4–1.
If A, B, and D are given vectors, prove the
distributive law for the vector cross product, i.e.,
A : (B + D) = (A : B) + (A : D).
SOLUTION
Consider the three vectors; with A vertical.
Note obd is perpendicular to A.
od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3
ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1
bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2
Also, these three cross products all lie in the plane obd since they are all
perpendicular to A. As noted the magnitude of each cross product is proportional to
the length of each side of the triangle.
The three vector cross products also form a closed triangle o¿b¿d¿ which is similar to
triangle obd. Thus from the figure,
A * (B + D) = (A * B) + (A * D)
(QED)
Note also,
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
D = Dx i + Dy j + Dz k
A * (B + D) = 3
i
Ax
Bx + Dx
j
Ay
By + Dy
k
Az 3
Bz + Dz
= [A y (Bz + Dz) - A z(By + Dy)]i
- [A x(Bz + Dz) - A z(Bx + Dx)]j
+ [A x(By + Dy) - A y(Bx + Dx)]k
= [(A y Bz - A zBy)i - (A x Bz - A z Bx)]j + (A x By - A y Bx)k
+ [(A y Dz - A z Dy)i - (A x Dz - A z Dx)j + (A x Dy - A y Dx)k
i
= 3 Ax
Bx
j
Ay
By
k
i
Az 3 + 3 Ax
Bz
Dx
= (A * B) + (A * D)
j
Ay
Dy
k
Az 3
Dz
(QED)
4–2.
Prove
the
triple
scalar
A # (B : C) = (A : B) # C.
product
identity
SOLUTION
As shown in the figure
Area = B(C sin u) = |B * C|
Thus,
Volume of parallelepiped is |B * C||h|
But,
|h| = |A # u(B * C)| = ` A # a
B * C
b`
|B * C|
Thus,
Volume = |A # (B * C)|
Since |(A * B) # C| represents this same volume then
A # (B : C) = (A : B) # C
(QED)
Also,
LHS = A # (B : C)
i
= (A x i + A y j + A z k) # 3 Bx
Cx
j
By
Cy
k
Bz 3
Cz
= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx)
= A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx
RHS = (A : B) # C
i
= 3 Ax
Bx
j
Ay
By
k
A z 3 # (Cx i + Cy j + Cz k)
Bz
= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(A xBy - A yBx)
= A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx
Thus, LHS = RHS
A # (B : C) = (A : B) # C
(QED)
4–3.
Given the three nonzero vectors A, B, and C, show that if
A # (B : C) = 0, the three vectors must lie in the same
plane.
SOLUTION
Consider,
|A # (B * C)| = |A| |B * C | cos u
= (|A| cos u)|B * C|
= |h| |B * C|
= BC |h| sin f
= volume of parallelepiped.
If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.
.
.
4–6.
The crane can be adjusted for any angle 0° … u … 90° and
any extension 0 … x … 5 m. For a suspended mass of
120 kg, determine the moment developed at A as a function
of x and u. What values of both x and u develop the
maximum possible moment at A? Compute this moment.
Neglect the size of the pulley at B.
x
9m
1.5 m
θ
A
SOLUTION
a + MA = - 12019.81217.5 + x2 cos u
= 5-1177.2 cos u17.5 + x26 N # m
= 51.18 cos u17.5 + x26 kN # m (Clockwise)
Ans.
The maximum moment at A occurs when u = 0° and x = 5 m.
Ans.
a + 1MA2max = 5-1177.2 cos 0°17.5 + 526 N # m
= - 14 715 N # m
= 14.7 kN # m (Clockwise)
Ans.
B
4–7.
Determine the moment of each of the three forces about
point A.
F1
F2
250 N 30
300 N
60
A
2m
3m
4m
SOLUTION
The moment arm measured perpendicular to each force from point A is
d1 = 2 sin 60° = 1.732 m
B
F3
d3 = 2 sin 53.13° = 1.60 m
Using each force where MA = Fd, we have
a + 1MF12A = - 25011.7322
Ans.
a + 1MF22A = - 30014.3302
= - 1299 N # m = 1.30 kN # m (Clockwise)
Ans.
a + 1MF32A = - 50011.602
= - 800 N # m = 800 N # m (Clockwise)
5
3
d2 = 5 sin 60° = 4.330 m
= - 433 N # m = 433 N # m (Clockwise)
4
Ans.
500 N
4–8.
Determine the moment of each of the three forces about point B.
F2 ⫽ 300 N
F1 ⫽ 250 N 30⬚
60⬚
A
2m
3m
SOLUTION
4m
The forces are resolved into horizontal and vertical component as shown in Fig. a.
For F1,
a + MB = 250 cos 30°(3) - 250 sin 30°(4)
= 149.51 N # m = 150 N # m d
Ans.
B
4
5
3
For F2,
F3 ⫽ 500 N
a + MB = 300 sin 60°(0) + 300 cos 60°(4)
= 600 N # m d
Ans.
Since the line of action of F3 passes through B, its moment arm about point B is
zero. Thus
MB = 0
Ans.
.
.
4–11.
The railway crossing gate consists of the 100-kg gate arm
having a center of mass at Ga and the 250-kg counterweight
having a center of mass at GW. Determine the magnitude
and directional sense of the resultant moment produced by
the weights about point A.
A
2.5 m
Ga
GW
1m
0.5 m
0.75 m
B
SOLUTION
0.25 m
+(MR)A = gFd; (MR)A = 100(9.81)(2.5 + 0.25) - 250(9.81)(0.5 - 0.25)
= 2084.625 N # m = 2.08 kN # m (Counterclockwise)
Ans.
4–12.
The railway crossing gate consists of the 100-kg gate arm
having a center of mass at Ga and the 250-kg counterweight
having a center of mass at GW. Determine the magnitude
and directional sense of the resultant moment produced by
the weights about point B.
A
2.5 m
SOLUTION
Ga
a +(MR)B = gFd; (MR)B = 100(9.81)(2.5) - 250(9.81)(0.5)
= 1226.25 N # m = 1.23 kN # m (Counterclockwise)
GW
1m
0.5 m
0.75 m
B
Ans.
0.25 m
4 –13.
The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point
A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the
same moment.
Given:
F
70 N
a
0.9 m
b
0.3 m
c
0.7 m
T
60 deg
Solution:
F sin T c F cos T a
(a)
MA
(b)
F C ( a)
MA
FC
MA
a
MA
73.9 N˜ m
Ans.
FC
82.2 N
Ans.
4 –14.
The force F acts on the end of the pipe at B. Determine the angles T ( 0° d T d 180° ) of
the force that will produce maximum and minimum moments about point A. What are the
magnitudes of these moments?
Given:
F
70 N
a
0.9 m
b
0.3 m
c
0.7 m
Solution:
MA
F sin T c F cos T a
For maximum moment
T max
MAmax
MAmin
dT
MA
c F cos T a F sin T
0
§c·
¸
© a¹
atan ¨
T max
F sin T max c F cos T max a
For minimum moment
T min
d
F sin T c F cos T a
MA
§ a ·
¸
©c¹
180 deg atan ¨
F c sin T min F ( a) cos T min
MAmax
37.9 deg
79.812 N˜ m
Ans.
Ans.
0
T min
MAmin
128 deg
Ans.
0 N˜ m
Ans.
4–15.
The Achilles tendon force of Ft = 650 N is mobilized when
the man tries to stand on his toes. As this is done, each of his
feet is subjected to a reactive force of Nf = 400 N.
Determine the resultant moment of Ft and Nf about the
ankle joint A.
Ft
5
A
200 mm
SOLUTION
Referring to Fig. a,
a +(MR)A = ©Fd;
(MR)A = 400(0.1) - 650(0.065) cos 5°
= - 2.09 N # m = 2.09 N # m (Clockwise)
Ans.
65 mm
100 mm
Nf
400 N
4–16.
The Achilles tendon force Ft is mobilized when the man tries
to stand on his toes. As this is done, each of his feet is
subjected to a reactive force of Nt = 400 N. If the resultant
moment produced by forces Ft and Nf about the ankle joint
A is required to be zero, determine the magnitude of Ff.
Ft
5
A
200 mm
SOLUTION
Referring to Fig. a,
a +(MR)A = ©Fd;
0 = 400(0.1) - F cos 5°(0.065)
F = 618 N
Ans.
65 mm
100 mm
Nf
400 N
4–17.
The total hip replacement is subjected to a force of
F = 120 N. Determine the moment of this force about the
neck at A and the stem at B.
120 N
15°
40 mm
A
SOLUTION
150°
Moment About Point A: The angle between the line of action of the load and the
neck axis is 20° - 15° = 5°.
a + MA = 120 sin 5°10.042
= 0.418 N # m
10°
(Counterclockwise)
Ans.
B
Moment About Point B: The dimension l can be determined using the law of sines.
55
l
=
sin 150°
sin 10°
l = 158.4 mm = 0.1584 m
Then,
a + MB = - 120 sin 15°10.15842
= - 4.92 N # m = 4.92 N # m (Clockwise)
Ans.
15 mm
4–18.
4m
The tower crane is used to hoist the 2-Mg load upward at
constant velocity. The 1.5-Mg jib BD, 0.5-Mg jib BC, and
6-Mg counterweight C have centers of mass at G1, G2, and
G3, respectively. Determine the resultant moment produced
by the load and the weights of the tower crane jibs about
point A and about point B.
G2
9.5m
B
G3
7.5 m
12.5 m
23 m
SOLUTION
Since the moment arms of the weights and the load measured to points A and B are
the same, the resultant moments produced by the load and the weight about points
A and B are the same.
a + (MR)A = (MR)B = ©Fd;
(MR)A = (MR)B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5)
- 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kN # m (Counterclockwise)
D
C
Ans.
A
G1
4–19.
The tower crane is used to hoist a 2-Mg load upward at constant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have
centers of mass at G1 and G2, respectively. Determine the
required mass of the counterweight C so that the resultant
moment produced by the load and the weight of the tower
crane jibs about point A is zero. The center of mass for the
counterweight is located at G3.
4m
G2
9.5m
B
G3
7.5 m
12.5 m
23 m
SOLUTION
a + (MR)A = ©Fd;
A
0 = MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5)
MC = 4966.67 kg = 4.97 Mg
D
C
Ans.
G1
4–20.
Determine the magnitude of the force F that
should be applied at the end of the lever such
that this force creates a clockwise moment M
about point O.
Given:
M
15 N m
I
60 deg
T
30 deg
a
50 mm
b
300 mm
Solution:
M
F
F cos T a b sin I
F sin T b cos I
M
cos T a b sin I
sin T b cos I
F
77.6 N
Ans.
4 –21.
Determine the angle T(0 <= T <= 90 deg) so that the force F develops a clockwise moment M
about point O.
Given:
F
M
100 N
20 N˜ m
I
a
60 deg
50 mm
b
300 mm
Solution:
Initial Guess
T
30 deg
Given
M
F cos T a b sin I
T
Find T
T
F sin T b cos I
28.6 deg
Ans.
4–22.
A
The tool at A is used to hold a power lawnmower blade
stationary while the nut is being loosened with the wrench.
If a force of 50 N is applied to the wrench at B in the direction
shown, determine the moment it creates about the nut at C.
What is the magnitude of force F at A so that it creates the
opposite moment about C?
13
12
F
5
400 mm
50 N
C
SOLUTION
B
c + MA = 50 sin 60°(0.3)
300 mm
MA = 12.99 = 13.0 N # m
a + MA = 0;
F = 35.2 N
- 12.99 + Fa
Ans.
12
b (0.4) = 0
13
Ans.
60
4–23.
The towline exerts a force of P = 4 kN at the end of the
20-m-long crane boom. If u = 30°, determine the
placement x of the hook at A so that this force creates a
maximum moment about point O. What is this moment?
B
P
4 kN
20 m
O
u
1.5 m
SOLUTION
Maximum moment, OB
A
x
BA
a + (MO)max = - 4kN(20) = 80 kN # m b
Ans.
4 kN sin 60°(x) - 4 kN cos 60°(1.5) = 80 kN # m
x = 24.0 m
Ans.
4–24.
The towline exerts a force of P = 4 k N at the end of the
20-m-long crane boom. If x = 25 m, determine the position u
of the boom so that this force creates a maximum moment
about point O. What is this moment?
B
P
4 kN
20 m
O
u
1.5 m
SOLUTION
Maximum moment, OB
x
BA
c + (MO)max = 4000(20) = 80 000 N # m = 80.0 kN # m
Ans.
4000 sin f(25) - 4000 cos f(1.5) = 80 000
25 sin f - 1.5 cos f = 20
f = 56.43°
u = 90° - 56.43° = 33.6°
Ans.
Also,
(1.5)2 + z2 = y 2
2.25 + z2 = y 2
Similar triangles
20 + y
25 + z
=
z
y
20y + y2 = 25z + z2
20(22.25 + z2) + 2.25 + z2 = 25z + z2
z = 2.260 m
y = 2.712 m
u = cos -1 a
2.260
b = 33.6°
2.712
A
Ans.
4–25 .
Determine the resultant moment of the forces about point A. Solve the problem first by
considering each force as a whole, and then by using the principle of moments.
Units Used:
kN
3
10 N
Given:
F1
250 N
a
2m
F2
300 N
b
3m
F3
500 N
c
4m
T1
60 deg
d
3
T2
30 deg
e
4
Solution Using Whole Forces:
Geometry
MA
D
atan §¨
d·
¸
©e¹
L
e
§a b d c·
¨
¸
e ¹ 2
2
©
e d
F1 ª¬( a)cos T 2 º¼ F2 ( a b) sin T 1 F3 L
MA
2.532 kN˜ m
Ans.
Solution Using Principle of Moments:
MA
F1 cos T 2 a F2 sin T 1 ( a b) F3
d
2
2
d e
e
c F3
2
2
( a b)
d e
MA
3
2.532 u 10 N˜ m
Ans.
4 – 26.
If the resultant moment about point A is M clockwise, determine the magnitude of F 3.
Units Used:
3
kN
10 N
Given:
M
4.8 kN˜ m a
2m
F1
300 N
b
3m
F2
400 N
c
4m
T1
60 deg
d
3
T2
30 deg
e
4
Solution:
Initial Guess
F3
1N
Given
M
F 1 cos T 2 a F 2 sin T 1 ( a b) F 3
F3
Find F3
F3
1.593 kN
e
§ d ·c F §
· ( a b)
3 ¨
¨ 2 2¸
¸
2
2
© d e ¹
© d e ¹
Ans.
4–27.
The connected bar BC is used to increase the lever arm of the
crescent wrench as shown. If the applied force is F = 200 N and
d = 300 mm, determine the moment produced by this force
about the bolt at A.
C
d
15⬚
30⬚
300 mm
B
SOLUTION
By resolving the 200-N force into components parallel and perpendicular to the box
wrench BC, Fig. a, the moment can be obtained by adding algebraically the moments
of these two components about point A in accordance with the principle of
moments.
a +(MR)A = ©Fd;
MA = 200 sin 15°(0.3 sin 30°) - 200 cos 15°(0.3 cos 30° + 0.3)
= -100.38 N # m = 100 N # m (Clockwise)
Ans.
A
F
4–28.
The connected bar BC is used to increase the lever arm of
the crescent wrench as shown. If a clockwise moment of
MA = 120 N # m is needed to tighten the bolt at A and the
force F = 200 N, determine the required extension d in
order to develop this moment.
C
d
15⬚
30⬚
300 mm
B
SOLUTION
A
By resolving the 200-N force into components parallel and perpendicular to the box
wrench BC, Fig. a, the moment can be obtained by adding algebraically the moments
of these two components about point A in accordance with the principle of moments.
a +(MR)A = ©Fd; -120 = 200 sin 15°(0.3 sin 30°) - 200 cos 15°(0.3 cos 30° + d)
d = 0.4016 m = 402 mm
Ans.
