cap 8, novena edc_text

8/17/2019 Cap 8, Novena Edc_text

1/166

CHAPTER

8


2/166


3/166

240

lb I

/i,

=

0.35

I

//fc

=

0.35

PROBLEM 8.1

Determine whether

the block

shown

is

in

equilibrium and find the magnitude

and

i* direction of

the friction force when

6-25°

and P

-

1

50 lb.

SOLUTION

Z4-Ot*»

r

Assume

equilibrium:

T

-25°

=

0:

F

+

(240 lb) sin

25°

-(1

50

lb)

cos

25°

=

^

F

=

+34.5

1 8

lb

F

=

34.5 1 8

lb

\

1

r

lS0

1V>

X

+/ZF

y

=

0:

N

-

(240

lb)

cos

25°

-

(1

50

lb)sin

25°

=

N

=

+280.9

lib

N

=

280.9

lib/

Maximum friction

force:

F

m

=p

3

N

=

0.35(280.91

lb)

=

98.3

19

lb

Block

is

in equilibrium ^

Since.

F


4/166

240 lb

I JJ

S

~ 0.35

I

W

=

0.25

PROBLEM 8.2

Determine

whether

the

block shown is in equilibrium

and find

the

magnitude

and

direction

of the friction force when 9

=

30°

and

P

~

30

lb.

SOLUTION

Assume equilibrium:

«*u

-30°

=

0:

F

+

(240

lb)

sin

30°

-(30

lb)cos

30°

=

F

=

-94.019

lb

F-

-•

0: N

~

(240

lb) cos

30°

-

(30

lb)

sin

30°

=

=

=

94.019 lb

\

-0

1

1/

30*

^3o°

N

=

+222.85

lb

N

=

=

222.85 lb

/

Maximum

friction

force:

=

0.35(222.85

lb)

=

77.998

lb

Since

F

is

\

and

F

> F

m>

Block

moves

down

^

Actual friction force: F

-

F

k

=

jU

k

N

=

0.25(222.85

lb)

F

=

55.7

1b\^

PROPRIETARY MATERIAL. © 2010

The

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inc.

All rights

reserved. No

pari

of

this Manual

maybe

displayed,

reproduced or distributed in

any

form

or

by

any means,

without the

prior

written

permission

of

the publisher,

or

used

beyond

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and

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If

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1208


5/166

MM)

N

;/

5

.

-0.20

/ij.

=

O.J5

PROBLEM

8.3

Determine

whether

the

block

shown is

in

equilibrium

and

find

the

magnitude

and direction

of

the

friction

force when

=

40°

and

p

=

400

N.

SOLUTION

Assume

equilibrium

:

leyf***

0CC#

.$

IS-

2£-

?7>

+/

ZF

y

=

:

iV

-

(800

N)

cos

25°

+

(400

N)sin

1

5°

=

JV

=

+621.5N

N-621..5NJ

^2^=0:

-^

+

(800N)sin

25°-(400N)cosI5°

=

F

=

+48.28 N

F

=

48.28

N\

Maximum,

friction force:

=--

0.20(62.1.5

N)

=

124.3

N

SinceF


6/166

\>m \

//,

0.20

//,

l)

15

PROBLEM

8.4

Determine whether

the

block

shown is

in

equilibrium

and

find

the

magnitude and

direction of the friction

force

when #

=

35°

and

P

=

200N.

SOLUTION

Assume

equilibrium :

£**»•

T/*-

+/ /-;

=0:

N

-

(800

lM)cos

25°

+

(200

N)sin

10°

=

N

'

=

690.3 N

N= 690.3

NJ

^

XR

=

0:

-F

+

(800

N)

sin

25°

-

(200

N)

cos

1

0°

=

Maximum friction force:

Since

F>F

m9

Friction

force:

F

=

14.1.1.3

N

=

(0.20)(690.3

N)

=

1.38.06 N

F

=

Mk

N

=

(0.15)(690.3

N)

=

1 03.547 N

F

=

141.13N\

Block moves

down

F

=

103.5

N\


7/166

SOON

p

s

=

0.20

ft

= 0-15

PROBLEM

8.5

Knowing

that

9

-

45°,

determine

the

range

of

values

of

P for

which

equilibrium

is maintained.

SOLUTION

To

start

block up

the

incline :

4

Us

=0.20

&

=

tan

1

0.20

=

1

1.31

QooN

V

/\-

y

?£,


8/166

It.

=

0.25

\

/

PROBLEM

8,6

Determine the

range

of values of

P

for

which

equilibrium

of

the

block

shown is maintained.

MX)

N

SOLUTION

FBD

block:

(Impending motion

down):

(Impending

motion

up):

££>oN

Sbo

/£>

(j)

s

-

tan

~

//

A

=

tan

0.25

P

=

(500

lb)

tan

(30°

-

tan

1

0.25)

=

143.03 lb

S&&N

P

Sbo

lb

P

=

(500

lb) tan

(30°

+ tan

'

0.25)

=

483.46 lb

Equilibrium is

maintained

for

143.0

lb


9/166

l.-.k

PROBLEM

8.7

Knowing

that the

coefficient

of friction

between the

1

5-kg block and

the

incline

is

ju

s

=0.25,

determine

(a)

the

smallest

value

of/-*

required to

maintain

the block

in equilibrium,

(b) the

corresponding

value

of/?.

SOLUTION

FBD block

(Impending

motion

downward):

(#)

AZote;

For

minimum

P,

So

and

m

W-(15kg)(9.81m/s

2

)

-147.150 N


10/166

fi

t

=

0.25

m

/it

= 030

:.m


11/166

3

s5

'

r

500 N

PROBLEM

8.9

The

coefficients

of friction

between

the

block

and the rail

are

//,

=

0.30

and

/i

k

=0.25.

Knowing

that

=

65°,

determine

the smallest value

of

P required

(a) to

start the block

moving

up the rail,

(b) to

keep

it

from

moving

down.

SOLUTION

(a)

To

start block

up the rail:

M

s

=0.30

(p

s

~

tan

0.30

= 16.70°

^

fe&tMsSfJO

5-OOfJ

hr

Force

triangle:

P

500 N

sin

51.70°

sin(18p°-

25°-

51.70°)

(b)

To

prevent block from

moving down:

X

5tort

50Ot*

<

>%

I&70

hr

Force

triangle:

P

500

N

sin

8.30°

sin

(

1 £

0°-

25°

-18.30°)

/>

=

403N

<

P

=

229N <

PROPRIETARY

MATERIAL.

©

2010 The

McGraw-Hill

Companies,

Inc.

All rights

reserved.

No part

of

this

Manual may

be

displayed,

reproduced

or distributed in

any

form

or by

any

means,

without

the

prior written

permission

of

the

publisher,

or

used

beyond

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distribution to teachers

and

educators

permitted by McGraw-Hill

for

their

individual

coarse

preparation.

If

you

are

a

student

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this .Manual,

you

are using

it withoutpermission.

1215


12/166

80

11

-I

M7

PROBLEM

8.10

The 80-lb

block

is

attached to

link

AB and rests

on

a

moving

belt.

Knowing

that

Lt

x

=0.25

and

fi

k

=0.20,

determine

the

magnitude of

the horizontal

force P

that

should

be applied to

the

belt

to

maintain

its

motion

(a) to

the

right, (b)

to the

left.

SOLUTION

We

note

that link

AB is

a

two-force

member, since

there

is motion between

belt

and

block

fi

k

=

0.20

and

&

=

tan

1

0.20

=

11.

