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 = A x i + A y j + Az k B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k A * (B + D) =
3
i Ax
j Ay
Bx + Dx
By + Dy
k Az Bz + Dz
3
= [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
3
i = Ax Bx
j Ay By
3 3
k i Az + Ax Bz Dx
= (A * B) + (A * D)
j Ay Dy
k Az Dz
3 (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,
` a
|h| = |A # u(B * C)| = A #
B * C |B * C|
b`
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)
= (A x i + A y j + A z k) # Bix
3
Cx
Bjy Cy
Bkz Cz
3
= 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
3
i = Ax Bx
j Ay By
k Az Bz
3
# (Cx i
+ Cy j + Cz k)
= 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 9 m
1.5 m θ
A
SOLUTION
1 21 2 5- 1177.2 cos 17.5 + 26 N # m
a + MA = - 120 9.81 7.5 + x cos u =
x
u
= 1.18 cos u 7.5 + x
kN # m (Clockwise)
5 1 26 The maximum moment at occurs when = 0° and = 5 - 1177.2 cos 0°17.5 + 526 N # m a+ 1 2 A
MA
u
x = 5 m.
Ans. Ans.
max
= - 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
m 2
m 3
4m
SOLUTION
he moment arm measured perpendicular to each force from point A is d1 = 2 sin 60° = 1.732 m
B
4
5 3
d2 = 5 sin 60° = 4.330 m
F3
d3 = 2 sin 53.13° = 1.60 m
Using each force where MA = Fd, we have
1 2
a + MF1
1 2
a + MF2
1 2
a + MF3
A
1 2
= - 250 1.732
= - 433 N # m = 433 N # m (Clockwise) A
= - 1299 N # m = 1.30 kN # m (Clockwise) A
Ans.
1 2
= - 300 4.330
Ans.
1 2
= - 500 1.60
= - 800 N # m = 800 N # m (Clockwise)
Ans.
500 N
4–8.
Determine the moment of each of the three forces about point
B.
F1
F2
250 N 30
300 N
60
A
m 2
m 3
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
For F , 2 a + MB = 300 sin 60°(0) + 300 cos 60°(4) = 600 N m d Since the line of action of F3 passes through zero.Thus
Ans.
=0
4
5 3
F3 500 N
#
MB
B
Ans. B,
its moment arm about point
B
is
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
0.5 m
0.75 m B
SOLUTION
+ (MR)A =
g
Fd;
0.25 m
(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 )
An s.
1m
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 a + (MR)B =
g
Ga
Fd;
(MR)B
= 100(9.81)(2.5) - 250(9.81)(0.5)
) = 1226.25 N # m = 1.23 kN # m (Counterclockwise
GW
0.5 m
0.75 m B
An s.
0.25 m
1m
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
60 deg
Solution:
(a)
MA F sin c F cos a
(b)
F C ( a) M A
FC
MA a
M A 73.9N m
Ans.
F C 82.2 N
Ans.
4 –14
.
The force F acts on the end of the pipe at B. Determine the angles ( 0° 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:
M A F sin c F cos a
For maximum moment
d d
max
MA c F cos a F sin 0
atan
c a
MAmax F sin max c F cos max a For minimum moment
min
37.9 deg
Ans.
MAmax 79.812N m
Ans.
128deg
Ans.
max
M A F sin c F cos a 0
a c
180 deg atan
MAmin F c sin min F ( a) cos min
min
MAmin 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)
65 mm
Ans.
100 mm
Nf
400 N
4–16.
The Achilles tendon force Ft is mobilized whenthe 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 F
= =
400(0.1) 618 N
-
F cos 5°(0.065) 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°.
1 2
a + MA = 120 sin 5° 0.04
10°
= 0.418 N # m (Counterclockwise)
Ans. B
Moment About Point B: The dimension l can be determined using the law of sines. l sin 150°
=
55 sin 10°
l
= 158.4 mm = 0.1584 m
Then,
1
2
a + MB = - 120 sin 15° 0.1584
= - 4.92 N # m = 4.92 N # m (Clockwise)
Ans.
15 mm
4–18.
The tower crane is used to hoist the 2-Mg load upward at constant velocity. The 1.5-Mg jib BD, 0.5-Mg ji b 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.
4m G2
9.5m B
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;
D
C G3
(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 ) 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
7.5 m
12.5 m
23 m
SOLUTION a + (MR)A = © Fd;
A
0 MC
D
C G3
= MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) = 4966.67 kg = 4.97 Mg 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 Nm
60 deg
30 deg
a 50 mm b 300 mm Solution:
M F cos a b sin F sin b cos
F
M cos a
b sin sin b cos
F 77.6 N
Ans.
4 –21
.
Determine the angle (0 <= <= 90 deg) so that the force about point O. Given: F
M
100 N 20 N m
60 deg
a
50 mm
b
300 mm
Solution: I nitial
Guess
30 deg
Given
a b sin F sin b cos
M
F cos
Find
28.6 deg
Ans.
F
develops a clockwise moment M
4–22.
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 atB 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
C
50 N
SOLUTION B
c + M A = 50 sin 60°(0.3) MA
= 12.99 = 13.0 N # m
a + M A = 0; F
300 mm
= 35.2 N
- 12.99 +
F
Ans.
a 1213 b (0.4) = 0 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
A
SOLUTION Maximum moment, OB
x
BA
= - 4kN(20) = 80 kN # m b a + (M ) 4 kN sin 60°(x) - 4 kN cos 60°(1.5) = 80 kN # m O max
x
= 24.0 m
Ans.
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
A
SOLUTION Maximum moment, OB
x
BA
= 4000(20) = 80 000 N # m = 80.0 kN # m
c + (MO)max
4000 sin f(25)
Ans.
- 4000 cos f(1.5) = 80 000
- 1.5 cos f = 20
25 sin f f
= 56.43°
u
= 90° - 56.43° = 33.6°
Ans.