F
4–29.
The connected bar BC is used to increase the lever arm of
the crescent wrench as shown. If a clockwise moment of
MA = 120 N # m is needed to tighten the nut at A and the
extension d = 300 mm, determine the required force F in
order to develop this moment.
C
d
15⬚
30⬚
300 mm
B
SOLUTION
A
By resolving force F into components parallel and perpendicular to the box wrench
BC, Fig. a, the moment of F can be obtained by adding algebraically the moments
of these two components about point A in accordance with the principle of
moments.
a +(MR)A = ©Fd;
-120 = F sin 15°(0.3 sin 30°) - F cos 15°(0.3 cos 30° + 0.3)
F = 239 N
Ans.
F
4–30.
A force F having a magnitude of F = 100 N acts along the
diagonal of the parallelepiped. Determine the moment of F
about point A, using M A = rB : F and M A = rC : F.
z
F
C
200 mm
rC
SOLUTION
F = 100 a
400 mm
B
-0.4 i + 0.6 j + 0.2 k
b
0.7483
F
600 mm
x
F = 5-53.5 i + 80.2 j + 26.7 k6 N
M A = rB * F = 3
i
0
-53.5
j
-0.6
80.2
i
-0.4
-53.5
j
0
80.2
k
0 3 = 5 -16.0 i - 32.1 k6 N # m
26.7
Ans.
Also,
M A = rC * F =
k
0.2 =
26.7
-16.0 i - 32.1 k N # m
rB
Ans.
A
y
4–31.
The force F = 5600i + 300j - 600k6 N acts at the end of
the beam. Determine the moment of the force about
point A.
z
A
x
SOLUTION
1.2 m
r = {0.2i + 1.2j} m
i
M O = r * F = 3 0.2
600
O
F
B
0.4 m
j
1.2
300
k
0 3
- 600
M O = { - 720i + 120j - 660k} N # m
0.2 m
Ans.
y
4–32.
Determine the moment produced by force FB about point O.
Express the result as a Cartesian vector.
z
A
6m
FC
420 N
FB
780 N
SOLUTION
Position Vector and Force Vectors: Either position vector rOA or rOB can be used to
determine the moment of FB about point O.
rOA = [6k] m
rOB = [2.5j] m
x
(0 - 0)i + (2.5 - 0)j + (0 - 6)k
2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2
R = [300j - 720k] N
Vector Cross Product: The moment of FB about point O is given by
i
M O = rOA * FB = 3 0
0
j
0
300
k
6 3 = [ -1800i] N # m = [- 1.80i] kN # m
- 720
Ans.
j
2.5
300
k
0 3 = [ -1800i] N # m = [ -1.80i] kN # m
- 720
Ans.
or
i
M O = rOB * FB = 3 0
0
2.5 m
C
3m
The force vector FB is given by
FB = FB u FB = 780 B
2m
O
B
y
4–33.
Determine the moment produced by force FC about point O.
.Express the result as a Cartesian vector
z
A
6m
FC
420 N
FB
780 N
SOLUTION
Position Vector and Force Vectors: Either position vector rOA or rOC can be used to
determine the moment of FC about point O.
rOA = {6k} m
x
The force vector FC is given by
(2 - 0)i + (- 3 - 0)j + (0 - 6)k
2(2 - 0)2 + (- 3 - 0)2 + (0 - 6)2
R = [120i - 180j - 360k] N
Vector Cross Product: The moment of FC about point O is given by
i
M O = rOA * FC = 3 0
120
j
0
- 180
k
6 3 = [1080i + 720j] N # m
-360
Ans.
j
-3
-180
k
0 3 = [1080i + 720j] N # m
-360
Ans.
or
i
M O = rOC * FC = 3 2
120
2.5 m
C
3m
rOC = (2 - 0)i + (- 3 - 0)j + (0 - 0)k = [2i - 3j] m
FC = FCuFC = 420 B
2m
O
B
y
4–34.
Determine the resultant moment produced by forces FB
and FC about point O. Express the result as a Cartesian
vector.
z
A
6m
FC
420 N
FB
780 N
SOLUTION
Position Vector and Force Vectors: The position vector rOA and force vectors FB and
FC, Fig. a, must be determined first.
rOA = {6k} m
2m
2.5 m
C
3m
FB = FB uFB = 780 B
FC = FCuFC = 420 B
(0 - 0)i + (2.5 - 0)j + (0 - 6)k
2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2
(2 - 0)i + (-3 - 0)j + (0 - 6)k
2(2 - 0)2 + (- 3 - 0)2 + (0 - 6)2
R = [300j - 720k] N
R = [120i - 180j - 360k] N
Resultant Moment: The resultant moment of FB and FC about point O is given by
MO = rOA * FB + rOA * FC
i
3
= 0
0
j
0
300
i
k
3
3
+ 0
6
120
- 720
j
0
- 180
= [- 720i + 720j] N # m
k
6 3
- 360
Ans.
x
O
B
y
4–35.
Using a ring collar the 75-N force can act in the vertical
plane at various angles u. Determine the magnitude of the
moment it produces about point A, plot the result of M
(ordinate) versus u (abscissa) for 0° … u … 180°, and
specify the angles that give the maximum and minimum
moment.
z
A
2m
1.5 m
SOLUTION
i
MA = 3 2
0
j
1.5
75 cos u
k
3
0
75 sin u
y
x
75 N
θ
= 112.5 sin u i - 150 sin u j + 150 cos u k
MA = 21112.5 sin u22 + 1-150 sin u22 + 1150 cos u22 = 212 656.25 sin2 u + 22 500
dMA
1
1
= 112 656.25 sin2 u + 22 5002- 2 112 656.25212 sin u cos u2 = 0
du
2
sin u cos u = 0;
u = 0°, 90°, 180°
Mmax = 187.5 N # m at u = 90°
Mmin = 150 N # m at u = 0°, 180°
Ans.
4–36.
The curved rod lies in the x–y plane and has a radius of 3 m.
If a force of F = 80 N acts at its end as shown, determine
the moment of this force about point O.
z
O
y
B
3m
45
3m
SOLUTION
A
1m
rAC = {1i - 3j - 2k} m
F
rAC = 2(1)2 + (-3)2 + (- 2)2 = 3.742 m
M O = rOC * F = 3
i
4
1
3.742 (80)
j
0
3
- 3.742
(80)
M O = {- 128i + 128j - 257k} N # m
k
3
-2
2
- 3.742
(80)
2m
x
Ans.
C
80 N
4–37.
The curved rod lies in the x–y plane and has a radius of 3 m.
If a force of F = 80 N acts at its end as shown, determine
the moment of this force about point B.
z
O
y
B
3m
45
SOLUTION
rAC = {1i - 3j - 2k} m
A
1m
rAC = 2(1) + (- 3) + (- 2) = 3.742 m
2
M B = rBA
2
2
i
* F = 3 3 cos 45°
1
3.742 (80)
M B = { -37.6i + 90.7j -
3m
F
j
k
3
(3 - 3 sin 45°)
0
3
2
- 3.742(80)
- 3.742(80)
155k} N # m
2m
x
Ans.
C
80 N
4–38.
Force F acts perpendicular to the inclined plane. Determine
the moment produced by F about point A. Express the
result as a Cartesian vector.
z
A
3m
F
400 N
3m
B
SOLUTION
x
Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF
is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here
rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m
rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [ -3i + 4j] m
Thus,
i
b = rCA * rCB = 3 0
-3
j
4
4
k
-3 3
0
= [12i + 9j + 12k] m2
Then,
uF =
12i + 9j + 12k
b
=
= 0.6247i + 0.4685j + 0.6247k
b
212 2 + 92 + 12 2
And finally
F = FuF = 400(0.6247i + 0.4685j + 0.6247k)
= [249.88i + 187.41j + 249.88k] N
Vector Cross Product: The moment of F about point A is
M A = rAC * F = 3
i
0
249.88
j
4
187.41
k
-3 3
249.88
= [1.56i - 0.750j - 1.00k] kN # m
Ans.
4m
C
y
4–39. Solve 4–38 if F ⫽ 500 N.
z
A
3m
F
500 N
3m
B
SOLUTION
x
C
4m
y
Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF
is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here
rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m
rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [ - 3k + 4j] m
Thus,
i
b = rCA * rCB = 3 0
-3
j
4
4
k
- 3 3 = [12i + 9j + 12k] m2
0
Then,
500 N
12i + 9j + 12k
b
=
= 0.6247i + 0.4685j + 0.6247k
uF =
b
2122 + 92 + 122
And finally
F = FuF = 500(0.6247i + 0.4685j + 0.6247k)
= [312.35i + 234.25j + 312.35k] N
Vector Cross Product: The moment of F about point B is
MB = rBC * F = 3
i
-3
312.35
j
4
234.25
k
0 3
312.35
= [1.25i + 0.937j - 1.95k] kN # m
Ans.
4–40.
The pipe assembly is subjected to the 80-N force. Determine
the moment of this force about point A.
z
A
400 mm
B
x
SOLUTION
300 mm
Position Vector And Force Vector:
200 mm
rAC = {(0.55 - 0)i + (0.4 - 0)j + ( -0.2 - 0)k} m
200 mm C
250 mm
= {0.55i + 0.4j - 0.2k} m
40
F = 80(cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N
30
= (44.53i + 53.07j - 40.0k} N
F
Moment of Force F About Point A: Applying Eq. 4–7, we have
MA = rAC * F
i
3
= 0.55
44.53
j
0.4
53.07
k
- 0.2 3
- 40.0
= { - 5.39i + 13.1j + 11.4k} N # m
Ans.
80 N
y
4–41.
The pipe assembly is subjected to the 80-N force. Determine
the moment of this force about point B.
z
A
400 mm
B
x
SOLUTION
300 mm
Position Vector And Force Vector:
200 mm
rBC = {(0.55 - 0) i + (0.4 - 0.4)j + (- 0.2 - 0)k} m
200 mm C
250 mm
= {0.55i - 0.2k} m
40
F = 80 (cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N
30
= (44.53i + 53.07j - 40.0k} N
F
Moment of Force F About Point B: Applying Eq. 4–7, we have
MB = rBC * F
i
3
= 0.55
44.53
j
0
53.07
k
- 0.2 3
- 40.0
= {10.6i + 13.1j + 29.2k} N # m
Ans.
80 N
y
4–42.
Strut AB of the 1-m-diameter hatch door exerts a force of
450 N on point B. Determine the moment of this force
about point O.
z
B
30°
0.5 m
SOLUTION
y
0.5 m
Position Vector And Force Vector:
30°
rOB = 510 - 02i + 11 cos 30° - 02j + 11 sin 30° - 02k6 m
x
= 50.8660j + 0.5k6 m
rOA = 510.5 sin 30° - 02i + 10.5 + 0.5 cos 30° - 02j + 10 - 02k6 m
= 50.250i + 0.9330j6 m
F = 450 ¢
F = 450 N
O
10 - 0.5 sin 30°2i + 31 cos 30° - 10.5 + 0.5 cos 30°24j + 11 sin 30° - 02k
210 - 0.5 sin 30°22 + 31 cos 30° - 10.5 + 0.5 cos 30°242 + 11 sin 30° - 022
= 5-199.82i - 53.54j + 399.63k6 N
Moment of Force F About Point O: Applying Eq. 4–7, we have
M O = rOB * F
= 3
i
0
-199.82
j
0.8660
-53.54
k
0.5 3
399.63
= 5373i - 99.9j + 173k6 N # m
Ans.
Or
M O = rOA * F
=
i
0.250
-199.82
j
0.9330
-53.54
k
0
399.63
=
373i - 99.9j + 173k N # m
Ans.
≤N
A
– .
Determine the moment of the force ) at A about point O. Express the result as a cartesian
vector.
Units Used:
3
N1 10 1
*LYHQ
) 13N1
D 6P
E 2.5P
F 3P
G 3P
H 8P
I 6P
J 4P
K 8P
6ROXWLRQ
§¨ E J ¸·
U$% ¨ F G ¸
¨K D ¸
©
¹
02 U2$ u )
§¨ E ¸·
U2$ ¨ F ¸
¨D ¸
© ¹
02
) )˜
§ 84 ·
¨ 8 ¸ N1˜ P
¨
¸
© 39 ¹
U$%
U$%
Ans.
– .
Determine the moment of the force ) at A about point P. Express the result as a cartesian
vector.
Units Used:
3
N1 10 1
*LYHQ
) 13N1
D 6P
E 2.5P
F 3P
G 3P
H 8P
I 6P
J 4P
K 8P
6ROXWLRQ
§¨ E J ¸·
U$% ¨ F G ¸
¨K D ¸
©
¹
02 U3$ u )
§¨ E I ¸·
U3$ ¨ F H ¸
¨ D ¸
©
¹
02
§ 116 ·
¨ 16 ¸ N1˜ P
¨
¸
© 135 ¹
) )˜
U$%
U$%
Ans.
4–45.
A force of F = 5 6i - 2j + 1k 6 kN produces a moment
of M O = 54i + 5j - 14k6 kN # m about the origin of
coordinates, point O. If the force acts at a point having an x
coordinate of x = 1 m, determine the y and z coordinates.
z
F
P
MO
z
d
y
O
1m
SOLUTION
y
MO = r * F
i
4i + 5j - 14k = 3 1
6
j
y
-2
x
k
z3
1
4 = y + 2z
5 = -1 + 6z
-14 = -2 - 6y
y = 2m
Ans.
z = 1m
Ans.
4–46.
The force F = 56i + 8j + 10k6 N creates a moment about
point O of M O = 5 -14i + 8j + 2k6 N # m. If the force
passes through a point having an x coordinate of 1 m,
determine the y and z coordinates of the point.Also, realizing
that MO = Fd, determine the perpendicular distance d from
point O to the line of action of F.
z
F
P
MO
z
d
y
O
1m
SOLUTION
y
i
3
- 14i + 8j + 2k = 1
6
j
y
8
k
z 3
10
x
- 14 = 10y - 8z
8 = -10 + 6z
2 = 8 - 6y
y = 1m
Ans.
z = 3m
Ans.
MO = 2( -14)2 + (8)2 + (2)2 = 16.25 N # m
F = 2(6)2 + (8)2 + (10)2 = 14.14 N
d =
16.25
= 1.15 m
14.14
Ans.
4–47.
Determine the magnitude of the moment of each of the
three forces about the axis AB. Solve the problem (a) using
a Cartesian vector approach and (b) using a scalar
approach.
z
F1 = 60 N
F2 = 85 N
F3 = 45 N
SOLUTION
A
a) Vector Analysis
B
Position Vector and Force Vector:
y
x
1.5 m
r1 = 5-1.5j6 m
r2 = r3 = 0
F1 = 5-60k6 N
F2 = 585i6 N F3 = 545j6 N
Unit Vector Along AB Axis:
uAB =
12 - 02i + 10 - 1.52j
212 - 022 + 10 - 1.522
= 0.8i - 0.6j
Moment of Each Force About AB Axis: Applying Eq. 4–11, we have
1MAB21 = uAB # 1r1 * F12
0.8
= 3 0
0
-0.6
-1.5
0
0
0 3
-60
= 0.831-1.521-602 - 04 - 0 + 0 = 72.0 N # m
Ans.
1MAB22 = uAB # 1r2 * F22
0.8
3
= 0
85
-0.6
0
0
0
03 = 0
0
Ans.
0
03 = 0
0
Ans.
