PROBLEM 3.87
A machine component is subjected to the forces shown, each
of which is parallel to one of the coordinate axes. Replace
these forces with an equivalent force–couple system at A.
SOLUTION
PROBLEM 3.88
A mechanic uses a crowfoot wrench to loosen a bolt at C.
The mechanic holds the socket wrench handle at Points A
and B and applies forces at these points. Knowing that
these forces are equivalent to a force–couple system at C
consisting of the force
(8 lb) + (4 lb)= −C ik
and the
couple
(360 lb·
C
=M
in.)i, determine the forces applied at
A and at B when
2 lb.
z
A=
SOLUTION
We have
:ΣF
+=ABC
or
: 8 lb
x xx
F AB+=
( 8 lb)
xx
BA=−+
(1)
:0
y yy
F ABΣ +=
or
yy
AB= −
(2)
: 2 lb 4 lb
zz
FBΣ +=
or
2 lb
z
B=
(3)
We have
//
:
C BC AC C
Σ ×+ ×=M r Br AM
8 0 2 8 0 8 lb in. (360 lb in.)
22
xy xy
BB AA
+ ⋅= ⋅
i jk i jk
i
or
(2 8 ) (2 16 8 16)
yy x x
BA B A− + −+ −ij
(8 8 ) (360 lb in.)
yy
BA++ = ⋅ki
From i–coefficient:
2 8 360 lb in.
yy
BA−= ⋅
(4)
j–coefficient:
2 8 32 lb in.
xx
BA−+ = ⋅
(5)
k–coefficient:
880
yy
BA+=
(6)
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
From Equations (2) and (4):
2 8( ) 360
yy
BB−− =
36 lb 36 lb
yy
BA= =
From Equations (1) and (5):
2( 8) 8 32
xx
AA−−+ =
1.6 lb
x
A=
From Equation (1):
(1.6 8) 9.6 lb
x
B=− +=−
(1.600 lb) (36.0 lb) (2.00 lb)= −+A ijk
(9.60 lb) (36.0 lb) (2.00 lb)=−+ +B ijk
consent of McGraw–Hill Education.
PROBLEM 3.89
In order to unscrew the tapped faucet A, a
plumber uses two pipe wrenches as shown.
By exerting a 40–lb force on each wrench, at
a distance of 10 in. from the axis of the pipe
and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from
rotating, and thus avoids loosening or further
tightening the joint between the pipe and the
tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the
vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C
equivalent to the two 40–lb forces when this
condition is satisfied.
SOLUTION
We first reduce the given forces to force–couple systems at A and B, noting that
| | | | (40 lb)(10 in.)
400 lb in.
AB
= =
= ⋅
MM
We now determine the equivalent force–couple system at C.
(40 lb)(1 cos ) (40 lb)sin
θθ
= −−R ij
(1)
(15 in.) [ (40 lb)cos (40 lb)sin ]
(7.5 in.) (40 lb)
400 400 600cos 600sin 300
(600 lb in.)sin (300 lb in.)(1 2cos )
R
C AB
θθ
θθ
θθ
= + + ×− −
+×
=+−− + +
= ⋅ + ⋅−
MMM k i j
ki
j ij
ij
(2)
(a) For no rotation about vertical, y component of
R
C
M
must be zero.
1 2cos 0
cos 1/2
θ
θ
−=
=
60.0
θ
= °
(b) For
60.0
θ
= °
in Eqs. (1) and (2),
(20.0 lb) (34.641 lb) ; (519.62 lb in.)
R
C
=−=⋅R i jM i
(20.0 lb) (34.6 lb) ; (520 lb in.)
R
C
=−=⋅R i jM i
consent of McGraw–Hill Education.
PROBLEM 3.90
Assuming θ = 60° in Prob. 3.89, replace the
two 40–lb forces with an equivalent force–
couple system at D and determine whether
the plumber’s action tends to tighten or
loosen the joint between (a) pipe CD and
elbow D, (b) elbow D and pipe DE.
Assume all threads to be right-handed.
