Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
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
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
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
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
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
AB
PAB
= +=
=
−
=
= −
P
ij
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
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
AB
PAB
P
PP
= +=
=
−
=
= −
P
ij
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
M PP
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
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
P
M
φ
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
OA
OA
= = +−
= +−
i jk
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 PC OA
PC OA
PC
T
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
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
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
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
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
AB
PAB
= +=
=
−
=
= −
P
ij
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
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
AB
PAB
P
PP
= +=
=
−
=
= −
P
ij
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
M PP
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
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
P
M
φ
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
OA
OA
= = +−
= +−
i jk
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 PC OA
PC OA
PC
T
T
PC
TPC
T
θ
= ⋅
= ⋅
= ⋅
=
=
Tλ
λλ
λ
or
( ) 0.973 lb
PC OA
T=
consent of McGraw-Hill Education.