1MAB23 = uAB # 1r3 * F32
0.8
= 3 0
0
-0.6
0
45
b) Scalar Analysis: Since moment arm from force F2 and F3 is equal to zero,
1MAB22 = 1MAB23 = 0
Ans.
Moment arm d from force F1 to axis AB is d = 1.5 sin 53.13° = 1.20 m,
MAB
1
= F1d = 60 1.20 = 72.0 N # m
Ans.
2m
F {6i 3j 10k} N
4–48. Determine the moment produced by force F about
the diagonal AF of the rectangular block. Express the result
as a Cartesian vector.
z
B
A
D
C
O
x
3m
1.5 m
G
3m
F
y
F {6i 3j 10k} N
4–49. Determine the moment produced by force F about
the diagonal OD of the rectangular block. Express the
result as a Cartesian vector.
z
B
A
D
C
O
x
3m
1.5 m
G
3m
F
y
4– 50.
Determine the magnitude of the moment of the force F about the base line CA of the tripod.
Given:
F
§ 50 ·
¨ 20 ¸ N
¨
¸
© 80 ¹
a
4m
b
2.5 m
c
1m
d
0.5 m
e
2m
f
1.5 m
g
2m
Solution:
rCA
§ g ·
¨e ¸ u
¨ ¸ CA
©0¹
rCA
rCA
rCD
§b g·
¨ e ¸
¨
¸
© a ¹
MCA
rCD u F ˜ uCA
MCA
226 N˜ m
Ans.
4–51. Determine the magnitude of the moment
produced by the force of F = 200 N about the hinged axis
(the x axis) of the door.
z
0.5 m
B
F 200 N
2m
15
A
1m
x
2.5 m
y
4–52 .
The lug nut on the wheel of the automobile is to be removed using the wrench and applying the
vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is
initially required to turn the nut. If the force F can be applied at A in any other direction, will it
be possible to turn the nut?
Given:
F
30 N
M
14 N˜ m
a
0.25 m
b
0.3 m
c
0.5 m
d
0.1 m
Solution:
2
Mx
F c b
Mx
12 N˜ m
Mx M
2
No
Ans.
For Mxmax, apply force perpendicular to the handle and the x-axis.
Mxmax
Fc
Mxmax
15 N˜ m
Mxmax ! M
Yes
Ans.
4 – 53 .
The lug nut on the wheel of the automobile is to be removed using the wrench and applying the
vertical force F . Assume that the cheater pipe AB is slipped over the handle of the wrench and
the F force can be applied at any point and in any direction on the assembly. Determine if this
force is adequate, provided a torque M about the x axis is initially required to turn the nut.
Given:
F1
14 N˜ m
M
30 N
a
0.25 m
b
0.3 m
c
0.5 m
Solution:
ac
c
Mx
F1
Mx
18 N˜ m
Mx ! M
2
2
c b
Ans.
Yes
Mxmax occurs when force is applied perpendicular to both the handle and the x-axis.
Mxmax
F1 ( a c)
Mxmax
22.5 N˜ m
Mxmax ! M
Yes
Ans.
d
0.1 m
4–54.
z
The board is used to hold the end of a four-way lug wrench in
the position shown when the man applies a force of F = 100
N. Determine the magnitude of the moment produced by this
force about the x axis. Force F lies in a vertical plane.
F
60⬚
x
SOLUTION
250 mm
Vector Analysis
Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine
the moment of F about the x axis.
rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m
The force vector F, Fig. a, can be written as
F = 100(cos 60°j - sin 60°k) = {50j - 86.60k} N
Knowing that the unit vector of the x axis is i, the magnitude of the moment of F
about the x axis is given by
Mx = i # rAB * F =
1
30
0
0
0.25
50
0
0 3
-86.60
= 1[0.25(-86.60) - 50(0)] + 0 + 0
= -21.7 N # m
Ans.
The negative sign indicates that Mx is directed towards the negative x axis.
Scalar Analysis
This problem can be solved by summing the moment about the x axis
Mx = ©Mx;
Mx = -100 sin 60°(0.25) + 100 cos 60°(0)
= -21.7 N # m
Ans.
250 mm
y
4–55.
z
The board is used to hold the end of a four-way lug wrench
in position. If a torque of 30 N # m about the x axis is
required to tighten the nut, determine the required magnitude of the force F that the man’s foot must apply on the
end of the wrench in order to turn it. Force F lies in a vertical plane.
F
60⬚
x
250 mm
SOLUTION
Vector Analysis
Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis.
rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m
The force vector F, Fig. a, can be written as
F = F(cos 60°j - sin 60°k) = 0.5Fj - 0.8660Fk
Knowing that the unit vector of the x axis is i, the magnitude of the moment of F
about the x axis is given by
Mx = i # rAB
1
3
* F = 0
0
0
0.25
0.5F
0
3
0
-0.8660F
= 1[0.25(-0.8660F) - 0.5F(0)] + 0 + 0
= -0.2165F
Ans.
The negative sign indicates that Mx is directed towards the negative x axis. The
magnitude of F required to produce Mx = 30 N # m can be determined from
30 = 0.2165F
F = 139 N
Ans.
Scalar Analysis
This problem can be solved by summing the moment about the x axis
Mx = ©Mx;
-30 = - F sin 60°(0.25) + F cos 60°(0)
F = 139 N
Ans.
250 mm
y
4–56.
The cutting tool on the lathe exerts a force F on the shaft
as shown. Determine the moment of this force about the
y axis of the shaft.
z
F
{6i
4j
7k} kN
SOLUTION
My = uy # (r * F)
0
= 3 30 cos 40°
6
30 mm
1
0
-4
40
0
30 sin 40° 3
-7
My = 276.57 N # mm = 0.277 N # m
y
Ans.
x
4–57.
z
The cutting tool on the lathe exerts a force F on the shaft as
shown. Determine the moment of this force about the x and
z axes.
F ⫽ {6i ⫺ 4j ⫺ 7k} kN
30 mm
40⬚
SOLUTION
y
Moment About x and y Axes: Position vectors rx and r z shown in Fig. a can be conveniently used in computing the moment of F about x and z axes respectively.
rx = {0.03 sin 40° k} m
rz = {0.03 cos 40°i} m
Knowing that the unit vectors for x and z axes are i and k respectively. Thus, the
magnitudes of moment of F about x and z axes are given by
1
Mx = i # rx * F = 3 0
6
0
0
-4
0
0.03 sin 40° 3
-7
= 1[0(-07) - (-4)(0.03 sin 40°)] - 0 + 0
= 0.07713 kN # m = 77.1 N # m
0
Mz = k # rz * F = 3 0.03 cos 40°
6
0
0
-4
1
03
7
= 0 - 0 + 1[0.03 cos 40°(-4) - 6(0)]
= - 0.09193 kN # m = - 91.9 N # m
Thus,
Mx = Mxi = {77.1i} N # m
Mz = Mzk = {- 91.9 k} N # m
Ans.
x
4–58. If F = 450 N, determine the magnitude of the
moment produced by this force about the x axis.
z
F
45⬚
A
B
60⬚
60⬚
100 mm
x
300 mm
150 mm
y
4–59. The friction at sleeve A can provide a maximum
resisting moment of 125 N # m about the x axis. Determine
the largest magnitude of force F that can be applied to the
bracket so that the bracket will not turn.
z
F
45⬚
A
B
60⬚
60⬚
100 mm
x
300 mm
150 mm
y
4–60. A vertical force of F = 60 N is applied to the
handle of the pipe wrench. Determine the moment that this
force exerts along the axis AB (x axis) of the pipe assembly.
Both the wrench and pipe assembly ABC lie in the x-y
plane. Suggestion: Use a scalar analysis.
z
A
y
500 mm
F
150 mm
B
x
45⬚
200 mm
C
.
4–61. Determine the magnitude of the vertical force F
acting on the handle of the wrench so that this force
produces a component of moment along the AB axis (x axis)
of the pipe assembly of (MA)x = 5 -5i6 N # m. Both the pipe
assembly ABC and the wrench lie in the x-y plane.
Suggestion: Use a scalar analysis.
z
A
y
500 mm
F
150 mm
B
x
45⬚
200 mm
C
.
4–62. Determine the magnitude of the moments of the
force F about the x, y, and z axes. Solve the problem (a) using
a Cartesian vector approach and (b) using a scalar approach.
z
A
4m
m
y
m
x
3m
C
2m
kN # m
kN # m
kN # m
.
B
F ⫽ {4i ⫹ 12j ⫺ 3k} kN
.
.
kN # m
.
kN # m
.
kN # m
.
4–63. Determine the moment of the force F about an axis
extending between A and C. Express the result as a
Cartesian vector.
z
A
4m
y
x
3m
C
2m
B
F ⫽ {4i ⫹ 12j ⫺ 3k} kN
m
m
m
kN..
kN # m
kN # m
kN # m
.
4–64 .
z
The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. If
FB = 150 N, determine the magnitude of the moment produced by this force about the y axis. Also, what is the magnitude of force FA in order to counteract this moment?
50 mm 50 mm
y
SOLUTION
Vector Analysis
Moment of FB About the y Axis: The position vector rCB, Fig. a, will be used to
determine the moment of FB about the y axis.
rCB = (-0.15 - 0)j + (0.05 - 0.05)j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m
FB = 150(cos 60°i - sin 60°k) = {75i - 129.90k} N
Knowing that the unit vector of the y axis is j, the magnitude of the moment of FB
about the y axis is given by
0
3 - 0.15
1
0
0
75
0
-0.2598 3
-129.90
= 0 - 1[-0.15(-129.90) - 75( -0.2598)] + 0
= - 38.97 N # m = 39.0 N # m
Ans.
The negative sign indicates that My is directed towards the negative y axis.
Moment of FA About the y Axis: The position vector rDA, Fig. a, will be used to
determine the moment of FA about the y axis.
rDA = (0.15 - 0)i + [-0.05 - (- 0.05)]j + (-0.2598 - 0)k = {0.15i - 0.2598k} m
Referring to Fig. a, the force vector FA can be written as
FA = FA(-cos 15°i + sin 15°k) = - 0.9659FAi + 0.2588FAk
Since the moment of FA about the y axis is required to counter that of FB about the
same axis, FA must produce a moment of equal magnitude but in the opposite sense
to that of FA.
Mx = j # rDA * FB
+ 0.38.97 = 3
0
0.15
- 0.9659 FA
1
0
0
0
- 0.2598 3
0.2588FA
+ 0.38.97 = 0 - 1[0.15(0.2588FA) - (-0.9659FA)(-0.2598)] + 0
FA = 184 N
Ans.
Scalar Analysis
This problem can be solved by first taking the moments of FB and then FA about
the y axis. For FB we can write
My = ©My;
My = -150 cos 60°(0.3 cos 30°) - 150 sin 60°(0.3 sin 30°)
= -38.97 N # m
Ans.
The moment of FA, about the y axis also must be equal in magnitude but opposite
in sense to that of FB about the same axis
My = ©My;
38.97 = FA cos 15°(0.3 cos 30°) - FA sin 15°(0.3 sin 30°)
FA = 184 N
300 mm
300 mm
30⬚
Ans.
30⬚
FA 135⬚
120⬚
A
Referring to Fig. a, the force vector FB can be written as
My = j # rCB * FB =
x
FB
B
4–65.
z
The wrench A is used to hold the pipe in a stationary position
while wrench B is used to tighten the elbow fitting.
Determine the magnitude of force FB in order to develop a
torque of 50 N # m about the y axis. Also, what is the required
magnitude of force FA in order to counteract this moment?
50 mm 50 mm
y
x
SOLUTION
300 mm
Vector Analysis
Moment of FB About the y Axis: The position vector rCB, Fig. a, will be used to
determine the moment of FB about the y axis.
rCB = (-0.15 - 0)i + (0.05 - 0.05)j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m
Referring to Fig. a, the force vector FB can be written as
FB = FB(cos 60°i - sin 60°k) = 0.5FBi - 0.8660FBk
Knowing that the unit vector of the y axis is j, the moment of FB about the y axis is
required to be equal to -50 N # m, which is given by
My = j # rCB * FB
0
-50i = † -0.15
0.5FB
1
0
0
0
-0.2598 †
-0.8660FB
-50 = 0 - 1[-0.15(-0.8660FB) - 0.5FB( -0.2598)] + 0
FB = 192 N
Ans.
Moment of F A About the y Axis: The position vector rDA, Fig. a, will be
used to determine the moment of FA about the y axis.
rDA = (0.15 - 0)i + [-0.05 - (- 0.05)]j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m
Referring to Fig. a, the force vector FA can be written as
FA = FA(-cos 15°i + sin 15°k) = -0.9659FAi + 0.2588FAk
Since the moment of FA about the y axis is required to produce a countermoment
of 50 N # m about the y axis, we can write
My = j # rDA * FA
50 = †
0
0.15
- 0.9659FA
1
0
0
0
-0.2598 †
0.2588FA
50 = 0 - 1[0.15(0.2588FA) - ( -0.9659FA)(-0.2598)] + 0
FA = 236 N # m
Ans.
Scalar Analysis
This problem can be solved by first taking the moments of FB and then FA about
the y axis. For FB we can write
My = ©My;
-50 = -FB cos 60°(0.3 cos 30°) - FB sin 60°(0.3 sin 30°)
FB = 192 N
Ans.
For FA, we can write
My = ©My;
300 mm
50 = FA cos 15°(0.3 cos 30°) - FA sin 15°(0.3 sin 30°)
FA = 236 N
Ans.
30⬚
30⬚
FA 135⬚
120⬚
A
FB
B
4– 66.
The wooden shaft is held in a lathe.
The cutting tool exerts force F on the
shaft in the direction shown.
Determine the moment of this force
about the x axis of the shaft. Express
the result as a Cartesian vector. The
distance OA is a.
Given:
a
25 mm
T
30 deg
F
§ 5 ·
¨ 3 ¸ N
¨ ¸
©8¹
Solution:
r
ª 0
«( a)cos T
«
¬ ( a)sin T
º
»
»
¼
i
§1·
¨0¸
¨ ¸
©0¹
Mx
ª¬ r u F ˜ iº¼ i
Mx
§ 0.211 ·
¨ 0 ¸ N˜ m
¨
¸
© 0 ¹
Ans.
4–67.
A twist of 4 N # m is applied to the handle of the screwdriver.
Resolve this couple moment into a pair of couple forces F
exerted on the handle and P exerted on the blade.
–F
–P
P
5 mm
SOLUTION
For the handle
MC = ©Mx ;
F10.032 = 4
F = 133 N
Ans.
For the blade,
MC = ©Mx ;
P10.0052 = 4
P = 800 N
Ans.
30 mm
F
4 N·m
4–68.
The ends of the triangular plate are subjected to three
couples. Determine the plate dimension d so that the
resultant couple is 350 N # m clockwise.
100 N
600 N
d
100 N
30°
600 N
SOLUTION
a + MR = ©MA ;
200 N
-350 = 2001d cos 30°2 - 6001d sin 30°2 - 100d
d = 1.54 m
Ans.
200 N
4–69.
The caster wheel is subjected to the two couples. Determine
the forces F that the bearings exert on the shaft so that the
resultant couple moment on the caster is zero.
500 N
F
A
40 mm
B
F
100 mm
SOLUTION
a + ©MA = 0;
45 mm
500(50) - F (40) = 0
F = 625 N
Ans.
50 mm
500 N
4–70. If u = 30°, determine the magnitude of force F so that
the resultant couple moment is 100 N # m, clockwise. The
disk has a radius of 300 mm.
300 N
300 mm
15⬚
⫺F
u
30⬚
F
30⬚
u
300 N
15⬚
4–71. If F = 200 N, determine the required angle u so that
the resultant couple moment is zero. The disk has a radius of
300 mm.