Also, (1.5)2 2.25
+
+ z
z2 2
=
= y
y2 2
Similar triangles 20
+
y
z
20y
+
25
=
y2
+
z
y
= 25z +
20( 22.25
+
z
= 2.260 m
y
= 2.712 m
u
= cos -1
2
z)
z2
+ 2.25 +
z2
a 2.260 b = 33.6° 2.712
= 25z +
z2
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
103 N
Given:
F 1 250 N
a 2m
F 2 300 N
b 3m
F 3 500 N
c 4m
1
60 deg
d 3
2
30 deg
e 4
Solution Using Whole Forces:
Geometry
atan
d e
d
L ab
e
c
e 2
2
e d
M A F 1 ( a) cos 2 F 2 ( a b) sin 1 F3 L
M A 2.532 kN m
Ans.
Solution Using Principle of Moments:
MA F 1 cos 2 a F2 sin 1 ( a b) F 3
d 2
2
d e
c F3
e 2
2
(a
b)
d e
3
MA 2.532 10 N m
Ans.
4 – 26
.
If the resultant moment about point A is M clockwise, determine the magnitude of F 3. Units Used: kN
103 N
Given:
M 4.8 kN m a 2 m F 1 300 N
b 3m
F 2 400 N
c 4m
1
60 deg
d 3
2
30 deg
e 4
Solution: Initial Guess
F3 1 N
Given
M F 1 cos 2 a F 2 sin 1 ( a b) F 3
F 3 Find F3
F 3 1.593 kN
Ans.
d 2
2
d e
c F 3
e 2
2
d e
( a b)
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 atA.
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)
An s.
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 sho wn. If a clockwise mom ent 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 A
of these two components about point moments.
a + (MR)A = © Fd;
- 120 =
in accordance with the principle of
F sin 15°(0.3 sin 30°) F
= 239 N
-
F cos 15°(0.3 cos 30°
+ 0.3) Ans.
F
4–30.
A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment ofF about point A, using M A = rB : F and M A = rC : F.
z
F
C
200 mm rC
SOLUTION F F
MA
= 100
400 mm
a - 0.4
i
+ 0.6 j + 0.2 k 0.7483
5
B
b
F
6
x
= - 53.5 i + 80.2 j + 26.7 k N = rB * F =
3
i 0 - 53.5
j
- 0.6 80.2
k 0 26.7
3
= - 16.0 i - 32.1 k N # m
5
6
Ans.
Also, i MA
= rC * F =
- 0.4 - 53.5
j 0 80.2
k 0.2 26.7
rB
600 mm
= - 16.0 i - 32.1 k N # m
Ans.
A
y
4–31.
5
6
The force F = 600i + 300 j - 600k N acts at the end of the beam. Determine the moment of the force about point A .
z
A
x
O
F
SOLUTION 1.2 m
r
= {0.2i + 1.2 j} m
3
i 0.2 600
B 0.4 m
j 1.2 300
k 0 - 600
3
MO
= r* F=
MO
= { - 720i + 120 j - 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
= [2.5 j] m
=
FB u FB
B
x
- 0)i + (2.5 - 0)j + (0 - 6)k = 780 2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2 (0
R=
[300 j
- 720k] N
Vector Cross Product: The moment of FB about point O is given by
MO
3
i
= rOA * FB = 0 0
j 0 300
k 6 - 720
j 2.5 300
k 0 - 720
3
= [ - 1800i] N # m = [ - 1.80i] kN # m
Ans.
= [ - 1800i] N # m = [ - 1.80i] kN # m
Ans.
or
MO
3
i
= rOB * FB = 0 0
2.5 m
3m
The force vector FB is given by FB
2m
C
3
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
rOC
= (2 - 0)i + ( - 3 - 0)j + (0 - 0)k = [2i - 3 j] m
=
FCuFC
= 420 B
x
(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 MO
=
rOA
*
FC
=
3
i 0 120
j 0 - 180
k 6 = [1080i + 720j] N m - 360
3
#
rOC
*
FC
=
3
i 2 120
j
=
-3 - 180
k 0 = [1080i + 720j] N m - 360
3
#
Ans.
or
MO
2.5 m
3m
The force vector FC is given by FC
2m
C
Ans.
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
- 0)i + (2.5 - 0) j + (0 - 6)k = 780 2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2
FC
=
FCu FC
= 420
B B
(0
(2
- 0)i + ( - 3 - 0) j + (0 - 6)k
2(2 - 0)2 + ( - 3 - 0)2 + (0 - 6)2
R= R=
[300 j
- 720k] N
[120i
- 180 j - 360k] N
Resultant Moment: The resultant moment of FB and FC about point O is given by MO
=
3
i 0 0
j 0 300
= rOA * FB + rOA * FC k 6 - 720
3 3 +
i 0 120
j 0 - 180
= [ - 720i + 720 j] N # m
k 6 - 360
3 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 2 m
1.5 m
SOLUTION
MA
MA dMA du
sin u cos u Mmax Mmin
3
i
j 1.5 75 cos u
= 2 0
k 0 75 sin u
3
75 N θ
= 112.5 sin u i - 150 sin u j + 150 cos u k = 2 112.5 sin u 2 + - 150 sin u 2 + 150 cos u
1
=
= 0;
y
x
2 1
1 12 656.25 sin2 u 2
2 1 2 = 212 656.25 sin + 22 5002 112 656.25212 sin cos 2 = 0
= 0°, 90°, 180°
Ans.
