solution a b -...

11–1.

SOLUTION

Ans.F = 2P cot u

-P(4L cos u du) - F(- L sin u du) = 0

dU = 0; -Pdy - Fdx = 0

y = L sin u, dy = L cos u du

x = L cos u, dx = - L sin u du

The scissors jack supports a load P. Determine the axialforce in the screw necessary for equilibrium when the jackis in the position Each of the four links has a length Lis pin-connected at its center. Points B and D can movehorizontally.

u.

C D

A B

P

u

2 and

2 2

4 4

2

11–4.

The spring has an unstretched length of 0.3 m. Determinethe angle for equilibrium if the uniform links each have amass of 5 kg.

u

SOLUTIONFree Body Diagram: The system has only one degree of freedom defined by theindependent coordinate When undergoes a positive displacement only thespring force and the weights of the links (49.05 N) do work.

Virtual Displacements: The position of points B, D and G are measured from thefixed point A using position coordinates and respectively.

(1)

(2)

(3)

Virtual–Work Equation: When points B, D and G undergo positive virtualdisplacements and the spring force that acts at point B doespositive work while the spring force that acts at point D and the weight of linkAC and CE (49.05 N) do negative work.

(4)

Substituting Eqs. (1), (2) and (3) into (4) yields

(5)

However, from the spring formula,Substituting this value into Eq. (5) yields

Since then

Ans.

and Ans.u = 85.4°

u = 15.5°

34.335 sin u - 576 sin u cos u + 144 cos u = 0

du Z 0,

134.335 sin u - 576 sin u cos u + 144 cos u2 du = 0

= 480 sin u - 120.Fsp = kx = 4003210.6 sin u2 - 0.34

134.335 sin u - 1.2Fsp cos u2 du = 0

21-49.05dyG2 + Fsp1dxB - dxD2 = 0dU = 0;

Fsp

FspdyG ,dxDdxB ,

yG = 0.35 cos u dyG = -0.35 sin udu

xD = 210.7 sin u2 - 0.1 sin u = 1.3 sin u dxD = 1.3 cos udu

xB = 0.1 sin u dxB = 0.1 cos udu

yG ,xDxB ,

Fsp

du,uu.0.1 m

0.6 m

C

A

BD

Ek = 400 N/m

θ θ

mass 50 kg

490.5

245.25 N

50 (9.81) N = 490.5 N

.

11–10.

The thin rod of weight W rest against the smooth wall andfloor. Determine the magnitude of force P needed to hold itin equilibrium for a given angle .

SOLUTION

Free-Body Diagram: The system has only one degree of freedom defined by theindependent coordinate . When undergoes a positive displacement , only theweight of the rod W and force P do work.

Virtual Displacements: The weight of the rod W and force P are located from thefixed points A and B using position coordinates and , respectively

(1)

(2)

Virtual-Work Equation: When points C and A undergo positive virtual displacementsand , the weight of the rod W and force F do negative work.

; (3)

Substituting Eqs. (1) and (2) into (3) yields

Since , then

Ans.P =W

2 cot u

Pl sin u -Wl

2 cos u = 0

du Z 0

aPl sin u -Wl

2 cos ub du = 0

-WdyC - PdyA = 0dU = 0

dxAdyC

dxA = - l sin uduxA = l cos u

dyC =12

cos uduyC =12

sin u

xAyC

duuu

u

A

l

B

P θ

11–11.

SOLUTIONFree–Body Diagram: Only force F and the weight of link AB (98.1 N) do work.

Virtual Displacement: Force F and the weight of link AB (98.1 N) are located fromthe top of the fixed link using position coordinates and Since the cord has aconstant length, l, then

(1)

Virtual–Work Equation: When and undergo positive virtual displacements ds1 (98.1 N) and force F do positive work and negative

work, respectively.

(2)

Substituting Eq. (1) into (2) yields

Since then

Ans.F =

- 4F = 0

ds1 Z 0,

1 - 4F2 ds1 = 0

1ds12 - F1ds22 = 0dU = 0;

s2s1

4s1 - s2 = l 4ds1 - ds2 = 0

s1 .s2

Determine the force F acting on the cord which is requiredto maintain equilibrium of the horizontal 20-kg bar AB.Hint: Express the total constant vertical length l of the cordin terms of position coordinates and The derivative ofthis equation yields a relationship between and d2.d1

s2.s1s2

s1

A B

F

and the weight of link ABds2 ,

196.2

196.2

196.2

49.1 N

20 (9.81) = 196.2 N

11–13.

Each member of the pin-connected mechanism has a massof 8 kg. If the spring is unstretched when determinethe angle for equilibrium. Set aM = # m.

k = >m ndu

u = 0°,

SOLUTION

Solving, Ans.

or Ans.u =

u =

47.088 cos u - 1 5 sin 2u + 0 = 0

Fs = (0.3 sin u) = in u

47.088 cos u - Fs(0.3 cos u) + = 0

[2(78.48)(0.15 cos u) + 78.48(0.3 cos u) - Fs(0.3 cos u) + du = 0

dU = 0; 2(78.48) dy1 + 78.48dy2 - Fsdy2 + du = 0

dy2 = 0.3cos u du

dy1 = 0.15 cos u du

y2 = 0.3 sin u

y1 = 0.15 sin u

AM

D

C

B300 mm

300 mm

300 mm

ku

u

100 N3000 N

100

100]

100

900s3000

3 10

42.7°

54.8°

11–14.

SOLUTION

Ans.k = 1.24 kN>m

2(78.48)(0.15 cos 25°) + 78.48(0.3 cos 25°) - 0.1268k(0.3 cos 25°) = 0

u = Fs = k(0.3 sin 25°) = 0.1268k

[2(78.48)(0.15 cos u) + 78.48(0.3 cos u) - Fs(0.3 cos u)] du = 0

dU = 0; 2(78.48)dy1 + 78.48dy2 - Fsdy2 = 0

dy1 = 0.15 cos u du, dy2 = 0.3 cos u du

y1 = 0.15 sin u, y2 = 0.3 sin u

Each member of the pin-connected mechanism has a massof 8 kg. If the spring is unstretched when determinethe required stiffness k so that the mechanism is inequilibrium when Set M = 0.u =

u = 0°,

AM

D

C

B300 mm

300 mm

300 mm

ku

u

25°.