Ans.
4–72.
Friction on the concrete surface creates a couple moment of
MO = 100 N # m on the blades of the trowel. Determine the
magnitude of the couple forces so that the resultant couple
moment on the trowel is zero. The forces lie in the horizontal plane and act perpendicular to the handle of the trowel.
–F
750 mm
F
MO
SOLUTION
Couple Moment: The couple moment of F about the vertical axis is
MC = F(0.75) = 0.75F. Since the resultant couple moment about the vertical axis is
required to be zero, we can write
(Mc)R = ©Mz;
0 = 100 - 0.75F
F = 133 N
Ans.
1.25 mm
4–73.
The man tries to open the valve by applying the couple forces
of F = 75 N to the wheel. Determine the couple moment
produced.
150 mm
150 mm
F
SOLUTION
a +Mc = ©M;
Mc = -75(0.15 + 0.15)
= -22.5 N # m = 22.5 N # m b
Ans.
F
4–74.
If the valve can be opened with a couple moment of 25 N # m, determine
the required magnitude of each couple force which must be applied
to the wheel.
150 mm
150 mm
F
SOLUTION
a +Mc = ©M;
-25 = -F(0.15 + 0.15)
F = 83.3 N
Ans.
F
.
4–76.
Three couple moments act on the pipe assembly. Determine the magnitude of M3 and the
bend angle T so that the resultant couple moment is zero.
Given:
T1
45 deg
M1
900 N˜ m
M2
500 N˜ m
Solution:
Initial guesses:
T
10 deg
M3
10 N˜ m
Given
6Mx = 0;
o
M1 M3 cos T M2 cos T 1
M3 sin T M2 sin T 1
n 6My = 0;
§ T ·
¨ ¸
© M3 ¹
Find T M3
0
0
T
32.9 deg
M3
651 N˜ m
Ans.
4–77. The cord passing over the two small pegs A and B of
the square board is subjected to a tension of 100 N.
Determine the required tension P acting on the cord that
passes over pegs C and D so that the resultant couple
produced by the two couples is 15 N # m acting clockwise.
Take u = 15°.
300 mm
C
B
u
30⬚
⫺P
100 N
300 mm
100 N
A
.
P
45⬚
30⬚
u
D
4–78. The cord passing over the two small pegs A and B of
the board is subjected to a tension of 100 N. Determine the
minimum tension P and the orientation u of the cord
passing over pegs C and D, so that the resultant couple
moment produced by the two cords is 20 N # m, clockwise.
300 mm
C
B
u
30⬚
⫺P
100 N
300 mm
100 N
.
P
45⬚
30⬚
A
u
D
4–79.
Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude
of the couple moment?
Given:
F
125 N
a
150 mm
b
150 mm
c
200 mm
d
600 mm
Solution:
M
§ c · §0·
¨a b¸ u ¨ 0 ¸
¨
¸ ¨ ¸
© 0 ¹ ©F¹
M
§ 37.5 ·
¨ 25 ¸ N˜ m
¨
¸
© 0 ¹
M
45.1 N˜ m
Ans.
4–80.
If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the
forces applied to the wrenches.
Given:
M
300 N˜ m
a
150 mm
b
150 mm
c
200 mm
d
600 mm
Solution:
Initial guess:
Given
F
1N
§ c · §0·
¨a b¸ u ¨ 0 ¸
¨
¸ ¨ ¸
© 0 ¹ ©F¹
M
F
Find ( F)
F
832.1 N
Ans.
4–81.
Two couples act on the cantilever beam. If F = 6 kN,
determine the resultant couple moment.
3m
3m
5 kN
F
5
4
3
A
SOLUTION
30
30
0.5 m
0.5 m
5
4
F
a) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, the
couple moments (Mc)1 and (Mc)2 produced by the 6-kN and 5-kN couples,
respectively, are given by
a + (MC)1 = 6 sin 30°(3) - 6 cos 30°(0.5 + 0.5) = 3.804 kN # m
4
3
a + (MC)2 = 5a b(0.5 + 0.5) - 5a b (3) = -9 kN # m
5
5
Thus, the resultant couple moment can be determined from
(MC)R = (MC)1 + (MC)2
= 3.804 - 9 = -5.196 kN # m = 5.20 kN # m (Clockwise)
Ans.
b) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, and
summing the moments of these force components about point A, we can write
a + (MC)R = ©MA ;
B
4
3
(MC)R = 5a b (0.5) + 5a b (3) - 6 cos 30°(0.5) - 6 sin 30°(3)
5
5
4
3
+ 6 sin 30°(6) - 6 cos 30°(0.5) + 5a b(0.5) - 5a b (6)
5
5
= -5.196 kN # m = 5.20 kN # m (Clockwise)
Ans.
3
5 kN
4–82.
Determine the required magnitude of force F, if the
resultant couple moment on the beam is to be zero.
3m
3m
5 kN
F
5
4
3
A
SOLUTION
30
B
30
0.5 m
0.5 m
5
4
F
By resolving F and the 5-kN couple into their x and y components, Fig. a, the couple
moments (Mc)1 and (Mc)2 produced by F and the 5-kN couple, respectively, are
given by
a + (MC)1 = F sin 30°(3) - F cos 30°(1) = 0.6340F
4
3
a + (MC)2 = 5a b (1) - 5a b(3) = -9 kN
5
5
The resultant couple moment acting on the beam is required to be zero. Thus,
(MC)R = (MC)1 + (MC)2
0 = 0.6340F - 9
F = 14.2 kN
Ans.
3
5 kN
4–83.
Express the moment of the couple acting on the pipe
assembly in Cartesian vector form. Solve the problem
(a) using Eq. 4–13, and (b) summing the moment of each
force about point O. Take F = 525k6 N.
z
O
y
300 mm
200 mm
F
150 mm
SOLUTION
B
(a) MC = rAB * (25k)
i
= 3 -0.35
0
j
- 0.2
0
x
k
0 3
25
F
400 mm
A
MC = {- 5i + 8.75j} N # m
Ans.
(b) MC = rOB * (25k) + rOA * ( -25k)
i
3
= 0.3
0
j
0.2
0
i
k
3
3
0 + 0.65
0
25
j
0.4
0
k
0 3
- 25
MC = (5 - 10)i + (-7.5 + 16.25)j
MC = {- 5i + 8.75j} N # m
200 mm
Ans.
4–84.
If the couple moment acting on the pipe has a magnitude of
400 N # m, determine the magnitude F of the vertical force
applied to each wrench.
z
O
y
300 mm
200 mm
F
150 mm
SOLUTION
B
M C = rAB * (Fk)
i
= 3 -0.35
0
j
- 0.2
0
k
03
F
x
F
400 mm
MC = { -0.2Fi + 0.35Fj} N # m
A
MC = 2(- 0.2F)2 + (0.35F)2 = 400
F =
400
2( - 0.2)2 + (0.35)2
= 992 N
200 mm
Ans.
4–85.
The gear reducer is subjected to the couple moments
shown. Determine the resultant couple moment and specify
its magnitude and coordinate direction angles.
z
M2 = 60 N · m
M1 = 50 N · m
30°
SOLUTION
x
Express Each Couple Moment as a Cartesian Vector:
M 1 = 550j6 N # m
M 2 = 601cos 30°i + sin 30°k2 N # m = 551.96i + 30.0k6 N # m
Resultant Couple Moment:
M R = ©M;
MR = M1 + M2
= 551.96i + 50.0j + 30.0k6 N # m
= 552.0i + 50j + 30k6 N # m
Ans.
The magnitude of the resultant couple moment is
MR = 251.962 + 50.02 + 30.02
= 78.102 N # m = 78.1 N # m
Ans.
The coordinate direction angles are
a = cos-1 a
51.96
b = 48.3°
78.102
Ans.
b = cos-1 a
50.0
b = 50.2°
78.102
Ans.
g = cos-1
30.0
78.102
Ans.
= 67.4°
y
4–86.
The meshed gears are subjected to the couple moments
shown. Determine the magnitude of the resultant couple
moment and specify its coordinate direction angles.
z
M2 = 20 N · m
20°
30°
SOLUTION
y
M 1 = 550k6 N # m
M 2 = 201-cos 20° sin 30°i - cos 20° cos 30°j + sin 20°k2 N # m
= 5-9.397i - 16.276j + 6.840k6 N # m
x
M1 = 50 N · m
Resultant Couple Moment:
M R = ©M;
MR = M1 + M2
= 5-9.397i - 16.276j + 150 + 6.8402k6 N # m
= 5-9.397i - 16.276j + 56.840k6 N # m
The magnitude of the resultant couple moment is
MR = 21-9.39722 + 1-16.27622 + 156.84022
= 59.867 N # m = 59.9 N # m
Ans.
The coordinate direction angles are
a = cos-1 a
-9.397
b = 99.0°
59.867
Ans.
b = cos-1 a
-16.276
b = 106°
59.867
Ans.
g = cos-1
56.840
59.867
Ans.
= 18.3°
4– 87.
If the resultant couple of the two
couples acting on the fire hydrant is
MR = { 15i + 30j} N ˜ m, determine
the force magnitude P.
Given:
a
0.2 m
b
0.150 m
M
§ 15 ·
¨ 30 ¸ N˜ m
¨
¸
© 0 ¹
F
75 N
Solution:
Initial guess
P
1N
Given
M
§ F a ·
¨ Pb ¸
¨
¸
© 0 ¹
P
Find ( P)
P
200 N
Ans.
4–88.
A couple acts on each of the handles of the minidual
valve. Determine the magnitude and coordinate direction
angles of the resultant couple moment.
z
35 N
25 N
60
SOLUTION
y
Mx = - 35(0.35) - 25(0.35) cos 60° = - 16.625
My = - 25(0.35) sin 60° = - 7.5777 N # m
|M| = 2(- 16.625) + (-7.5777) =
2
2
18.2705 = 18.3 N # m
175 mm
175 mm
x
25 N
Ans.
a = cos-1 a
- 16.625
b = 155°
18.2705
Ans.
b = cos-1 a
- 7.5777
b = 115°
18.2705
Ans.
g = cos-1 a
0
b = 90°
18.2705
Ans.
35 N
4–89.
Determine the resultant couple moment of the two couples
that act on the pipe assembly. The distance from A to B is
d = 400 mm. Express the result as a Cartesian vector.
z
{35k} N
B
250 mm
d
{ 50i} N
C
30
{ 35k} N
SOLUTION
350 mm
Vector Analysis
x
Position Vector:
rAB = {(0.35 - 0.35)i + (-0.4 cos 30° - 0)j + (0.4 sin 30° - 0)k} m
= { -0.3464j + 0.20k} m
Couple Moments: With F1 = {35k} N and F2 = { - 50i} N, applying Eq. 4–15, we have
(M C)1 = rAB * F1
i
= 30
0
j
-0.3464
0
k
0.20 3 = {- 12.12i} N # m
35
(M C)2 = rAB * F2
i
= 3 0
- 50
j
- 0.3464
0
k
0.20 3 = { - 10.0j - 17.32k} N # m
0
Resultant Couple Moment:
M R = ©M;
M R = (M C)1 + (M C)2
= {- 12.1i - 10.0j - 17.3k}N # m
Scalar Analysis: Summing moments about x, y, and z axes, we have
(MR)x = ©Mx ;
(MR)x = -35(0.4 cos 30°) = - 12.12 N # m
(MR)y = ©My ;
(MR)y = -50(0.4 sin 30°) = -10.0 N # m
(MR)z = ©Mz ;
(MR)z = -50(0.4 cos 30°) = -17.32 N # m
Express M R as a Cartesian vector, we have
M R = { - 12.1i - 10.0j - 17.3k} N # m
Ans.
A
{50i} N
y
4–90.
Determine the distance d between A and B so that the
resultant couple moment has a magnitude of MR = 20 N # m.
z
{35k} N
B
250 mm
d
{ 50i} N
C
30
{ 35k} N
SOLUTION
350 mm
Position Vector:
x
rAB = {(0.35 - 0.35)i + (-d cos 30° - 0)j + (d sin 30° - 0)k} m
= {- 0.8660d j + 0.50d k} m
Couple Moments: With F1 = {35k} N and F2 = {- 50i} N, applying Eq. 4–15, we have
(M C)1 = rAB * F1
i
3
= 0
0
j
- 0.8660d
0
k
0.50d 3 = { - 30.31d i} N # m
35
(M C)2 = rAB * F2
i
= 3 0
-50
j
-0.8660d
0
k
0.50d 3 = {- 25.0d j - 43.30d k} N # m
0
Resultant Couple Moment:
M R = ©M;
M R = (M C)1 + (M C)2
= {- 30.31d i - 25.0d j - 43.30d k} N # m
The magnitude of M R is 20 N # m, thus
20 = 2(-30.31d)2 + (- 25.0d)2 + (43.30d)2
d = 0.3421 m = 342 mm
Ans.
A
{50i} N
y
4–91.
If F = 80 N, determine the magnitude and coordinate
direction angles of the couple moment. The pipe assembly
lies in the x–y plane.
z
F
300 mm
300 mm
F
x
200 mm
SOLUTION
It is easiest to find the couple moment of F by taking the moment of F or –F about
point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be
determined first.
200 mm
rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m
rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m
The force vectors F and –F can be written as
F = {80 k} N and - F = [ -80 k] N
Thus, the couple moment of F can be determined from
i
M c = rAB * F = 3 0.1
0
j
0.5
0
k
0 3 = [40i - 8j] N # m
80
or
i
Mc = rBA * -F = 3 - 0.1
0
j
- 0.5
0
k
0 3 = [40i - 8j] N # m
- 80
The magnitude of Mc is given by
Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + (- 8)2 + 02 = 40.79 N # m = 40.8 N # m
The coordinate angles of Mc are
a = cos
-1
b = cos
-1
g = cos
-1
¢
¢
¢
Mx
40
≤ = cos ¢
≤ = 11.3°
M
40.79
My
M
Mz
M
-8
≤ = 101°
40.79
Ans.
0
≤ = 90°
40.79
Ans.
≤ = cos ¢
≤ = cos ¢
Ans.
Ans.
300 mm
y
4–92.
If the magnitude of the couple moment acting on the pipe
assembly is 50 N # m, determine the magnitude of the couple
forces applied to each wrench. The pipe assembly lies in the
x–y plane.
z
F
300 mm
300 mm
F
x
200 mm
SOLUTION
It is easiest to find the couple moment of F by taking the moment of either F or –F
about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must
be determined first.
rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m
rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m
The force vectors F and –F can be written as
F = {Fk} N and - F = [ -Fk]N
Thus, the couple moment of F can be determined from
i
M c = rAB * F = 3 0.1
0
j
0.5
0
k
0 3 = 0.5Fi - 0.1Fj
F
The magnitude of Mc is given by
Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F
Since Mc is required to equal 50 N # m,
50 = 0.5099F
F = 98.1 N
Ans.
200 mm
300 mm
y
4–93.
If F = 5100k6 N, determine the couple moment that acts on
the assembly. Express the result as a Cartesian vector.
Member BA lies in the x–y plane.
z
O
O
B
F
60
SOLUTION
x
2
f = tan a b - 30° = 3.69°
3
-1
200 mm
–F
150 mm
r1 = { -360.6 sin 3.69°i + 360.6 cos 3.69°j}
200 mm
= { -23.21i + 359.8j} mm
u = tan-1 a
2
b + 30° = 53.96°
4.5
r2 = {492.4 sin 53.96°i + 492.4 cos 53.96°j}
= {398.2i + 289.7j} mm
Mc = (r1 - r2) * F
i
= 3 - 421.4
0
j
70.10
0
k
0 3
100
Mc = {7.01i + 42.1j} N # m
y
300 mm
Ans.