31°

(a)

Belt moves to

right

Free

body:

Block

Force

triangle:

R 80

lb

Free

body:

Belt

sin

120°

sin

48.69°

R

=

92.23

lb

+~XF

x

=

0: P

-

(92.23

lb)sin

11.31°

P

=

18.089 lb

(/>)

Belt

moves

to

left

Free body:

Block

Force

triangle:

R

801b

sin

60°

sin

108.69°

#

=

73.139 lb

Free

body:

Belt

±*

2/^

=

0: (73.139 lb)sin

11.31°

-P-0

P

=

14.344

lb

&>'*>

P

=

1 8.09

lb

—

-

^1

ffOlb

f~/K?

0

^JMTnA

8

P

=

14.34

lb

PROPRIETARY

MATERIAL.

€>

2010 The


Inc. All

rights reserved. No part

of

this Manual may be

displayed,

reproduced

or

distributed in any

form

or by

any means, without the

prior written permission

of

the

publisher,

or

used

beyond the

limited

distribution to teachers and

educatorspermitted by

McGraw-Hill

for

their

individual

course

preparation.

If

you

area

student using

this

Manual,

you

are using

it

without

permission.

1216


13/166

M.HP

PROBLEM

8.11

The

coefficients

of

friction

are

//,

-

0.40

and

,u

k

=

0.30

between

all

surfaces

of

contact.

Determine

the

smallest

force

P

required

to

start

the

30-kg

block

moving

if

cable

AB

(a)

is

attached

as

shown,

(b)

is

removed.

SOLUTION

(«)

Free

body:

20-kg

block

W

=(20

kg)(9.81m/s

2

)-

196.2

N

I'\

=

//,

A/,

-

0.4(1

96.2

N)

=

78.48

N

-±-£F =

0:

r-^=0 7-

=

^

=78.48

N

Free

body:

30-kg block

^

2

=

(30

kg)(9.81

m/s

2

)

=

294.3

N

7V

2

=

1

96.2

N

+

294.3

N =

490,5

N

F

2

=ji

s

N

2

=

0.4(490.5

N)

=

196.2

N

P

=

78.48

N

+

1

96.2 N

+ 78.48

N

=

353.2

N

(h)

Free

body:

Both

blocks

Blocks

move

together

«

/

-(50kg)(9.81

m/s

2

)

=

490.5N

-±-£F

=

0:

P~F

=

P =

Us

N =

0.4(490.5

N)

=

.1

96.2

N

T

r^^

39-.V*//

.4

In,

P-353.N—

-^

^y

=

#»*i>y

£f

/*•**

fvv*#*-r//

P

=

196.2

N

—

^

^wTI^'ff '

°/°

10

T

'

e McGraw

-

Hi

Com

P

a»ies

»

'»-

AH rights

reserved.

*o

part

of

this

Manual

tnay

be

displayed

tepwducedotdistnbutedmanyjorm

or by

any

means,

without

the prior

mitten

permission

of

the

publisher,

or used

beyond

the

limited

1217


14/166

PROBLEM

8.12

The

coefficients

of

friction are

fi

s

=

0.40

and

{i

k

-

0.30

between

all

surfaces of

contact.

Determine

the

smallest

force

P

required

to

start

the

30-kg

block

moving

if

cableAB

(a)

is

attached

as

shown, {b)

is

removed.

SOLUTION

(a)

Free

bodv: 20-kg

block

W

=

(20kg)(9.81

m/s

2

)

=

196.2N

F,

~fi

s

N

x

=

0.4(196.2

M)

=

78.48

N

)Nj*J9SA*t

T

z:

ft*

J9A.*N\

F,

u^r

±»SF

=

0: 7--F,

=0

7

=

F,

=78.48

N

Free bodv:

30-kg

block

fr

2

=

(30

kg)(9.8

1

m/s

2

)

-

294.3

N

N

2

=

196.2

N

+

294.3

N

=

490.5

N

F

2

=

//,JV

2

=0.4(490.5

N)

=

196.2

N

J^IF

=

0:

P~F ~F

2

^0

P

=

78.48 M

+

196.2 N

=

274.7 N

(b)

Free

bodv:

Both

blocks

Blocks

move

together

*

4

—

*-./-*

»,

3&«W

.4

tMi

P

=

275

N

^

=

(50kg)(9.81m/s

2

)

=

490.5

N

,+_EF

=

0:

P-F

= Q

P

=

jU

x

N

=

0.4(490.5

N)

=

196.2

N

Stf

*•£

P

f*


15/166

PROBLEM

8.13

Three

4-kg

packages

A,

B,

and

C

are

placed

on

a

conveyor

belt

that

is

at

rest.

Between

the

belt

and

both

packages

A

and

C

the

coefficients

of

friction

are

/^

=

0.30

and

//

A

.

=

0.20; between

package

B

and

the

belt

the

coefficients

are

//,

=

0.10

and

//^0.08.

The

packages

are

placed

on

the

belt

so that

they

are

in

contact

with

each

other

and at

rest.

Determine

which,

if

any,

of

the

packages

will

move

and

the

friction

force

acting

on

each

package.

SOLUTION

Consider

C

bv

itself

Assume

equilibrium

+\-LF

y

=0:

tf

c

-JFcos15°

=

N

C

=W cos

15°

=

0.966

W

+/ /


16/166

PROBLEM

8.13

(Continued)

Consider

A

and

B together

:

Assume

equilibrium

Thus,

F

A

=F

B

=

0.259

W

N

A

*N

B

=0.966W

F

A

+

F

B

=2(Q259W)

=

0.5\W

(F

A

\,

HF

B

)

m

=0.3N

A

+0AN

B

=03S6W

F

A

=/i

k

N

A

=

0.2(0.966)(4)(9.81)

A

and

B

move

A

F,

=7.58N/

<

F

B

^ftNs

=

0.08(0.966)(4)(9.81)

F„=3.03N/

<

PROPRIETARY

MATERIAL,

©

20 1

The

McGraw-Hill

Companies,

Inc.

All

rights

reserved. No

part

of

this

Manual

rnay be

displayed

reproduced

or

distributed in any

farm

or by

any

means,

without the

prior

written

permission

of

the

publisher, or

used

beyond the

limited

distribution to

teachers

and

educators

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McGraw-Hill

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their

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course

preparation.

If

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using

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Manual,

you

are

using

it

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permission.

1220


17/166

PROBLEM

8.14

Solve

Problem

8.13

assuming that

package

B

is

placed

to

the

right

of

both packages

A

and

C.

PROBLEM

8.13

Three

4-kg

packages A, B,

and C

are

placed on

a

conveyor

belt that is

at rest.

Between

the

belt

and

both

packages A

and

C

the coefficients

of

friction

are

fi

s

-

0.30

and

jA.

k

-

0.20;

between

package

B and

the

belt,

the coefficients

are

^-0.10

and

//

A

=

0.08.

The

packages

are placed

on the

belt so

that they

are

in contact

with

each other

and

at

rest.

Determine

which,

if any,

of the

packages

will

move

and the friction

force

acting

on

each

package.

SOLUTION

Consider

package B

by itself

:

Assume

equilibrium

+\

~LF

y

=

:

N

B

~W cos

1

5°

=

N

K

=

W

cos

15°

=

0.966

W

VLF

x

=0:

F

B

-Wsm\5°

=

F

B

=

Wsin

15°

=

0.259

W

But

F

m

=M,N

a

=

0.1

0(0.966

W)

~Q.0966W

Thus,

F

B

>

F

m

. Package

B would

move

if

alone.