PROBLEM 3.89 In order to unscrew the
tapped faucet A, a plumber uses two pipe
wrenches as shown. By exerting a 40–lb
force on each wrench, at a distance of 10 in.
from the axis of the pipe and in a direction
perpendicular to the pipe and to the wrench,
he prevents the pipe from rotating, and thus
avoids loosening or further tightening the
joint between the pipe and the tapped elbow
C. Determine (a) the angle θ that the
wrench at A should form with the vertical if
elbow C is not to rotate about the vertical,
(b) the force–couple system at C equivalent
to the two 40–lb forces when this condition
is satisfied.
SOLUTION
The equivalent force–couple system at C for
60
θ
= °
was obtained in the solution to Prob. 3.89:
(20.0 lb) (34.641 lb)
(519.62 lb in.)
R
C
= −
= ⋅
Ri j
Mi
The equivalent force–couple system at D is made of R and
R
D
M
where
/
(519.62 lb in.) (25.0 in.) [(20.0 lb) (34.641 lb) ]
(519.62 lb in.) (500 lb in.)
RR
D C CD
=+×
= ⋅+ × −
= ⋅− ⋅
MMr R
i ji j
ik
Equivalent force-couple at D:
(20.0 lb) (34.6 lb) ; (520 lb in.) (500 lb in.)
R
C
= − = ⋅− ⋅R i jM i k
R
consent of McGraw–Hill Education.
R
R
PROBLEM 3.91
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force–couple
system at A consisting of
(30 N) + +
yz
RR= −R i jk
and
(12 N · m) .
R
A
= −Mi
(b) Find the corresponding
values of
y
R
and
.
z
R
(c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a) Equivalence requires
:Σ=+F RBC
or
(30 N) ( )
yz xyz
RR B CCC− + + =− +− + +i jk k i jk
Equating the i coefficients:
: 30 N or 30 N
xx
CC−=− =i
Also,
//
:R
A A BA CA
Σ = ×+ ×M M r Br C
or
(12 N m) [(0.2 m) (0.15 m) ] ( )
(0.4 m) [ (30 N) ]
yz
B
CC
− ⋅ = + ×−
+ ×− + +
i i jk
i i jk
Equating coefficients:
: 12 N m (0.15 m) or 80 N
: 0 (0.4 m) or 0
: 0 (0.2 m)(80 N) (0.4 m) or 40 N
yy
zz
BB
CC
CC
− ⋅=− =
= =
=−=
i
k
j
(80.0 N) (30.0 N) (40.0 N)=− =−+B kC i k
(30 N) (80 N) [( 30 N) (40 N) ]
yz
RR− + + =− +− +i jk k i k
Equating coefficients:
:0
y
R=j
0
y
R=
: 80 40
z
R=−+k
or 40.0 N
z
R= −
z
when the slot in the head of the screw is vertical.
consent of McGraw–Hill Education.
PROBLEM 3.92
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine the magnitude and the point of
application of the resultant of the four wind forces
when
1 fta=
and
12 ft.b=
SOLUTION
We have
Equivalence then requires
: 105 90 160 50
z
FRΣ − −− −=−
or
405 lbR=
: (5 ft)(105 lb) (1 ft)(90 lb) (3 ft)(160 lb)
(5.5 ft)(50 lb) (405 lb)
x
M
y
Σ −+
+=−
or
2.94 fty= −
: (5.5 ft)(105 lb) (12 ft)(90 lb) (14.5 ft)(160 lb)
(22.5 ft)(50 lb) (405 lb)
y
M
x
Σ ++
+=−
or
12.60 ftx=
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.
consent of McGraw–Hill Education.
PROBLEM 3.93
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine a and b so that the point of application
of the resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then requires that
G
ΣM
of the applied system of forces also be zero. Then
at
: : ( 3) ft (90 lb) (2 ft)(105 lb)
(2.5 ft)(50 lb) 0
x
GM aΣ −+ × +
+=
or
0.722 fta=
: (9 ft)(105 ft) (14.5 ) ft (90 lb)
(8 ft)(50 lb) 0
y
MbΣ − − −×
+=
or
20.6 ftb=
consent of McGraw–Hill Education.