A
4–94.
If the magnitude of the resultant couple moment is
15 N # m, determine the magnitude F of the forces applied
to the wrenches.
z
O
O
B
x
2
f = tan a b - 30° = 3.69°
3
-1
200 mm
–F
150 mm
200 mm
= { -23.21i + 359.8j} mm
2
b + 30° = 53.96°
4.5
r2 = {492.4 sin 53.96°i + 492.4 cos 53.96°j}
= {398.2i + 289.7j} mm
Mc = (r1 - r2) * F
i
3
= - 421.4
0
j
70.10
0
k
03
F
Mc = {0.0701Fi + 0.421Fj} N # m
Mc = 2(0.0701F)2 + (0.421F)2 = 15
F =
15
2(0.0701) 2 + (0.421)2
= 35.1 N
Ans.
Also, align y¿ axis along BA.
Mc = -F(0.15)i¿ + F(0.4)j¿
15 = 2(F(-0.15))2 + (F(0.4))2
F = 35.1 N
y
300 mm
r1 = { -360.6 sin 3.69°i + 360.6 cos 3.69°j}
u = tan-1 a
F
60
SOLUTION
Ans.
A
4–95.
If F1 = 100 N, F2 = 120 N and F3 = 80 N, determine the
magnitude and coordinate direction angles of the resultant
couple moment.
z
–F4 ⫽ [⫺150 k] N
0.3 m
0.2 m
0.2 m
0.3 m
0.2 m
F1
0.2 m – F1
x
SOLUTION
Couple Moment: The position vectors r1, r2, r3, and r4, Fig. a, must be determined
first.
r1 = {0.2i} m
r2 = {0.2j} m
0.2 m
r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k
= {0.1837i + 0.1837j - 0.15k} m
The force vectors F1, F2, and F3 are given by
F2 = {120k} N
F3 = {80i} N
Thus,
M 1 = r1 * F1 = (0.2i) * (100k) = {-20j} N # m
M 2 = r2 * F2 = (0.2j) * (120k) = {24i} N # m
M 3 = r3 * F3 = (0.2j) * (80i) = {-16k} N # m
M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m
Resultant Moment: The resultant couple moment is given by
(M c)R = ∑M c;
(M c)R = M 1 + M 2 + M 3 + M 4
= (-20j) + (24i) + (-16k) + (27.56i-27.56j)
= {51.56i - 47.56j - 16k} N # m
The magnitude of the couple moment is
(M c)R = 2[(M c)R]x2 + [(M c)R]y 2 + [(M c)R]z 2
= 2(51.56)2 + (-47.56)2 + ( -16)2
= 71.94 N # m = 71.9 N # m
Ans.
The coordinate angles of (Mc)R are
a = cos -1 a
b = cos -1 a
g = cos -1 a
[(Mc)R]x
51.56
b = cos a
b = 44.2°
(Mc)R
71.94
[(Mc)R]y
(Mc)R
[(Mc)R]z
(Mc)R
b = cos a
Ans.
-47.56
b = 131°
71.94
Ans.
-16
b = 103°
71.94
Ans.
b = cos a
F2
– F3
r3 = {0.2j} m
From the geometry of Figs. b and c, we obtain
F1 = {100k} N
– F2 0.2 m
F3
30⬚
F4 ⫽ [150 k] N
y
4–96.
Determine the required magnitude of F1, F2, and F3
so that the resultant couple moment is (Mc)R =
[50 i - 45 j - 20 k] N # m.
z
–F4 ⫽ [⫺150 k] N
0.3 m
0.2 m
0.2 m
Couple Moment: The position vectors r1, r2, r3, and r4, Fig. a, must be determined
first.
r1 = {0.2i} m
r2 = {0.2j} m
F1
0.2 m – F1
x
– F2 0.2 m
r3 = {0.2j} m
F2
– F3
From the geometry of Figs. b and c, we obtain
0.2 m
r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k
F3
= {0.1837i + 0.1837j - 0.15k} m
The force vectors F1, F2, and F3 are given by
F1 = F1k
F2 = F2k
F3 = F3i
Thus,
M 1 = r1 * F1 = (0.2i) * (F1k) = -0.2 F1j
M 2 = r2 * F2 = (0.2j) * (F2k) = 0.2 F2i
M 3 = r3 * F3 = (0.2j) * (F3i) = -0.2 F3k
M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m
Resultant Moment: The resultant couple moment required to equal
(M c)R = {50i - 45j - 20k} N # m. Thus,
(M c)R = ©M c;
0.3 m
0.2 m
SOLUTION
(M c)R = M 1 + M 2 + M 3 + M 4
50i - 45j - 20k = (-0.2F1j) + (0.2F2i) + ( -0.2F3k) + (27.56i - 27.56j)
50i - 45j - 20k = (0.2F2 + 27.56)i + (-0.2F1 - 27.56)j - 0.2F3k
Equating the i, j, and k components yields
50 = 0.2F2 + 27.56
F2 = 112 N
Ans.
-45 = -0.2F1 - 27.56
F1 = 87.2 N
Ans.
-20 = -0.2F3
F3 = 100 N
Ans.
30⬚
F4 ⫽ [150 k] N
y
4–97.
Replace the force and couple system by an equivalent force
and couple moment at point O.
y
3m
8 kN m
P
O
3m
4 kN
6 kN
4m
5m
+ ©F = ©F ;
:
Rx
x
FRx = 6a
5
b - 4 cos 60°
13
A
4m
FRy = 6 a
12
b - 4 sin 60°
13
= 2.0744 kN
FR = 2(0.30769)2 + (2.0744)2 = 2.10 kN
u = tan-1 c
2.0744
d = 81.6° a
0.30769
a + MO = ©MO ;
60
5
= 0.30769 kN
+ c ©FRy = ©Fy ;
12
13
SOLUTION
MO = 8 - 6a
Ans.
Ans.
5
12
b(4) + 6a b(5) - 4 cos 60°(4)
13
13
MO = -10.62 kN # m = 10.6 kN # m b
Ans.
x
4–98.
Replace the force and couple system by an equivalent force
and couple moment at point P.
y
3m
8 kN m
P
O
3m
4 kN
6 kN
4m
5m
+ ©F = ©F ;
:
Rx
x
FRx = 6a
5
b - 4 cos 60°
13
A
4m
FRy = 6a
12
b - 4 sin 60°
13
= 2.0744 kN
FR = 2(0.30769)2 + (2.0744)2 = 2.10 kN
u = tan-1 c
2.0744
d = 81.6° a
0.30769
a + MP = ©MP ;
60
5
= 0.30769 kN
+ c ©FRy = ©Fy ;
12
13
SOLUTION
MP = 8 - 6a
Ans.
Ans.
5
12
b(7) + 6a b(5) - 4 cos 60°(4) + 4 sin 60°(3)
13
13
MP = -16.8 kN # m = 16.8 kN # m b
Ans.
x
4–99.
Replace the force system acting on the beam by an
equivalent force and couple moment at point A.
3 kN
2.5 kN 1.5 kN 30
5
3
4
B
A
2m
SOLUTION
4
FRx = 1.5 sin 30° - 2.5a b
5
+ F = ©F ;
:
Rx
x
= - 1.25 kN = 1.25 kN ;
3
FRy = - 1.5 cos 30° - 2.5 a b - 3
5
+ c FRy = ©Fy ;
= - 5.799 kN = 5.799 kN T
Thus,
FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN
Ans.
and
u = tan
a + MRA = ©MA ;
-1
¢
FRy
FRx
≤ = tan
-1
a
5.799
b = 77.8° d
1.25
Ans.
3
MRA = -2.5a b (2) - 1.5 cos 30°(6) - 3(8)
5
= -34.8 kN # m = 34.8 kN # m (Clockwise)
Ans.
4m
2m
4–100.
Replace the force system acting on the beam by an
equivalent force and couple moment at point B.
3 kN
2.5 kN 1.5 kN 30
5
3
4
B
A
2m
SOLUTION
4
FRx = 1.5 sin 30° - 2.5a b
5
+ F = ©F ;
:
Rx
x
= -1.25 kN = 1.25 kN ;
3
FRy = -1.5 cos 30° - 2.5a b - 3
5
+ c FRy = ©Fy ;
= -5.799 kN = 5.799 kN T
Thus,
FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN
Ans.
and
u = tan
-1
¢
FRy
FRx
a + MRB = ©MRB ;
≤ = tan
-1
a
5.799
b = 77.8° d
1.25
Ans.
3
MB = 1.5cos 30°(2) + 2.5a b (6)
5
= 11.6 kN # m (Counterclockwise)
Ans.
4m
2m
4–101.
Replace the loading system acting on
the beam by an equivalent resultant
force and couple moment at point O.
Given:
F1
200 N
F2
450 N
M
200 N˜ m
a
0.2 m
b
1.5 m
c
2m
d
1.5 m
T
30 deg
Solution:
FR
§ sin T
§0·
¨
¸
¨
F1 1 F2 cos T
¨ ¸
¨
©0¹
© 0
FR
§ 225 ·
¨ 190 ¸ N
¨
¸
© 0 ¹
·
¸
¸
¹
FR
294 N
MO
§ b c · ª § 0 ·º § b · ª § sin T
¨ a ¸ u «F ¨ 1 ¸» ¨ a ¸ u «F ¨ cos T
¨
¸ « 1¨ ¸» ¨ ¸ « 2 ¨
© 0 ¹ ¬ © 0 ¹¼ © 0 ¹ ¬ © 0
MO
§ 0 ·
¨ 0 ¸ N˜ m
¨
¸
© 39.6 ¹
Ans.
·º
§0·
¸» M¨ 0 ¸
¸»
¨ ¸
© 1 ¹
¹¼
Ans.
.
.
.
.
.
.
4–104. The system of four forces acts on the roof truss.
Determine the equivalent resultant force and specify its
location along AB, measured from point A.
200
1 kNlb
30⬚
1.375
275kN
lb 14 m
ft
B
1.5
kNlb 14 m
300
ft
0.75
150kN
lb 41 ft
m
A
↓
+FRx = ΣFx;
FRx = 1 sin 30° = 0.5 kN
↓
30⬚
+FRy = ΣFy;
FRy = 0.75 + 1.5 + 1.375 + 1 cos 30° = 4.491 kN
FR =
(0.5) 2 + (4.491) 2 = 4.52 kN
FRy = 4.491 kN
Ans.
⎛ 0.5 ⎞
␪ = tan–1 ⎜
= 6.35°
⎝ 4.491 ⎟⎠
␾ = 30° – 6.35° = 23.6°
哭
+ MRA = ΣMA;
Ans.
4.491 (d) = 1 (1.5) + 2 (1.375) + 3 cos 30° (1)
d = 1.52 m
FRx = 0.5 kN
Ans.
FRx = 0.5 kN
FRy = 4.491 kN
4–105.
Replace the force system acting on the frame by a resultant
force and couple moment at point A.
5 kN
3
13
5
1m
4m
4
B
C
5m
SOLUTION
Equivalent Resultant Force: Resolving F1, F2, and F3 into their x and y components, Fig. a, and summing these force components algebraically along the x and
y axes, we have
+ ©(F ) = ©F ;
:
R x
x
+ c (FR)y = ©Fy;
4
5
(FR)x = 5 a b - 3 a b = 2.846 kN :
5
13
3
12
(FR)y = - 5a b - 2 - 3 a b = -7.769 kN = 7.769 kN T
5
13
The magnitude of the resultant force FR is given by
FR = 2(FR)x2 + (FR)y2 = 22.8462 + 7.7692 = 8.274 kN = 8.27 kN
Ans.
The angle u of FR is
u = tan-1 c
(FR)y
(FR)x
d = tan-1 c
7.769
d = 69.88° = 69.9° c
2.846
Ans.
Equivalent Couple Moment: Applying the principle of moments and summing the
moments of the force components algebraically about point A, we can write
a + (MR)A = ©MA;
3
4
12
5
(MR)A = 5 a b(4) - 5 a b(5) - 2(1) - 3a b (2) + 3 a b (5)
5
5
13
13
= -9.768 kN # m = 9.77 kN # m (Clockwise)
Ans.
3 kN
2 kN
A
1m
5
D
12
4–106.
Replace the force system acting on the bracket by a resultant
force and couple moment at point A.
450 N
600 N
B
30⬚
45⬚
0.3 m
SOLUTION
A
Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y
axes, we have
+ ©(F ) = ©F ;
:
R x
x
(FR)x = 450 cos 45° - 600 cos 30° = -201.42 N = 201.42 N
+ c (FR)y = ©Fy;
(FR)y = 450 sin 45° + 600 sin 30° = 618.20 N
0.6 m
;
c
The magnitude of the resultant force FR is given by
FR = 2(FR)x2 + (FR)y2 = 2201.4 2 + 618.202 = 650.18 kN = 650 N
Ans.
The angle u of FR is
u = tan - 1 c
(FR)y
(FR)x
d = tan - 1 c
618.20
d = 71.95° = 72.0° b
201.4
Ans.
Equivalent Resultant Couple Moment: Applying the principle of moments, Figs. a
and b, and summing the moments of the force components algebraically about
point A, we can write
a + (MR)A = ©MA;
(MR)A = 600 sin 30°(0.6) + 600 cos 30°(0.3) + 450 sin 45°(0.6) - 450 cos 45°(0.3)
= 431.36 N # m = 431 N # m (Counterclockwise)
Ans.
4–107.
z
A biomechanical model of the lumbar region of the human
trunk is shown. The forces acting in the four muscle groups
consist of FR = 35 N for the rectus, FO = 45 N for the
oblique, FL = 23 N for the lumbar latissimus dorsi, and
FE = 32 N for the erector spinae. These loadings are
symmetric with respect to the y–z plane. Replace this system
of parallel forces by an equivalent force and couple moment
acting at the spine, point O. Express the results in Cartesian
vector form.
FO
FR
FO
FL
FL
O
15 mm
45 mm
x
FR = ©Fz ;
FR = {2(35 + 45 + 23 + 32)k } = {270k} N
MROx = ©MOx ;
MR O = [-2(35)(0.075) + 2(32)(0.015) + 2(23)(0.045)]i
MR O = {-2.22i} N # m
FE
FE
75 mm
SOLUTION
FR
Ans.
Ans.
50 mm
30 mm
40 mm
y
4–108.
Replace the two forces acting on the post by a resultant
force and couple moment at point O. Express the results in
Cartesian vector form.
z
A
C
FD
SOLUTION
FD = FDuCD = 7C
(0 - 0)i + (6 - 0)j + (0 - 8)k
(0 - 0) + (6 - 0) + (0 - 8)
2
2
2
S = [3j - 4k] kN
(2 - 0)i + (- 3 - 0)j + (0 - 6)k
(2 - 0)2 + (- 3 - 0)2 + (0 - 6)2
S = [2i - 3j - 6k] kN
The resultant force FR is given by
FR = πF; FR = FB + FD
= (3j - 4k) + (2i - 3j - 6k)
= [2i - 10k] kN
Ans.
Equivalent Resultant Force: The position vectors rOB and rOC are
rOB = {6j} m
rOC = [6k] m
Thus, the resultant couple moment about point O is given by
(M R)O = ©M O;
(M R)O = rOB * FB + rOC * FD
i
= 30
0
j
6
3
i
k
0 3 + 30
2
-4
= [- 6i + 12j] kN # m
j
0
-3
8m
2m
D
3m
x
k
6 3
-6
Ans.
5 kN
7 kN
6m
Equivalent Resultant Force: The forces FB and FD, Fig. a, expressed in Cartesian
vector form can be written as
FB = FBuAB = 5C
FB
O
6m
B
y
4–109.
z
Replace the force system by an equivalent force and couple
moment at point A.