25°;

11–15.

SOLUTION

Ans.F =M

2a sin u

-M du + F (2a sin u)du = 0

dU = 0; -M du - F dx = 0

x = 2a cos u, dx = -2a sin u du

The service window at a fast-food restaurant consists of glassdoors that open and close automatically using a motor whichsupplies a torque M to each door. The far ends, A and B,move along the horizontal guides. If a food tray becomesstuck between the doors as shown, determine the horizontalforce the doors exert on the tray at the position u. M A

a a a a

C B DM

u u

11–16.

If the spring is unstretched when , the mass of thecylinder is 25 kg, and the mechanism is in equilibrium when

, determine the stiffness k of the spring. Rod ABslides freely through the collar at A. Neglect the mass of therods.

SOLUTION

Free-Body Diagram: When undergoes a positive virtual angular displacement of ,the dash line configuration shown in Fig. a is formed. We observe that only the springforce and the weight W of the cylinder do work when the virtual displacement takes place.

Virtual Displacement: The position of the point B at which and act isspecified by the position coordinates and , measured from the fixed point C.

(1)

(2)

Virtual–Work Equation: In this case must be resolved into its horizontal andvertical component, i.e. and . Since and Wand act towards the negative sense of their corresponding virtual displacements, theirwork is negative. However, does positive work since it acts towards the positivesense of its corresponding virtual displacement. Thus,

(3)

Substituting , Eqs. (1) and (2) into Eq. (3), we have

Using the indentity , the above equation canbe rewritten as

Since , then

(4)

Applying the law of cosines to the geometry shown in Fig. b, we have

Thus, the stretch of the spring is given by

= 0.1171 m

x = 20.5625 - 0.54 cos 45° - 20.5625 - 0.54 cos 30°

= 20.5625 - 0.54 cos u

AB = 20.452 + 0.62 - 2(0.45)(0.6) cos u

Fsp =245.25 sin u

cos (u - f)

110.3625 sin u - 0.45Fsp cos (u - f) = 0

du Z 0

du A110.3625 sin u - 0.45Fsp cos (u - f) B = 0

cos (u - f) = cos u cos f + sin u sin f

du A110.3625 sin u - 0.45Fsp( cos u cos f + sin u sin f) B = 0

-Fsp cos f(0.45 cos udu) + Fsp sin f(-0.45 sin udu) - 245.25(-0.45 sin udu) = 0

W = 25(9.81) = 245.25 N

-Fsp cos fdxB + Fsp sin fdyB + (-WdyB) = 0dU = 0;

(Fsp)y

(Fsp)x(Fsp)y = Fsp sin f(Fsp)x = Fsp cos fFsp

dyB = -0.45 sin uduyB = 0.45 cos u

dxB = 0.45 cos uduxB = 0.45 sin u

yBxB

FspWD

Fsp

duu

u = 45°

u = 30°

A

B

C

0.6 m

0.45 m

k

u

11–16. (continued)

The magnitude of computed using the spring force formula is therefore

The angle at can be obtained by referring to the geometry shown in Fig. c.

Substituting and the results for and into Eq. (4), we have

Ans. k = 1484 N>m = 1.48 kN>m

0.1171 k = 245.25 sin 45°

cos(45° - 41.53°)

fFspu = 45°

f = 41.53°

tan f =0.6 - 0.45 cos 45°

0.45 sin 45°

u = 45°f

Fsp = kx = 0.1171 k

Fsp

0.5 m

M � 600 N�m

k � 300 N�m/rad

0.5 m

A

B

Cu

11–17.

If the spring has a torsional stiffness of and it is unstretched when , determine the angle when the frame is in equilibrium.

SOLUTION

Free-Body Diagram: When undergoes a positive virtual angular displacement of, the dash line configuration shown in Fig. a is formed. We observe that only

couple moment M and the torque developed in the torsional spring do workwhen the virtual displacement takes place. The magnitude of can be computedusing the spring force formula,

Virtual Displacement: Since , then

(1)

Virtual–Work Equation: Since M and act towards the negative sense of theircorresponding angular virtual displacements, their work is negative. Thus,

; (2)

Substituting , , and Eq. (1) into Eq. (2), we have

Since , then

Ans.u = 1.071 rad = 61.4°

-600 + 600(p - 2u) = 0

du Z 0

du[-600 + 600(p - 2u)] = 0

-600du - 2[300(p - 2u)](-du) = 0

Msp = 300(p - 2u)M = 600 N # m

-Mdu + 2(-Mspda) = 0dU = 0

Msp

da = -du

a =p

2- u

Msp = k(2a) = 300 c2ap

2- ub d = 300(p - 2u)

Msp

Msp

du

u

uu = 90°k = 300 N # m>rad

11–18.

A 5-kg uniform serving table is supported on each side bypairs of two identical links, and , and springs . Ifthe bowl has a mass of , determine the angle where thetable is in equilibrium. The springs each have a stiffness of

and are unstretched when . Neglectthe mass of the links.

u = 90°k = >m

u1 kgCECDAB

SOLUTIONFree - Body Diagram: When undergoes a positive virtual angular displacement of

, the dash line configuration shown in Fig. a is formed. We observe that only thespring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do workwhen the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula,

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp arespecified by the position coordinates and xC, respectively. Here,are measured from the fixed point B while xC is measured from the fixed point D.

(1)

(2)

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense oftheir corresponding virtual displacement, their work is negative. Thus,

(4)

Substituting , ,

, Eqs. (1), (2), and (3) into Eq. (4), we have

Since , then

Solving the above equation,

Ans.