Consider

a

ll

packages

together

:

Assume

equilibrium.

In a

manner

similar

to above, we

find

N

A

=N

B

=N

C

=0.966W

F

/f

=F

li

=F

c

=

0.259 1¥

F

A

+

F

B

+F

C

=3(0.259W)

=

0.777W

Bui

and

=

0.30(0.966

W)

=

0.290

W

(F

B

)

m

=0.10(0.966^)

=

0.Q966W

PROPRIETARY

MATERIAL.

©

2010

The McGraw-Hill

Companies,

Inc. All

tights

reserved.

No part

of

this

Manual

may

be displayed,

reproduced

or

distributed in

any

form or

by

any

means,

without

the

prior

written permission

of

the publisher, or

used beyond the

limited

distribution

to teachers

and

educators

permitted

by

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for

their

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course

preparation.

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you are a student

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are using

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1221


18/166

PROBLEM 8.14

(Continued)

Thus,

(F

A

)

m

+

(F

c

\„

+

(F

B

)

m

=

2(0.290

WO

+

0.0966

W

=

0.677W

and we

note

that

F

A

+

F

B

+

F

c

>

{F

A

),„

+

(F

c

)

m

+

(F

B

),„

All

packages

move

A

,2-

F

A

~F

c

={i

k

N

=

0.20(0.966)(4

kg)(9.81

m/s

z

)

=

7.58N

F

B

^fi

k

N

=

0.08(0.966)(4kg)(9.81

m/s

2

)

=

3.03N

F^

=

F

c

=

7.58 N

/

;

¥

B

=

3.03

N

/

^

PROPRIETARY

MATERIAL. ©

2010 The

McGraw-Hill

Companies,

Inc.

All rights

reserved. M>

/«w7

o////«-

Manual may

be

displayed,

reproduced or

distributed in

any

form

or by

any means,

without

the

prior

written

permission

of

the

publisher,

or used

beyond the

limited

distribution

to

teachers

and

educators

permittedby

McGraw-Hill

for

their

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preparation.

If

you are

a

student using

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are using it

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1222


19/166

V

ip

=

2.66667

(a) All

casters

locked.

Impending

slip:

F

B

^fi

a

N

B

So

ZF

y

=0:

N

A

+N

B

-W

=

N

A

+

N

B

^W

.0

hi.—

*j

^

=

120 lb

//,

-0.3

P

=

0.3(120

lb)

or

P

=

36.01b-

(P = 0.3^.

OK)

(&)

Casters

at

A

tree,

so

Impending

slip:

^=0

Fb^&Ni

L/


20/166

PROBLEM

8.15

(Continued)

(lM

A

=i): (32

in/)P +

(12

m.)W

-(24 m.)N

B

=0

8P

+

3W

-

6—

=

P

=

0.25

W

0,3

(P

=

0.25^

<

Z^ip

OK)

P

=

0.25(1

20

lb)

or

P

=

30.0

lb— <

(c)

Casters

at

i?

free,

so

P

B

=

impending

slip:

F

A

~

}A

S

N

A

ZF

X

^():

P-F

A

=0

P

=

F

A

=ti

s

N

A

N

=JL^JL

A

M,

0.3

XM

B

-

0:

(12

in.)

W

-

(32

in.)P

-

(24

m.)N

A

=

3JF-8P-6

—

=

0.3

P

=

0.1

07.1

43^

=

1

2.8572

(P


21/166


22/166

::,';;.-^

E

-

:

::^

PROBLEM

8.17

The

cylinder shown

is of weight

W and radius

r, and

the

coefficient of

static

friction

fi

s

is the same

at A

and

B.

Determine

the

magnitude of

the

largest

couple M that

can

be

applied

to

the

cylinder

if

it is

not to

rotate.

SOLUTION

FBD

cylinder:

For maximum

M>

motion

impends

at both

A

and

B

or

and

h

=

&Nb

-TF

X

=

0:

N

A

-F

B

=Q

N

A

=F

B

=ti

t

N

B

Fa=Ma

=

/£**

|L/v

=

0:

N

B

+F

A

-W

=

N

B

W

1

+

rf

Fb

-

^

1+ft

2

F

A

_

A

2

*

1+//

2

XM

C

'-

=

0:

M-r(F

A

+ F

B

)

=

M

—A ii

4-

ir\~

^

A/,

+

ft

PROPRIETARY

MATERIAL

© 2010

The

McGraw-Hill

Companies, Inc.

All rights

reserved.

A'o

ywf

o//Akt

Manual may be

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reproduced

or distributed

in any

form

or

by

any means,

without the prior

written permission

of

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1226


23/166

PROBLEM

8,18

The

cylinder

shown

is

of

weight

W and radius

r. Express

in

terms

W

and

r

the

magnitude

of

the

largest

couple

M that

can be applied

to

the

cylinder

if

it is

not

to

rotate,

assuming

the

coefficient

of

static

friction

to be

(a)

zero

at A

and

0.30 at B

y

(b)

0.25

at

,4

and

0.30 at

B.

SOLUTION

FB.D

cylinder:

For

maximum

M,

motion impends

at both

A

and

B

f

a=Ma

n

a

f

b=Mb

n

b

—

2/^=0:

N

A

-F

B

=0

oi-

and

N

A

=F

B

=/i

B

N

B

\ZF

y

=0:

N

B

+ F

A

-W

=

Q

N

b

{\

+ li

a

li

b

)~W

1

N,

\+MaMb

W

Mb

1

+

MaMb

pA-^

A

^B-r

hBs

~

w

1

+

MaMb

{XM

c

=0:

M~r(F

A

+F

B

)

=

Q

M

-

Wrfi,

1

+

MaMb

(a) For//

4

-0

and

ji

b

-030:

(b) For

jU

A

=

0.25

and

//

£

=0.30:

M

=

Q.300Wr

*4

M

=

QM9Wr

<

PROPRIETARY

MATERIAL.

©

2010 The

McGraw-Hill

Companies,

Inc.

AH

rights

reserved.

No part

of

this

Manual may

be displayed,

reproduced

or distributed

in any

form or by

any

means, without

the prior

written

permission

of

the

publisher, or used

beyond the

limited

distribution

to teachers

and

educators

permitted

by

McGraw-Hill

for

their

individual

coursepreparation.

If

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astudent

using

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mi


24/166

,«

s

-

0.-10

A0*k-

fl

k

=

0.30

.150 inin

|

J

300 rum

..n-

SOLUTION

Free

body:

Drum

Free

body: Left

arm

ABL

OJC-m'S_

Free body:

Right

arm

PER

iii

150

mm

iiT

300 mm

PROBLEM

8.19

The

hydraulic

cylinder shown

exerts a force

of

3

kN

directed

to

the

right

on

Point B

and

to

the

left

on

Point E.

Determine

the magnitude

of the couple

M required

to

rotate the

drum

clockwise

at

a

constant speed.

50 mm

\

150 mm

250 mm

+)IM

C

=

0: M

-

(0.25

m)(F

L

+

P

R

)

=

.M

=

(0.25m)(F

/

,

+ F

w

)

Since

drum

is

rotating

+)

tM

A

=

: (3

k.N)(0. 1 5

m)

+

P

L

(0.

1

5

m)

-

N

L

(0.45

m)

=

0.45

kN

•

rn +

(0.3

JV,, )(().

