978-0133915426 Chapter 11 Part 3

1159

*11–36.

Determine the angle for equilibrium and investigate the

stability at this position. The bars each have a mass of 3 kg

and the suspended block Dhas a mass of 7 kg. Cord DC has

a total length of 1 m.

u

SOLUTION

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(–Wsin u–2W

Dcos u)

u=12.1°

tan u=W

2W

D

=3(9.81)

14(9.81) =0.2143

dV

du

=l(Wcos u–2W

Dsin u)=0

=Wl sin u–W

Dl(3 –2 cos u)

V=2Wy1–W

Dy2

y2=l+2l(1 –cos u)=l(3 –2 cos u)

y1=l

l=500 mm

500 mm

A

C

500 mm

uu

1160

11–37.

Determine the angle

u

for equilibrium and investigate the

stability at this position. The bars each have a mass of 10 kg

and the spring has an unstretched length of 100 mm.

500 mm

AC

500 mm

k 1.5 kN/m

uu

SOLUTION

du

11–37. Continued

1162

11–38.

The two bars each have a mass of 8 kg. Determine the

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

equilibrium when

u=60°.

The spring has an unstretched

length of 1 m. Investigate the stability of the system at the

equilibrium position. 1.5 m

1.5 m

B

A

k

u

SOLUTION

du

11–38 Continued

2

V

2

V

2

11–39.

SOLUTION

However, for a smallangle ,.Thus,

The elastic potentialenergy of the torsional spring canbe computed using

,where .Thus,

The totalpotentialenergy of the systemis

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.

W=8k

3L

–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

=-

3WL

2u+4ku=ua–3WL

2+4kb

V

V=V

g+V

e=3WL

2a1–u2

2b+2ku2

V

g=1

2k(2u)2=2ku2

b=2uV

e=1

2kb2

V

g=Wy =Wa3

2Lba1–u2

2b=3WL

2a1–u2

2b

cos u1–u2

2

u

A

k

L

2

L

2

A spring with a torsional stiffness k is attached to the pin

at B. It is unstretched when the rod assembly is in the

vertical position. Determine the weight W of the block that

results in neutral equilibrium. Hint: Establish the potential

energy function for a small angle u. i.e., approximate sin u L 0,

and .

cos uL1–u2>2

*11–40.

SOLUTION

P

otential Function: First, we must determine the center of gravity of the cylinder.By

referring to

Fig. a,

W

ith reference to the datum, Fig. a, the gravitational potential energy of the cylinder

is positive since its center of gravity is located above the datum.

Here,

T

hus,

Equilibrium Configuration:

Taking the first derivative of V,

Equilibrium requires

.Thus,

Stability:

The second derivative of Vis

To

have neutral equilibrium at ,.Thus,

Ans.

d=12h;2(–12h)2–4(3)(6h2)

2(3) =0.5858h=0.586h

6h2–12hd +3d2=0

–W

B

6h2–12hd +3d2

4(3h–d)

R

cos 0° =0

d2V

d2u2u=0°

=0u=0°

d2V

d2u

=-W

B

6h2–12hd +3d2

4(3h–d)

R

cos u

sin u=0u=0°

–W

B

6h2–12hd +3d2

4(3h–d)

R

sin u=0

dV

du

=0

dV

du

=-W

B

6h2–12hd +3d2

4(3h–d)

R

sin u

V=V

g=Wy =W

B

6h2–12hd +3d2

4(3h–d)

R

cos u

y

=(y–d)cos u=

B

6h2–d2

4(3h–d)–d

R

cos u=

B

6h2–12hd +3d2

4(3h–d)

R

cos u

h

rpr2h–1

3rpr2d

A conical hole is drilled into the bottom of the cylinder, and

it

is then supported on the fulcrum at A. Determine the

minimum

distance din order for it to remain in stable

equilibrium.

11–41.

The uniform rod has a mass of 100 kg. If the spring is

unstretched when

u=60°

, determine the angle

u

for

equilibrium and investigate the stability at the equilibrium

position. The spring is always in the horizontal position due

to the roller guide at B2 m

k  500 N/m

A

B

2 m

u

SOLUTION

u=24.62°

u=24.6°

11–42.