1
u
= 187.5 N # m at u = 90° = 150 N # m at u = 0°, 180°
- 12
2
2
u
u
u
+ 22 500
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
= {1i - 3 j - 2k} m
rAC
= 2(1)2 + ( - 3)2 + ( - 2)2 = 3.742 m
MO
= rOC * F =
MO
3
i 4 1 3.742 (80)
A
1m
rAC
F
j 0
-
3 3.742 (80)
= { - 128i + 128 j - 257k} N # m
k
-2 -
2 3.742 (80)
3
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 - 3 j - 2k} m
rAC
= 2(1)2 + ( - 3)2 + ( - 2)2 = 3.742 m
MB
A
1m
F
i MB
3m
j
45° (3 - 3sin45°) = rBA * F = 3 cos 1 3 (80) - 3.742 3.742 (80)
3
= { - 37.6i + 90.7 j - 155k} N # m
k
0
2 (80) - 3.742
3
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 * r,F a. Here BCig. rAC
= (0 - 0)i + (4 - 0) j + (0 - 3)k = [4 j - 3k] m
rBC
= (0 - 3)i + (4 - 0) j + (0 - 0)k = [ - 3i + 4 j] m
Thus,
b
= rCA * rCB =
3
i 0 -3
j 4 4
k
-3 0
3
= [12i + 9 j + 12k] m2 Then, uF
=
b b
12i
=
+ 9 j + 12k
212 2 + 92 + 12 2
= 0.6247i + 0.4685 j + 0.6247k
And finally F
= FuF = 400(0.6247i + 0.4685 j + 0.6247k) = [249.88i + 187.41 j + 249.88k] N
Vector Cross Product: The moment of F about point A is
MA
= rAC * F =
3
i 0 249.88
j 4 187.41
k
-3 249.88
3
= [1.56i - 0.750 j - 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 y
4m
Force Vector:Since force F is perpendicular to the inclined plane, its unit vector uF ise qual to the unit vector of the cross product, b = rAC * r,F a. Here BCig. rAC
= (0 - 0)i + (4 - 0) j + (0 - 3)k = [4 j - 3k] m
rBC
= (0 - 3)i + (4 - 0)j + (0 - 0)k = [ - 3k + 4 j] m
Thus,
= rCA * rCB =
b
3
i 0 -3
j 4 4
k
3
- 3 = [12i + 9 j + 12k] m2 0
Then, 500 N
uF
=
b b
=
+ 9 j + 12k = 0.6247i + 0.4685 j + 0.6247k 2122 + 92 + 122 12i
And finally F
= FuF = 500(0.6247 i + 0.4685 j + 0.6247k) = [3123. 5 i + 2342. 5 j + 3123. 5 k] N
Vector Cross Product: The moment of F about point B is
MB
= rBC * F =
3
i
-3 3123 . 5
j 4 2342 . 5
k 0 3123 . 5
3
= [1.25 i + 0.937 j - 1.95 k] 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 + 0.4 j - 0.2k} m F
40
= 80(cos 30° sin 40°i + cos 30° cos 40° j - sin 30°k) N
30
= (44.53i + 53.07 j - 40.0k} N
F
Moment of ForceF About Point A:Applying Eq. 4–7, we have MA
= rAC * F
3
i
= 0.55 44.53
j 0.4 53.07
k
- 0.2 - 40.0
3
= { - 5.39i + 13.1 j + 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 F
40
= 80 (cos 30° sin 40°i + cos 30° cos 40° j - sin 30°k) N
30
= (44.53i + 53.07 j - 40.0k} N
F
Moment of ForceF About Point B:Applying Eq. 4–7, we have MB
= rBC * F
3
i
= 0.55 44.53
j 0 53.07
k
- 0.2 - 40.0
3
= {10.6i + 13.1 j + 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
F = 450 N O
SOLUTION
y 0.5 m
Position Vector And Force Vector: rOB
30°
510 - 02 + 11 cos 30° - 02 + 11 sin 30° - 02 6 m 5 6 = 510.5 sin 30° - 02 + 10.5 + 0.5 cos 30° - 02 + 10 - 02 6 m = 50.250 + 0.9330 6 m 10 - 0.5 sin 30°2 + 31 cos 30° - 10.5 + 0.5 cos 30°24 + 11 sin 30° - 02 ≤ N = 450 ¢ 210 - 0.5 sin 30°2 + 31 cos 30° - 10.5 + 0.5 cos 30° 24 + 11 sin 30° - 02 = 5 - 199.82 - 53.54 + 399.63 6 N =
i
j
k
x
= 0.8660 j + 0.5k m
rOA
i
i
F
j
k
j
i
j
2
i
j
k
2
2
k
Moment of ForceF About Point O:Applying Eq. 4–7, we have MO
= rOB * F =
3
i 0 - 199.82
j 0.8660 - 53.54
k 0.5 399.63
= 373i - 99.9 j + 173k N # m
5
6
3 Ans.
Or MO
= rOA * F =
i 0.250 - 199.82
j 0.9330 - 53.54
k 0 399.63
= 373i - 99.9 j + 173k N # m
Ans.
A
–
.
Determine the moment of the force at A about point O. Express the result as a cartesian vector. Units Used:
103
13
6
2.5
3
3
8
6
4
8
84
8
39
Ans.
–
.
Determine the moment of the force at A about point P . Express the result a s a cartesian vector. Units Used:
103
13
6
2.5
3
3
8
6
4
8
116
16
135
Ans.
4–45.
5
6
A force of F = 6i - 2 j + 1k kN produces a moment of M O = 4i + 5 j - 14k kN m about the ori gin 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.
5
6
#
z
F
P MO z
d
y
O 1m
SOLUTION MO
4i
y
= r* F
3
i
j y -2
+ 5 j - 14k = 1 6
4
=
5
= - 1 + 6z
y
x
3
k z 1
+ 2z
- 14 = - 2 - 6y y
= 2m
Ans.
z
= 1m
Ans.
4–46.
5
6
The force F = 6i + 8 j + 10k N creates a moment about point O of M O = - 14i + 8 j + 2k N m. the Ifforce passes through a point having an x coordinate of 1 m, determine they 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.