1

5

m)

-

N

L

(0.45

m)

=

0.4057V,

=

0.45

N

Substitute

for./*)

and F

R

into

(1):

,

-~i.il I

kN

F)

j

^0.3/V

A

=

0.3(1. 11.1.

kN)

=

0.3333

kN

+)XM

D

=0:

(3

kN)(0.15

m)~F

R

(0.15

m)-/V

/f

(0.45

m)

=

0.45 kN

•

m

-

(0.3JV

#

)(0.

1 5

m)

-

N

H

(0.45

m)

-

0A95N

R

-

0.45

7V

W

=0.9091

kN

Fjt=Mk

N

R=

0.3(0.909

IkN)

=

0.2727

kN

M

-

(0.25 m)(0.333

kN

+

0.2727

kN)

M

=

0.1515

kN-m

(1)

(2)

(3)

M

=

151.5

N-m

^

PROPRIETARY

MATERIAL

© 2010

The

McGraw-Hill

Companies, Inc.

All rights

reserved. No

part

of

this Manual

may be

displayed,

reproduced

or distributed

in

any

form

or

by

any

means,

without

the

prior written

permission

of

the

publisher,

or used beyond

the

limited

distribution to

teachers

and

educators

permitted by

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individual

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1228


25/166

150

nun]

1

|

m

300 IT

M

s

-

O.-iO

//A

=

0.30

?*,/>

J.50

nun

m—*

250 in

mm

PROBLEM

8.20

1

°

*

'

A

couple

M of magnitude

100

N

•

m is applied to

the drum

as

|

ft

I shown.

Determine

the

smallest force that must

be

exerted

by

the

'soo

mm

hydraulic cylinder

on

joints

B and E if the drum

is not

to

rotate.

SOLUTION

Free

body:

Drum

Free body:

Left

arm

A

BL

AtfB-ZLjpF

Free

body: Right arm PER

+J

LM

C

=

0:

1

00

N

•

m

-

(0.25

m)(F

L

+ F

R

)

=

F,

+

.F

ft

=400N

Since

motion impends

F

r

=jU,Hh~QAN

r

+)

XM

A

=

0:

7/(0.

1

5 m) + F

L

(0.

15m)-

7V

L

(0.45

m)

=

0.

J

57'

+

(0.4/V

A

)(0.

1 5

m)

-

/V

A

(0.45

m)

=

0.397V/

=

0. 1

5f; JV

A

=

0.384627'

F

L

=

0.47V,,

=0.4(0.384627)

F

7

=0.153857'

^

ZM

A>

=

0:

r(0. 1

5

m)

-

F„

(0.

15m)- yV„

(0.45

m)

-

0.

1

57'

-

(0.4

/V

fi

)(().

1 5

m)

-

N

R

(0.45

m)

-

0.5

N

R

=

0. 1

57';

N

R

-

0.294

1

27*

F

K

=

0AN

R

=

0.4(0.294

127')

7^=0.11765.7

Substitute

for f,

and

F

R

into

Eq.

(

1

):

0.1

53857'

+

0.11

7657

=

400

7

=

1473.3 N

(1)

(2)

(3)

7'

=

1.473 kN

^

PROPRIETARY

MATERIAL.

©

2010

The

McGraw-Hill

Companies,

Inc.

All rights

reserved. No pari

of

this

Manual

may

be displayed,

reproduced or

distributed

in

any

form

or

by

any means,

without the prior

written

permission

of

the publisher,

or used beyond

the

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distribution

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educators

permitted

by

McGraw-Hill

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their

individual

course

preparation.

If

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are a student using

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yon

are

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1229


26/166

PROBLEM 8.21

A

6.5-m

ladder AB

leans against

a

wall

as

shown.

Assuming that

the

coefficient

of

static friction

//

s

is

zero at

B,

determine

the

smallest value

of

fl

s

at A

for

which

equilibrium

is

maintained.

6 m

\*~

z.o

m +\

SOLUTION

Free body:

Ladder

Three-force body.

Line

of

action

ofA

must

pass

through Z), whereW and

B intersect.

fas'*

}

/.«-*

At A:

M

x

- tan

#

6 m

0.2083

//.

-0.208

<

PROPRIETARY

MATERIAL

© 2010 The

McGraw-Hill Companies, Inc.

All


of

this Manual

may

be

displayed,

reproduced

or

distributed

in any

form

or

by

any

means, without the

prior

written

permission

of

the publisher, or used

beyond

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limited

distribution to

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1230


27/166

/\

PROBLEM 8.22

A

6.5-m ladder

AB leans

against a wall

as shown.

Assuming

that the coefficient

of

static

friction

ft

is

the same

at A

and B,

determine

the smallest

value of

ft

for

which

equilibrium is

maintained.

6 m

SOLUTION

Free

body: Ladder

Motion impending:

Fa-

Ma

+)ZM

A

=Q:

W(].25m)~N

B

(6m)~

ftA^(2.5m)

=

W .

//

M

}.25W

N

~

:

6

+

2.5/4

(I)

Jr&

*-

«/•/

o/V/ha-

Manual

may

be

displayed,

reproduced

or

distributed

in

any

form or

by any

means,

without the prior written

permission

of

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1231


28/166

PROBLEM

8.23

End A of

a

slender,

uniform rod

of length

L

and

weight W

bears

on

a

surface

as

shown, while

end

B

is supported by

a cord

BC.

Knowing

that

the

coefficients of

friction

are ju

s

=

0.40

and

ju

k

=

0.30,

determine (a)

the

largest

value of

for which

motion

is

impending, (b)

the

corresponding

value

of

the

tension

in the

cord.

SOLUTION

c

Free-body

diagram

/

Three-force

body.

Line of action

of

R

must

pass

through D, where

T

and

R

intersect.

/

Motion

impends:

/R

A

tm\tp

s

=0.4

&

=21.80°

.#

5*«

(a) SinceBG

=

GA, it

follows

that BD

-

DC

and

AD

bisects ABAC

~

+

d>

=

90°

2

A

-

+

21.8° =

90°

2

=

136.4°

^

p^^.

4

s

*je/,&°

~7-9>*

(b) Force

triangle (right

triangle):

T~Wcos2\.S°

fLz,s

r

=

0.928^

^

PROPRIETARY

MATERIAL,

©

2010 The


Inc. Ail

rights reserved.

No part

of

this

Manual

may

be

displayed,


any

form

or by any

means, without

the

prior

written

permission

of

the

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or

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preparation.

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Manual,

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are

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iI

withoutpermission.

1232


29/166

PROBLEM

8.24

End A

of

a

slender,

uniform

rod

of length

L

and

weight

W

bears on a

surface

as

shown,

while

end

B

is supported

by a

cord

BC.

Knowing

that

the

coefficients

of

friction are

fi

s

—

0.40 and

fi

k

-

0.30,

determine (a)

the

largest value

of

for

which

motion

is impending,

(b)

the

corresponding

value

of

the tension

in

the

cord.

SOLUTION

Free-body

diagram

Rod AB is

a

three-force

body.

Thus,

line

of action ofR

must pass

through D,

where

W

and T

intersect.

Since AG

-

GB, CD

=

DB

and the

medianAD

of the

isosceles

triangle ABC bisects the angle

0.

(a) Thus,

Since motion impends,

< >

s

=

tan

1

0.40

=

21.80*

= 2^=2(21.8°)

(p)

Force

triangle :

This

is

a

right triangle.

T

=

rVsm0

x

=

Wsm2)..S

c

=

42.6°

<

fUs

PROPRIETARY MATERIAL,

©

2010

The


Inc. AH


of

this

Manual

may

be

displayed,


any

form

or

by

any means, without the

prior

written

permission

of

the

publisher, or

used

beyond

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distribution

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and

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1233


30/166

36

in.

