978-0073398167 Chapter 3 Solution Manual Part 11

PROBLEM 3.97

For the shift lever shown, determine the magnitude and the direction of the smallest

force P that has a 210-lb ⋅ in. clockwise moment about B.

SOLUTION

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,

1

8 in.

tan 22 in.

19.98

aθ

−

=

= °

and

minB

M dP=

Where

/

22

(8 in.) (22 in.)

23.409 in.

AB

dr=

= +

=

Then

min

210 lb in.

23.409 in.

P⋅

=

8.97 lb

=

min 8.97 lb=P

19.98° 

PROBLEM 3.98

Consider the volleyball net

shown. Determine the angle

formed by guy wires AB and

AC.

SOLUTION

First note:

2 22

( 6.5) ( 8) (2) 10.5 ft

(0) ( 8) (6) 10 ft

AB

AC

= − +− + =

= +− + =

and

(6.5 ft) (8 ft) (2 ft)

(8 ft) (6 ft)

AB

AC

=− −+

=−+

ijk

jk

C

By definition,

( )( )cosAB AC AB AC

θ

⋅=

C C

or

( 6.5 8 2 ) ( 8 6 ) (10.5)(10)cos

( 6.5)(0) ( 8)( 8) (2)(6) 105cos

θ

− − + ⋅− + =

− +− − + =

ijk jk

or

cos 0.72381

θ

=

or

43.6

θ

= °



PROBLEM 3.99

A crane is oriented so that the end of the 25–m boom AO lies in

the yz plane. At the instant shown, the tension in cable AB is

4 kN. Determine the moment about each of the coordinate axes

of the force exerted on A by cable AB.

SOLUTION

22

()()

(25 m) (15 m)

20 m

OC OA AC= −

= −

=

(15 m) (20 m)

A= +r jk

22

(2.5 m) (15 m) 15.2069 m

2.5 15

(4 kN) 15.2069

(0.65760 kN) (3.9456 kN)

AB

PAB

= +=

=

−

=

= −

P

ij



Using Eq. (3.11):

0 15 20

0.65760 3.9456 0

(78.912 kN m) (13.1520 kN m) (9.8640 kN m)

OA

O

=×=

=⋅ ⋅⋅

i jk

M rP

M i+ j- k

But

Ox y z

MMM=++M i jk

Therefore,

78.9 kN m, 13.15 kN m, 9.86 kN m

xy z

MM M= ⋅= ⋅=−⋅



PROBLEM 3.100

The 25–m crane boom AO lies in the yz plane. Determine the

maximum permissible tension in cable AB if the absolute value

of moments about the coordinate axes of the force exerted on A

by cable AB must be

|Mx| ≤ 60 kN∙m, |My| ≤ 12 kN∙m, |Mz| ≤ 8 kN∙m

22

()()

(25 m) (15 m)

20 m

OC OA AC= −

= −

=

(15 m) (20 m)

A

= +r jk

22

(2.5 m) (15 m) 15.2069 m

2.5 15

( ) 15.2069

(0.164399 kN) (0.98639 kN)

AB

PAB

P

PP

= +=

=

−

=

= −

P

ij



Using Eq. (3.11):

0 15 20

0.163499 0.98639 0

(19.7278 ) (3.2700 ) (2.4525 )

OA

O

PP

P PP

=×=

−

=

i jk

M rP

M i+ j- k

But

60 kN m: 19.7278 60 3.04 kN

12 kN m: 3.270 12 3.67 kN

8 kN m: 2.4525 8 3.15 kN

x

y

z

M PP

⋅ <≤

<⋅ < ≤

<

PROBLEM 3.101

A single force P acts at C in a direction perpendicular to

the handle BC of the crank shown. Determine the moment

Mx of P about the x–axis when θ = 65°, knowing that

My = −15 N · m and Mz = −36 N · m.