5
#
6
z
F
P MO z
d
y
O 1m
SOLUTION
- 14i + 8 j + 2k =
3
i 1 6
j y 8
k z 10
y
3
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 14.14
= 1.15 m
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
= - 1.5 j m
r2
= r3 = 0
F1
= - 60k N
5 6
F2
= 85i N F3 = 45 j N
5 6
5 6
Unit Vector Along AB Axis: uAB
12 - 02 + 10 - 1.52 1 2 + 10 - 1.52 i
=
2 2 - 0
j
2
2
= 0.8i - 0.6 j
Moment of Eac h Force About AB Axis: Applying Eq. 4–11, we have
1 2 MAB
1
= uAB # r1 * F1
1
=
3
2
- 0.6 - 1.5
0.8 0 0
0 0 - 60
0
3
= 0.8 - 1.5 - 60 - 0 - 0 + 0 = 72.0 N # m
1 2 MAB
2
31 21 2 4 #1 * 2
= uAB r2 =
1 2 MAB
3
3 3
F2
- 0.6
0.8 0 85
Ans.
0 0 0
=0
Ans.
0 0 0
=0
Ans.
3 2 3
0 0
= uAB # r3 * F3 =
1
- 0.6
0.8 0 0
0 45
b) Scalar Analysis:Since moment arm from force F2 and F3 is equal to zero,
1 2 =1 2 MAB
2
MAB
3
Moment arm d from force F1 to axis AB is d MAB
1
=
F1d
=0
Ans.
= 1.5 sin 53.13° = 1.20 m,
= 60 1.20 = 72.0 N # m
Ans.
2 m
4–48. Determine the moment produced by force F about the diagonal AF of the rectangular block. Express the result as a Cartesian vector.
F {6i 3 j 10k} N z B
A D
C O
x
3m
1.5 m
G
3m F
y
4–49. Determine the moment produced by forceF about the diagonal OD of the rectangular block. Express the result as a Cartesian vector.
F {6i 3j 10k} N 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
a 4m b 2.5 m c 1m d 0.5 m e 2m f 1.5 m g 2m
Solution:
g rCA
e 0
b g
r uCA
CA rCA
rCD
e a
MCA rCD F uCA
M CA 226N 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 M
30 N 14 N m
a
0.25 m
b
0.3 m
c
0.5 m
d
0.1 m
Solution: F
2
Mx
c
Mx
12N m
Mx
M
2
b
No
Ans.
For Mxmax, apply force perpendicular to the handle and the x-axis. M xmax
Fc
M xmax
15N m
M xmax
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
M
30 N
14 N m
a
0.25 m
b
0.3 m
c
0.5 m
Solution:
Mxmax
F1
a
Mx
Mx
18N m
Mx
M
c
2
c
c
2
b
Ans.
Yes
occurs when force is applied perpendicular to both the handle and the x-axis. Mxmax
F 1( a
Mxmax
22.5 N m
Mxmax
M
c)
Yes
Ans.
d
0.1 m
4–54.
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.
z
F
60
250 mm y
SOLUTION
x
250 mm
Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to deter mine the moment of F about the x axis. rAB
= (0.25 - 0.25)i + (0.25 - 0) j + (0 - 0)k = {0.25 j} m
The force vector F, Fig. a, can be written as F
= 100(cos 60° j - sin 60°k) = {50 j - 86.60k} N
Knowing that the unit vector of the about the x axis is given by Mx
3
1
= i # rAB * F = 0 0
0 0.25 50
x
axis is i, the magnitude of the moment of F
0 0 - 86.60
3
= 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.
4–55.
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.
#
z
F
60
250 mm y
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 = {0.25 j} m
The force vector F, Fig. a, can be written as F
= F(cos 60° j - sin 60°k) = 0.5F j - 0.8660Fk
Knowing that the unit vector of the about the x axis is given by 1
Mx
= i # rAB * F = 30 0
0 0.25 0.5F
x
axis is i, the magnitude of the moment of F
0 3 0 - 0.8660F
= 1[0.25( - 0.8660F) - 0.5F(0)] + 0 + 0 = - 0.2165F
Ans. M
x
#
The negative sign indicates that xM isxdirected negative from axis. The magnitude of F required to produce m can bethe determined = 30 N towards 30 F
= 0.2165F = 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) + = 139 N
F
F cos 60°(0)
Ans.
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
My
=
uy
=
3
# (r * F)
0 30cos40° 6
30 mm
1 0 -4
0 30sin40° -7
= 276.57 N # mm = 0.277 N # m
3
40 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 rz 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 magnitudes of moment of F about x and z axes are given by
Mx
=
#
i rx
*
F
3
1
= 0 6
0 0 -4
0 0.03sin40° -7
k
respectively. Thus, the
3
= 1[0( - 07) - ( - 4)(0.03 sin 40°)] - 0 + 0 = 0.07713 kN # m = 77.1 N # m
Mz
=
#
k rz
*
F
3
0
0
= 0.03cos40° 6
0
-4
1 0 7
3
= 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
If F = 450 N, determine the magnitude of the moment produced by this force about the x axis. 4–58.
z
F
45
A
B
60 60
100 mm
x
300 mm
150 mm
y
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. 4–59.
#
z
F
45
A
B
60 60
100 mm
x
300 mm
150 mm
y
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.
4–60.
z
A y 500 mm
F
150 mm
B 45
x
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 = - 5i N m. Both the pipe assembly ABC and the wrench lie in the x - y plane. Suggestion: Use a scalar analysis.
5 6
z
#
A y 500 mm
F
150 mm
B 45
x
200 mm
C
.
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.
4–62.
z
A
4m
m
y
m
x
3
m
C
2m
#
kN m
#
kN m
.
B
F {4i 12 j 3k} kN
.
#
.
#
.
kN m
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 12 j 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 tighte n the elbow fitti ng. 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
x
300 mm 300 mm 30
FA
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 My
3
0
1 0 0
= j # rCB * FB = - 0.15 75
0
- 0.2598 -129.90
3
= 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.