I)i

PROBLEM 8.25

A window

sash weighing 1 lb

is

normally

supported by

two

5-lb

sash

weights.

Knowing that

the

window remains

open

after

one

sash

cord

has

broken, determine

the

smallest possible

value

of

the coefficient

of

static

friction.

(Assume that

the

sash

is

slightly smaller than

the

frame

and will

bind

only at Points

A

and

D.)

SOLUTION

FBD window: r

=

=

51b

=

0: N

A

-N

D

~0

t

U

1

r

*rf

A

-

*

N

A

-

N

D

i

2T

in. ~~\8j*.-4*-

»*it-

Impending motion:

F

A

=

/'A

1

h/

Fd

=

MsN

D

1

V

ZM

D

=

0:

(\%in.W-(21m.)N

A

-(36in.)F

4

W^=~N

A

+2

Ms

N

A

3

+

4

A

=

^

=

101b

\*F

y

=

0: F

A

~W +

T

+

F

D

=0

W

F

A

+F

D

=W-T

=

^

Now

Fa+Fd=M,W

a

+N

d

)

~2/i,N

a

Then

W

„

IW

—

-

2n

-

2

'3

+

4/1,

or

ft

=0.750

<

PROPRIETARY

MATERIAL.

©

2010 The

McGraw-Hill

Companies, Inc. All

rights

reserved. /Vb

part

of

this Manual

may

be displayed,

reproduced

or distributed in any

form

or by

any means,

without

the prior written permission

of

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used

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1234


31/166

300

N

105

nun

-

3.15

mm

'

rsoo

r;

360

mm

PROBLEM

8.26

A

500-N concrete

block

is

to be lifted

by

the pair

of

tongs shown.

Determine

the

smallest allowable value

of the coefficient

of

static

friction

between

the

block

and

the

tongs atF and

G.

SOLUTION

Free

body:

Members CA,

AR,

BD

By

symmetry

:

C

v

=£>

;

=-(500)

=

250N

Since

CA

is

a

two-force

member,

a

250

N

90 mm

75

mm

75

mm

C

r

=300N

&r=0:

D

x

-C

x

O

=

300

N

lOEvrtVH

Free body: Tong

DBF

+)

£M

£

=

:

(300

N)(J 05

mm)

+

(250

N)(l

35

mm)

+

(250

N)(l 57,5 mm)

-

F

v

(360

mm)

=

F

=

+290.625

N

Minimum

value of

jU

s

:

F„

Ms

250

N

F

290.625

N

560

1

»2STo

M

f*25oti

_ii


32/166

in

\:

50

ii)

PROBLEM 8.27

The

press

shown

is used

to emboss a

small

seal at E.

Knowing

that

the coefficient of

static

friction between the vertical guide

and

the

embossing

die

D

is 0.30, determine

the

force exerted by

the die

on

the

seal.

SOLUTION

Free

body:

MemberABC

_«_

—

.

—

—

>i

8

«»&&>'

+)

TM

A

=

0: F

BD

cos

20°(4)

+

F

BD

sin

20°(6.9282)

-(50

lb)(4

+

15.4548)

=

7^

=

158.728 lb

Free body:

Die

D

(t>

s

=

tan

-1

fi

s

.-

tan

1

0.3

16.6992°

Force

triangle:

D .58.7281b

sin

53.301°

sin

106.6992'

D

=

132.869

lb

On seal:

D

=

132.9

lb

\<

PROPRIETARY

MATERIAL.

CO 2010

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rights reserved. No

part

of

this Manna may be displayed,

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form

or

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any means,

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1236


33/166

PROBLEM

8.28

1.00

mm

;

/

.';

.eljs

100 mm

i«

Fi

e

The

1

00-mm-radius cam shown is

used

to

control

the motion of the

plate CD.

Knowing that

the

coefficient of static

friction

between

the

cam and

the plate

is 0.45

and

neglecting friction

at the roller

supports,

determine (a)

the

force

P

required

to maintain

the

motion

of the

plate,

knowing

that

the

plate

is 20 mm

thick,

(/>)

the

largest

thickness

of the

plate

for

which the

mechanism

is

self

locking (i.e.,

for which the

plate

cannot, be

moved however

large the force P

may be).

SOLUTION

Free body:

Cam

Q**6C/V

Impending

motion:

RS'r>&

F

=

ft

s

N

)ZM

A

=

0: QR-NRsin0+

Qt

s

N)Rcos0

=

O

Q

sm&-jU

s

cos*

Free body: Plate

XF

X

=0

P

=

ft

s

N

Geometry

in AABD

with

#

=

100 mm

and

d

=

20

mm

R~d

(1)

(2)

cos

=

R

80 mm

100

mm

=

0.8

sin

-Vl-

cos

2

=

0.6

P

T

\

R

\

\

iff

F-Hs*

1

Y~T

1

PROPRIETARY

MATERIAL.

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Companies, Inc. All

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reserved. No

pari

of

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Manual may

be.

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or distributed in any

form

or

by

any

means, without

the prior written permission

of

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used beyond

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1237


34/166

PROBLEM

8.28

(Continued)

(fl)

Eq.

(1)

using

g

=

60N

and

^

=0.45

A'~

6

°

N

0.6

-(0.45)(0.8)

=

6

°

=

250N

0.24

Eq.

(2)

P

=

//

A

,W

=

(0.45)(250N)

/>

=

112.5N

<

(b)

For

P

=

00,

Af

=

=


35/166

PROBLEM

8.29

A slender

rod

of

length

L

is

lodged between

peg C and the vertical

wall

and

supports

a

load

P

at

end

A,

Knowing

that

the coefficient of

static

friction is

0.20

at

both B

and

C, find

the

range

of values of

the

ratio

LI

a

for which equilibrium

is

maintained.

SOLUTION

We shall first assume that

the

motion

ofend

B

is

impending

upward. The friction

forces

at B

and

C

will have

the values and

directions

indicated

in

the

FB

diagram.

+

)IM

W

=0:

PLsm$~N

r

sin0

PI

-±~

SF

V

=

: N

c

cos

+

///V

c

sin

~

N

B

=

+1

ZF

=--

:

tf

c

sin

-

juN

c

cos

-

///V

B

-

P

=

(1)

(2)

(3)

Multiply

Eq.

(2)

by

//

and

subtract

from

Eq.

(3):

N

c

(sin

(9

-

(X

cos

0)

-

//JV

C

(cos

+

//

sin

0)

-

P

=

P

=

A'

c

[sin

0(1

-

//

2

)

-

2/1

cos

0]

Substitute

for N

c

from

Eq.

(1)

and

solve

for a/L:

L

sin

2

0[(1

-

//

2

)sin 0-2//

cos0]

(4)

Making

0-35°

and

//

=

0.20 in

Eq.

(4):

-

=

sin

2

35°[(1

-

0.04)sin

35°

-

2(0.20)

cos

35°]

=

0.07336

-

=

13.63

<

a

PROPRIETARY

MATERIAL.

©

2010 The

McGraw-Hill

Companies, Inc.

Ail

rights

reserved.

No part

of

this Manual

may

be

displayed,

reproduced or distributed

in

any

form

or by

any

means, without

the prior

written

permission

of

the

publisher,

or

used beyond

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to teachers


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for

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coursepreparation.

If

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are

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1239


36/166

PROBLEM

8.29

(Continued)

Assuming

now

that

the motion

at B

is impending

downward,

we

should

reverse

the direction of F# and

F

(

.

in.

the FB diagram

.