PROBLEM 3.94
A concrete foundation mat of 5–m radius supports four equally spaced
the four loads.
SOLUTION
Have:
:
ACDE
Σ +++=F FFFF R
( ) ( ) ( ) ( )
100 kN 125 kN 25kN 75 kN=− − −−R j jjj
( )
325 kN= − j
(325 kN)= = −RP j
or
325 kNR=
Have:
( )
( )
( )
:
x CC EE G
M F z F z RzΣ +=
( )
125 kN (4 m) (75 kN)(4 m)−+
( )
325 kN G
z=
0.61539 m
G
z∴=−
or
0.615 m
G
z= −
( ) ( )
( )
:
z AA DD G
M F x F x RxΣ +=
( ) ( )
100 kN (4 m) (25 kN) 4 m−
( )
325 kN
G
x= −
or
0.923 m
G
x=−
consent of McGraw–Hill Education.
PROBLEM 3.95
Three children are standing on a
5 5-m×
raft. If the
weights of the children at Points A, B, and C are 375 N,
260 N, and 400 N, respectively, determine the magnitude
and the point of application of the resultant of the three
weights.
SOLUTION
We have
:
ABC
Σ ++=FFF F R
(375 N) (260 N) (400 N)
(1035 N)
−− − =
−=
jjjR
jR
or 1035 NR=
We have
: () () () ()
x AA BB CC D
M Fz Fz Fz RzΣ ++=
(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (1035 N)( )
D
z++ =
3.0483 m
D
z=
or 3.05 m
D
z=
We have
: () () () ()
z AA BB CC D
M Fx Fx Fx RxΣ ++=
375 N(1m) (260 N)(1.5 m) (400 N)(4.75 m) (1035 N)()
D
x++ =
2.5749 m
D
x=
or 2.57 m
D
x=
consent of McGraw–Hill Education.
PROBLEM 3.88
A mechanic uses a crowfoot wrench to loosen a bolt at C.
The mechanic holds the socket wrench handle at Points A
and B and applies forces at these points. Knowing that
these forces are equivalent to a force–couple system at C
consisting of the force
(8 lb) + (4 lb)= −C ik
and the
couple
(360 lb·
C
=M
in.)i, determine the forces applied at
A and at B when
2 lb.
z
A=
SOLUTION
We have
:ΣF
+=ABC
or
: 8 lb
x xx
F AB+=
( 8 lb)
xx
BA=−+
(1)
:0
y yy
F ABΣ +=
or
yy
AB= −
(2)
: 2 lb 4 lb
zz
FBΣ +=
or
2 lb
z
B=
(3)
We have
//
:
C BC AC C
Σ ×+ ×=M r Br AM
8 0 2 8 0 8 lb in. (360 lb in.)
22
xy xy
BB AA
+ ⋅= ⋅
i jk i jk
i
or
(2 8 ) (2 16 8 16)
yy x x
BA B A− + −+ −ij
(8 8 ) (360 lb in.)
yy
BA++ = ⋅ki
From i–coefficient:
2 8 360 lb in.
yy
BA−= ⋅
(4)
j–coefficient:
2 8 32 lb in.
xx
BA−+ = ⋅
(5)
k–coefficient:
880
yy
BA+=
(6)
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
From Equations (2) and (4):
2 8( ) 360
yy
BB−− =
36 lb 36 lb
yy
BA= =
From Equations (1) and (5):
2( 8) 8 32
xx
AA−−+ =
1.6 lb
x
A=
From Equation (1):
(1.6 8) 9.6 lb
x
B=− +=−
(1.600 lb) (36.0 lb) (2.00 lb)= −+A ijk
(9.60 lb) (36.0 lb) (2.00 lb)=−+ +B ijk
consent of McGraw–Hill Education.