Mo men t o f 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 0.2588FA
3
+ 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°)
= - 38.97 N # m
- 150 sin 60°(0.3 sin 30°) 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
= FA cos 15°(0.3 cos 30°) = 184 N
FA sin 15°(0.3 sin 30°)
Ans.
135
120
A
Referring to Fig. a, the force vector FB can be written as
30
F
B
B
4–65.
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?
z
50 mm 50 mm
#
y
x
SOLUTION
300 mm 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
=
- sin 60°k) = 0.5FBi - 0.8660FBk
FB(cos 60°i
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
#
= j # rCB * FB
My
†
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 th e y Ax is :The posit ion vec tor rDA, Fig. a, wil l 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
=
50 FA
†
0 0.15 - 0.9659FA
1 0 0
0
- 0.2598 0.2588FA
†
= 0 - 1[0.15(0.2588FA) - ( - 0.9659FA)( - 0.2598)] + 0 = 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 = 192 N
FB sin 60°(0.3 sin 30°)
Ans.
For FA, we can write My
= © My;
50 FA
= FA cos 15°(0.3 cos 30°) = 236 N
FA sin 15°(0.3 sin 30°)
Ans.
30
FA
30
135
120
A
F
B
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
30 deg
F
3 N
5
8
Solution:
r
0
( a) cos
( a) sin
1 i
0
0
0.211 Mx
r F i
i
Mx
0 0
N m
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 ;
1 2=4
F 0.03
F
= 133 N
Ans.
For the blade, MC
= © Mx ;
1 2=4
P 0.005
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 ;
1
- 350 = 200 d
d cos 30°
= 1.54 m
2 - 6001
2 - 100
d sin 30°
200 N
d 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;
500(50) F
= 625 N
45 mm
- F (40) = 0 Ans.
50 mm 500 N
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. 4–70. If
#
300 N 300 mm
15
F
u
30 F
30
u 300 N
15
F = 200 N, determine the required angle u so that the resultant couple moment is zero. The dis khas a radius of 300 mm.
4–71. If
Ans.
4–72.
Friction on the concrete surface creates a couple mom ent of MO = 100 N m on the blade s of the trowe l. Determi ne the magnitude of the couple forces so that the resultant couple moment on the trowel is zero . The forces lie in the hor izontal 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.
150mm
150mm
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 openedwith a couplemomentof25 N m , determine the required magnitude of each couple force which must be applied to the wheel.
150mm
150mm
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 M 3 and the bend angle so that the r esultant couple moment is zero.
Given:
45 deg
1
M1 900 N m M2 500 N m
Solution: Initial guesses:
10 deg
M3 10 N m
Given
M = 0;
M1 M3 cos M 2 cos 1 0
M3 sin M2 sin 1 0
x
M = 0; y
M3
Find M3
32.9 deg
M3
651N 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 pegsA 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
ab 0
0
F
37.5 M
25 N m
0
M
45.1N 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:
F 1N
Initial guess:
Given
c
0
ab 0
0
F
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
B
0.5 m
30
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 a + (MC)2 = 5
a 35 b (0.5 + 0.5) - 5 a 45 b (3) = - 9 kN # m
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 A, we can write summing the moments of these force components about point
a 35 b (0.5) + 5 a 45 b (3) - 6 cos 30°(0.5) - 6 sin 30°(3) 3 4 + 6 sin 30°(6) - 6 cos 30°(0.5) + 5 a b (0.5) - 5 a b (6) 5 5
a + (MC)R = © MA ;
(MC)R = 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
0.5 m
30
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 a + (MC)2 = 5
a 35 b (1) - 5 a 45 b (3) = - 9 kN
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) su mming the mo ment of ea ch force about point O. Take F = 25k N.
z
5 6
O y
300 mm 200 mm F
150 mm
SOLUTION B
(a) MC = rAB * (25k) i
= - 0.35
3 3
j
k
- 0.2
0
0 0 25 MC = { - 5i + 8.75j} N m
#
x
F
400 mm 200 mm
3
A
Ans.
(b) MC = rOB * (25k) + rOA * ( - 25k) i
= 0.3 0
j
0.2 0
k
3 3
i
0 + 0.65 25 0
j
k
0.4 0
0 - 25
MC
= (5 - 10)i + (-7.5 + 16.25)j
MC
= { - 5i + 8.75j} N # m
3 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 MC
B
= rAB * (Fk)
3
i
= - 0.35 0
j
k
- 0.2
0
0
F
3
x
200 mm
MC
= { - 0.2Fi + 0.35F j} N # m
MC
= 2( - 0.2F)2 + (0.35F)2 = 400
F
=
400
2( - 0.2)2 + (0.35)2
F
400 mm
= 992 N
A
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: M1
= 50j N # m
M2
= 60 cos 30°i + sin 30°k N # m = 51.96i + 30.0k N # m
5 6 1
2
Resultant Couple Moment: MR
= © M;
MR
5
6
= M1 + M2 = 51.96i + 50.0j + 30.0k N # m
5 = 552.0
i
6
+ 50j + 30k N # m
6
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
b
= cos-1
51.96 a 78.102 b = 48.3° 50.0
Ans.
= 50.2°
Ans.
= 67.4°
Ans.
78.102
g
= cos-1
a 30.0 b 78.102
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 = 2 0 N· m
20° 30°
SOLUTION M1 M2
y
= 50k N # m
5 6 = 201 - cos 20° sin 30° - cos 20° cos 30° + sin 20° 2 N # m i
j
= - 9.397i - 16.276j + 6.840k
5
= © M;
MR
x
6
Resultant Couple Moment: MR
k
N#m
M1 = 50 N · m
= M1 + M2 = - 9.397i - 16.276j + 50 + 6.840 k N # m
5 5
1
26
= - 9.397i - 16.276j + 56.840k N # m
6
The magnitude of the resultant couple moment is MR
1
2 1
2 1
2
= 2 - 9.397 2 + - 16.276 2 + 56.840 = 59.867 N # m = 59.9 N # m
2
Ans.