SOLUTION

See the solution to Prob. 3.53 for the derivation of the following equations:

(0.2) sin( )

x

MP

θφ

= +

(1)

tan

z

y

M

φ

=

(4)

22

4

yz

P MM= +

(5)

Substituting for known data gives:

22

36 N m

tan 15 N m

67.380

4 ( 15) ( 36)

156.0 N

0.2 m(156.0 N)sin(65 67.380 )

23.047 N m

x

P

M

φ

−⋅

=−⋅

= °

= − +−

=

= °+ °

= ⋅

23.0 N m

x

M= ⋅



PROBLEM 3.102

While tapping a hole, a machinist applies the horizontal

forces shown to the handle of the tap wrench. Show that

these forces are equivalent to a single force, and specify, if

possible, the point of application of the single force on the

handle.

SOLUTION

Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one

of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-

couple.

We have

2.9 lb 2.65 lb 0.25 lb,

B

F=−=

where the 2.65-lb force is part of the couple. Combining the two

parallel forces,

couple

(2.65 lb)[(3.2 in. 2.8 in.)cos25 ]

14.4103 lb in.

M= +°

= ⋅

and

couple 14.4103 lb in.= ⋅M

A single equivalent force will be located in the negative z direction.

Based on

: 14.4103 lb in. [(0.25 lb)cos 25 ]( )

B

MaΣ − ⋅= °

63.600 in.a=

F′

(0.25 lb)(cos 25 sin 25 )= °+ °ik

F′

(0.227 lb) (0.1057 lb)= +ik

and is applied on an extension of handle BD at a

distance of 63.6 in. to the right of B. 

PROBLEM 3.103

A 500–N force is applied to a bent plate as shown. Determine

(a) an equivalent force–couple system at B, (b) an equivalent

system formed by a vertical force at A and a force at B.

SOLUTION

(a) Force–couple system at B

(500 N)sin30 (500 N)cos30

(250.0 N) (433.01 N)

= −

F ik

Fik



(0.3 0.175 ) (250 433.01 )

86.153 N m

B

=− ×−

=−⋅

Mi ji j

k

The equivalent force–couple system at B is

500 N

B=F

60.0°

86.2 N m

B

= ⋅M



(b) Require

Equivalence requires

86.153 N m (0.125 m); (689.22 N)

or ; (250 N) (433.01 N) (689.22 N)

BB

MM∑=Σ

⋅= =

===−−

AA j

F A+B B F-A B i j j

(250 N) (1122.23 N) 1149.74 N=−=Bi j

77.4°

1150 N=B

77.4°



689 N = ↑A



PROBLEM 3.104

A force and a couple are applied as shown to the end of a

cantilever beam. (a) Replace this system with a single force

F applied at Point C, and determine the distance d from C to

a line drawn through Points D and E. (b) Solve part a if the

directions of the two 360-N forces are reversed.

SOLUTION

PROBLEM 3.105

Slider P can move along rod OA. An elastic cord PC is attached

to the slider and to the vertical member BC. Knowing that the

distance from O to P is 6 in. and that the tension in the cord is

3 lb, determine (a) the angle between the elastic cord and the

rod OA, (b) the projection on OA of the force exerted by cord

PC at Point P.

SOLUTION

First note

22 2

(12) (12) ( 6) 18 in.

OA = + +− =

Then

1(12 12 6 )

18

1(2 2 )

3

OA

= = +−

= +−

i jk

λ

Now

1

6 in. ( )

3

OP OP OA= ⇒=

The coordinates of Point P are (4 in., 4 in., −2 in.)

so that

(5 in.) (11 in.) (14 in.)PC =++ijk

C

and

222

(5) (11) (14) 342 in.

PC = ++ =

(a) We have

( )cos

OA

PC PC

θ

⋅=

C λ

or

1

(5 11 14 ) (2 2 ) 342 cos

3

θ

+ + ⋅ +−=i j k i jk

or

1

cos [(5)(2) (11)(2) (14)( 1)]

3 342

0.32444

θ

= + +−

=

or

71.1

θ

= °



(b) We have

()

cos

(3 lb)(0.32444)

OA

PC PC OA

PC OA

PC

T

PC

TPC

T

θ

= ⋅

=

Tλ

λλ

λ

or

( ) 0.973 lb

PC OA

T=



consent of McGraw–Hill Education.

