PROBLEM 3.59
Shafts A and B connect the gear box to the wheel
assemblies of a tractor, and shaft C connects it to the
engine. Shafts A and B lie in the vertical yz plane,
while shaft C is directed along the x axis. Replace
the couples applied to the shafts by a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
1200sin20 1200cos20 410.42 1127.63
900sin 20 900cos20 307.82 845.72
840
The single equivalent couple is the sum of the three moments
(840 lb ft) (102.60 lb ft) (1973.35 lb ft
A
B
C
=− + =−+
=+=+
= −
=− ⋅− ⋅+ ⋅
M j k jk
M j k jk
Mi
Mi j
AA
AA
)k
2150 lb ftM= ⋅
axis
z
0.39119 0.047921 0.91899
cos 0.39119
cos 0.047921
cos 0.91899
x
y
M
θ
θ
θ
==−− +
= −
= −
=
Mi jkλ
z
113.0 92.7 23.2
xy
θ θθ
= °=°=°
consent of McGraw–Hill Education.
PROBLEM 3.60
If
20 lb,P=
replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.78:
16-lb force:
1(480 lb in.)M=−⋅k
40-lb force:
28 5[(10 lb in.) (30 lb in.) (15 lb in.) ]M= ⋅+ ⋅+ ⋅i jk
P
20 lb=
3
(30 in.) (20 lb)
(600 lb in.)
C
MP= ×
= ×
= ⋅
r
ik
j
123
22 2
(480) 8 5 (10 30 15 ) 600
(178.885 lb in.) (1136.66 lb in.) (211.67 lb in.)
(178.885) (113.66) (211.67)
1169.96 lb in.
M
=++
=− + ++ +
= ⋅+ ⋅− ⋅
= ++
= ⋅
MM M M
k i jk j
i jk
1170 lb in.M= ⋅
axis
0.152898 0.97154 0.180921
cos 0.152898
cos 0.97154
cos 0.180921
x
y
z
M
θ
θ
θ
== +−
=
=
= −
Mij kλ
81.2 13.70 100.4
xy z
θθ θ
=°= °= °
consent of McGraw–Hill Education.
PROBLEM 3.61
A 30–lb vertical force P is applied at A to the bracket shown, which is held
by screws at B and C. (a) Replace P with an equivalent force-couple
system at B. (b) Find the two horizontal forces at B and C that are
equivalent to the couple obtained in part a.
SOLUTION
(a)
(30 lb)(5 in.)
150.0 lb in.
B
M=
= ⋅
30.0 lb=F
,
150.0 lb in.
B
= ⋅M
(b)
150 lb in. 50.0 lb
3.0 in.
BC ⋅
= = =
50.0 lb=B
;
50.0 lb=C
consent of McGraw–Hill Education.
PROBLEM 3.62
The force P has a magnitude of 250 N and is applied at the end C of
a 500–mm rod AC attached to a bracket at A and B. Assuming
α
30°=
and
60°,
β
=
replace P with (a) an equivalent force–couple system at B,
(b) an equivalent system formed by two parallel forces applied at
A and B.
SOLUTION
PROBLEM 3.63
Solve Problem 3.62, assuming
25°.
αβ
= =
PROBLEM 3.62 The force P has a magnitude of 250 N and is applied
at the end C of a 500–mm rod AC attached to a bracket at A and B.
Assuming
α
30°=
and
60°,
β
=
replace P with (a) an equivalent force–
couple system at B, (b) an equivalent system formed by two parallel
forces applied at A and B.
SOLUTION
(a) Equivalence requires
: or 250 N
BB
Σ= =FF P F
25.0°
: (0.3 m)[(250 N)sin50 ] 57.453 N m
BB
MΣ =− °=− ⋅M
The equivalent force–couple system at B is
250 N
B
=F
25.0°
57.5 N m
B= ⋅M
(b) We require
Equivalence requires
(0.3 m)[(250 N)sin50 ]
[(0.2 m)sin50 ]
375 N
B AE
M dQ
Q
Q
= °
= °
=
Adding the forces at B:
375 N
A
=F
25.0°
625 N
B=F
25.0°
consent of McGraw–Hill Education.
PROBLEM 3.64
A 260–lb force is applied at A to the rolled–steel section shown. Replace that
force with an equivalent force–couple system at the center C of the section.
SOLUTION
22
(2.5 in.) (6.0 in.) 6.50 in.AB = +=
2.5 in. 5
sin 6.5 in. 13
6.0 in. 12
cos 22.6
6.5 in. 13
α
αα
= =
= = = °
sin cos
5 12
(260 lb) (260 lb)
13 13
(100.0 lb) (240 lb)
FF
αα
=−−
=−−
=−−
F ij
ij
ij
/
(2.5 4.0 ) ( 100.0 240 )
400 600
(200 lb in.)
C AC
= ×
= + ×− −
= −
=−⋅
Mr F
ij i j
kk
k
260 lb
=F
67.4°;
200 lb in.
C= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.65
A dirigible is tethered by a cable attached to its cabin at B.