Ans.u = .1°

-7.3575 + 25 sin u = 0

cos u = 0 u = 90°

cos u(-7.3575 + 25 sin u) = 0

-7.3575 cos u + 25 sin u cos u = 0

du Z 0

du A -7.3575 cos u + 25 sin u cos u B= 0

-4.905(0.25 cos udu) - 24.525(0.25 cos udu) - u(-0.25 sin udu) = 0

Fsp = u N

Wt = a52b(9.81) = 24.525 NWb = a

12b(9.81) = 4.905 N

dU = 0; -WbdyGb+ A -WtdyGt

B + A -FspdxC B = 0

xC = 0.25 cos u dxC = -0.25 sin udu

yGt= 0.25 sin u + a dyGt

= 0.25 cos udu

yGb= 0.25 sin u + b dyGb

= 0.25 cos udu

yGband yGt

yGb, yGt

,

Fsp = kx = A u B = u N.

du

u

A C k

250 mm

250 mm 150 mm

150 mm

BD

E

u u

400 N

400 0.25 cos 100 cos

100 cos

100 cos

17

11–19.

SOLUTIONFree - Body Diagram: When undergoes a positive virtual angular displacement of

, the dash line configuration shown in Fig. a is formed. We observe that only thespring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do workwhen the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, .

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp arespecified by the position coordinates and xC, respectively. Here,are measured from the fixed point B while xC is measured from the fixed point D.

(1)

(2)

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense oftheir corresponding virtual displacement, their work is negative. Thus,

(4)

Substituting , ,

, Eqs. (1), (2), and (3) into Eq. (4), we have

Since , then

When , then

Ans.k =117.72sin 60°

= 1 6 N>m

u =

k =117.72

sin u

-7.3575 cos u + 0.0625k sin u cos u = 0

du Z 0

du A -7.3575 cos u + 0.0625k sin u cos u B = 0

-4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(-0.25 sin udu) = 0

Fsp = 0.25k cos u N

Wt = a52b(9.81) = 24.525 NWb = a

12b(9.81) = 4.905 N

dU = 0; -WbdyGb+ A -WtdyGt

B + A -FspdxC B = 0

xC = 0.25 cos u dxC = -0.25 sin udu

yGt= 0.25 sin u + a dyGt

= 0.25 cos udu

yGb= 0.25 sin u + b dyGb

= 0.25 cos udu

yGband yGt

yGb, yGt

,

Fsp = kx = k A0.25 cos u B = 0.25 k cos u

du

u

A 5-kg uniform serving table is supported on each side bytwo pairs of identical links, and , and springs . Ifthe bowl has a mass of and is in equilibrium when

, determine the stiffness of each spring.The springsare unstretched when . Neglect the mass of the links.u = 90°

ku =1 kg

CECDAB

A C k

250 mm

250 mm 150 mm

150 mm

BD

E

u u

60°

60°

3

11–20. The “Nuremberg scissors” is subjected to ahorizontal force of . Determine the angle forequilibrium. The spring has a stiffness of andis unstretched when .u = 15°

k = 15 kN>muP = 600 N

P

200 mm

200 mm

A

CD

E

Bk

u

11–21.horizontal force of . Determine the stiffness k ofthe spring for equilibrium when . The spring isunstretched when .u = 15°

u = 60°P = 600 N

P

200 mm

200 mm

A

CD

E

Bk

u

The “Nuremberg scissors” is subjected to a

11–22.

The dumpster has a weight W and a center of gravity at G.Determine the force in the hydraulic cylinder needed tohold it in the general position .u

SOLUTION

Ans.F = aW(a + b - d tan u)

acb2a2 + c2 + 2a c sin u

F(a2 + c2 + 2a c sin u)- 12 ac cos u du - W(a + b) cos u du + Wd sin u = 0

dU = 0; Fds - Wdy = 0

dy = (a + b) cos u du - d sin u

y = (a + b) sin u + d cos u

ds = (a2 + c2 + 2a c sin u)- 12 ac cos u du

= 2a2 + c2 + 2a c sin u

s = 2a2 + c2 - 2a c cos (u + 90°)θ

b dG

a

c

du

du

11–23.

The crankshaft is subjected to a torque of Determine the horizontal compressive force F applied tothe piston for equilibrium when u = 60°.

M = 50 N #m.

SOLUTION

For

Ans.

Ans.F = 512 N

(-50 + 0.09769F)du = 0

dx = -0.09769 du

u = 60°, x = 0.4405 m

dU = 0; -50du - Fdx = 0

0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx

(0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u)

100 mm400 mm

F

M

u

11–24.

SOLUTION

(1)

From Eq. (1)

Ans.F =50020.04 cos2u + 0.6

(0.2 cos u + 20.04 cos2u + 0.6) sin u

x =0.2cos u + 20.04 cos2u + 0.6

2

x =0.2cos u;20.04 cos2u + 0.6

2, since20.04 cos2u + 0.6 7 0.2 cos u

x2 - 0.2x cos u - 0.15 = 0

F =50(2x - 0.2cos u)

0.2x sin u

-50du - Fa 0.2x sin u0.2 cos u - 2x

b du = 0, du Z 0

dU = 0; -50du - Fdx = 0

dx = a 0.2x sin u0.2 cos u - 2x

b du0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx

(0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u)

The crankshaft is subjected to a torque ofDetermine the horizontal compressive force F and plotthe result of F (ordinate) versus (abscissa) for0° … u … 90°.

u

M = 50 N # m.

100 mm400 mm

F

M

u

11–25.

SOLUTION

Ans.m = 100 kg

[m(9.81) - 981]du = 0

-m(9.81)(0.625)a 11.25b(-2du) - 981(1.25)a 2

2.5bdu = 0

-m(9.81)(0.625 cos f df) - 981( 1.25 cos u du) = 0

dU = 0; -m(9.81)dy1 - 981dy2 = 0

dy2 = 1.25 cos u du

dy1 = 0.625 cos f df

y2 = 1.25 sin u

y1 = a1.252b sin f

df = -2du

0.75df = -1.5du

1.25a0.751.25b df = -2.5a1.5

2.5bdu

1.25 sin f df = -2.5 sin u du

3 - 1.25 cos f = 2.5 cos u

x = 1.25 cos f; 3 - x = 2.5 cos u

Rods AB and BC have centers of mass located at theirmidpoints. If all contacting surfaces are smooth and BC hasa mass of 100 kg, determine the appropriate mass of ABrequired for equilibrium.