PROBLEM 3.97

For the shift lever shown, determine the magnitude and the direction of the smallest

force P that has a 210-lb ⋅ in. clockwise moment about B.

SOLUTION

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,

1

8 in.

tan 22 in.

19.98

aθ

−

=

= °

and

minB

M dP=

Where

/

22

(8 in.) (22 in.)

23.409 in.

AB

dr=

= +

=

Then

min

210 lb in.

23.409 in.

P⋅

=

8.97 lb

=

min 8.97 lb=P

19.98° 

PROBLEM 3.98

Consider the volleyball net

shown. Determine the angle

formed by guy wires AB and

AC.

SOLUTION

First note:

2 22

( 6.5) ( 8) (2) 10.5 ft

(0) ( 8) (6) 10 ft

AB

AC

= − +− + =

= +− + =

and

(6.5 ft) (8 ft) (2 ft)

(8 ft) (6 ft)

AB

AC

=− −+

=−+

ijk

jk

C

By definition,

( )( )cosAB AC AB AC

θ

⋅=

C C

or

( 6.5 8 2 ) ( 8 6 ) (10.5)(10)cos

( 6.5)(0) ( 8)( 8) (2)(6) 105cos

θ

− − + ⋅− + =

− +− − + =

ijk jk

or

cos 0.72381

θ

=

or

43.6

θ

= °



PROBLEM 3.99

A crane is oriented so that the end of the 25–m boom AO lies in

the yz plane. At the instant shown, the tension in cable AB is

4 kN. Determine the moment about each of the coordinate axes

of the force exerted on A by cable AB.

SOLUTION

22

()()

(25 m) (15 m)

20 m

OC OA AC= −

= −

=

(15 m) (20 m)

A= +r jk

22

(2.5 m) (15 m) 15.2069 m

2.5 15

(4 kN) 15.2069

(0.65760 kN) (3.9456 kN)

AB

PAB

= +=

=

−

=

= −

P

ij



Using Eq. (3.11):

0 15 20

0.65760 3.9456 0

(78.912 kN m) (13.1520 kN m) (9.8640 kN m)

OA

O

=×=

=⋅ ⋅⋅

i jk

M rP

M i+ j- k

But

Ox y z

MMM=++M i jk

Therefore,

78.9 kN m, 13.15 kN m, 9.86 kN m

xy z

MM M= ⋅= ⋅=−⋅



PROBLEM 3.100

The 25–m crane boom AO lies in the yz plane. Determine the

maximum permissible tension in cable AB if the absolute value

of moments about the coordinate axes of the force exerted on A

by cable AB must be

|Mx| ≤ 60 kN∙m, |My| ≤ 12 kN∙m, |Mz| ≤ 8 kN∙m

22

()()

(25 m) (15 m)

20 m

OC OA AC= −

= −

=

(15 m) (20 m)

A

= +r jk

22

(2.5 m) (15 m) 15.2069 m

2.5 15

( ) 15.2069

(0.164399 kN) (0.98639 kN)

AB

PAB

P

PP

= +=

=

−

=

= −

P

ij



Using Eq. (3.11):

0 15 20

0.163499 0.98639 0

(19.7278 ) (3.2700 ) (2.4525 )

OA

O

PP

P PP

=×=

−

=

i jk

M rP

M i+ j- k

But

60 kN m: 19.7278 60 3.04 kN

12 kN m: 3.270 12 3.67 kN

8 kN m: 2.4525 8 3.15 kN

x

y

z

M PP

⋅ <≤

<⋅ < ≤

<

PROBLEM 3.101

A single force P acts at C in a direction perpendicular to

the handle BC of the crank shown. Determine the moment

Mx of P about the x–axis when θ = 65°, knowing that

My = −15 N · m and Mz = −36 N · m.