The coordinate direction angles are a
= cos-1
b
= cos
g
= cos-1
-1
9.397 a -59.867 b = 99.0°
Ans.
- 16.276
a 59.867 b = 106° 56.840 59.867
= 18.3°
Ans. Ans.
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
15 M
30 0
N m
F 75 N
Solution:
P 1N
Initial guess
Given
F a M
Pb 0
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 Mx
= - 35(0.35) - 25(0.35) cos 60° = -16.625
My
= - 25(0.35) sin 60° = - 7.5777 N # m
y 175 mm
x
#
|M| = 2( - 16.625)2 + ( - 7.5777)2 = 18.2705 = 18.3 N m a
= cos-1
b
= cos-1
g
= cos-1
16.625 a -18.2705 b = 155° 7.5777 a -18.2705 b = 115° 0 a 18.2705 b = 90°
175 mm
25 N
Ans. Ans.
Ans.
Ans.
35 N
4–89.
Determine the resultant couple moment of the two couples that act on the pipe assembly.The distance fromA to B is d = 400 mm. Express the result asa 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
3 3
i
j
= 0
k
- 0.3464
0
0
3
#
0.20 = { - 12.12i} N m 35
(M C)2 = rAB * F2
=
i
j
0 - 50
- 0.3464 0
k
3
#
0.20 = { - 10.0j - 17.32k} N m 0
Resultant Couple Moment: MR
= © M;
MR
= (M C)1 + (M C)2
# = { - 12.1i - 10.0j - 17.3k}N m
Ans.
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 MR
= { - 12.1i - 10.0j - 17.3k} N # m
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: rAB
x
= {(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
3
i
j
= 0
k
- 0.8660d
0
0
3
#
0.50d = { - 30.31d i} N m 35
(M C)2 = rAB * F2
=
3
i
j
0 - 50
- 0.8660d
k
0
3
#
0.50d = { - 25.0d j - 43.30d k} N m 0
Resultant Couple Moment: MR
= © M;
MR
= (M C)1 + (MC)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. 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
200 mm
The force vectorsF 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 Mc
3
i
j
= rAB * F = 0 .1 0.5 0
0
k
3
#
0 = [40i - 8 j] N m 80
or Mc
3
i
= rBA * - F = - 0.1 0
j
- 0.5 0
k
3
#
0 = [40i - 8 j] N m - 80
The magnitude ofMc 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 M
My M
Mz M
cos
40 40.79
11.3°
Ans.
cos
8 40.79
101°
Ans.
cos
0 40.79
90°
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 vectorsF and –F can be written as F = {Fk} N and - F = [ - Fk]N Thus, the couple moment of F can be determined from Mc
3
i
j
= rAB * F = 0.1
0.5 0
0
k
3
0 = 0.5Fi - 0.1F j 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.
5 6
If F = 100k N, determine the couple moment that acts on the assembly. Express the result as a Cartesi an vector. Member BA lies in the x–y plane.
z O O
B
60
F
SOLUTION x
f
= tan-1 r1
a 23 b - 30° = 3.69°
200 mm 300 mm
150 mm
= { - 360.6 sin 3.69°i + 360.6 cos 3.69°j}
200 mm
= { - 23.21i + 359.8j} mm u
= tan-1 r2
a 4.52 b + 30° = 53.96°
= {492.4 sin 53.96°i + 492.4 cos 53.96°j} = {398.2i + 289.7j} mm
Mc
= (r1 - r2) * F
3
i
= - 421.4 Mc
y
–F
0
j
k
70.10 0
0 100
= {7.01i + 42.1j} N # m
3 Ans.
A
4–94.
If the magnitude of the resultant couple moment is 15 N m, determine the magnitude F of the force s applie d to the wrenches.
z
#
O O
B
60
SOLUTION f
= tan-1 r1
x
a 23 b - 30° = 3.69°
F
200 mm 300 mm
y
–F
150 mm
= { - 360.6 sin 3.69°i + 360.6 cos 3.69° j}
200 mm
= { - 23.21i + 359.8j} mm
a 4.52 b + 30° = 53.96°
u
= tan-1
r2
= {492.4 sin 53.96°i + 492.4 cos 53.96° j} = {398.2i + 289.7j} mm
Mc
= (r1 - r2) * F
3
i
= - 421.4 Mc
0
j
k
70.10 0
0
3
= {0.0701Fi + 0.421F j} N # m
Mc = 2(0.0701F)2 F
F
=
+ (0.421F)2 = 15
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
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 F1
0.2 m
x
SOLUTION Couple Moment:The position vectors r1, r2, r3, and r4, Fig. a, must be determined first. r1
= {0.2i} m
= {0.2j} m
r2
r3
= {0.2j} m
= 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 vectorsF1, F2, and F3 are given by
= {100k} N
F1
F2
= {120k} N
F3
= {80i} N
Thus, M1 M2 M3 M4
= r1 * F1 = (0.2i) * (100k) = { - 20 j} N # m = r2 * F2 = (0.2j) * (120k) = {24i} N # m
= r3 * F3 = (0.2j) * (80i) = { - 16k} N # m
= 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 = M1 + M2 + M3 + M4
= ( - 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 [((
Mc)R]x
a
= cos -1
b
= cos -1
g
= cos -1
b = cos a 51.56 b = 44.2° 71.94
Mc)R
[(Mc)R]y (Mc)R [(Mc)R]z
aa (
Mc)R
bb
= cos
- 47.56
= 131° 71.94 - 16 = cos = 103° 71.94
aa b b
0.2 m F2
– F3
From the geometry of Figs.b and c, we obtain r4
– F2
Ans. Ans. Ans.