The

same

result

may

be

obtained

by making

—

35°

and

//

= -0.20

in

Eq.(4):

-

=

sin

2

35°[(1

-

0.04)sin

35°

-

2(-0.20)cos35°]

-0.2889

L

a

=

3.461

20 iO

The

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Companies, Inc.

All

rights reserved.

No part

of

this Manual may be

displayed,

reproduced

or

distributed in

any

form

or by

any

means, without

the

prior

written

permission

of

the


the

limited

distribution

to

teachers andeducatorspermitted by McGraw-Hill

for

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If

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1240


37/166

i&C

PROBLEM

8.30

The

50-1

b

plate

ABCD

is

attached at A

and

D

to

collars

that can

slide on

the

vertical rod. Knowing that

the

coefficient of

static friction

is

0.40

between

both

collars

and the rod,

determine whether

the

plate

is

in

equilibrium in

the position

shown

when the

magnitude of the

vertical

force

applied atE

is

(a) P

-

0,

(b) P

-

20 lb.

SOLUTION

(a)

P

=

+)ZM

D

=0:

JV

/(

(2ft)-(501b)(3ft)

=

A.

'

ffl

.W

f

L

I

-*t

«

£

£F

V

=0:

N

A

=

75 Ib

N

D

=

N

A

=

75 lb

.i

,

P

t&

&

/=

.

3

ft

**

ofi

>

-t>

+\LF

y

=0:

/^+F

D

-501b

=

F

A

+F

D

=5Q\b

But:

(F

A )m

=

Ma

=

0-40(75

lb)

=

30

lb

(^))„,=^^/>-0.40(751b)

=

301b

Thus;

(^)*+(^L=60Ib

and

(^A)

m

HF

D

)

m

>F

A

+F

Plate is in equilibrium

M

(b) P

=

201b

+)ZM

D

=0:

JV,

(2

ft)

-

(50

lb)(3 ft)

+

(20

Ib)(5 ft)

=

N

A

=

251b

£F

V

=0:

N

D

=N

A

=25\b

+|2^=0:

P

A

+F

D

-

50

lb

+ 20

lb

=

^+7^,

=30

lb

But:

(Fa)*

=

/^=0.4(251b)

=

101b

(FiX^MsNn

=0.4(25

lb)

=

101b

Thus:

(KX

+

(F

)„,=

20 lb

and

^

+

^>(^A,+(^)

W,

Plate moves downward

-^

PROPRIETARY

MATERIAL.

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2010 The

McGraw-Hill

Companies, Inc. All

rights

reserved. JVo

part

of

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may

be

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in

any

form

or by

any

means, without

the

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written

permission

of

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publisher, or used

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1241


38/166

v-;

«

PROBLEM 8.31

In Problem 8.30,

determine the

range

of

values of

the magnitude

P of

the

vertical

force

applied atE

for which

the plate will move downward.

PROBLEM

8.30

The

50-lh

plate ABCD

is

attached at A.

and

D to

collars

that can slide

on the vertical rod.

Knowing

that

the

coefficient of

static

friction

is

0.40

between

both

collars

and

the

rod,

determine

whether

the

plate is

in equilibrium in

the position shown

when

the magnitude of

the

vertical force applied atE

is

{a) P

-

0, (/?)

P

=

20

lb.

TV,,

=75

lb

-2.5

But:

Plate

moves

I if:

or

(2)

30lb


39/166

PROBLEM

8.31 (Continued)

But:

(^),,,=(^),,,=/0^

=

0.40(2.5P~75)

=

P~30\b

Plate

moves

I

if:

F

A

+

F„

>

(F,

),„

+

(/y)

;

„

50-P>(P~30)

+

(P-30)

/><

—

=

36.7

lb


40/166

60

nun

PROBLEM 8.32

A

pipe of diameter 60 mm

is

gripped

by the

stillson

wrench shown. Portions

AB

and

DE

of the

wrench

are

rigidly attached to

each

other,

and

portion

CF is

connected by

a pin at

D. If

the

wrench is to

grip the pipe

and

be

self-locking,

determine the

required minimum

coefficients

of

friction at.

A

and

C.

SOLUTION

FBD ABD:

Q

2M

=

0:

(1

5

mm^

-(110 mm)F

/f

=

Impending

motion:

F

A

~ju

A

N

A

HO

mm

Then

or

15

—

11

0//^

=0

^=0.13636

FBD

pipe:

ZF

X

=():

F

A

-D

x

=0

D

X

=

F

A

ZF

V

=Q:

N

c

-N

A

=0

N

C

=N

A

FBD

DF:

(^

ZM

F

=

: (550

mm)F

c

-

(1

5 mm)N

c

-

(500

mm)D

x

-

Impending motion:

F

c

=f.t

c

N

c

Then

F,

550//

r

-15

=

500^-

But

So

F

A

N,

=

N

,

and

-^--u.=

0.

1

3636

t A

N

^

t*A

550//

c

=

15

+

500(0. 1

3636)

5£i0iv>iw

//

f;

=0.1512

<

PROPRIETARY

MATERIAL CO

2010

The


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rights reserved.

No

pari

of

this Manual

may

be displayed,

reproduced

or distributed

in

any form or by

any means, without the prior

written

permission

of

the

publisher,

or

used beyond

the

limited

distribution to

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and

educators

permitted

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coursepreparation.

If

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1244


41/166

15

o

it

HSI

GO

mm

50

Him

500

mm

PROBLEM

8.33

Solve Problem

8.32

assuming

that

the

diameter of the

pipe

is 30

mm.

PROBLEM

8.32

A pipe

of diameter 60 mm is gripped

by the

stillson wrench

shown.

Portions

AB

and

DE

of the wrench

are

rigidly attached to each other,

and

portion CF

is connected

by

a

pin

at

/).

If

the wrench is

to

grip

the

pipe

and

be self-locking,

determine

the required

minimum

coefficients

of

friction at

A

and

C.

SOLUTION

FBD

ABD:

ZM

D

=

:

(15

mm)W

,

-

(80

mm)F

A

=

Impending

motion:

F

A

~

jU

A

N

A

Then

1

5

mm

-

(80

mm)//

A

=

-*EF

v

-0:

F

A

-D

X

=Q

So that

D

x

=

F

A

=

0.

1

875W,

FBD pipe:

fsF^O:

N

C

-N

A

=Q

Nc=N

A

FBD

DF:

Impending motion:

£.M

/;

, =

o : (550 mm)F

c

-

(1

5

mm)iY

c

-

(500

mm)/)

(

=

F

c

={i

€

N

c

N.

But

N

.,

=

jV

c

(from pipe

F5Z>)

so

and

550/^-15

=

500(0.1875)'

J

A'

N

c

*c

1

m>y>

500fn»

PROPRIETARY

MATERIAL.

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2010

The

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Companies,

Inc.

All rights

reserved.

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of

this Manual may he displayed,

reproduced

or distributed in

any

form

or

by

any means,

without

the

prior

written permission

of

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or

used

beyond

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limited


andeducatorspermittedby

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preparation.

If

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1245


42/166

PROBLEM

8.34

A

A

10-ft beam, weighing

1200

lb, is to be

moved

to the left onto

the

B

p

platform. A

horizontal force P

is applied

to

the

dolly,

which is

•*x*-

C

^

^_\*t*mmm

mounted on

frict

ion less

wheels.

The coefficients

of

friction

between

2ft

-'-:--':-. -----

all surfaces

are

//

v

~0.30

and ju

k

-

0.25,

and initially j-2ft.