SOLUTION

See the solution to Prob. 3.53 for the derivation of the following equations:

(0.2) sin( )

x

MP

θφ

= +

(1)

tan

z

y

M

φ

=

(4)

22

4

yz

P MM= +

(5)

Substituting for known data gives:

22

36 N m

tan 15 N m

67.380

4 ( 15) ( 36)

156.0 N

0.2 m(156.0 N)sin(65 67.380 )

23.047 N m

x

P

M

φ

−⋅

=−⋅

= °

= − +−

=

= °+ °

= ⋅

23.0 N m

x

M= ⋅



PROBLEM 3.102

While tapping a hole, a machinist applies the horizontal

forces shown to the handle of the tap wrench. Show that

these forces are equivalent to a single force, and specify, if

possible, the point of application of the single force on the

handle.

SOLUTION

Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one

of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-

couple.

We have

2.9 lb 2.65 lb 0.25 lb,

B

F=−=

where the 2.65-lb force is part of the couple. Combining the two

parallel forces,

couple

(2.65 lb)[(3.2 in. 2.8 in.)cos25 ]

14.4103 lb in.

M= +°

= ⋅

and

couple 14.4103 lb in.= ⋅M

A single equivalent force will be located in the negative z direction.

Based on

: 14.4103 lb in. [(0.25 lb)cos 25 ]( )

B

MaΣ − ⋅= °

63.600 in.a=

F′

(0.25 lb)(cos 25 sin 25 )= °+ °ik

F′

(0.227 lb) (0.1057 lb)= +ik

and is applied on an extension of handle BD at a

distance of 63.6 in. to the right of B. 

PROBLEM 3.103

A 500–N force is applied to a bent plate as shown. Determine

(a) an equivalent force–couple system at B, (b) an equivalent

system formed by a vertical force at A and a force at B.

SOLUTION

(a) Force–couple system at B

(500 N)sin30 (500 N)cos30

(250.0 N) (433.01 N)

= −

F ik

Fik



(0.3 0.175 ) (250 433.01 )

86.153 N m

B

=− ×−

=−⋅

Mi ji j

k

The equivalent force–couple system at B is

500 N

B=F

60.0°

86.2 N m

B

= ⋅M



(b) Require

Equivalence requires

86.153 N m (0.125 m); (689.22 N)

or ; (250 N) (433.01 N) (689.22 N)

BB

MM∑=Σ

⋅= =

===−−

AA j

F A+B B F-A B i j j

(250 N) (1122.23 N) 1149.74 N=−=Bi j

77.4°

1150 N=B

77.4°



689 N = ↑A



PROBLEM 3.104

A force and a couple are applied as shown to the end of a

cantilever beam. (a) Replace this system with a single force

F applied at Point C, and determine the distance d from C to

a line drawn through Points D and E. (b) Solve part a if the

directions of the two 360-N forces are reversed.

SOLUTION

PROBLEM 3.105

Slider P can move along rod OA. An elastic cord PC is attached

to the slider and to the vertical member BC. Knowing that the

distance from O to P is 6 in. and that the tension in the cord is

3 lb, determine (a) the angle between the elastic cord and the

rod OA, (b) the projection on OA of the force exerted by cord

PC at Point P.

SOLUTION

First note

22 2

(12) (12) ( 6) 18 in.

OA = + +− =

Then

1(12 12 6 )

18

1(2 2 )

3

OA

= = +−

= +−

i jk

λ

Now

1

6 in. ( )

3

OP OP OA= ⇒=

The coordinates of Point P are (4 in., 4 in., −2 in.)

so that

(5 in.) (11 in.) (14 in.)PC =++ijk

C

and

222

(5) (11) (14) 342 in.

PC = ++ =

(a) We have

( )cos

OA

PC PC

θ

⋅=

C λ

or

1

(5 11 14 ) (2 2 ) 342 cos

3

θ

+ + ⋅ +−=i j k i jk

or

1

cos [(5)(2) (11)(2) (14)( 1)]

3 342

0.32444

θ

= + +−

=

or

71.1

θ

= °



(b) We have

()

cos

(3 lb)(0.32444)

OA

PC PC OA

PC OA

PC

T

PC

TPC

T

θ

= ⋅

=

Tλ

λλ

λ

or

( ) 0.973 lb

PC OA

T=



consent of McGraw–Hill Education.