0.2 m
F3
30
0.3 m
0.2 m – F1
F4 y
[150 k] N
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
SOLUTION
F1
Couple Moment:The position vectors r1, r2, r3, and r4, Fig. a, must be determined first.
0.2 m
x
– F2
r1
= {0.2i} m
r2
= {0.2 j} m
r3
= {0.2 j} m
F2 – F3
From the geometry of Figs. b and c, we obtain r4
0.2 m
0.2 m
= 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45° j - 0.3 sin 30°k
F3
= {0.1837i + 0.1837 j - 0.15k} m The force vectors F1, F2, and F3 are given by F1
=
F1k
F2
=
F2k
F3
=
F3i
Thus, M1
= r1 * F1 = (0.2i) * (F1k) = - 0.2 F1 j
M2
= r2 * F2 = (0.2j) * (F2k) = 0.2 F2i
M3
= r3 * F3 = (0.2j) * (F3i) = - 0.2 F3k
= r4 * F4 = (0.1837i + 0.1837 j - 0.15k) * (150k) = {27.56i - 27.56 j} N # m
M4
Resu ltan t Mome nt:The resultant couple moment required to equal (M ) = {50i - 45 j - 20k} N m. Thus,
#
c R
(M c)R
= © M c;
(M c)R
= M1 + M2 + M3 + M4 50i
- 45 j - 20k = ( - 0.2F1 j) + (0.2F2i) + ( - 0.2F3k) + (27.56i - 27.56 j)
50i
- 45 j - 20k = (0.2F2 + 27.56)i + ( - 0.2F1 - 27.56) j - 0.2F3k
Equating the i, j, and k components yields F2
= 112 N
Ans.
- 45 = - 0.2F1 - 27.56
F1
= 87.2 N
Ans.
- 20 = - 0.2F3
F3
= 100 N
Ans.
50
= 0.2F2 + 27.56
30
0.3 m
0.2 m – F1
F4 y
[150 k] N
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 6 kN
4 kN 4m
5m 13
SOLUTION
+ © FRx = © Fx ;
:
FRx
=6
a 135 b - 4 cos 60°
4m
FRy
= 6 12 - 4 sin 60°
a 13 b
= 2.0744 kN FR
u
=
2(0.30769)2
= tan-1
+ (2.0744)2 = 2.10 kN
2.0744 c 0.30769 d = 81.6°
a + MO = © MO ;
60
A
= 0.30769 kN + c © FRy = © Fy ;
12 5
Ans. Ans.
a
a 1213 b (4) + 6 a 135 b (5) - 4 cos 60°(4)
MO
=8- 6
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 6 kN
4 kN 4m
5m 13
SOLUTION
+ © FRx = © Fx ;
:
FRx
=6
a 135 b - 4 cos 60°
4m
FRy
= 6 12
a 13 b - 4 sin 60°
= 2.0744 kN FR u
=
2(0.30769)2 +
= tan-1
(2.0744)2
2.0744 c 0.30769 d = 81.6°
a + MP = © MP ;
60
A
= 0.30769 kN + c © FRy = © Fy ;
12 5
= 2.10 kN
Ans. Ans.
a
a 1213 b (7) + 6 a 135 b (5) - 4 cos 60°(4) + 4 sin 60°(3)
MP
=8-6
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
m
SOLUTION
+
:
FRx
= © Fx ;
FRx
= 1.5 sin 30° - 2.5
a 45 b
= - 1.25 kN = 1.25 kN + c FR = © Fy ;
FRy
y
;
3 2.5 5
= - 1.5 cos 30° -3 = - 5.799 kN = 5.799 kN T
ab
Thus, FR
=
2F 2R
x
+
F 2Ry
=
21.252 +
5.7992
= 5.93 kN
Ans.
and u
= tan - 1
a + MRA = © MA ;
¢ ≤= FRy
FRx
MRA
a
5.799 tan - 1 1.25
= - 2.5
b = 77.8°
d
Ans.
a 35 b (2) - 1.5 cos 30°(6) - 3(8)
= - 34.8 kN # m = 34.8 kN # m (Clockwise)
Ans.
2
m4 m 2
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
m
SOLUTION
+
:
FRx
= © Fx ;
FRx
= 1.5 sin 30° - 2.5
a 45 b
= - 1.25 kN = 1.25 kN + c FR = © Fy ;
FRy
y
= - 1.5 cos 30° - 2.5
;
3
-3
a5b
= - 5.799 kN = 5.799 kN T Thus, FR
=
2F 2R
u
= tan - 1
x
+
F 2Ry
=
21.252 +
5.7992
= 5.93 kN
Ans.
and
¢ ≤= FRy
FRx
a + M RB = © MRB ;
a
5.799 tan - 1 1.25
MB
b = 77.8°
= 1.5cos 30°(2) + 2.5
d
Ans.
a 35 b (6)
= 11.6 kN # m (Counterclockwise )
Ans.
2
m4 m 2
4–101
.
Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given:
F 1 200 N F 2 450 N M 200 N m a 0.2 m b 1.5 m c 2m d 1.5 m
30 deg
Solution:
sin
0
F 1 1 F2 cos
FR
0
0
225 FR
190 N
0
FR
294 N
Ans.
b c 0 b sin MO
MO
F1 1
a F2 cos
a 0
0 0
0
0
39.6
N m
0
0
M 0 1
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 1 4m ft
B
1.5 kNlb 14 m 300 ft 0.75 150kN lb 4 1 ft m
A
30
↓+FRx
= ΣFx;
FRx = 1 sin 30° = 0.5 kN
↓ +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 = 6.35° = tan–1 4.491 = 30° – 6.35° = 23.6°
+ 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.