F2
{100i
100j
50k} N
F1
4m
F3
SOLUTION
FR = ©F;
8m
FR = F1 + F2 + F3
1m
= 1300 + 1002i + 1400 - 1002j + 1-100 - 50 - 5002k
= 5400i + 300j - 650k6 N
The position vectors are rAB = 512k6 m and rAE = 5 -1j6 m.
M RA = ©M A ;
M RA = rAB * F1 + rAB * F2 + rAE * F3
i
3
= 0
300
j
0
400
k
i
3
3
12
+ 0
-100
100
i
+ 0
0
j
-1
0
k
0
-500
=
{ 500k} N
-3100i + 4800j N # m
j
0
-100
k
12 3
-50
Ans.
{300i
400j
100k} N
4–110.
z
The belt passing over the pulley is subjected to forces F1 and
F2, each having a magnitude of 40 N. F1 acts in the -k direction.
Replace these forces by an equivalent force and couple
moment at point A. Express the result in Cartesian vector form.
Set u = 0° so that F2 acts in the -j direction.
␪
r ⫽ 80 mm
y
300 mm
A
SOLUTION
x
FR = F1 + F2
FR = {-40j - 40 k} N
Ans.
M RA = ©(r * F)
i
3
= -0.3
0
j
0
-40
F2
F1
k
i
3
3
0.08 + - 0.3
0
0
MRA = {-12j + 12k} N # m
j
0.08
0
k
0 3
-40
Ans.
4–111.
The belt passing over the pulley is subjected to two forces F1
and F2, each having a magnitude of 40 N. F1 acts in the - k
direction. Replace these forces by an equivalent force and
couple moment at point A. Express the result in Cartesian
vector form. Take u = 45°.
z
r
80 mm
y
300 mm
A
SOLUTION
x
F R = F1 + F 2
= -40 cos 45°j + ( -40 - 40 sin 45°)k
F2
FR = {- 28.3j - 68.3k} N
Ans.
rAF1 = {- 0.3i + 0.08j} m
rAF2 = -0.3i - 0.08 sin 45°j + 0.08 cos 45°k
= {- 0.3i - 0.0566j + 0.0566k} m
MRA = (rAF1 * F1) + (rAF2 * F2)
i
= 3 - 0.3
0
j
0.08
0
i
k
0 3 + 3 -0.3
0
- 40
j
- 0.0566
-40 cos 45°
MRA = {- 20.5j + 8.49k} N # m
k
0.0566 3
-40 sin 45°
Ans.
Also,
MRAx = ©MAx
MRAx = 28.28(0.0566) + 28.28(0.0566) - 40(0.08)
MRAx = 0
MRAy = ©MAy
MRAy = -28.28(0.3) - 40(0.3)
MRAy = -20.5 N # m
MRAz = ©MAz
MRAz = 28.28(0.3)
MRAz = 8.49 N # m
MRA = {- 20.5j + 8.49k} N # m
Ans.
F1
4–112.
Handle forces F1 and F2 are applied to the electric drill.
Replace this force system by an equivalent resultant force
and couple moment acting at point O. Express the results in
Cartesian vector form.
F2 ⫽ {2j ⫺ 4k} N
z
F1 ⫽ {6i ⫺ 3j ⫺ 10k} N
0.15 m
0.25 m
0.3 m
SOLUTION
FR = ©F;
O
FR = 6i - 3j - 10k + 2j - 4k
x
= {6i - 1j - 14k} N
Ans.
M RO = ©M O ;
i
M RO = 3 0.15
6
j
0
-3
k
i
0.3 3 + 3 0
-10
0
j
-0.25
2
k
0.3 3
-4
= 0.9i + 3.30j - 0.450k + 0.4i
= {1.30i + 3.30j - 0.450k} N # m
Ans.
Note that FRz = -14 N pushes the drill bit down into the stock.
(MRO)x = 1.30 N # m and (MRO)y = 3.30 N # m cause the drill bit to bend.
(MRO)z = -0.450 N # m causes the drill case and the spinning drill bit to rotate
about the z-axis.
y
4–113.
Replace the force at A by an equivalent force and couple moment at point O.
Given:
F
375 N
a
2m
b
4m
c
2m
d
1m
T
30 deg
Solution:
Fv
MO
§ sin T
¨
F cos T
¨
© 0
·
¸
¸
¹
§ a ·
¨ b ¸uF
¨ ¸ v
©0¹
Fv
MO
§ 187.5 ·
¨ 324.76 ¸ N
¨
¸
© 0 ¹
·
§ 0
¨ 0
¸ N˜ m
¨
¸
© 100.481 ¹
Ans.
Ans.
4–114 .
Replace the force at A by an equivalent force and couple moment at point P .
Given:
F
375 N
a
2m
b
4m
c
2m
d
1m
T
30 deg
Solution:
Fv
MP
§ sin T
¨
F cos T
¨
© 0
·
¸
¸
¹
§ a c ·
¨ bd ¸uF
¨
¸ v
© 0 ¹
Fv
MP
§ 187.5 ·
¨ 324.76 ¸ N
¨
¸
© 0 ¹
§ 0 ·
¨ 0 ¸ N˜ m
¨
¸
© 736.538 ¹
Ans.
Ans.
4–115.
Determine the magnitude and direction T
of force F and its placement d on the
beam so that the loading system is
equivalent to a resultant force FR acting
vertically downward at point A and a
clockwise couple moment M.
Units Used:
kN
3
10 N
Given:
F1
5 kN
a
3m
F2
3 kN
b
4m
FR
12 kN
c
6m
M
50 kN˜ m
e
7
f
24
Solution:
Initial guesses:
Given
F
1 kN
T
§ e · F F cos T
¨ 2 2¸ 1
© e f ¹
d
30 deg
2m
0
§ f · F F sin T F
2
¨ 2 2¸ 1
© e f ¹
F R
f
§
·
¨ 2 2 ¸ F1 a F sin T ( a b d) F2 ( a b)
© e f ¹
§F·
¨T ¸
¨ ¸
©d¹
Find F T d
F
4.427 kN
T
71.565 deg
M
d
3.524 m
Ans.
4–116.
Determine the magnitude and direction T of force F and its placement d on the beam so that the
loading system is equivalent to a resultant force F R acting vertically downward at point A and a
clockwise couple moment M.
Units Used:
3
kN
10 N
Given:
F1
5 kN
a
3m
F2
3 kN
b
4m
FR
10 kN
c
6m
45 kN˜ m e
M
7
f
24
F
1 kN
Solution:
Initial guesses:
Given
T
§ e · F F cos T
¨ 2 2¸ 1
© e f ¹
d
30 deg
1m
0
§ f · F F sin T F
2
¨ 2 2¸ 1
© e f ¹
F R
f
§
·
¨ 2 2 ¸ F1 a F sin T ( a b d) F2 ( a b)
© e f ¹
§F·
¨T ¸
¨ ¸
©d¹
Find F T d
F
2.608 kN
T
57.529 deg
M
d
2.636 m
Ans.
4–117.
700 N
Replace the loading acting on the beam by a single resultant
force. Specify where the force acts, measured from end A.
450 N
30
300 N
60
B
A
2m
4m
SOLUTION
+ F = ©F ;
:
Rx
x
FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N
+ c FRy = ©Fy ;
FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N
;
T
F = 2(-125)2 + ( -1296)2 = 1302 N
Ans.
1296
b = 84.5°
125
Ans.
u = tan-1 a
c + MRA = ©MA ;
d
1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 3000
x = 8.51 m
Ans.
3m
3000 N m
4–118.
Replace the loading acting on the beam by a single resultant
force. Specify where the force acts, measured from B.
700 N
450 N
30
300 N
60
B
A
2m
4m
SOLUTION
+ F = ©F ;
:
Rx
x
FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N
+ c FRy = ©Fy ;
FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N
F = 2(- 125)2 + ( - 1296)2 = 1302 N
u = tan-1 a
1296
b = 84.5° d
125
c + MRB = ©MB ;
;
T
Ans.
Ans.
1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 3000
x = 2.52 m (to the right)
Ans.
3m
3000 N m
4–119. The system of parallel forces acts on the top of the
Warren truss. Determine the equivalent resultant force of the
system and specify its location measured from point A.
2 kN
1 kN
500 N
500 N
1m
A
.
.
1m
500 N
1m
1m
4–120.
Replace the loading on the frame by a single resultant force.
Specify where its line of action intersects member AB,
measured from A.
300 N
250 N
1m
C
2m
3m
5
B
4
D
2m
400 N m
60
SOLUTION
+ ©F = F ;
:
x
Rx
3m
500 N
4
FRx = -250a b - 500(cos 60°) = -450 N = 450 N ;
5
A
+ c ©Fy = ©Fy ;
FRy
3
= -300 - 250a b - 500 sin 60° = -883.0127 N = 883.0127 N T
5
FR = 2(- 450)2 + (- 883.0127)2 = 991 N
u = tan-1 a
883.0127
b = 63.0° d
450
a + MRA = ©MA ;
Ans.
Ans.
3
4
450y = 400 + (500 cos 60°)(3) + 250a b(5) - 300(2) - 250a b(5)
5
5
y =
800
= 1.78 m
450
Ans.
3
4–121.
Replace the loading on the frame by a single resultant force.
Specify where its line of action intersects member CD,
measured from end C.
300 N
250 N
1m
C
2m
3m
5
B
4
D
2m
400 N m
60
SOLUTION
+ ©F = F ;
:
x
Rx
+ c ©Fy = ©Fy ;
FRx
FRy
3
= -300 - 250a b - 500 sin 60° = -883.0127 N = 883.0127 N T
5
FR = 2(- 450)2 + (- 883.0127)2 = 991 N
u = tan-1 a
3m
500 N
4
= -250a b - 500(cos 60°) = -450 N = 450 N ;
5
A
Ans.
883.0127
b = 63.0° d
450
c + MRA = ©MC ;
3
883.0127x = - 400 + 300(3) + 250 a b(6) + 500 cos 60°(2) + (500 sin 60°)(1)
5
x =
2333
= 2.64 m
883.0127
Ans.
3
4–122 .
Replace the force and couple system by an equivalent force and couple moment at point O.
Units Used :
3
kN 10 N
Given:
M 30kN ˜ m
T 60deg
a 3m
f 12
b 3m
g 5
c 4m
F1 8kN
d 4m
F2 6kN
e 6m
Solution:
§g ·
§ cos ( T ) ·
¨ f ¸ F ˜ ¨ sin ( T ) ¸
2¨
¸
2
2¨ ¸
f g ©0 ¹
© 0 ¹
F1
FR FR
§ 0.077 ·
¨ 2.188 ¸ kN
¨
¸
© 0.000 ¹
FR
2.19 kN
Ans.
§ g ·º § 0 · ª § cos ( T ) ·º
§ 0 · § c · ª F
1
¨
¸
¨
¸
«
¨ f ¸» ¨ d ¸ u «F ˜ ¨ sin ( T ) ¸»
0 e u
MO ¨ ¸ ¨ ¸ « 2 2 ¨ ¸» ¨ ¸ « 2 ¨
¸»
© M ¹ © 0 ¹ ¬ f g © 0 ¹¼ © 0 ¹ ¬ © 0 ¹¼
MO
§
¨
¨
©
0.00
0.00
6.9 2
·
¸ kN˜ m
¸
¹
Ans.
4 –123 .
Replace the force and couple system by an equivalent force and couple moment at point P.
Units Used :
3
kN 10 N
Given:
M 30kN ˜m
T 60deg
a 3m
f 12
b 3m
g 5
c 4m
F1 8kN
d 4m
F2 6kN
e 6m
Solution:
§g ·
§ cos ( T ) ·
¨ f ¸ F ˜ ¨ sin ( T ) ¸
2¨
¸
2
2¨ ¸
f g ©0 ¹
© 0 ¹
F1
FR FR
§ 0.077
¨ 2.188
¨
© 0.000
·
¸ kN
¸
¹
FR
2.19 kN
Ans.
§ g ·º § b · ª § cos ( T ) ·º
§ 0 · § c b · ª F
1
¨
¸
¨
¸
«
¨ f ¸» ¨ d ¸ u «F ˜ ¨ sin ( T ) ¸»
0 MP u
e
¨ ¸ ¨
¸ « 2 2 ¨ ¸» ¨ ¸ « 2 ¨
¸»
© M ¹ © 0 ¹ ¬ f g © 0 ¹¼ © 0 ¹ ¬ © 0 ¹¼
MP
§
¨
¨
©
·
¸
0.000 kN˜ m
¸
0.358 ¹
0.000
Ans.
4–124.
Replace the force system acting on the post by a resultant
force, and specify where its line of action intersects the post
AB measured from point A.
0.5 m
B
1m
500 N
5
3
0.2 m
4
30
1m
SOLUTION
300 N
Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y
components, Fig. a. Summing these force components algebraically along the x and
y axes,
+ (F ) = ©F ;
:
R x
x
4
(FR)x = 250 a b - 500 cos 30° - 300 = -533.01 N = 533.01 N ;
5
+ c (FR)y = ©Fy;
3
(FR)y = 500 sin 30° - 250a b = 100 N c
5
1m
A
The magnitude of the resultant force FR is given by
FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N
Ans.
The angle u of FR is
u = tan
-1
B
(FR)y
(FR)x
R = tan
-1
c
100
d = 10.63° = 10.6° b
533.01
Ans.
Location of the Resultant Force: Applying the principle of moments, Figs. a and b,
and summing the moments of the force components algebraically about point A,
a +(MR)A = ©MA;
4
3
533.01(d) = 500 cos 30°(2) - 500 sin 30°(0.2) - 250 a b (0.5) - 250 a b(3) + 300(1)
5
5
d = 0.8274 mm = 827 mm
Ans.
250 N
4–125.
Replace the force system acting on the post by a resultant
force, and specify where its line of action intersects the post
AB measured from point B.
0.5 m
B
1m
500 N
5
3
0.2 m
4
30
1m
SOLUTION
300 N
Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y
components, Fig. a. Summing these force components algebraically along the x and
y axes,
1m
4
(FR)x = 250 a b - 500 cos 30° - 300 = -533.01N = 533.01 N ;
5
+ ©(F ) = ©F ;
:
R x
x
3
(FR)y = 500 sin 30° - 250a b = 100 N c
5
+ c (FR)y = ©Fy;
The magnitude of the resultant force FR is given by
FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N
Ans.
The angle u of FR is
u = tan
-1
B
(FR)y
(FR)x
R = tan
-1
c
100
d = 10.63° = 10.6° b
533.01
Ans.
Location of the Resultant Force: Applying the principle of moments, Figs. a and b,
and summing the moments of the force components algebraically about point B,
a +(MR)B = ©Mb;
3
-533.01(d) = -500 cos 30°(1) - 500 sin 30°(0.2) - 250 a b (0.5) - 300(2)
5
d = 2.17 m
Ans.
A
250 N
4–126. Replace the force and couple moment system
acting on the overhang beam by a resultant force, and
specify its location along AB measured from point A.
30 kN
26 kN
30⬚
13
12
A
0.3 m
5
45 kN⭈m
0.3 m
B
2m
1m
1m
2m
4–127.
The tube supports the four parallel forces. Determine the
magnitudes of forces FC and FD acting at C and D so that
the equivalent resultant force of the force system acts
through the midpoint O of the tube.
FD
z
600 N
D
FC
A
400 mm
SOLUTION
Since the resultant force passes through point O, the resultant moment components
about x and y axes are both zero.