Knowing that the

top

surface of the

dolly

is

slightly higher than

the

platform,

determine

the

force

P

required

to

start

moving

the

beam.

(Hint: The

beam is supported

atA and/).)

SOLUTION

FBD

beam:

Sf+

i

Zoo

lb

Fa

Ma

ir*~

S^-r

F

M

+

)

EM

A

=

: JV„

(8

it)

-

(1

200

lb)(5

ft)

=

N

i}

=750Ib

+|

XF

Y

=

:

N

A

-\

200

+

750

=

N

/f

=450

lb

C

?

A,=/'A

=

0.3(450)

=

135.0

lb

(/^)

IH

=//,AT

D

=

0.3(750)

=

225

lb

Since

(F

A

)

<

(F

n

)

,

sliding

first

impends

atA with

FBD

dolly:

From FBD of

dolly:

^=(^A,=i35Ib

+-Z/7-0:

F

A

-F

D

~0

F

D

=F

A

=135.0

lb

++2F

X

=

0: F

D

~P

=

p

=

F

D

=

135.0

lb

F

ft

[

PROPRIETARY

MATERIAL. ©

2010 The McGraw-Hill

Companies,

Inc.

All rights reserved. No part

of

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displayed,

reproduced or

distributed in

any

form

or

by

any

means, without the prior

written

permission

of

the


the

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course

preparation.

If

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1246


43/166

PROBLEM

8.35

(#)

Show

that

the

beam of

Problem

8.34 cannot

be

moved

if

the

top

surface

of the

dolly is slightly

lower than

the platform,

(b)

Show

that

the

beam

can

be moved if two

1.75-lb

workers stand

on the

beam

at B

and determine

how

far

to

the left the

beam

can

be

moved.

PROBLEM

8.34

A.

10-ft

beam,

weighing 1200

lb,

is

to

be

moved

to

the

left

onto

the

platform. A

horizontal

force

P

is

applied to

the

dolly,

which is

mounted on

frictionless

wheels.

The

coefficients

of

friction

between

all surfaces

are

ju

s

-

0.30 and

fi

k

=

0.25,

and

initially

x

-

2

ft.

Knowing

that

the

top

surface of the

dolly is

slightly

higher than

the platform,

determine

the

force

P

required

to

start

moving

the

beam.

(Hint; The

beam

is supported

at A

and D.)

SOLUTION

(a)

Beam alone

+)£M

C

=

0:

N

ff

(S

ft)

-

(1200

lb)(3

ft)

=

N

fi

=450Ib

+|

IF

t

,

=

0:

N

r

+ 450-1

200

=

N

c

=

750 lb

(F

c

),„

=

jU

x

N

c

=

0.3(750)

=

225

lb

(^A^/'A^

0.3(450)

=

135

lb

Since

(F

B

)

m

< (F

c

)

,

sliding first

impends

at

5,

and

F

t

=*5C

&

*&

He

4.

Beam

cannot

be moved

A

(b)

Beam

with

workers standing

at B

X

5**| 1200

&

iSTofr

5

-s

J-*

F^

~S

«.

+)IM

c

=0:

JV

fl

(1

-

jc)

-

(1

200X5

j)-

350(1

0-*)

=

+)XM

B

=

0:

(1200)(5)-

W

c

(10~-x)

-

*c

iV,

9500-1

550*

10-Jc

6000

lO-x

PROPRIETARY

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Manual

may

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distributed in

any

form

or by

any

means,

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written

permission

of

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publisher, or used

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1247


44/166

PROBLEM

8.35 (Continued)

Check

that beam

starts moving

for

x

-

2

ft:

x,

9500-1550(2)

nnn

„

For

x

=

2 ft:

N

B

=

^

=

800

lb

B

10-2

Nc

=

-^

=

750

lb

L

10-2

(^c)*=A^c=

0-3(750)

=

225

lb

(/^L-//.,^^

0.3(800)

=

240 lb

Since (F

c

),„

<

(^),„,

sliding

first impends at C,

Beam

moves

A

How

far

does

beam

move?

Beam

will

stop

moving

when

6000

1500

But

F

c

=fii

k

N

c

=0.2$

and

(F

B

)

m

=M,N

B

=030

10-x

10

-

x

9500-1550* 2850-465*

10-

jc

10-x

Setting

F

c

=

(F

B

)

m

: 1

500

-

2850

-

465*

x

=

2.90 ft

<

(Note: We

have assumed

that,

once

started,

motion is

continuous and

uniform

(no

acceleration).)

PROPRIETARY

MATERIAL.

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or

by

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means,

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.1248


45/166

ifh i

*

M

100

mm

k

100

mm

PROBLEM

8.36

;

Knowing

that

the

coefficient of

static friction

between

the collar

and

the rod

H

c

is

0.35, determine

the range

ofvalues

ofP for

which

equilibrium is

maintained

m when -

50°

and

M

~

20 N

m.

SOLUTION

Free

body member AB:

BC is

a

two-force

member.

+)

ZM

A

-

0:

20

N

•

m

-

F

liC

cos

50°(0.

1

m)

=

F

ac

=31

1.145

N

Motion

of

C

impending

upward

:

-i*

£/?

=0:

(3

1 1

.

1

45

N)cos

50°

-

N

=

TV

=

200

N

+JEF

;

,

=0:

(311.145 N)sin

50°-

P-(0.35)(200N)

=

Motion,

of

C

impending

downward

:

±»E/r

=:

=

308.35

N

<

168.4 N<

308

N

^

PROPRIETARY

MATERIAL

©

2010 The

McGraw-Hill

Companies,

Inc.

All rights

reserved.

No

part

of

this Manual

may be displayed,

reproduced

or distributed

in

any

form

or

by

any

means,

without

the prior

written

permission

of

the

publisher,

or used beyond

the

limited

distribution

to teachers

and

educators

permitted

by McGraw-Hill

for

their

individual

course

preparation.

If

you

area

student

using

this Manual,

you

arc

using it without

permission.

1249


46/166

JT~

100

mm

•

M

-100

mm-

PROBLEM

8.37

Knowing

that

the

coefficient

of

static

friction

between

the

collar and

the

rod

is 0.40,

determine

the

range

of

values

of

M

for

which

equilibrium is

maintained

when $

=

60°

and

P

=

200

N.

SOLUTION

Free body

member

AB:

BC

is

a

two-force

member.

+)

~LM

A

=0:

M

-

F

BC

cos

60°(0.

i

m)

=

A**

M

=

0.05F

BC

Motion ofC

impending

upward :

+~TF

x

=

0: F

BC

cos60°-N~0

N~0.SF

BC

+\ XF

V

=

0:

F

HC

sin

60°

-

200N

-

(0.40)(0.5F

BC

)

=

(1)

Pz&Ofl

F

BC

=

300.29

N

Eq.

(1):

M

=

0.05(300.29)

M

=

15.014N-m


47/166

PROBLEM

8.38

The

slender

rod

AB

of

length

/

=

600

mm

is

attached

to a collar

at

B

and

rests

on

a small

wheel

located

at a

horizontal

distance

a

=

80 mm from

the

vertical

rod

on

which

the

collar

slides.

Knowing

that

the coefficient

of

static

friction

between

the

collar

and

the

vertical

rod is

0.25

and

neglecting

the

radius

of the

wheel,

determine

the

range

ofvalues

of

P

for

which

equilibrium

is

maintained

when

Q

=

100

N

and

-

30°.