If the tension in the cable is 1040 N, replace the force
exerted by the cable at B with an equivalent system formed
by two parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of
α
with the vertical.
Then for equivalence,
: (1040 N)sin30 sin sin
x AB
F FF
αα
Σ °= +
(1)
: (1040 N)cos30 cos cos
y AB
F FF
αα
Σ − °=− −
(2)
Dividing Equation (1) by Equation (2),
( )sin
(1040 N)sin30
(1040 N)cos30 ( )cos
AB
AB
FF
FF
α
α
+
°=
− °− +
Simplifying yields
30 .
α
= °
Based on
: [(1040 N)cos30 ](4 m) ( cos30 )(10.7 m)
CA
MFΣ °= °
388.79 N
A
F=
or
389 N
A=F
60.0°
Based on
: [(1040 N)cos30 ](6.7 m) ( cos30 )(10.7 m)
AC
MFΣ− ° = °
651.21 N
C
F=
C=F
consent of McGraw–Hill Education.
PROBLEM 3.66
A force and couple act as shown on a square plate of side a = 25 in. Knowing
that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by
a single force applied at a point located (a) on line AB, (b) on line AC. In each
case determine the distance from A to the point of application of the force.
SOLUTION
Replace the given force–couple system with an equivalent force-
couple system at A.
(60 lb)(cos50 ) 38.567 lb
x
P= °=
(60 lb)(sin50 ) 45.963 lb
y
P= °=
(45.963 lb)(25 in.) (40 lb)(25 in.)
Ay
M P a Qa= −
= −
149.075 lb in.= ⋅
(a) Equating moments about A gives:
149.075 lb in. (45.963 lb)
3.24 in.
x
x
⋅=
=
60.0 lb=P
50.0°; 3.24 in. from A
(b)
149.075 lb in. (38.567 lb)
3.87 in.
y
y
⋅=
=
60.0 lb=P
50.0°; 3.87 in. below A
consent of McGraw–Hill Education.
PROBLEM 3.67
An eccentric, compressive 250–kN force P is applied to the end
of a column. Replace P with an equivalent force–couple system
at G.
SOLUTION
Have
( )
: 250 kNΣ− =F jF
or
( )
250 kN= −Fj
Also have
:
GP
Σ ×=M r PM
0.030 0 0.060 kN m =
0 250 0
−⋅
−
i jk
M
( ) ( )
15 kN m 7.5 kN m∴= ⋅ + ⋅Mi k
or
( ) ( )
15.00 kN m 7.50 kN m= ⋅+ ⋅M ik
consent of McGraw–Hill Education.
PROBLEM 3.59
Shafts A and B connect the gear box to the wheel
assemblies of a tractor, and shaft C connects it to the
engine. Shafts A and B lie in the vertical yz plane,
while shaft C is directed along the x axis. Replace
the couples applied to the shafts by a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
1200sin20 1200cos20 410.42 1127.63
900sin 20 900cos20 307.82 845.72
840
The single equivalent couple is the sum of the three moments
(840 lb ft) (102.60 lb ft) (1973.35 lb ft
A
B
C
=− + =−+
=+=+
= −
=− ⋅− ⋅+ ⋅
M j k jk
M j k jk
Mi
Mi j
AA
AA
)k
2150 lb ftM= ⋅
axis
z
0.39119 0.047921 0.91899
cos 0.39119
cos 0.047921
cos 0.91899
x
y
M
θ
θ
θ
==−− +
= −
= −
=
Mi jkλ
z
113.0 92.7 23.2
xy
θ θθ
= °=°=°
consent of McGraw–Hill Education.
PROBLEM 3.60
If
20 lb,P=
replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.78:
16-lb force:
1(480 lb in.)M=−⋅k
40-lb force:
28 5[(10 lb in.) (30 lb in.) (15 lb in.) ]M= ⋅+ ⋅+ ⋅i jk
P
20 lb=
3
(30 in.) (20 lb)
(600 lb in.)
C
MP= ×
= ×
= ⋅
r
ik
j
123
22 2
(480) 8 5 (10 30 15 ) 600
(178.885 lb in.) (1136.66 lb in.) (211.67 lb in.)
(178.885) (113.66) (211.67)
1169.96 lb in.
M
=++
=− + ++ +
= ⋅+ ⋅− ⋅
= ++
= ⋅
MM M M
k i jk j
i jk
1170 lb in.M= ⋅
axis
0.152898 0.97154 0.180921
cos 0.152898
cos 0.97154
cos 0.180921
x
y
z
M
θ
θ
θ
== +−
=
=
= −
Mij kλ
81.2 13.70 100.4
xy z
θθ θ
=°= °= °
consent of McGraw–Hill Education.
PROBLEM 3.61
A 30–lb vertical force P is applied at A to the bracket shown, which is held
by screws at B and C. (a) Replace P with an equivalent force-couple
system at B. (b) Find the two horizontal forces at B and C that are
equivalent to the couple obtained in part a.
SOLUTION
(a)
(30 lb)(5 in.)
150.0 lb in.
B
M=
= ⋅
30.0 lb=F
,
150.0 lb in.