PROBLEM 3.89
In order to unscrew the tapped faucet A, a
plumber uses two pipe wrenches as shown.
By exerting a 40–lb force on each wrench, at
a distance of 10 in. from the axis of the pipe
and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from
rotating, and thus avoids loosening or further
tightening the joint between the pipe and the
tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the
vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C
equivalent to the two 40–lb forces when this
condition is satisfied.
SOLUTION
We first reduce the given forces to force–couple systems at A and B, noting that
| | | | (40 lb)(10 in.)
400 lb in.
AB
= =
= ⋅
MM
We now determine the equivalent force–couple system at C.
(40 lb)(1 cos ) (40 lb)sin
θθ
= −−R ij
(1)
(15 in.) [ (40 lb)cos (40 lb)sin ]
(7.5 in.) (40 lb)
400 400 600cos 600sin 300
(600 lb in.)sin (300 lb in.)(1 2cos )
R
C AB
θθ
θθ
θθ
= + + ×− −
+×
=+−− + +
= ⋅ + ⋅−
MMM k i j
ki
j ij
ij
(2)
(a) For no rotation about vertical, y component of
R
C
M
must be zero.
1 2cos 0
cos 1/2
θ
θ
−=
=
60.0
θ
= °
(b) For
60.0
θ
= °
in Eqs. (1) and (2),
(20.0 lb) (34.641 lb) ; (519.62 lb in.)
R
C
=−=⋅R i jM i
(20.0 lb) (34.6 lb) ; (520 lb in.)
R
C
=−=⋅R i jM i
consent of McGraw–Hill Education.
PROBLEM 3.90
Assuming θ = 60° in Prob. 3.89, replace the
two 40–lb forces with an equivalent force–
couple system at D and determine whether
the plumber’s action tends to tighten or
loosen the joint between (a) pipe CD and
elbow D, (b) elbow D and pipe DE.
Assume all threads to be right-handed.
PROBLEM 3.89 In order to unscrew the
tapped faucet A, a plumber uses two pipe
wrenches as shown. By exerting a 40–lb
force on each wrench, at a distance of 10 in.
from the axis of the pipe and in a direction
perpendicular to the pipe and to the wrench,
he prevents the pipe from rotating, and thus
avoids loosening or further tightening the
joint between the pipe and the tapped elbow
C. Determine (a) the angle θ that the
wrench at A should form with the vertical if
elbow C is not to rotate about the vertical,
(b) the force–couple system at C equivalent
to the two 40–lb forces when this condition
is satisfied.
SOLUTION
The equivalent force–couple system at C for
60
θ
= °
was obtained in the solution to Prob. 3.89:
(20.0 lb) (34.641 lb)
(519.62 lb in.)
R
C
= −
= ⋅
Ri j
Mi
The equivalent force–couple system at D is made of R and
R
D
M
where
/
(519.62 lb in.) (25.0 in.) [(20.0 lb) (34.641 lb) ]
(519.62 lb in.) (500 lb in.)
RR
D C CD
=+×
= ⋅+ × −
= ⋅− ⋅
MMr R
i ji j
ik
Equivalent force-couple at D:
(20.0 lb) (34.6 lb) ; (520 lb in.) (500 lb in.)
R
C
= − = ⋅− ⋅R i jM i k
R
consent of McGraw–Hill Education.
PROBLEM 3.91
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force–couple
system at A consisting of
(30 N) + +
yz
RR= −R i jk
and
(12 N · m) .