2 m1 m

0.75 m

1.5 mA

B

C

11–26.degree-of-freedom system is expressed by the relation

, where y is given in meters,determine the equilibrium positions and investigate thestability at each position.

V = (3y3 + 2y2 - 4y + 50) J

If the potential energy for a conservative one-

11–27.

SOLUTION

Ans.

Ans.

Ans.d2V

dx2 = 4 7 0 Stablex = 0.167 m,

d2V

dx2 = -4 6 0 Unstablex = 0,

d2V

dx2 = 48x - 4

x = 0 and x = 0.167 m

124x - 42x = 0

dV

dx= 24x2 - 4x = 0

V = 8x3 - 2x2 - 10

If the potential function for a conservative one-degree-of-f erehw si metsys modeer x isgiven in meters, determine the positions for equilibrium andinvestigate the stability at each of these positions.

V = 18x3 - 2x2 - 102 J,

11–28.

If the potential function for a conservative one-degree-of-f erehw si metsys modeer

determine the positions for equilibrium andinvestigate the stability at each of these positions.0° 6 u 6 180°,

(12 sin 2u + 15 cos u) J,V =

SOLUTION

Choosing the angle

Ans.

and

Ans.

Ans.

Ans.d2V

du2 = 57.2 7 0 Stableu = 145°,

d2V

du2 = -57.2 6 0 Unstableu = 34.6°,

d2V

du2 = -48 sin 2u - 15 cos u

u = 145°

u = 34.6°

0° 6 u 6 180°

48 sin2 u + 15 sin u - 24 = 0

2411 - 2 sin2 u2 - 15 sin u = 0

24 cos 2u - 15 sin u = 0dV

du= 0;

V = 12 sin 2u + 15 cos u

11–29. If the potential energy for a conservative one-degree-of-freedom system is expressed by the relation

, determine theequilibrium positions and investigate the stability at each position.

0° … u … 90°V = (24 sin u + 10 cos 2u) J,

.

.

.

.

11–30.

SOLUTION

For equilibrium:

Ans.

and

Ans.

Stability:

Ans.

Ans.

Ans.d2V

du2 = 15 7 0, Stableu = 90°,

d2V

du2 = -24.4 6 0, Unstableu = 141°,

d2V

du2 = -24.4 6 0, Unstableu = 38.7°,

d2V

du2 = -40 cos 2u - 25 sin u

u = cos-1 0 = 90°

u = sin-1a2540b = 38.7° and 141°

1-40 sin u + 252 cos u = 0

dV

du= -20 sin 2u + 25 cos u = 0

V = 10 cos 2u + 25 sin u

If the potential function for a conservative one-degree-of-f erehw si metsys modeer

determine the positions for equilibrium andinvestigate the stability at each of these positions.0° 6 u 6 180°,

V = 110 cos 2u + 25 sin u2 J,

11–31.

Determine the angle for equilibrium and investigate thestability of the mechanism in this position. The spring has astiffness of and is unstretched when The block A has a mass of 40 kg. Neglect the mass ofthe links.

SOLUTION

Potential Function: With reference to the datum, Fig. a, the gravitational potentialenergy of block A is positive since its center of gravity is located above the datum.Here, m, where b is a constant. Thus,

The eleastic potential energy of spring BF can be computed using , where

. Thus,

The total potential energy of the system is

Equilibrium Configuration: Taking the first derivative of V, we have

Equilibrium requires . Thus,

Ans.

Ans.

Stability: We can write . Thus, the second

derivative of V is

At

Stable Ans.

At

Unstable Ans.= -201.10 6 0

u = 35.54°, d2V

d2u2u= 35.54°

= -176.58 sin 35.54° - 303.75 cos 71.09°

= 127.17 7 0

u = 90°, d2V

d2u2u= 0°

= -176.58 sin 90° - 303.75 cos 180° = 127.17 7 0

d2V

d2u= -176.58 sin u - 303.75 cos 2u

dV

du= 176.58 cos u - 151.875 sin 2u

u = 35.54° = 35.5°

176.58 - 303.75 sin u = 0

cos u = 0 u = 90°

cos u(176.58 - 303.75 sin u) = 0

176.58 cos u - 303.75 cos u sin u = 0

dV

du= 0

dV

du= 176.58 cos u - 303.75 cos u sin u

V = Vg + Ve = 176.58 sin u + 151.875 cos2 u + 392.4 b

Ve =12

(1500)(0.45 cos u)2 = 151.875 cos2 u

s = 0.45 cos u m

Ve =12

ks 2

Vg = mgy = 40(9.81)(0.45 sin u + b) = 176.58 sin u + 392.4 b

y = (0.45 sin u + b)

u = 90°.k = 1.5 kN>m

u

450 mm

600 mm

C

B D

E

F

kA

uu

11–32.

The spring of the scale has an unstretched length of a. Determine the angle for equilibrium when a weight Wis supported on the platform. Neglect the weight of the members. What value W would be required to keep the scale in neutral equilibrium when u = 0°?

u

SOLUTIONPotential Function: The datum is established at point A. Since the weight W isabove the datum, its potential energy is positive. From the geometry, the springstretches and .

Equilibrium Position: The system is in equilibrium if .

Solving,

Ans.

Stability: To have neutral equilibrium at .

Ans.W = 2kL

d2V

du22u - 0°

= 4kL2 cos 0° - 2WLcos 0° = 0

d2V

du2 = 4kL2 cos 2 u - 2WL cos u

u = 0°, d2V

du22u - 0°

= 0

u = 0° or u = cos-1a W2kLb

dV

du= 2kL2 sin 2u-2WLsin u = 0

dV

du= 4kL2 sin u cos u - 2WL sin u = 0

dV

du= 0

= 2kL2 sin 2u + 2WL cos u

=12

(k)(2 L sin u)2 + W(2L cos u)

=12kx2 + Wy

V = Ve + Vg

y = 2Lcos ux = 2L sin u

k

L

W

LL

L

a

uu

� .

The uniform beam has mass M. If thecontacting surfaces are smooth, determinethe angle �� for equilibrium and investigatethe stability of the beam when it is in thisposition. The spring has an unstretchedlength of �.Units Used:

kN 103 N�

Given:

M 200 kg� k 1.2kNm

�

� 0.5 m�l 2 m�

Solution:

V M gl2

��

sin �� 12

k l cos �� 2��

�Vd

dV'� M g

l2

��

cos �� k l cos �� l sin �� 0�

2�Vd

d

2V''� M� g

l2

��

sin �� k l2 sin �� 2� k l cos �� l cos ��

There are 2 equilibrium points

Guess � 30 deg� Given M gl2

��


�1 Find �� 1 36.4 deg�

Guess � 60 deg� Given M gl2

��


�2 Find �� 2 62.3 deg�

Check Stability

V''1 M� gl2

��

sin �1� k l2 sin �1� 2� k l cos �1� �� l cos �1� ��

V''2 M� gl2

��

sin �2� k l2 sin �2� 2� k l cos �2� �� l cos �2� ��

V''1 1.624� kN m�� Unstable V''2 1.55 kN m�� Stable

Ans.

Ans.

Ans.

–

11–36.

Determine the angle for equilibrium and investigate thestability at this position. The bars each have a mass of 3 kgand the suspended block D has a mass of 7 kg. Cord DC hasa total length of 1 m.

u

SOLUTION

Ans.

Ans.= -70.2 6 0 Unstable

u = 12.1°, d2V

du2 = 0.5[-3(9.81) sin 12.1° - 14(9.81) cos 12.1°]

d2V

du2 = l(-W sin u - 2WD cos u)

u = 12.1°

tan u =W

2WD=

3(9.81)

14(9.81)= 0.2143

dV

du= l(W cos u - 2WD sin u) = 0

= Wl sin u - WDl(3 - 2cos u)

V = 2Wy1 - WDy 2

y2 = l + 2l(1 - cos u) = l(3 - 2 cos u)

y1 =l

2sin u

l = 500 mm

500 mm

500 mm

A

D

C

500 mm

u u

11 37.

Each of the two springs has anunstretched length �. Determine the massM of the cylinder when it is held in theequilibrium position shown, i.e., y = a.

Given:

a 1 m�

b 500 mm�

� 500 mm�

k 200Nm

�

Solution:

V 2k2

y2 b2� �� 2 M g y��

dVdy

2k y2 b2� �� y

y2 b2�

��

�

M g��

Set y a�

Guess M 1 kg�

Given

2k y2 b2� �� y

y2 b2�

��

�

M g� 0�

M Find M( )� M 22.5 kg� Ans.

–

11–38.

A homogeneous block rests on top of the cylindricalsurface. Derive the relationship between the radius of thecylinder, r, and the dimension of the block, b, for stableequilibrium. Hint: Establish the potential energy functionfor a small angle , i.e., approximate , and

.cos u L 1 - u2>2sin u L 0u

SOLUTIONPotential Function: The datum is established at point O. Since the center of gravityfor the block is above the datum, its potential energy is positive. Here,

.

[1]

For small angle , and . Then Eq. [1] becomes

Equilibrium Position: The system is in equilibrium if

Stability: To have stable equilibrium, .

Ans.b 6 2r

ar -b

2b 7 0

d2V

du22u=0°

= War -b

2b 7 0

d2V

du22u=0°

7 0

dV

du= War -

b

2bu = 0 u = 0°

dV

du= 0.

= Wa ru2

2-bu2

4+ r +

b

2b

V = WB ar +b

2b ¢1 -

u2

2≤ + ru2R

u = 1 -u2

2sin u = uu

V = Wy = W c ar +b

2b cos u + ru sin u d

y = ar +b

2b cos u + ru sin u

b

r

b

11 39.

The uniform rod AB has a mass M. Ifspring DC is unstretched at � = 90 deg,determine the angle � for equilibrium andinvestigate the stability at the equilibriumposition. The spring always acts in thehorizontal position due to the roller guideat D.

Units Used:

kN 103 N�

Given:

M 80 kg� a 1 m�

k 2kNm

� b 2 m�

Solution:

V M ga b�

2��

�

sin �� 12

k a cos �� 2��

V'�

Vdd

� M ga b�

2��

�

cos �� k2

a2 sin 2��

V'' 2�Vd

d

2� M� g

a b�2

��

�

sin �� k a2 cos 2��

Equilibrium

Guess � 30 deg� Given M ga b�

2��

�

cos �� k2

a2 sin 2�� 0� �1 Find ��

Guess � 70 deg� Given M ga b�

2��

�

cos �� k2

a2 sin 2�� 0� �2 Find ��

Check Staibility

V''1 M� ga b�

2��

�

sin �1� k a2 cos 2�1� ��

V''2 M� ga b�

2��

�

sin �2� k a2 cos 2�2� ��

�1 36.1 deg� V''1 1.3� kN m�� Unstable

�2 90.0 deg� V''2 0.82 kN m�� Stable

Ans.

Ans.

–

11–40.

A spring with a torsional stiffness is attached tothe pin at. It is unstretched when the rod assembly is in the vertical

position. Determine the weight of the block that resultsin neutral equilibrium. Hint: Establish the potential energyfunction for a small angle , i.e., approximate , and

.

SOLUTION

Potential Function: With reference to the datum, Fig. a, the gravitational potential

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 . Thus, .

However, for a small angle , .Thus,

The elastic potential energy of the torsional spring can be computed using

, where . Thus,


Equilibrium Configuration: Taking the first derivative of , we have

Equilibrium requires . Thus,

Stability: The second derivative of is

To have neutral equilibrium at , . Thus,

Ans.