B
= ⋅M
(b)
150 lb in. 50.0 lb
3.0 in.
BC ⋅
= = =
50.0 lb=B
;
50.0 lb=C
consent of McGraw–Hill Education.
PROBLEM 3.62
The force P has a magnitude of 250 N and is applied at the end C of
a 500–mm rod AC attached to a bracket at A and B. Assuming
α
30°=
and
60°,
β
=
replace P with (a) an equivalent force–couple system at B,
(b) an equivalent system formed by two parallel forces applied at
A and B.
SOLUTION
PROBLEM 3.63
Solve Problem 3.62, assuming
25°.
αβ
= =
PROBLEM 3.62 The force P has a magnitude of 250 N and is applied
at the end C of a 500–mm rod AC attached to a bracket at A and B.
Assuming
α
30°=
and
60°,
β
=
replace P with (a) an equivalent force–
couple system at B, (b) an equivalent system formed by two parallel
forces applied at A and B.
SOLUTION
(a) Equivalence requires
: or 250 N
BB
Σ= =FF P F
25.0°
: (0.3 m)[(250 N)sin50 ] 57.453 N m
BB
MΣ =− °=− ⋅M
The equivalent force–couple system at B is
250 N
B
=F
25.0°
57.5 N m
B= ⋅M
(b) We require
Equivalence requires
(0.3 m)[(250 N)sin50 ]
[(0.2 m)sin50 ]
375 N
B AE
M dQ
Q
Q
= °
= °
=
Adding the forces at B:
375 N
A
=F
25.0°
625 N
B=F
25.0°
consent of McGraw–Hill Education.
PROBLEM 3.64
A 260–lb force is applied at A to the rolled–steel section shown. Replace that
force with an equivalent force–couple system at the center C of the section.
SOLUTION
22
(2.5 in.) (6.0 in.) 6.50 in.AB = +=
2.5 in. 5
sin 6.5 in. 13
6.0 in. 12
cos 22.6
6.5 in. 13
α
αα
= =
= = = °
sin cos
5 12
(260 lb) (260 lb)
13 13
(100.0 lb) (240 lb)
FF
αα
=−−
=−−
=−−
F ij
ij
ij
/
(2.5 4.0 ) ( 100.0 240 )
400 600
(200 lb in.)
C AC
= ×
= + ×− −
= −
=−⋅
Mr F
ij i j
kk
k
260 lb
=F
67.4°;
200 lb in.
C= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.65
A dirigible is tethered by a cable attached to its cabin at B.
If the tension in the cable is 1040 N, replace the force
exerted by the cable at B with an equivalent system formed
by two parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of
α
with the vertical.
Then for equivalence,
: (1040 N)sin30 sin sin
x AB
F FF
αα
Σ °= +
(1)
: (1040 N)cos30 cos cos
y AB
F FF
αα
Σ − °=− −
(2)
Dividing Equation (1) by Equation (2),
( )sin
(1040 N)sin30
(1040 N)cos30 ( )cos
AB
AB
FF
FF
α
α
+
°=
− °− +
Simplifying yields
30 .
α
= °
Based on
: [(1040 N)cos30 ](4 m) ( cos30 )(10.7 m)
CA
MFΣ °= °
388.79 N
A
F=
or
389 N
A=F
60.0°
Based on
: [(1040 N)cos30 ](6.7 m) ( cos30 )(10.7 m)
AC
MFΣ− ° = °
651.21 N
C
F=
C=F
consent of McGraw–Hill Education.
PROBLEM 3.66
A force and couple act as shown on a square plate of side a = 25 in. Knowing
that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by
a single force applied at a point located (a) on line AB, (b) on line AC. In each
case determine the distance from A to the point of application of the force.
SOLUTION
Replace the given force–couple system with an equivalent force-
couple system at A.
(60 lb)(cos50 ) 38.567 lb
x
P= °=
(60 lb)(sin50 ) 45.963 lb
y
P= °=
(45.963 lb)(25 in.) (40 lb)(25 in.)
Ay
M P a Qa= −
= −
149.075 lb in.= ⋅
(a) Equating moments about A gives:
149.075 lb in. (45.963 lb)
3.24 in.
x
x
⋅=
=
60.0 lb=P
50.0°; 3.24 in. from A
(b)
149.075 lb in. (38.567 lb)
3.87 in.
y
y
⋅=
=
60.0 lb=P
50.0°; 3.87 in. below A
consent of McGraw–Hill Education.
PROBLEM 3.67
An eccentric, compressive 250–kN force P is applied to the end
of a column. Replace P with an equivalent force–couple system
at G.
SOLUTION
Have
( )
: 250 kNΣ− =F jF
or
( )
250 kN= −Fj
Also have
:
GP
Σ ×=M r PM
0.030 0 0.060 kN m =
0 250 0
−⋅
−
i jk
M
( ) ( )
15 kN m 7.5 kN m∴= ⋅ + ⋅Mi k
or
( ) ( )
15.00 kN m 7.50 kN m= ⋅+ ⋅M ik
consent of McGraw–Hill Education.