Mechanical Engineering Chapter 11 Homework When All The Weights Undergo Positive Virtual

1123

11–1.

SOLUTION

Free Body Diagram: The tension in cable AC can be determined by releasing

cable AC .The system has only one degree of freedom defined by the independent

coordinate .When undergoes a positive displacement ,only FAC and the weight

of lamp (10 lb force) do work.

Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B

using position coordinates yAand xA.

duuu

Use the method of virtual work to determine the tensions in

cable AC.The lamp weighs 10 lb.

B

A

C

45°30°

FAC =7.32

lb

11–2.

SOLUTION

x=Lcos u,dx=-Lsin udu

The scissors jack supports a load P. Determine the axial

force in the screw necessary for equilibrium when the jack

is in the position Each of the four links has a length Land

is pin-connected at its center.Points Band Dcan move

horizontally.

u.

CD

AB

P

u

11–3.

*11–4.

T

h

e

di

s

k

h

as a we

i

g

h

t of 10

lb

an

d

i

s su

bj

ecte

d

to a vert

i

ca

l

force and a couple moment

Determine the disk’s rotation if the end of the spring

w

raps around the periphery of the disk as the disk turns.

The spring is originally unstretched.

u

M=8lb#ft.P=8lb

SOLUTION

Since

W

Ans:

11–5.

The punch press consists of the ram R, connecting rod AB,

and a flywheel. If a torque of

M=75 N #m

is applied to the

flywheel, determine the force F applied at the ram to hold

the rod in the position

u=60°.

SOLUTION

Free Body Diagram. The system has only one degree of freedom, defined by the

independent coordinate

u

. When

u

undergoes a positive angular displacement

du

as

shown in Fig. a, only force F and couple moment M do work.

Virtual Displacement. The position of force F is measured from fixed point O by

position coordinate

xA

. Applying the law of cosines by referring to Fig. b,

Differentiating the above expression,

Virtual–Work Equation. When point A undergoes a positive virtual displacement,

and the ywheel undergoes positive virtual angular displacement

du

, both F and M

do negative work.

0.4 x

A

sin

u

0.4 cos u

–

2x

A

F+M=0

F

200 mm

M

B

R

A

600 mm

u

1128

11–6.

The flywheel is subjected to a torque of

M=75 N #m

.

Determine the horizontal compressive force F and plot the

result of F (ordinate) versus the equilibrium position

u

(abscissa) for

0°…u…180°.

SOLUTION

u

du

Virtual Displacement. The position of force F is measured from fixed point O by

position coordinate

xA

. Applying the law of cosines by referring to Fig. b,

Differentiating the above expression,

Virtual–Work Equation. When point A undergoes a positive virtual displacement

and the ywheel undergoes positive virtual angular displacement

du

, both F and M

Substituting Eq. (2) into (3)

Here,

M=75 N #m.

Then

Using Eq. (1) and (4), the following tabulation can be computed. Subsequently, the

graph of F vs

u

shown in Fig. c can be plotted

F

200 mm

M

B

R

A

600 mm

u

11–7.

SOLUTION

Free Body Diagram: The system has only one degree of freedom defined by the

independent coordinate .When undergoes a positive displacement ,only the

spring forces Fsp,the weight of the block (50 lb),the weights of the links (10 lb) and

the couple moment Mdo work.

duuu

Whenthe50-lb uniform block compresses the

twoverticalsprings 4in. Iftheuniform linksAB and CD

eachweigh10 lb,determinethe magnitudeoftheapplied

couplemoments Mneededto maintain equilibrium when

u=20°.

u

=20°,

A

BD

C

MM

1ft

4ft

1ft

k2lb/in.k2lb/in.

1ft

2ft

uu

Ans:

1130

*11–8.

The bar is supported by the spring and smooth

collar that allows the spring to be always perpendicular to

the bar for any angle If the unstretched length of the

spring is determine the force Pneeded to hold the bar in

the equilibrium position Neglect the weight of the bar.u.

l0,

u.

SOLUTION

Ans:

1131

11–9.

SOLUTION

x=4 cos uy=4sin u,

The 4-ft members of the mechanism are pin-connected at

their centers.If vertical forces act at Cand

Eas shown,determine the angle for equilibrium. The

spring is unstretched when Neglect the weight of

the members.

u=45°.

u

P

1=P

2=30 lb

P1P2

k= 200 lb/ft

CE

B

2ft

Ans:

11–10.

The thin rod of weight Wrest against the smooth wall and

floor.Determine the magnitude of force Pneeded to hold it

in equilibrium for a given angle

SOLUTION

Free-Body Diagram: The system has only one degree of freedom defined by the

independent coordinate .When undergoes a positive displacement ,only the

weight of the rod Wand force Pdo work.

Virtual Displacements: The weight of the rod Wand force Pare located from the

fixed points Aand Busing position coordinates and ,respectively

Virtual-Work Equation:When points Cand Aundergo positive virtual displacements

and ,the weight of the rod Wand force Fdo negative work.

dxA

dyC

xA

yC

duuu

A

l

B

P

θ

u.

11–11.

If each of the three links of the mechanism have a mass of

4 kg, determine the angle

u

for equilibrium. The spring,

which always remains vertical, is unstretched when

u=0°.

SOLUTION

u

W

Virtual Displacement. The positions of the weights

W

1, W

2, W

3

and spring force

Fsp are measured from fixed point A using position coordinates

y1, y2, y3

and

y4

respectively

y1=0.1 sin

u

d

y1=0.1 cos

u

du (1)

y2=0.2 sin

y2=0.2 cos

y3=0.1 sin

y3=0.1 cos

y4=0.5 sin

u

d

y4=0.2 cos

u

du (4)

Virtual Work Equation. When all the weights undergo positive virtual displacement,

all of them do positive work. However, Fsp does negative work when its undergoes

W

sp

(0.4 W cos

u

–0.2 F

sp

cos

u

+M)

du

=0

Since d

≠0,

then

200 mm

C

D

A

M  30 N fi m

k  3 kN/m

B

u

0.4 W cos

1134

11–11. Continued

Solve numerically:

Note:

u=23.8°

is a stable equilibrium, while

u=72.3°

is an unstable one.

Ans:

*11–12.

The disk is subjected to a couple moment M. Determine the

disk’s rotation

u

required for equilibrium. The end of the

spring wraps around the periphery of the disk as the disk

turns. The spring is originally unstretched.

SOLUTION

Free Body Diagram. The system has only one degree of freedom, defined by the

independent coordinate

u

. When

u

undergoes a positive angular displacement

du

as

shown in Fig. a, only the spring force Fsp and couple moment M do work.

du

k  4 kN/m

M  300 N fi m

0.5 m

Ans:

1136

11–13.

A

5

-kg uniform serving table is supported on each side by

pairs of two identical links, and ,and springs .If

the bowl has a mass of , determine the angle where the

table is in equilibrium. The springs each have a stiffness of

and are unstretched when .Neglect

the mass of the links.

u=90°k=200 N>m

u1kg

CECDAB

SOLUTION

Free – Body Diagram: When undergoes a positive virtual angular displacement of

,the dash line configuration shown in Fig.aisformed. We observe that only the

Virtual Displacement: The position of points of application of Wb,Wt,and Fsp are

specified by the position coordinates and xC,respectively.Here,

are measured from the fixed point Bwhile xCis measured from the fixed point D.

(1)

cos u(–7.3575 +12.5 sin u)=0

–7.3575 cos u+12.5sin ucos u=0

yGb=0.25 sin u+bdyGb=0.25 cos udu

yGband yGt

yGb,yGt,

du

u

ACk

250 mm

250 mm 150 mm

150 mm

B

D

E

uu

1137

11–14.

SOLUTION

Free – Body Diagram: When undergoes a positive virtual angular displacement of

,the dash line configuration shown in Fig.aisformed. We observe that only the

spring force Fsp,the weight Wtofthe table, and the weight Wbofthe bowl do work

Virtual Work Equation: Since Wb,Wt,and Fsp act towards the negative sense of

their corresponding virtual displacement, their work is negative.Thus,

(4)

dU=0;–W

bdyGb+

A

–W

tdyGt

B

+

A

–F

spdxC

B

=0

du

u

A

5

-kg uniform serving table is supported on each side by

two pairs of identical links, and ,and springs .If

the bowl has a mass of and is in equilibrium when

,determine the stiffness of each spring.The springs

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

ku=45°

1kg

CECDAB

ACk

250 mm

250 mm 150 mm

150 mm

B

D

E

uu

11–15.

SOLUTION

x=2acos u,dx=-2asin udu

Theservice windowat a fast-foodrestaurantconsists ofglass

doorsthat openand close automatically usingamotor which

supplies atorqueMto eachdoor. Thefarends, Aand B,

movealongthehorizontalguides.Ifafoodtraybecomes

stuckbetweenthedoorsasshown,determinethehorizontal

force thedoorsexertonthetrayat thepositionu.

MA

aa

CB

D

M

uu

1139

*11–16.

The members of the mechanism are pin connected. If a

vertical force of 800 N acts at A, determine the angle

u

for

equilibrium. The spring is unstretched when

u=0°.

Neglect

the mass of the links.

1 m

800 N

1 m

k  6 kN/m

D

B

A

u

SOLUTION

Free Body Diagram. The system has only one degree of freedom, defined by the

independent coordinate

u

. When

u

undergoes a positive angular displacement

du

as

shown in Fig. a, only spring force Fsp and force P do work.

u=23.58°=23.6°

u=23.6°

1140

11–17.

When

u=30°

, the 25-kg uniform block compresses the two

horizontal springs 100 mm. Determine the magnitude of the

applied couple moments M needed to maintain equilibrium.

Take

k=3 kN>m

and neglect the mass of the links.

A

B

D

C

M

300 mm

200 mm

100 mm

50 mm

k

u

SOLUTION

Free Body Diagram. The system has only one degree of freedom, defined by the

independent coordinate

u

. When

u

undergoes a positive angular displacement

du

as

shown in Fig. a, only spring force Fsp, the weight of the block W, and couple moment

M do work.

Virtual–Work Equation. When Fsp, W and M undergo their respective positive

virtual displacement, all of them do negative work. Thus

Since du

≠0,

then

11–18.

SOLUTION

Free – Body Diagram: When undergoes a positive virtual angular displacement of

, the dash line configuration shown in Fig. ais a formed. We observe that only the

spring force Fsp acting at points Aand Band the force Pdo work when the virtual

displacements take place.The magnitude of Fsp can be computed using the spring

force formula,

Virtual Displacement: The position of points Aand Bat which Fsp acts and point C

at which force Pacts are specified by the position coordinates yA,yB, and yC,

measured from the fixed point E, respectively.

(1)

(2)

Since , then

Ans.u=41.2°

–2400(sin u–0.2588) +960 =0

cos udu Z0

yB=3(0.2 sin u)dyB=0.6 cos udu

yA=0.2 sin udyA=0.2 cos udu

du

u

The “Nuremberg scissors” is subjected to a horizontal force

of . Determine the angle for equilibrium. The

spring has a stiffness of and is unstretched

when .u=15°

k=15 kN>m

uP=600 N

P

200 mm

200mm

A

C

D

E

B

k

u

Ans:

1142

11–19.

The “Nuremberg scissors” is subjected to a horizontal force

of . Determine the stiffness kof the spring for

equilibrium when .The spring is unstretched when

.u=15°

u=60°

P=600 N

SOLUTION

Free – Body Diagram: When undergoes a positive virtual angular displacement of

, the dash line configuration shown in Fig. ais formed. We observe that only the

spring force Fsp acting at points Aand Band the force Pdo work when the virtual

displacements take place.The magnitude of Fsp can be computed using the spring

force formula.

du

u

P

200 mm

200mm

A

C

D

E

B

k

u

k=9.88

kN>m