Each bar has a mass per length of

m0

. Determine the angles

u

and f at which they are suspended in equilibrium. The

contact at A is smooth, and both are pin connected at B.B

3

2

uf

l

A

SOLUTION

11–43.

The truck has a mass of 20 Mg and a mass center at G.

Determine the steepest grade along which it can park

without overturning and investigate the stability in this

position.

u

SOLUTION

G

3.5 m

1169

*11–44.

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 the

angles and and the dimensions shown. All members are

pin connected.

fu

W

2

W

1

,

W

2

ba

af

f

u

SOLUTION

where

is a constant and

Ans.W

2=W

1ab

absin u

cos f

dV

du

=W

1bsin u–W

2acos (90° –u–g)

=-W

1bcos u+W

2asin (90° –u–g)

V=-W

1y1+W

2y2

f=(90° –u–g)g

y2=asin f=asin (90° –u–g)

y1=bcos u

1170

11–45.

A 3-lb weight is attached to the end of rod ABC. If the rod

is supported by a smooth slider block at Cand rod BD,

determine the angle for equilibrium. Neglect the weight of

the rods and the slider.

u

SOLUTION

Thus,

Ans.

or,

Ans.u=33.0°

A

36 –16 sin2u

B

1

2=6.667 cos u

u=0°

sin u=0

Wc4–0.8333(36 –16 sin2u)–1

2(32 cos u)dsin udu=0

dU=0; Wdy=0

dy=4 sin udu+1.667a1

2b(36 –16 sin2u)–1

2(–32 sin ucos u)du

y=-4 cos u+1.667 236 –16 sin2u

x+4 cos u+y=2.667x

6

x

=16

(x+4 cos u)+y

x=2(6)2–(4 sin u)2=236 –16 sin2u

6 in.

θ

10 in.

4 in.

B

C

D

Ans:

u=0°

u=33.0°

11–46.

If the uniform rod OA has a mass of 12 kg, determine the

mass mthat will hold the rod in equilibrium when

Point Cis coincident with Bwhen OA is horizontal. Neglect

the size of the pulley at B.

u=30°.

Potential Function: The datum is established at point O. Since the center of gravity

of the rod and the block are above the datum, their potential energy is positive.

dna,ereH

Equilibrium Position: The system is in equilibrium if

At

Ans.m=5.29 kg

dV

du

`

u=30°

=-

29.43mcos 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

=-9.81mc–1

2110 –6 sin u2–1

21–6 cos u2d +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 u2

V=V

g=W

1y1+W

2y2

y2=0.5 sin um.y1=3–l=33–1210 –210 –6 sin u24 m

l=lAB –lA¿B=210 –210 –6 sin u

lAB =21

2+32=210 m

lA¿B=21

m

C

B

3m

1172

11–47.

SOLUTION

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 60

d2V

du2=-mgd cos u

sin u=0u=0°

dV

du

=-mgd sin u=0

dV

du

=0

V=V

g=Wy =mg(r+dcos u)

The cylinder is made of two materials such that it has a mass

of mand a center of gravity at point G. Show that when G

lies above the centroid Cof the cylinder, the equilibrium is

unstable.

C

G

a

r

1173

*11–48.

SOLUTION

Take the positive root

Ans.L=8 in.

L=20 ;2(–20)2–4(5)(–160)

10

0=5L2–20L–160

0=

2a16

12 b(5) –aL

2–2ba2L

12 b(5)

a16

12 b(5) +a2L

12 b(5)

The bent rod has a weight of 5 lb ft. A pivot of negligible

size is attached at its center Aand the rod is balanced as

shown. Determine the length Lof its vertical segments so

that it remains in neutral equilibrium. Neglect the thickness

of the rod.

8 in. 8 in.

2 in.

LL

A

Ans:

L=8 in.

1174

11–49.

T

h

e tr

i

angu

l

ar

bl

oc

k

of we

i

g

h

t Wrests on t

h

e smoot

h

corners which are a distance aapart. If the block has three

equal sides of length d, determine the angle for

equilibrium.

u

SOLUTION

Require, Ans.

or

Ans.u=cos–1d

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 ucos u–0.5 sin ucos u)d=0

V

=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

G60

60

u

Ans:

u=0°

u=cos–1

a

d

4ab