©Mx = 0;
FD(0.4) + 600(0.4) - FC(0.4) - 500(0.4) = 0
FC - FD = 100
©My = 0;
(1)
500(0.2) + 600(0.2) - FC(0.2) - FD(0.2) = 0
FC + FD = 1100
(2)
Solving Eqs. (1) and (2) yields:
FC = 600 N
FD = 500 N
Ans.
O
500 N
C
400 mm
x
z
B
200 mm
200 mm y
4–128.
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point P. Express the results in Cartesian vector form.
Units Used:
3
kN
10 N
Given:
F
§8·
¨ 6 ¸ kN
¨ ¸
©8¹
M
§ 20 ·
¨ 70 ¸ kN˜ m
¨
¸
© 20 ¹
a
3m
b
3m
e
5m
c
4m
f
6m
d
6m
g
5m
Solution:
FR
F
MR
§ f ·
¨ e ¸uF
M
¨
¸
©d g¹
FR
§8·
¨ 6 ¸ kN
¨ ¸
©8¹
MR
§ 46 ·
¨ 66 ¸ kN˜ m
¨
¸
© 56 ¹
Ans.
4–129.
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point Q. Express the results in Cartesian vector form.
Units Used:
3
kN
10 N
Given:
§8·
¨ 6 ¸ kN
F
¨ ¸
©8¹
§ 20 ·
¨ 70 ¸ kN˜ m
M
¨
¸
© 20 ¹
a
3m
b
3m
e
5m
c
4m
f
6m
d
6m
g
5m
Solution:
FR
F
MR
§0·
¨ ¸
M e uF
¨ ¸
©g¹
FR
§8·
¨ 6 ¸ kN
¨ ¸
©8¹
MR
§ 10 ·
¨ 30 ¸ kN˜ m
¨
¸
© 20 ¹
Ans.
4–130.
z
The building slab is subjected to four parallel column
loadings. Determine the equivalent resultant force and
specify its location (x, y) on the slab. Take F1 = 30 kN,
F2 = 40 kN.
20 kN
F1
50 kN
F2
SOLUTION
+ c FR = ©Fz;
x
FR = -20 - 50 - 30 - 40 = -140 kN = 140 kN T
(MR)x = ©Mx;
(MR)y = ©My;
8m
Ans.
-140y = -50(3) - 30(11) - 40(13)
y = 7.14 m
Ans.
140x = 50(4)+ 20(10) + 40(10)
x = 5.71 m
4m
3m
Ans.
6m
2m
y
4–131.
The building slab is subjected to four parallel column
loadings. Determine the equivalent resultant force and
specify its location (x, y) on the slab. Take F1 = 20 kN,
F2 = 50 kN.
z
20 kN
F1
50 kN
F2
SOLUTION
+ TFR = ©Fz;
x
FR = 20 + 50 + 20 + 50 = 140 kN
MR y = ©My;
MR x = ©Mx;
8m
Ans.
140(x) = (50)(4) + 20(10) + 50(10)
x = 6.43 m
Ans.
- 140(y) = -(50)(3) - 20(11) - 50(13)
y = 7.29 m
4m
3m
Ans.
6m
2m
y
4–132.
If FA = 40 kN and FB = 35 kN, determine the magnitude
of the resultant force and specify the location of its point of
application (x, y) on the slab.
z
30 kN
0.75 m
FB 2.5 m
90 kN
20 kN
2.5 m
0.75 m
FA
0.75 m
SOLUTION
x
3m
Equivalent Resultant Force: By equating the sum of the forces along the z axis to
the resultant force FR, Fig. b,
+ c FR = ©Fz;
3m
- FR = -30 - 20 - 90 - 35 - 40
FR = 215 kN
Ans.
Point of Application: By equating the moment of the forces and FR, about the x and
y axes,
(MR)x = ©Mx;
- 215(y) = -35(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - 40(6.75)
y = 3.68 m
(MR)y = ©My;
Ans.
215(x) = 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75)
x = 3.54 m
Ans.
0.75 m
y
4–133.
If the resultant force is required to act at the center of the
slab, determine the magnitude of the column loadings FA
and FB and the magnitude of the resultant force.
z
30 kN
0.75 m
FB 2.5 m
90 kN
20 kN
2.5 m
0.75 m
FA
0.75 m
x
SOLUTION
3m
3m
Equivalent Resultant Force: By equating the sum of the forces along the z axis to
the resultant force FR,
+ c FR = ©Fz;
- FR = -30 - 20 - 90 - FA - FB
FR = 140 + FA + FB
(1)
Point of Application: By equating the moment of the forces and FR, about the x and
y axes,
(MR)x = ©Mx;
- FR(3.75) = -FB(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - FA(6.75)
FR = 0.2FB + 1.8FA + 132
(MR)y = ©My;
(2)
FR(3.25) = 30(0.75) + 20(0.75) + 90(3.25) + FA(5.75) + FB(5.75)
FR = 1.769FA + 1.769FB + 101.54
(3)
Solving Eqs.(1) through (3) yields
FA = 30 kN
FB = 20 kN
FR = 190 kN
Ans.
0.75 m
y
4–134.
Replace the two wrenches and the force, acting on the pipe
assembly, by an equivalent resultant force and couple
moment at point O.
100 N · m
300 N
z
C
SOLUTION
O
0.5 m
Force And Moment Vectors:
45°
F3 = 5100j6 N
F2 = 2005cos 45°i - sin 45°k6 N
M 1 = 5100k6 N # m
M 2 = 1805cos 45°i - sin 45°k6 N # m
= 5127.28i - 127.28k6 N # m
Equivalent Force and Couple Moment At Point O:
FR = F1 + F2 + F3
= 141.42i + 100.0j + 1300 - 141.422k
= 5141i + 100j + 159k6 N
Ans.
The position vectors are r1 = 50.5j6 m and r2 = 51.1j6 m.
M RO = r1 * F1 + r2 * F2 + M 1 + M 2
i
= 30
0
j
0.5
0
i
0
+
141.42
k
0 3
300
j
1.1
0
k
0
-141.42
+ 100k + 127.28i - 127.28k
=
122i - 183k N # m
200 N
180 N · m
= 5141.42i - 141.42k6 N
M RO = ©M O ;
B
0.8 m
x
F1 = 5300k6 N
FR = ©F;
A
0.6 m
Ans.
100 N
y
4–135. Replace the parallel force system acting on
the plate by a resultant force and specify its location on the
x–z plane.
z
0.5 m
1m
2 kN
5 kN
1m
1m
y
3 kN
0.5 m
x
4–136.
The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the
magnitude F of the couple forces so that the system can be simplified to a wrench acting at
point C.
Given:
a
b
0.6 m
0.8 m
c
0.25 m
d
0.7 m
e
0.3 m
f
0.3 m
g
0.5 m
h
0.25 m
P
60 N
Q
40 N
Solution:
Initial Guess
F
1N
MC
1 N˜ m
Given
§¨ MC ·¸
¨ 0 ¸
¨ 0 ¸
©
¹
§ F ·
¨
¸
© MC ¹
ª P ( c h) º ª 0
« 0
»« 0
«
» «
¬ 0
¼ ¬ F ( e Find F MC
MC
º § a · § Q ·
» ¨b¸ u ¨ 0 ¸
» ¨ ¸ ¨ ¸
f) ¼ © 0 ¹ © 0 ¹
30 N˜ m
F
53.3 N
Ans.
4–137.
Replace the three forces acting on the plate by a wrench.
Specify the magnitude of the force and couple moment for
the wrench and the point P(x, y) where its line of action
intersects the plate.
z
FB
FA
{800k} N
{500i} N
A
B
y
P
x
y
x
6m
4m
SOLUTION
C
FR = {500i + 300j + 800k} N
FR = 2(500)2 + (300)2 + (800)2 = 990 N
FC
Ans.
uFR = {0.5051i + 0.3030j + 0.8081k}
MRx¿ = ©Mx¿;
MRx¿ = 800(4 - y)
MRy¿ = ©My¿;
MRy¿ = 800x
MRz¿ = ©Mz¿;
MRz¿ = 500y + 300(6 - x)
Since MR also acts in the direction of uFR,
MR (0.5051) = 800(4 - y)
MR (0.3030) = 800x
MR (0.8081) = 500y + 300(6 - x)
MR = 3.07 kN # m
Ans.
x = 1.16 m
Ans.
y = 2.06 m
Ans.
{300j} N
–.
The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent
resultant location, measured from point O.
*LYHQ
Z 30
Z 50
1
P
1
P
D 1P
E 1.5P
F 0.5P
*XHVVHV
5 11
G 1P
6ROXWLRQ
*LYHQ
Z˜ E Z˜ F = 5
·
§ E·
§F
Z˜ E˜ ¨ D ¸ Z˜ F˜ ¨ E D¸ = G˜ 5
2
2
©
¹
©
¹
§5 ·
¨ ¸ )LQG ( 5 G)
©G ¹
5
70 1
G
0.107 P
Ans.
4–139.
Replace the distributed loading with an equivalent resultant
force, and specify its location on the beam measured from
point O.
3 kN/m
O
3m
SOLUTION
Loading: The distributed loading can be divided into two parts as shown in Fig. a.
Equations of Equilibrium: Equating the forces along the y axis of Figs. a and b,
we have
+ T FR = ©F;
FR =
1
1
(3)(3) + (3)(1.5) = 6.75 kN T
2
2
Ans.
If we equate the moment of FR, Fig. b, to the sum of the moment of the forces in
Fig. a about point O, we have
a + (MR)O = ©MO;
1
1
(3)(3)(2) - (3)(1.5)(3.5)
2
2
x = 2.5 m
- 6.75(x) = -
Ans.
1.5 m
4–140.
Replace the loading by an equivalent force and couple
moment acting at point O.
200 N/m
O
4m
SOLUTION
Equivalent Force and Couple Moment At Point O:
+ c FR = ©Fy ;
FR = - 800 - 300
= - 1100 N = 1.10 kN T
a+ MRO = ©MO ;
Ans.
MRO = - 800122 - 300152
= - 3100 N # m
= 3.10 kN # m (Clockwise)
Ans.
3m
4–141.
Replace the loading by an equivalent resultant force and couple moment acting at point A.
Units Used:
3
kN
10 N
Given:
w1
600
w2
600
a
2.5 m
b
2.5 m
N
m
N
m
Solution:
FR
MRA
w1 a w2 b
w1 a§¨
a b·
¸
© 2 ¹
FR
MRA
0N
3.75 kN˜ m
Ans.
Ans.
.
.
4–143.
The masonry support creates the loading distribution
acting on the end of the beam. Simplify this load to a single
resultant force and specify its location measured from
point O.
0.3 m
O
1 kN/m
2.5 kN/m
SOLUTION
Equivalent Resultant Force:
+ c FR = ©Fy ;
FR = 1(0.3) +
1
(2.5 - 1)(0.3) = 0.525 kN c
2
Ans.
Location of Equivalent Resultant Force:
a + 1MR2O = ©MO ;
0.5251d2 = 0.30010.152 + 0.22510.22
d = 0.171 m
Ans.
.
.
4–145.
Replace the distributed loading by an equivalent resultant force, and specify its location on the
beam, measured from the pin at C.
Units Used:
3
kN 10 N
Given:
w 8000
N
m
a 5m
b 5m
T 30deg
Solution :
F R w˜ a w˜ b
2
F R˜ x = w˜ a˜
a w˜ b §
b·
˜ ¨a ¸
2
2 ©
3¹
w˜ a˜
x
a
2
FR
§
2 ©
w˜ b
FR
˜ ¨a 60 kN
Ans.
b·
¸
3¹
x
3.89 m
Ans.
4–146.
Replace the distributed loading with an equivalent resultant
force, and specify its location on the beam measured from
point A.
w0
w0
A
B
L
––
2
SOLUTION
Loading: The distributed loading can be divided into two parts as shown in Fig. a.
The magnitude and location of the resultant force of each part acting on the beam
are also shown in Fig. a.
Resultants: Equating the sum of the forces along the y axis of Figs. a and b,
+ TFR = ©F;
FR =
1
1
1
L
L
w a b + w0 a b = w0L T
2 0 2
2
2
2
Ans.
If we equate the moments of FR, Fig. b, to the sum of the moment of the forces in
Fig. a about point A,
a + (MR)A = ©MA;
1
L L
L 2
1
1
- w0L(x) = - w0 a b a b - w0 a b a Lb
2
2
2
6
2
2
3
x =
5
L
12
Ans.
L
––
2
4 –147.
The beam supports the distributed load caused by the sandbags. Determine the resultant force
on the beam and specify its location measured from point A.
Units Used:
3
kN
10 N
Given:
w1
1.5
w2
1
w3
2.5
kN
m
kN
m
kN
m
a
3m
b
3m
c
1.5 m
Solution:
FR
w1 a w2 b w3 c
MA
w1 a
MA
a
2
w2 b§¨ a 45.563 kN˜ m
©
FR
b·
§
¸ w3 c¨ a b ©
2¹
d
MA
FR
11.25 kN
Ans.
c·
¸
2¹
d
4.05 m
Ans.
4–148.
If the soil exerts a trapezoidal distribution of load on the
bottom of the footing, determine the intensities w1 and w2 of
this distribution needed to support the column loadings.
1m
50 kN
2.5 m
SOLUTION
3.5 m
1m
w2
Loading: The trapezoidal reactive distributed load can be divided into two parts
as shown on the free-body diagram of the footing, Fig. a. The magnitude and location measured from point A of the resultant force of each part are also indicated in
Fig. a.
Equations of Equilibrium: Writing the moment equation of equilibrium about
point B, we have
a + ©MB = 0; w2(8) ¢ 4 -
8
8
8
8
≤ + 60 ¢ - 1 ≤ - 80 ¢ 3.5 - ≤ - 50 ¢ 7 - ≤ = 0
3
3
3
3
w2 = 17.1875 kN>m = 17.2 kN>m
Ans.
Using the result of w2 and writing the force equation of equilibrium along the
y axis, we obtain
+ c ©Fy = 0;
80 kN
60 kN
1
(w - 17.1875)8 + 17.1875(8) - 60 - 80 - 50 = 0
2 1
w1 = 30.3125 kN>m = 30.3 kN>m
Ans.
w1
4–149.
Determine the length b of the triangular load and its position a on the beam such that the
equivalent resultant force is zero and the resultant couple moment is M clockwise.
Units Used:
3
kN
10 N
Given:
kN
w1
4
w2
M
8 kN˜ m
m
2.5
c
9m
a
1m
kN
m
Solution:
Initial Guesses:
1
Given
2
1
2
§a·
¨ ¸
©b¹
w1 b w1 b§¨ a ©
Find ( a b)
1
2
b
w2 c
2b ·
1m
0
1
2c
¸ w2 c
3¹ 2
3
§a·
¨ ¸
©b¹
M
§ 1.539 ·
¨
¸m
© 5.625 ¹
Ans.
4–150.
Replace the loading by an equivalent force and couple
moment acting at point O.
6 kN/m
500 kN m
O
7.5 m
SOLUTION
+ c FR = ©Fy ;
FR = -22.5 - 13.5 - 15.0
= -51.0 kN = 51.0 kN T
a + MRo = ©Mo ;
Ans.
MRo = -500 - 22.5(5) - 13.5(9) - 15(12)
= - 914 kN # m
= 914 kN # m (Clockwise)
Ans.
4.5 m
15 kN
4–151.
Replace the loading by a single resultant force, and specify
the location of the force measured from point O.
6 kN/m
500 kN m
O
7.5 m
SOLUTION
Equivalent Resultant Force:
+ c FR = ©Fy ;
- FR = -22.5 - 13.5 - 15
FR = 51.0 kN T
Ans.