Note: The equilibrium configuration of the system at is stable if

and is unstable if .W > 8k3L

ad2V

du2 < 0bW < 8k3L

ad2V

du2 > 0b

u = 0°

W =8k3L

-3WL

2+ 4k = 0

d2V

du22u= 0°

= 0u = 0°

d2V

du2 = -3WL

2+ 4k

V

u = 0°

ua -3WL

2+ 4kb = 0

dV

du= 0

dV

du= -

3WL2

u + 4ku = ua -3WL

2+ 4kb

V

V = Vg + Ve =3WL

2 a1 -

u2

2b + 2ku2

Vg =12

k(2u)2 = 2ku2

b = 2uVe =12

kb2

Vg = Wy = Wa32

Lb a1 -u2

2 b =

3WL2

a1 -u2

2b

cos u � 1 -u2

2u

y =L

2 cos u + L cos u =

32

L cos uu

cos u L 1 - u2>2sin u L 0u

WB

k

C

A

Bk

L2

L2

L2

11 41.

Each bar has a mass per length of m0. Determine the angles �and � at which they are suspended in equilibrium. The contactat A is smooth, and both are pin con-nected at B.

Solution:

� �� atan12

��

�

V3l2

� m03l4

��

�

cos �� l m0l2

��

cos �� l2

m0 l cos �� l4

sin ��

�

��

�Vd

d

9m0 l2

8sin �� m0 l2 sin ��

m0 l2

8cos �� 0�

Guess � 10 deg� � 10 deg�

Given � �� atan12

��

�98

sin �� sin �� 18

cos �� 0�

�

�

��

�

Find � ��

�

��

�

9.18

17.38��

�

deg� Ans.

–

11–42.

The small postal scale consists of a counterweight connected to the members having negligible weight.Determine the weight that is on the pan in terms of theangles and and the dimensions shown. All members arepin connected.

fu

W2

W1,W2

W1

b a

af

f

uSOLUTION

where is a constant and

Ans.W2 = W1abab

sin ucos f

dV

du= W1b sin u - W2 a cos (90° - u - g)

= -W1b cos u + W2 a sin (90° - u - g)

V = -W1y1 + W2y2

f = (90° - u - g)g

y2 = a sin f = a sin (90° - u - g)

y1 = b cos u

11–43.

If the spring has a torsional stiffness k = 300 and isunwound when , determine the angle for equilibriumif the sphere has a mass of 20 kg. Investigate the stability atthis position. Collar C can slide freely along the vertical guide.Neglect the weight of the rods and collar C.

SOLUTION

Potential Function: With reference to the datum, Fig. a, the gravitational potentialenergy of the sphere is positive its center of gravity is located above the datum. Here,

. Thus,

The elastic potential energy of the torsional spring can be computed using ,

where . Thus,


Equilibrium Configuration: Taking the first derivative of V, we have

Equilibrium requires

Solving by trial and error, we obtain

Ans.

Stability: The second derivative of V is

At the equilibrium configuration of

Ans.d2V

du22u= 76.78°

= -44.14 sin 38.39° + 300 = 272.58 7 0 Stable

u = 76.78°,

d2V

du2 = -44.145 sin (u/2) + 300

u = 1.340 rad = 76.78° = 76.8°

88.29 cos (u/2) + 300u - 150p = 0

dV

du= 0, Thus,

dV

du= 88.29 cos (u/2) + 300u - 150p

V = Vg + Ve = 176.58 sin (u/2) + 150u2 - 150pu + 37.5p2

Ve =12

300 ¢p2

- u≤2

= 150u2 - 150pu + 37.5p2

b =p

2- u

Ve =12

kb2

Vg = mgy = 20(9.81)(0.9 sin u>2) = 176.58 sin (u>2)

y = (0.3 sin (u>2) + 0.6 sin (u>2))m = (0.9 sin (u/2))

u = 90°N # m>rad

300 mm

300 mm

300 mm

D

B

A

C

ku

11–44.

The truck has a mass of 20 Mg and a mass center at G.Determine the steepest grade along which it can parkwithout overturning and investigate the stability in thisposition.

u

SOLUTIONPotential Function: The datum is established at point A. Since the center of gravityfor the truck is above the datum, its potential energy is positive. Here,

.


Since ,

Ans.

Stability:

Thus, the truck is in unstable equilibrium at Ans.u = 23.2°

d2V

du22u=23.20°

= W(-1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0

d2V

du2 = W(-1.5 sin u - 3.5 cos u)

u = 23.20° = 23.2°

1.5 cos u - 3.5 sin u = 0

W Z 0

dV

du= W(1.5 cos u - 3.5 sin u) = 0

dV

du= 0

V = Vg = Wy = W(1.5 sin u + 3.5 cos u)

y = (1.5 sin u + 3.5 cos u) m

G

u

3.5 m

1.5 m1.5 m

11–45.

SOLUTIONPotential Function: The datum is established at point A. Since the center of gravityof the cylinder is above the datum, its potential energy is positive. Here,

.

Equilibrium Position: The system is in equilibrium if .

Stability:

Thus, the cylinder is in unstable equilibrium at (Q.E.D.)u = 0°

d2V

du22u=0°

= -mgd cos 0° = -mgd 6 0

d2V

du2 = -mgd cos u

sin u = 0 u = 0°

dV

du= -mgd sin u = 0

dV

du= 0

V = Vg = Wy = mg(r + d cos u)

y = r + d cos u

The cylinder is made of two materials such that it has a massof m and a center of gravity at point G. Show that when Glies above the centroid C of the cylinder, the equilibrium isunstable.

C

Ga

r

11–46.

If each of the three links of the mechanism has a weight W,determine the angle for equilibrium. The spring, whichalways remains vertical, is unstretched when u = 0°.

u

SOLUTION

Assume , so .