R
A
= −Mi
(b) Find the corresponding
values of
y
R
and
.
z
R
(c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a) Equivalence requires
:Σ=+F RBC
or
(30 N) ( )
yz xyz
RR B CCC− + + =− +− + +i jk k i jk
Equating the i coefficients:
: 30 N or 30 N
xx
CC−=− =i
Also,
//
:R
A A BA CA
Σ = ×+ ×M M r Br C
or
(12 N m) [(0.2 m) (0.15 m) ] ( )
(0.4 m) [ (30 N) ]
yz
B
CC
− ⋅ = + ×−
+ ×− + +
i i jk
i i jk
Equating coefficients:
: 12 N m (0.15 m) or 80 N
: 0 (0.4 m) or 0
: 0 (0.2 m)(80 N) (0.4 m) or 40 N
yy
zz
BB
CC
CC
− ⋅=− =
= =
=−=
i
k
j
(80.0 N) (30.0 N) (40.0 N)=− =−+B kC i k
(30 N) (80 N) [( 30 N) (40 N) ]
yz
RR− + + =− +− +i jk k i k
Equating coefficients:
:0
y
R=j
0
y
R=
: 80 40
z
R=−+k
or 40.0 N
z
R= −
z
when the slot in the head of the screw is vertical.
consent of McGraw–Hill Education.
PROBLEM 3.92
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine the magnitude and the point of
application of the resultant of the four wind forces
when
1 fta=
and
12 ft.b=
SOLUTION
We have
Equivalence then requires
: 105 90 160 50
z
FRΣ − −− −=−
or
405 lbR=
: (5 ft)(105 lb) (1 ft)(90 lb) (3 ft)(160 lb)
(5.5 ft)(50 lb) (405 lb)
x
M
y
Σ −+
+=−
or
2.94 fty= −
: (5.5 ft)(105 lb) (12 ft)(90 lb) (14.5 ft)(160 lb)
(22.5 ft)(50 lb) (405 lb)
y
M
x
Σ ++
+=−
or
12.60 ftx=
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.
consent of McGraw–Hill Education.
PROBLEM 3.93
Four signs are mounted on a frame spanning a
highway, and the magnitudes of the horizontal
wind forces acting on the signs are as shown.
Determine a and b so that the point of application
of the resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then requires that
G
ΣM
of the applied system of forces also be zero. Then
at
: : ( 3) ft (90 lb) (2 ft)(105 lb)
(2.5 ft)(50 lb) 0
x
GM aΣ −+ × +
+=
or
0.722 fta=
: (9 ft)(105 ft) (14.5 ) ft (90 lb)
(8 ft)(50 lb) 0
y
MbΣ − − −×
+=
or
20.6 ftb=
consent of McGraw–Hill Education.
PROBLEM 3.94
A concrete foundation mat of 5–m radius supports four equally spaced
the four loads.
SOLUTION
Have:
:
ACDE
Σ +++=F FFFF R
( ) ( ) ( ) ( )
100 kN 125 kN 25kN 75 kN=− − −−R j jjj
( )
325 kN= − j
(325 kN)= = −RP j
or
325 kNR=
Have:
( )
( )
( )
:
x CC EE G
M F z F z RzΣ +=
( )
125 kN (4 m) (75 kN)(4 m)−+
( )
325 kN G
z=
0.61539 m
G
z∴=−
or
0.615 m
G
z= −
( ) ( )
( )
:
z AA DD G
M F x F x RxΣ +=
( ) ( )
100 kN (4 m) (25 kN) 4 m−
( )
325 kN
G
x= −
or
0.923 m
G
x=−
consent of McGraw–Hill Education.
PROBLEM 3.95
Three children are standing on a
5 5-m×
raft. If the
weights of the children at Points A, B, and C are 375 N,
260 N, and 400 N, respectively, determine the magnitude
and the point of application of the resultant of the three
weights.
SOLUTION
We have
:
ABC
Σ ++=FFF F R
(375 N) (260 N) (400 N)
(1035 N)
−− − =
−=
jjjR
jR
or 1035 NR=
We have
: () () () ()
x AA BB CC D
M Fz Fz Fz RzΣ ++=
(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (1035 N)( )
D
z++ =
3.0483 m
D
z=
or 3.05 m
D
z=
We have
: () () () ()
z AA BB CC D
M Fx Fx Fx RxΣ ++=
375 N(1m) (260 N)(1.5 m) (400 N)(4.75 m) (1035 N)()
D
x++ =
2.5749 m
D
x=
or 2.57 m
D
x=
consent of McGraw–Hill Education.