SOLUTION

For motion

of

collar

at

B

impending

upward:

+)£M

tf

=0:

Qls'mO

Ca

C

=

Q

sin#

n

a

sin

2

-

sin

2

cos

F

x

=0:

N^Ccos0

=

Q

+\£F

y

=0:

P

+

Q-Csm0-^

S

N

=

O

P

+

Q~Q\L\\n'0-

Ms

Q

fn

a

j

sin

2

6cos

6>

=

\UJ

P~Q

/

.

2

—

si

n

0(sin

-

//

cos

0)

~~

1

a

(1)

Substitute

data:

P

=

(100N)

600

mm

.?„„„,.

sin

2

30°(sin

30°

-

0.25

cos

30°)

-

80 mm

p

=

-46.84

N

(Pis

directed )

P =

-46.8N


48/166

PROBLEM

8.38

(Continued)

For

motion of

collar,

impending

downward :

F

=

/<

V

AT

In Eq.

(1)

we

substitute

-fi

s

for//,.

/

•

2

p

=

Q

J°

=

(100N)

—

sirr

0(sin +

fi

g

costf)

-

a

600mm

rJr

2

ono

80mm

'

sin

2

30°(sin

30°

+

0.25

cos

6)

-

1

For

equilibrium:

/^

+34.34

N


49/166

W-= 10 I

'.

'----=-:£.-

'-/J

X

«'=

lOtbl

PROBLEM

8.39

Two

10-lb

blocks

/*

and

B

are

connected

by

a slender

rod of

negligible

weight.

The

coefficient,

of

static

friction

is

0.30 between

all

surfaces

of

contact, and

the

rod forms

an angle =

30°.

with

the

vertical,

(a)

Show

that

the

system

is

in

equilibrium

when

P

=

0.

(b) Determine

the

largest

value

of

P

for

which

equilibrium

is

maintained.

SOLUTION

FBD

block

B:

(a)

Since

P

=

2.69 lb

to

initiate

motion,

(/;)

For P

mBKf

motion

impends

at

both surfaces:

equilibrium

exists

with P-Q

A

to

ib

Block

B:

Impending

motion:

£F

=0:

JV

fl

~10lb-F^cos30°

=

R

JV

B

=I0Ib

+—

2

AB

Solving

Eqs.(l)

and

(2):

FBD

block

A:

Then

F

B

^ju,N

B

=0.3N

fi

-2F

r

=0:

F

B

-F

AB

sin30°^0

F

AB

=2F

B

=Q.6N

B

N

l{

=

10

lb

+—

(0.6A/

/}

)

=

20.8166

ib

F

AB

=0.6N

B

=12.4900

lb

Block

/(

:

—

X,F

X

=

:

F

w

sin

3

0°

-

JV,

(

-

Impending

motion:

N

A

=

-F

(/i

=-(12.4900

lb)

=

6.2450

Ib

F

A

~,u

s

N

A

=0.3(6,2450

lb)

=

1.8735

lb

'

TF

y

=

:

F,

+

F

AB

cos

30°

-

P

-

1

lb

=

P

=

F

A+^~F

AB

-H)\h

=

1.8735 lb

+^(12.4900

Ib)-10

lb

=

2.69

lb

(1)

(2)

•3d

f\

F,

.A

-

be displayed,

reproduced

or

distributed

in

any

form

or

by

any

means,

without

the

prior

written

permission

of

the

publisher,

or

used beyond

the limited

distribution

to teachers

andeducators

permitted

by McGraw-Hill

for

their

individual

coursepreparation.

Ifvou are a

student

using

this

Manual,

you

are

using it without

permission.

1253


50/166

-J—

•i a

*d

8ft

vC

>

ft—

^-6

ft

—

PROBLEM

8.40

Two

identical

uniform

boards,

each of

weight

40

lb,

are

temporarily

leaned

against

each

other

as

shown.

Knowing

that

the

coefficient of

static

friction

between all

surfaces is 0.40,

determine

(a) the

largest

magnitude

of

the

force

P

for which

equilibrium

will be

maintained,

(h)

the

surface

at

which

motion will

impend.

SOLUTION

Board FBDs:

(I)

W

Assume

impending

motion at C, so

FBD

II:

(JM

B

=

0:

(6

ft)N

c

~(

8

Wc

(

3

ft

X

40 lb

>

a

°

[6

ft -0.4(8

ft)F

c

=(3 ft)(40

lb)

or

and

N

c

=42.857

lb

F

C

=QAN

C

=

17.143

lb

—

Lf;=0:

N

B

~F

c

^0

Check

for

motion at

.5:

FBD

I:

W

B

=F

C

=17.143

lb

f

£M

;)

=

0: -7^

-

40 lb +

N

c

-

7^=^

-40 lb

=

2.857 lb

h_

=

2

-

857

b

=

0.

1

67 <

//,

,

OK,

no

motion.

N

B

17.143 1b

(

ZM

A

=

: (8

fi)N

B

+

(6

ft)F

B

-

(3

ft)(P

+

40

lb)

=

(8

ft)(17.143 lb) +

(6

ft)(2.857

lb)

^

fe

3ft

1.1.4291b

PROPRIETARY

MATERIAL. ©

2010

The

McGraw-Hill

Companies,

Inc.

All

rights reserved.

No

part

of

this

Manual may be

displayed

reproduced

or

distributed in anyform

or by

any

means, without

the

prior written

permission

of

the

publisher,

or

used beyond

the limited

distribution to

teachers and

educators

permitted by

McGraw-Hill

for

their

individual

course

preparation.

If

you

are

a

student

using

this

Manual,

you are

using it

without

permission.

1254


51/166

PROBLEM

8.40

(Continued)

Check

for

slip

atA (unlikely

because

off):

—

2^=0: F

A

-N

B

=Q

or

F

A

=N

B

=17.143

lb

|zF

y

-0:

A^-P-40

1b

+

.F

B

=0

or

A^

=

1

1 .429 lb

+ 40

lb

-

2.857

lb

Then

£l

=

2™U Uo.353<

A

48.572

lb

17.3.43

1b

iV^

48.572

lb

OK, no

slip

:=>

assumption is

correct.

Therefore

(«)

/^

=

11.43

lb

^

(^)

Motion

impends

at

C

-^

PROPRIETARY

MATERIAL

©

2010

The

McGraw-Hill

Companies, Inc. All

rights reserved.

JVo /«*/•/

of

this Manual

may be displayed,

reproduced

or distributed

in

any

form

or

by any

means, without

the

prior

written

permission

of

the

publisher,

or used beyond

the limited

distribution

to teachers


by McGraw-Hill

for

their

individual course

preparation.

If

you are

a

student

using

this Manual,

you are

using it

without

permission.

1255


52/166

PROBLEM

8.41

Two identical

5-ft-long

rods

connected

by

a

pin at B

are

placed

between

two

walls

and

a

horizontal

surface

as

shown.

Denoting

by

fi

a

the

coefficient

of static

friction at

A,

B,

and C,

determine

the

smallest value

of

fi

s

for

which

equilibrium

is

maintained.

SOLUTION

Sense

of

impending

motion:

i:

.

**

^W*^

+)TM

B

=0:

2W~3N

A

-4u

s

N

/t

=0

4

+

4^)

£'V

N

M

=W-fi>N

A

(i)

(3)

(5)

£M

„

=

:

1

..W

-

4W

C

+

3//

4

./V

c

=

JW-

1.5W

(4-3/1.)

L/y

Nsc^W

+

MsNc

ZF

cap 8, novena edc_text

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