Ans.

or

Ans.u = 90°

u = sin-1 a4Wkab

4W - ka sin u = 0

cos u Z 0u 6 90°

(W - ka sin u)a cos u du + Wa cos u du + W(2a)cos u du = 0

dU = 0; (W - Fs)dy1 + Wdy2 + Wdy3 = 0

Fs = ka sin u

y3 = 2a + 2a sin u dy3 = 2a cos u du

y2 = 2a + a sin u dy2 = a cos u du

y1 = a sin u dy1 = a cos u du

2 a

k

2 a

a

a

u

11–47.

If the uniform rod OA has a mass of 12 kg, determine themass m that will hold the rod in equilibrium when Point C is coincident with B when OA is horizontal. Neglectthe size of the pulley at B.

u = 30°.

SOLUTIONGeometry: Using the law of cosines,

Potential Function: The datum is established at point O. Since the center of gravityof the rod and the block are above the datum, their potential energy is positive.

dna,ereH


At

Ans.m = 5.29 kg

dV

du`u=30°

= - 29.43m cos 30°

210 - 6 sin 30°+ 58.86 cos 30° = 0

u = 30°,

= - 29.43m cos u

210 - 6 sin u+ 58.86 cos u

dV

du= -9.81m c - 1

2110 - 6 sin u2- 1

21-6 cos u2 d + 58.86 cos u

dV

du`u=30°

= 0.

= 29.43m - 9.81m1210 - 210 - 6 sin u2 + 58.86 sin u

= 9.81m33 - 1210 - 210 - 6 sin u24 + 117.7210.5 sin u2V = Vg = W1y1 + W2y2

y2 = 0.5 sin u m.y1 = 3 - l = 33 - 1210 - 210 - 6 sin u24 m

l = lAB - lA¿B = 210 - 210 - 6 sin u

lAB = 212 + 32 = 210 m

lA¿B = 212 + 32 - 2112132 cos190° - u2 = 210 - 6 sin u

m

�

1 mA

C

O

B

3 m

11–48.

The triangular block of weight W rests on the smoothcorners which are a distance a apart. If the block has threeequal sides of length d, determine the angle forequilibrium.

u

SOLUTION

Require, Ans.

or

Ans.u = cos-1 d

4aba

-0.5774 d -a

sin 60° (-2 cos u) = 0

u = 0°sin u = 0

dV

du= Wc( -0.5774 d) sin u -

a

sin 60° (-1.5 sin u cos u - 0.5 sin u cos u)d = 0

V

y = cos u - AF

= Wy

=a

sin 60° (0.75 cos2u - 0.25 sin2u)

AF =a

sin 60° (sin (60° + u)) sin (60° - u)

AD =a

sin 60° (sin (60° + u))

AD

sin a=

a

sin 60°

AF = AD sin f = AD sin (60° - u)

d

a

G 60

60

u

d

32

11–49.

SOLUTION

(Q.E.D)cos usin3 u

=a

2r

ra cos usin2 u

b =a

2sin u

dV

du= 2Wa -r csc u cot u +

a

2sin ub = 0

V = 2War csc u -a

2cos ub

Two uniform bars, each having a weight W, are pin-connected at their ends. If they are placed over a smoothcylindrical surface, show that the angle for equilibriummust satisfy the equation cos u>sin3 u = a>2r.

u��a a

r

11-51. The uniform bar AB 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

K = 100 N/m

1 m

1 m

1

1

1 1 100 (1 – 1sinθ)

1 m

1 m

–50

[–50 + 100(1 – sinθ)](1cosθ δθ) = 0

50 – 100 sinθ = 0

.

K = 100 N/m 1 m

1 m

A

B

1m

1m

W

A

y

F1B

K = 100 N/m

1 m

1 m

1 m

1 m

11-51

2

2

2

150(1sin ) (100)(1 1sin )2

50cos 100(1 1sin )( 1cos )

Require, 0

50cos 100(1 1sin )cos 0

cos 0 or 50 100(1 sin ) 0

90 , or 30 Ans.

50sin 100(1 1sin )(1sin ) 100( 1cos )( 1cos

y

V

dVd

dVd

d Vd

θ

θ θ

θ θ θθ

θ

θ θ θ

θ θ

θ θ

θ θ θ θ θθ

=

= + −

= + − −

=

− − =

= − − =

= =

= − + − + − −

22

2

2

2

2

2

)

50sin 100(1 1sin )sin 100cos

90 , 50 0 Unstable Ans.

30 , 75 0 S table Ans.

d Vd

d Vd

d Vd

θ θ θ θθ

θθ

θθ

= − + − +

= = − <

= = <

1sin

K = 100 N/m 1 m

1 m

A

B 1m

1m

W

A

y

F1B

11–53.

The uniform right circular cone having a mass m issuspended from the cord as shown. Determine the angle at which it hangs from the wall for equilibrium. Is the conein stable equilibrium?

u

SOLUTION

Ans.

Stable Ans.

u = 9.46°, d2V

du2 = 1.52 a mg 7 0

d2V

du2 = - a - 3a2

cos u -a

4sin ub mg

u = 9.46°

tan u = 0.1667

3 sin u = 0.5 cos u

dV

du= - a -3a

2sin u +

a

4cos ub mg = 0

V = - a3a2

cos u +a

4sin ubmg

a

a

a

u

u

11–46. The assembly shown consists of a semicylinder anda rectangular block. If the block weighs 8 lb and thesemicylinder weighs 2 lb, investigate the stability when theassembly is resting in the equilibrium position. Set 4 in.

h

100 mm

250 mm

Wb = 40 N

50 mm

d = 42.441 mm

100 mm

yb = (100 + 50 cos �) mm

ys = (100 – 42.441 cos �) mm

Ws = 10 N

11–54. The assembly shown consists of a semicylinder and a rectangular block. If the block weighs 40 N and the semicylinder weighs 10 N, investigate the stability when the assembly is resting in the equilibrium position. Set h = 100 mm.

d = 4(100)

3π = 42.441 mm

V = Vg = 10(100 – 42.441 cos �) + 40(100 + 50 cos �)

dV

d� = 424.41 sin � – 2000 sin � = 0

sin � = 0

� = 0° (equilibrium position)

d V

d

2

2� = 424.41 cos � – 2000 cos �

At � = 0°, d V

d

2

2� = –1575.6 < 0 Unstable Ans.

