Mechanical Engineering Chapter 11 Homework Pearson Education Inc Upper Saddle River Nj

1143

*11–20.

The crankshaft is subjected to a torque of

Determine the horizontal compressive force Fapplied to

the piston for equilibrium when u=60°.

M=50 N #m.

SOLUTION

(0.4)2=(0.1)2+x2–2(0.1)(x)(cos u)

100 mm

400 mm

F

M

u

Ans:

11–21.

SOLUTION

(1)

From Eq. (1)

x=0.2 cos u;20.04 cos2u+0.6

x2–0.2xcos u–0.15 =0

(0.4)2=(0.1)2+x2–2(0.1)(x)(cos u)

Thecrankshaft is subjected to atorque of

Determine the horizontal compressive force Fand plot

the result of F(ordinate) versus (abscissa) for

0° …u…90°.

u

M=50 N

#

m.

100 mm

400 mm

F

M

u

11–22.

Ans:

11–23.

4in. 4in. x

A

B

CG

E

D

2in.

F

Determine the weight of block required to balance the

differential lever when the 20-lb load Fisplaced on the pan.

The lever is in balance when the load and block are not on

the lever. Take .x=12 in

G

SOLUTION

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

about point B,the dash line configuration shown in Fig.aisformed. We observe

that only the weight WGofblock Gand the weight WFofload Fdowork when the

virtual displacements take place.

du

Ans:

1147

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*11–24.

4in. 4in. x

A

B

CG

E

D

2in.

F

SOLUTION

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

about point B,the dash line configuration shown in Fig.aisformed. We observe

that only the weight WGofblock Gand the weight WFofload Fdowork when the

virtual displacements take place.

Virtual Displacement: Since is very small,the vertical virtual displacement of

block Gand load Fcan be approximated as

Virtual Work Equation: Since WGacts towards the positive sense of its

corresponding virtual displacement, its work is positive.However,force WFdoes

negative work since it acts towards the negative sense of its corresponding virtual

displacement.Thus,

dyG

du

If the load weighs 20 lb and the block weighs

2lb,determine its position for equilibrium of the

differential lever. The lever is in balance when the load and

block are not on the lever.

x

GF

Ans:

1148

11–25.

The dumpster has a weight Wand a center of gravity at G.

Determine the force in the hydraulic cylinder needed to

hold it in the general position .u

SOLUTION

=2a2+c2+2acsin u

s=2a2+c2–2accos (u+90°)

θ

bd

G

a

Ans:

1149

11–26.

The potential energy of a one-degree–of-freedom system is

defined by where xis

in ft.Determine the equilibrium positions and investigate the

stability for each position.

SOLUTION

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

dV

dx =60x2–20x–25

V=(20x3–10x2–25x–10) ft#lb,

Ans:

11–27.

If the potential function for a conservative one-degree-of-

fe

rehw si metsys modeer

determine the positions for equilibrium and

investigate the stability at each of these positions.

0° 6u6180°,

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

SOLUTION

24 cos 2u–15 sin u=0

dV

du

=0;

V=12 sin 2u+15 cos u

*11–28.

SOLUTION

dV

dx

=24x2–4x=0

V=8x3–2x2–10

If the potential function for a conservative one-degree-of-

f erehw si metsys modeer xis

given in meters, determine the positions for equilibrium and

investigate the stability at each of these positions.

V=18x3–2x2–102J,

Ans:

x=0.167 m

2

11–29.

SOLUTION

For equilibrium:

Stability:

d2V

du2=-40 cos 2u–25 sin u

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-

fe

rehw si metsys modeer

determine the positions for equilibrium and

investigate the stability at each of these positions.

0° 6u6180°,

V=110 cos 2u+25 sin u2J,

Ans:

u=38.7° unstable

u=90° stable

1153

11–30.

If t

h

e potent

i

a

l

energy for a conservat

i

ve one-

d

egree-of-

freedom system is expressed by the relation

, where xis given in feet,

determine the equilibrium positions and investigate the

stability at each position.