Location of Equivalent Resultant Force:
a + (MR)O = ©MO ;
- 51.0(d) = -500 - 22.5(5) - 13.5(9) - 15(12)
d = 17.9 m
Ans.
4.5 m
15 kN
4–152. Replace the loading by an equivalent resultant
force and couple moment at point A.
150kN/m
lb/ft
50
lb/ft
1 kN/m
B
4 ftm
1.2
ft m
61.8
2 kN/m
100
lb/ft
60⬚
A
F3 = 1.2 kN
F1 = 0.9 kN
F2 = 1.8 kN
0.9 m
0.6 m
FRx = 2.338 kN
1
F1 =
(1.8) (1) = 0.9 kN
2
F2 = (1.8) (1) = 1.8 kN
F3 = (1.2) (1) = 1.2 kN
+
→ FRx
= ΣFx;
+↓FRy = ΣFy;
FR =
FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN
FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN
(2.338) 2 + (2.55) 2 = 3.460 kN
⎛ 2.55 ⎞
␪ = tan–1 ⎜
= 47.5°
⎝ 2.338 ⎟⎠
Ans.
MRA = 0.9 (0.6) + 1.8 (0.9) + 1.2 (1.8 cos 60° + 0.6)
= 3.96 kN · m
哭
哭
+ MRA = ΣMA;
Ans.
Ans.
(1.8 cos 60° + 0.6) m
FRy = 2.55 kN
4–153. Replace the loading by an equivalent resultant
force and couple moment acting at point B.
150
kN/m
lb/ft
lb/ft
150kN/m
B
4 ftm
1.2
6 ft m
1.8
100
lb/ft
2 kN/m
60⬚
A
F1 =
1
(1.8) (1) = 0.9 kN
2
F2 = (1.8) (1) = 1.8 kN
F3 = (1.2) (1) = 1.2 kN
+
→ FRx
= ΣFx;
+↓FRy = ΣFy;
FR =
FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN
FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN
(2.338) 2 + (2.55) 2 = 3.460 kN
⎛ 2.55 ⎞
␪ = tan–1 ⎜
= 47.5°
⎝ 2.338 ⎟⎠
Ans.
MRB = 0.9 cos 60° (1.2 cos 60° + 1.2)
+ 0.9 sin 60° (1.2 sin 60°)
+ 1.8 cos 60° (0.9 cos 60° + 1.2)
+ 1.8 sin 60° (0.9 sin 60°) + 1.2 (0.6)
MRB = 5.04 kN · m
哭
哭
+ MRB = ΣMB;
Ans.
Ans.
4–154.
Replace the distributed loading by an equivalent resultant
force and specify where its line of action intersects member
AB, measured from A.
200 N/m
100 N/m
B
C
6m
5m
SOLUTION
+ ©FRx = ©Fx ;
;
FRx = 1000 N
+ TFRy = ©Fy ;
FRy = 900 N
FR = 2(1000)2 + (900)2 = 1345 N
FR = 1.35 kN
u = tan-1 c
Ans.
900
d = 42.0° d
1000
a + MRA = ©MA ;
200 N/m
Ans.
1000y = 1000(2.5) - 300(2) - 600(3)
y = 0.1 m
Ans.
A
4–155.
Replace the distributed loading by an equivalent resultant
force and specify where its line of action intersects member
BC, measured from C.
200 N/m
100 N/m
B
C
6m
5m
SOLUTION
+ ©FRx = ©Fx ;
;
FRx = 1000 N
+ TFRy = ©Fy ;
FRy = 900 N
FR = 2(1000)2 + (900)2 = 1345 N
FR = 1.35 kN
u = tan-1 c
Ans.
900
d = 42.0° d
1000
a + MRC = ©MC ;
200 N/m
Ans.
900x = 600(3) + 300(4) - 1000(2.5)
x = 0.556 m
Ans.
A
4–156.
Replace the distributed loading with an equivalent resultant
force, and specify its location on the beam measured from
point A.
w
p x)
––
w ⫽ w0 sin (2L
x
A
SOLUTION
L
Resultant: The magnitude of the differential force dFR is equal to the area of the
element shown shaded in Fig. a. Thus,
dFR = w dx = ¢ w0 sin
p
x ≤ dx
2L
Integrating dFR over the entire length of the beam gives the resultant force FR.
L
+T
FR =
LL
dFR =
L0
¢ w0 sin
L
2w0L
2w0L
p
p
cos
T
x ≤ dx = ¢x≤ ` =
p
p
2L
2L
0
Ans.
Location: The location of dFR on the beam is xc = x measured from point A. Thus,
the location x of FR measured from point A is given by
L
x =
LL
xcdFR
LL
=
dFR
L0
x ¢ w0 sin
p
x ≤ dx
2L
2w0L
p
4w0L2
=
2L
p2
=
2w0L
p
p
w0
Ans.
4–157.
Replace the distributed loading with an equivalent resultant
force, and specify its location on the beam measured from
point A.
w
10 kN/m
1 (⫺x2 ⫺ 4x ⫹ 60) kN/m
w ⫽ ––
6
A
B
6m
SOLUTION
Resultant: The magnitude of the differential force dFR is equal to the area of the
element shown shaded in Fig. a. Thus,
dFR = w dx =
1
( - x2 - 4x + 60)dx
6
Integrating dFR over the entire length of the beam gives the resultant force FR.
6m
FR =
+T
LL
dFR =
L0
6m
1
1
x3
(-x2 - 4x + 60)dx = B - 2x2 + 60x R `
6
6
3
0
= 36 kN T
Ans.
Location: The location of dFR on the beam is xc = x, measured from point A. Thus
the location x of FR measured from point A is
6m
x =
LL
xcdFR
LL
=
dFR
= 2.17 m
L0
1
x B (- x2 - 4x + 60) R dx
6
36
6m
=
L0
6m
x4
4x3
1
1
+ 30x2 ≤ `
¢- (-x3 - 4x2 + 60x)dx
6
4
3
0
6
=
36
36
Ans.
x
4–158.
Determine the equivalent resultant force
and couple moment at point O.
Units Used:
2
3
kN
10 N
Given:
a
3m
wO
w ( x)
3
kN
m
wO §¨
x·
2
¸
© a¹
Solution:
FR
´a
µ w ( x) dx
¶0
FR
3 kN
MO
´a
µ w ( x) ( a x) dx
¶0
MO
2.25 kN˜ m
Ans.
Ans.
4–159.
Wet concrete exerts a pressure distribution along the wall
of the form. Determine the resultant force of this
distribution and specify the height h where the bracing strut
should be placed so that it lies through the line of action of
the resultant force. The wall has a width of 5 m.
p
4m
p
SOLUTION
Equivalent Resultant Force:
h
z
+ F = ©F ;
:
R
x
-FR = - LdA = 4m
L0
8 kPa
wdz
a20z2 b A 103 B dz
1
FR =
L0
= 106.67 A 103 B N = 107 kN ;
z
Ans.
Location of Equivalent Resultant Force:
z
z =
LA
zdA
=
dA
LA
zwdz
L0
z
L0
wdz
4m
zc A 20z2 B (103) ddz
1
=
L0
4m
A 20z2 B (103)dz
1
L0
4m
c A 20z2 B (10 3) ddz
3
=
L0
4m
A 20z2 B (103)dz
1
L0
= 2.40 m
Thus,
h = 4 - z = 4 - 2.40 = 1.60 m
Ans.
1
(4 z /2) kPa
4–160.
Replace the loading by an equivalent force and couple
moment acting at point O.
w
1
––
w = (200x 2 ) N/m
x
O
9m
SOLUTION
Equivalent Resultant Force And Moment At Point O:
+ c FR = ©Fy ;
x
FR = -
LA
dA = -
9m
FR = -
L0
wdx
A 200x2 B dx
1
L0
= - 3600 N = 3.60 kN T
Ans.
x
a + MRO = ©MO ;
MRO = -
L0
xwdx
9m
= -
1
L0
9m
= -
x A 200x2 B dx
A 200x2 B dx
3
L0
= - 19 440 N # m
= 19.4 kN # m (Clockwise)
Ans.
600 N/m
.
.
4–162.
Determine the equivalent resultant force of the distributed
loading and its location, measured from point A. Evaluate
the integral using Simpson’s rule.
w
w
5x
(16
x2)1/2 kN/m
5.07 kN/m
2 kN/m
A
3m
SOLUTION
4
FR =
L
wdx =
1
L0
45x + (16 + x2)2 dx
FR = 14.9 kN
4
L0
Ans.
4
x dF =
1
L0
(x) 45x + (16x + x2 )2 dx
= 33.74 kN # m
x =
33.74
= 2.27 m
14.9
Ans.
x
B
1m
4–163.
Determine the resultant couple moment of the two couples
that act on the assembly. Member OB lies in the x-z plane.
z
y
A
400 N
O
500 mm
150 N
SOLUTION
x
45°
For the 400-N forces:
600 mm
M C1 = rAB * 1400i2
i
= 3 0.6 cos 45°
400
j
-0.5
0
k
- 0.6 sin 45° 3
0
B
400 mm
C
= - 169.7j + 200k
150 N
For the 150-N forces:
M C2 = rOB * 1150j2
i
= 3 0.6 cos 45°
0
j
0
150
k
-0.6 sin 45° 3
0
= 63.6i + 63.6k
M CR = M C1 + M C2
M CR =
63.6i - 170j + 264k N # m
Ans.
400 N
4–164.
The horizontal 30-N force acts on the handle of the wrench.
What is the magnitude of the moment of this force about the
z axis?
z
O
x
Position Vector And Force Vectors:
rBA = { -0.01i + 0.2j} m
rOA = [( - 0.01 - 0)i + (0.2 - 0)j + (0.05 - 0)k} m
= { -0.01i + 0.2j + 0.05k} m
F = 30(sin 45°i - cos 45° j) N
= [21.213i - 21.213j] N
Moment of Force F About z Axis: The unit vector along the z axis is k. Applying
Eq. 4–11, we have
Mz = k # (rBA * F)
0
= 3 - 0.01
21.213
0
0.2
- 21.213
1
03
0
= 0 - 0 + 1[(- 0.01)(- 21.213) - 21.213(0.2)]
= -4.03 N # m
Ans.
Or
Mz = k # (rOA * F)
0
= 3 - 0.01
21.213
0
0.2
- 21.213
1
0.05 3
0
= 0 - 0 + 1[(- 0.01)(- 21.213) - 21.213(0.2)]
= -4.03 N # m
The negative sign indicates that Mz, is directed along the negative z axis.
Ans.
A
45
45
10 mm
50 mm
SOLUTION
30 N
200 mm
B
y
4–165.
z
The horizontal 30-N force acts on the handle of the wrench.
Determine the moment of this force about point O. Specify
the coordinate direction angles a, b, g of the moment axis.
200 mm
B
10 mm
50 mm
O
SOLUTION
x
Position Vector And Force Vectors:
rOA = {(- 0.01 - 0)i + (0.2 - 0)j + (0.05 - 0)k} m
= {-0.01i + 0.2j + 0.05k} m
F = 30(sin 45°i - cos 45°j) N
= {21.213i - 21.213j} N
Moment of Force F About Point O: Applying Eq. 4–7, we have
MO = rOA * F
i
= 3 -0.01
21.213
j
0.2
-21.213
k
0.05 3
0
= {1.061i + 1.061j - 4.031k} N # m
= {1.06i + 1.06j - 4.03k} N # m
Ans.
The magnitude of MO is
MO = 21.0612 + 1.0612 + (-4.031)2 = 4.301 N # m
The coordinate direction angles for MO are
a = cos - 1 a
1.061
b = 75.7°
4.301
Ans.
b = cos - 1 a
1.061
b = 75.7°
4.301
Ans.
g = cos - 1 a
-4.301
b = 160°
4.301
Ans.
A
30 N
45⬚
45⬚
y
4–166.
The forces and couple moments that are exerted
on the toe and heel plates of a snow ski are
Ft = 5- 50i + 80j - 158k6N, M t = 5- 6i + 4j + 2k6N # m, and
Fh = 5- 20i + 60j - 250k6 N, M h = 5- 20i + 8j + 3k6 N # m,
respectively. Replace this system by an equivalent force and
couple moment acting at point P. Express the results in
Cartesian vector form.
z
P
Fh
Ft
O
Mh 800 mm
Mt
120 mm
y
x
SOLUTION
FR = Ft + Fh = { -70i + 140j - 408k}
i
MRP = 3 0.8
- 20
j
0
60
i
k
0 3 + 3 0.92
-50
-250
j
0
80
Ans.
N
k
0 3 + ( - 6i + 4j + 2k) + (-20i + 8j + 3k)
-158
MRP = (200j + 48k) + (145.36j + 73.6k) + (- 6i + 4j + 2k) + ( -20i + 8j + 3k)
MRP = {- 26i + 357.36j + 126.6k} N # m
MRP = {- 26i + 357j + 127k} N # m
Ans.
.
.
.
.
.
4–170.
Determine the moment of the force Fc about the door hinge
at A. Express the result as a Cartesian vector.
z
C
1.5 m
2.5 m
FC
250 N
a
SOLUTION
Position Vector And Force Vector:
30
A
rAB = {[-0.5 - (- 0.5)]i + [0 - ( - 1)]j + (0 - 0)k} m = {1j} m
FC = 250 §
[ -0.5 - (-2.5)]i +
{0 - [ - (1 + 1.5 cos 30°)]}j + (0 - 1.5 sin 30°)k
[ -0.5 - (-2.5)]2 +
B {0 - [ - (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°)2
a
x
Moment of Force Fc About Point A: Applying Eq. 4–7, we have
MA = rAB * F
i
0
159.33
j
1
183.15
k
0 3
- 59.75
= {- 59.7i - 159k}N # m
1m
0.5 m
¥N
= [159.33i + 183.15j - 59.75k]N
= 3
B
Ans.
y
4–171.
Determine the magnitude of the moment of the force Fc
about the hinged axis aa of the door.
z
C
1.5 m
2.5 m
FC
250 N
a
SOLUTION
Position Vector And Force Vectors:
30
A
rAB = {[- 0.5 - (- 0.5)]i + [0 - ( -1)]j + (0 - 0)k} m = {1j} m
FC = 250 §
{- 0.5 - (- 2.5)]i +
{0 - [ - (1 + 1.5 cos 30°)]}j + (0 - 1.5 sin 30°)k
[- 0.5 - (- 2.5)]2 +
B{0 - [ - (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°)2
a
x
Moment of Force Fc About a - a Axis: The unit vector along the a – a axis is i.
Applying Eq. 4–11, we have
Ma - a = i # (rAB * FC)
1
0
159.33
0
1
183.15
0
0 3
-59.75
= 1[1(- 59.75) - (183.15)(0)] - 0 + 0
= -59.7 N # m
The negative sign indicates that Ma - a is directed toward the negative x axis.
Ma - a = 59.7 N # m
1m
0.5 m
¥N
= [159.33i + 183.15j - 59.75k] N
= 3
B
Ans.
y
4–172.
The ends of the triangular plate are
subjected to three couples. Determine
the magnitude of the force F so that
the resultant couple moment is M
clockwise.
Given:
F1
600 N
F2
250 N
a
1m
T
40 deg
M
400 N˜ m
Solution:
Initial Guess
Given
F1
F
1N
§ a ·F aF
¨
¸
2
© 2 cos T ¹
§ a ·
¨
¸
© 2 cos T ¹
M
F
Find ( F)
F
830 N
Ans.
4–173. The tool is used to shut off gas valves that are
difficult to access. If the force F is applied to the handle,
determine the component of the moment created about the
z axis of the valve.
z
0.25 m
F ⫽ {⫺60i ⫹ 20j ⫹ 15k} N
0.4 m
.
30⬚
x
y