11–47. The 2-lb semicylinder supports the block which hasa specific weight of . Determine the height hof the block which will produce neutral equilibrium in theposition shown.

80 lb ft3

h

100 mm

250 mm

Wb = [13.5 (10–6) (h) (200) (250)] N

d = 42.441 mm

100 mm

ys = (100 – 42.441 cos �) mm

Ws = 10 N

yb = 100 +2

cosh

�⎛⎝⎜

⎞⎠⎟

mm

11–55. The 10-N (� 1-kg) semicylinder supports the block which has a specific weight of � = 13.5 kN/m3. Determine the height h of the block which will produce neutral equilibrium in the position shown.

d = 4(100)

3π = 42.441 mm

V = Vg = 10(100 – 42.441 cos �)

+ 13.51

10(200)(250)

6

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

h 100 +2

cosh

�⎛⎝⎜

⎞⎠⎟

dV

d� = 424.41 sin � – 0.3375 h2 sin � = 0

sin � = 0

� = 0° (equilibrium position)

d V

d

2

2� = 424.41 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-of-freedom system is expressed by therelation V = (ax3 + bx2 + cx + d), determine the equilibrium positions and investigate the stability ateach position.

Given: a 4N

m2�� b 1�

Nm

�� c 3� N�� d 10N m��

Solution:

V a x3� bx2

� c x�� d�=

Required Position :

dVdx

3 a� x2� 2 b� x�� c�= 0=

x12� b� 4 b2

� 4 3 a�( )� c��

2 3� a��

x1 0.590 m�

x22� b� 4 b2

� 4 3� a� c��

2 3� a��

x2 0.424� m�

Stability :

d2V

dx2V''= 6 a� x� 2 b��=

At x x1�� V''1 6 a� x1� 2 b�� V''1 12.2Nm

� V''1 0� Stable

At x x2�� V''2 6 a� x2� 2 b�� V''2 12.2�Nm

� V''2 0� Unstable

Ans.

Ans.

Ans.

Ans.

–

11–22. Determine the weight of block required tobalance the differential lever when the 20-lb load F isplaced on the pan.The lever is in balance when the load andblock are not on the lever. Take .12 in

100 mm 100 mm x

A

B

C G

ED

50 mm

F

11–57. Determine the weight of block G required to balance the differential lever when the 100 N (� 10 kg) load F is placed on the pan. The lever is in balance when the load and block are not on the lever. Take x = 300 mm.

Free – Body Diagram: When the lever undergoes a virtual angular displacement of �� about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place.

Virtual Displacement: Since �yG is very small, the vertical virtual displacement of block G and load F can be approximated as

�yG = (300 + 100) �� = 400 (1)

�yF = 50 �� (2)

Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus,

�U = 0; WG �yG + (–WF �yF) = 0 (3)

Substituting WG = 100 N and Eqs. (1) and (2) into Eq. (3),

WG(400 ��) – 100(50 ��) = 0

��(400 WG – 5000) = 0

Since �� ≠ 0, then

400 WG – 5000 = 0

WG = 12.5 N Ans.

��

11–23. If the load weighs 20 lb and the block weighs2 lb, determine its position for equilibrium of thedifferential lever. The lever is in balance when the load andblock are not on the lever.

100 mm 100 mm x

A

B

C G

ED

F

50 mm

11–58. If the load F weighs 100 N and the block G weighs 10 N, determine its position x for equilibrium of the differential lever. The lever is in balance when the load and block are not on the lever.

Free – Body Diagram: When the lever undergoes a virtual angular displacement of �� about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place.

Virtual Displacement: Since �yG is very small, the vertical virtual displacement of block G and load F can be approximated as

�yG = (100 + x) �� (1)

�yF = 50 �� (2)

Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus,

�U = 0; WG �yG + (–WF �yF) = 0 (3)

Substituting WF = 100 N, WG = 10 N, Eqs. (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 thelever. Determine the horizontal force Fon the piston for equilibrium.

Solution:

�s 2 l ��

x 2 l cos ��

�x 2� l sin ��

�U P� �s F�x�� 0�

P� 2l�� F2l sin �� 0�

F P csc �� Ans.

–

11–6 . uniform serving table is supported on eachside by pairs of two identical links, and , and springs

. If the bowl has a mass of , determine the angle where the table is in equilibrium. The springs each have astiffness of and are unstretched when .Neglect the mass of the links.

u = 90°k = 200 N>m

u1 kgCECDAB

A C k

250 mm

250 mm 150 mm

150 mm

BD

E

u u

0 A 5-kg

11– .side by two pairs of identical links, and , and springs

. If the bowl has a mass of and is in equilibrium when, determine the stiffness of each spring.The springs

are unstretched when . Neglect the mass of the links.u = 90°ku = 45°

1 kgCECDAB

A C k

250 mm

250 mm 150 mm

150 mm

BD

E

u u

61 A 5-kg uniform serving table is supported on each

11

Rods AB and BC have centers ofmass located at their midpoints. If allcontacting surfaces are smooth andBC has mass mBC determine theappropriate mass mAB of AB requiredfor equilibrium.

Given:

mBC 150 kg�

a 0.8 m�

b 1 m�

c 2 m�

d 1.6 m�

Solution:

Use � as the independent variable

Define L1 a2 b2�� L2 c2 d2�� atandc

��

� � atanab

��

�

Then L1 cos �� L2 cos �� b c�� L1� sin �� L2 sin �� 0�

Thus ��L2 sin �� L1 sin ��

��

��

Also y1L12

sin �� y1L12

cos �� L2� sin �� cot ��

2��

��

y2L22

sin �� y2L22

cos ��

�U mAB� g�y1 mBC g�y2�� g mABL2 sin �� cot ��

2��

�

mBCL22

cos ��

�

��

�� 0�

mAB mBC tan �� cot �� mAB 150 kg�

62.

Ans.

–

solution a b -...

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