V=(4x3–x2–3x+10) ft #lb

SOLUTION

Equilibrium Position:

Stability:

d2V

dx2=24x–2

V=4x3–x2–3x+10

Ans:

11–31.

k100 N/m

400 mm

D

C

B

A

u

SOLUTION

=(0.4)22(1 –cos u)

s=2(0.4)2+(0.4)2–2(0.4)2cos u

The uniform link AB has a mass of 3 kg and is pin connected

at both of its ends.The rod BD,having negligible weight,

passes through a swivel block at C.If the spring has a

stiffness of and is unstretched when ,

determine the angle for equilibrium and investigate the

stability at the equilibrium position. Neglect the size of the

swivel block.

u

u=0°k=100 N>m

Ans:

1155

*11–32.

The spring of the scale has an unstretched length

of a.Determine the angle for equilibrium when a weight W

is supported on the platform. Neglect the weight of the

members.What value Wwould be required to keep

the scale in neutral equilibrium when u=0°?

u

SOLUTION

Potential Function: The datum is established at point A. Since the weight Wis

above the datum, its potential energy is positive.From the geometry, the spring

stretches and .

Equilibrium Position: The system is in equilibrium if .

dV

du

=4kL2sin ucos u–2WL sin u=0

dV

du

=0

=1

2kx2+Wy

V=V

e+V

g

y=2Lcos ux=2Lsin u

k

L

W

LL

L

a

u

1156

11–33.

The uniform bar has a mass of 80 kg. Determine the angle

u

for equilibrium and investigate the stability of the bar when

it is in this position. The spring has an unstretched length

when

u=90°.

4 m

k  400 N/m

A

B

u

SOLUTION

Potential Function. The Datum is established through point A, Fig. a. Since the

center of gravity of the bar is above the datum, its potential energy is positive. Here,

y=2 sin

u and the spring stretches

x=4(1 –sin

u

) m.

Thus,

V=V

e

+V

g

=

1

2

kx2+Wy

Equilibrium Position. The bar is in equilibrium of

dV

du

=0

dV

du

=6400 sin u cos u–4830.4 cos u=0

dV

du

2

du

Stability. The equilibrium conguration is stable if d

2

V

du

270, unstable if d

2

V

du

260

and neutral if d

2

V

du

2=0.

At

u=90°

,

u=90°

Ans:

11–34.

The uniform bar AD has a mass of 20 kg. If the attached

spring is unstretched when

u=90°

, determine the angle

u

for equilibrium. Note that the spring always remains in the

vertical position due to the roller guide. Investigate the

stability of the bar when it is in the equilibrium position.

D

1 m

0.5 m

k  2 kN/m

C

B

A

u

SOLUTION

Potential Function. The Datum is established through point A, Fig. a. Since the

center of gravity of the bar is below the datum, its potential energy is negative. Here,

y=0.75 cos

u and the spring stretches

x=0.5 cos u.

Thus,

V=V

e

+V

g

Equilibrium Position. The bar is in equilibrium if

dV

du

=0.

dV

du

u=72.88°=72.9°

=–500 sin u cos u+147.15 sin u=0

Using the trigonometry identity

sin 2

u

=2 sin

u

cos

u

,

dV

du

=–250 sin 2u+147.15 sin u

du

Stability. The equilibrium conguration is stable if d

2

V

du

270, unstable if d

2

V

du

260

and neutral if d

2

V

du

2=0.

11–35.

The two bars each have a weight of 8 lb. Determine the

required stiffness k of the spring so that the two bars are in

equilibrium when u=30°. The spring has an unstretched

length of 1 ft.

SOLUTION

V=2(8)(1 sin u)+1

2

k(4 cos u–1)2

dV

du

=16 cos u+k(4 cos u–1)(–4 sin u)

2 ft

B

AC

θ

2 ft

k

Ans: