PROBLEM 3.142* (Continued)
Substituting
?
160
( 42 18 8 ) ( 3 ) 0
11
− + − ⋅ −− + =i j k ij k
PROBLEM 3.143*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y–axis and applied respectively at A and B.
SOLUTION
Express the forces at A and B as
xz
xz
AA
BB
= +
= +
Aik
Bik
Then, for equivalence to the given force system,
PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force
passes through a given point while the other force lies in a given plane.
SOLUTION
PROBLEM 3.144* (Continued)
Equation (6):
1()
x yz
A xR M
y
= −
Equation (1):
1()
xx y z
B R xR M
y
=−−
1()
y
y
PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given
point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the
constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes
PROBLEM 3.145* (Continued)
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that
0,
y
B≠
so that from Equation (4),
0.z=
To obtain one possible solution, arbitrarily let
0.
x
A=
0
yy zz
AB AB+=
(7)′
Then Equation (2) can be written
yy
A RB= −
Equation (3) can be written
zz
BA= −
Equation (6) can be written
y
y
aA
xB
= −
Substituting into Equation (5)′,
or
Then from Equations (2), (8), and (3),
22 2
22 2 22 2
23 2
22 2 22 2
2
22 2
y
z
z
a R RM
AR
aR M aR M
M a R aR M
AaR aR M aR M
aR M
BaR M
=−=
++
=−=−
++
=+
PROBLEM 3.145* (Continued)
In summary,
22 2
()
RM M aR
aR M
= −
+
A jk
2
22 2
()
aR aR M
aR M
= +
+
B jk
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is
applied at a given point.
PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
SOLUTION
PROBLEM 3.146* (Continued)
Case 1: Let
0z=
to satisfy Equation (4).
Now Equation (2):
yy
A RB
λ
= −
Equation (3):
zz
BA
λ
= −
Substitution into Equation (2):
2
1
yy y
z
z
y
zy
M
R BB
aR
aR
BaR M
λ
λ
λ
λλ
=−+
=−
Then
PROBLEM 3.146* (Continued)
Case 2: Let
0
y
B=
to satisfy Equation (4).
Now Equation (2):
y
R
A
λ
=
Substitution into Equation (5):
or
xz
zx y
yy
M
MzR xR x z R
λλλλ λ
λλ
=− −− − =
This last expression is the equation for the line of action of force B.
In summary,
PROBLEM 3.147
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving it
into horizontal and vertical components. (b) Using the result of
part (a), determine the perpendicular distance from O to the
line of action of P.
SOLUTION
(0.2 m)cos 40
0.153209 m
(0.2 m)sin 40
0.128558 m
x
y
= °
=
= °
=
PROBLEM 3.148
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION
(a) Slope of line:
0.875 m 5
1.90 m 0.2 m 12
EC = =
+
Then
12 ()
13
ABx AB
TT=