l l

1 2cos cosθ θ= +Cy l l

2 21 12 2C B CV Py kx kx= + +

or ( ) ( )22 21 2 1 1 2

1cos cos sin sin sin2

θ θ θ θ θ = + + + + V Pl kl

For small values of 1θ and 2 :θ

2 21 1 2 2 1 1 2 2

1 1sin , sin , cos 1 , cos 12 2

θ θ θ θ θ θ θ θ≈ ≈ ≈ − ≈ −

Then ( )2 2

22 21 21 1 2

11 12 2 2θ θ θ θ θ

= − + − + + + V Pl kl

and ( )21 1 1 2

1θ θ θ θ

θ∂ = − + + + ∂V Pl kl

( )22 1 2

2θ θ θ

θ∂

= − + +∂

V Pl kl

2 22 2

2 21 2

2 θ θ∂ ∂

= − + = − +∂ ∂

V VPl kl Pl kl

22

1 2θ θ∂

=∂ ∂

V kl

PROBLEM 10.97 CONTINUED

Stability Conditions for stability (see page 583).

For ( )1 21 2

0: 0 condition satisfiedV Vθ θθ θ∂ ∂

= = = =∂ ∂

22 2 2

2 21 2 1 2

0V V Vθ θ θ θ

∂ ∂ ∂− < ∂ ∂ ∂ ∂

Substituting, ( ) ( )( )22 22 0− − + − + <kl Pl kl Pl kl

2 4 2 2 3 2 43 2 0− + − <k l P l Pkl k l

2 2 23 0− + >P klP k l

Solving, 3 5 3 5 or 2 2− +

< >P kl P kl

or 0.382 or 2.62P kl P kl< >

22

21

0: 2 0V Pl klθ∂

> − + >∂

or 12

P kl<

2

222

0: 0V Pl klθ∂

> − + >∂

or P kl<

Therefore, all conditions for stable equilibrium are satisfied when

0 0.382P kl≤ <

PROBLEM 10.98 Solve Problem 10.97 knowing that 400 mml = and 1.25 kN/m.k =

SOLUTION

From the analysis of Problem 10.98 with

400 mm and 1.25 kN/ml k= =

( )( )0.382 0.382 1250 N/m 0.4 m 191 N< = =P kl

0 191.0 NP≤ <

PROBLEM 10.99 Bar ABC of length 2a and negligible weight is hinged at C to a drum of radius a as shown. Knowing that the constant of each spring is k and that the springs are undeformed when 1 2 0,θ θ= = determine the range of values of P for which the equilibrium position 1 2 0θ θ= = is stable.

SOLUTION

Have ( ) ( ) ( )2 22 1 2 1 2

1 1 sin sin 2 cos cos2 2

θ θ θ θ θ= + + + +V k a k a a P a a

Then ( )21 2 1 1

1sin sin cos 2 sinθ θ θ θ

θ∂

= + −∂

V ka Pa

21 1 2 1

1 sin 2 cos sin 2 sin2

θ θ θ θ = + −

ka Pa

and ( )2

21 1 2 12

1cos 2 sin sin 2 cosθ θ θ θ

θ∂

= − −∂

V ka Pa

2

21 2

1 2cos cosθ θ

θ θ∂

=∂ ∂

V ka

Also ( )2 22 1 2 2 2

2sin sin cos sinθ θ θ θ θ

θ∂

= + + −∂

V ka ka Pa

2 22 1 2 2 2

1sin cos sin 2 sin2

θ θ θ θ θ = + + −

ka ka Pa

and ( )2

2 21 2 2 22

2sin sin cos 2 cosθ θ θ θ

θ∂

= + − + −∂

V ka ka Pa

When 1 2 0θ θ= =

22

1 1 2 20 0

θ θ θ θ∂ ∂ ∂

= = =∂ ∂ ∂ ∂

V V Vka

2 22 2 2 2

2 21 2

2 2θ θ∂ ∂

= − = + − = −∂ ∂

V Vka Pa ka ka Pa ka Pa

PROBLEM 10.99 CONTINUED

Apply Equations 10.24 1

0: condition satisfiedVθ∂

=∂

2

0: condition satisfiedVθ∂

=∂

( ) ( )( )22 2 2 22 2 2

2 21 2 1 2

0: 2 2 0θ θ θ θ

∂ ∂ ∂− < − − − < ∂ ∂ ∂ ∂

V V V ka ka Pa ka Pa

or ( )( )2 2 2 2 0− − − <k a ka P ka P

Expanding 2 2 2 22 5 2 0− + − <k a ka kaP P

or 2 2 22 5 0− + >P kaP k a

or 5 17 5 17 and 4 4

− +< >P ka P ka

or 0.21922 and 2.2808< >P ka P ka

Also 2 2

2 22 2

1 20: 2 0 or 0: 2 0δ δ

δθ δθ> − > > − >

V Vka Pa ka Pa

or 1 or 22

< <P ka P ka

∴ For stable equilibrium when 1 2 0:θ θ= =

0 0.219≤ <P ka

PROBLEM 10.100 Solve Problem 10.99 knowing that 10 lb/in. and 14 in.k a= =

SOLUTION

From the solution to Problem 10.99, with 10 lb/in. and 14 in.k a= =

0 0.21922≤ <P ka

or ( )( )0 0.21922 10 lb/in. 14 in.≤ <P

or 0 30.7 lb≤ <P