3 kN 5 kN
2 kN 13
5
3
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 compo nents algebraic ally along the x and y axes, we have
+ © (FR)x = © Fx;
(FR)x
:
+ c (FR)y = © Fy;
(FR)y
A
a 45 b - 3 a 135 b = 2.846 kN 3 12 = - 5 a b - 2 - 3 a b = - 7.769 kN = 7.769 kN T 5 13 =5
:
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 (( )) d = tan c 7.769 d = 69.88° = 69.9° 2.846 FR
y
FR
x
-1
Ans.
c
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;
(MR)A
=5
3 5
ab
(4)
-5
4 5
ab
(5)
- 2(1) - 3
12 13
a b
= - 9.768 kN # m = 9.77 kN # m (Clockwise)
(2)
5
+3
13
a b
A ns.
(5)
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
+ © (FR)x = © Fx; (FR)x = 450 cos 45° - 600 cos 30° = - 201.42 N = 201.42 N
:
+ c (FR)y = © Fy;
(FR)y
0.6 m
;
= 450 sin 45° + 600 sin 30° = 618.20 N c
The magnitude of the resultant force FR is given by FR
=
2(FR)x2 +
(FR)y2
=
2201.42 +
618.202
= 650.18 kN = 650 N
Ans.
The angle u of FR is u
= tan - 1
c (( )) d = tan c 618.20 d = 71.95° = 72.0° 201.4 FR
y
FR
x
-1
b
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 )
An s.
4–107.
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 Nfor the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erec tor spinae . These loading s 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.
z
FO FR
FO
MROx
;
= © MOx
;
{2(35
+
45
+
23
+
32)k }
+
2(32)(0.015)
MR O
=
[ - 2(35)(0.075)
MR O
=
{ - 2.22i} N # m
=
{270k} N +
50 mm
30 mm 40 mm
y
x =
FL
15 mm 45 mm
FR
FL
O
SOLUTION = © Fz
FE FE
75 mm
FR
FR
Ans.
2(23)(0.045)]i Ans.
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
6m
Equivalent Resultant Force: The forces FB and FD, Fig. a, expressed in Cartesian vector form can be written as (0
FB
= FBuAB =
5
FD
= FDuCD =
7
-
0)i
(0
-
0)
2
+
(6
(2
-
0)i
+
(-3
-
0) j
+
(0
-
6)k
(2
-
0)2
+
(-3
-
0)2
+
(0
-
6)2
C
+
(6
-
0) j
-
2
0)
(0
+
-
8)k
-
2
(0
+
8)
[3 j
=
SS
=
-
[2i
4k] kN
-
3j
-
= πF;
FR
=
FB
+
x
6k] kN
FD =
(3 j
-
4k)
=
[2i
-
10k] kN
+
(2i
-
3j
-
6k) Ans.
Equivalent Resultant Force: The position vectors rOB and rOC are rOB
=
{6 j} 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
=
=
+
3
rOC
i 0 0
[ - 6i
*
j 6 3 +
FD k 0 -4
3 3 +
i 0 2
12 j] kN # m
j 0 -3
k 6 -6
5 kN
8m 2m
D
The resultant force FR is given by FR
FB
7 kN
3 Ans.
O
3m 6m
B y
4–109.
Replace the force system by an equivalent force and couple moment at point A.
z
F2
{100i
100j
50k} N
F1
4m
F3 {
SOLUTION
500 k} N 8m
FR
= © F;
FR
= = =
F1
+
+
F2
15300 400
+
i
+
F3
100 i 300 j
= ©MA ;
M RA
rAB
=
3
=
-
-
*
i 0 300 i 0 0
j
+
-
-
-
500 k
2
k
=
=
+
+
2 650 1400 6 N1002 1 100 50 512 6 m and 5 1 6 m.
The position vectors are rAB M RA
1m
k
F1
+
rAE
rAB
*
F2
j 0 400
k 12 - 100
j
k 0 - 500
-1
0
- 3100i +
=
+
rAE
3 3 +
4800 j N # m
-
i 0 100
j
*
F3 j 0 - 100
k 12 - 50
3 Ans.
{300i
400j
100k} N
4–110.
The belt passing over the pulley is subjected to forces F1 and F2, eachha ving a magnitude of 40 N. F1 actsinth e - k direction. Replace these forces by an equivalent force and couple momentat point A. Express theresult in Cartesianvector form. Set u = 0° sot hat F2 actsi nt he - j direction.
z
r
80 mm
y
300 mm
A
SOLUTION FR
=
F1
FR
=
{ - 40 j
M RA
=
3
x
F2
= © (r *
=
MRA
+
-
Ans.
F)
i
j 0 - 40
- 0.3
0
{ - 12 j
40 k} N
+
F2
F1
k 0.08 0
12k} N # m
3 3 +
i - 0.3
0
j 0.08 0
k 0 - 40
3 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
FR
= F1 + F2
FR
= { - 28.3 j - 68.3k} N
= - 40 cos 45° j + ( - 40 - 40 sin 45°)k
F2
Ans.
rAF1
= { - 0.3i + 0.08 j} m
rAF2
= - 0.3i - 0.08 sin 45° j + 0.08 cos 45°k = {- 0.3i - 0.0566 j + 0.0566k} m
MRA
= (rAF1 * F1) + (rAF2 * F2)
3
i
= - 0.3 MRA
0
j 0.08 0
k 0 - 40
3 3
i
+ - 0.3 0
j
- 0.0566 - 40 cos 45°
= { - 20.5 j + 8.49k} N # m
k 0.0566 - 40 sin 45°
3 Ans.
Also, MRAx
= © MA
MRAx
= 28.28(0.0566) + 28.28(0.0566) - 40(0.08)
M
x
=0
RAx
MRAy
= © MA
MRAy
= - 28.28(0.3) - 40(0.3)
MRAy
= - 20.5 N # m
y
MRAz
= © MA
MRAz
= 28.28(0.3)
MRAz
MRA
z
= 8.49 N # m = - 20.5 + 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
0.15 m
F1
{6i 3j 10k} N
0.25 m
0.3 m
SOLUTION FR
O
= © F; FR = 6i - 3 j - 10k + 2 j - 4k
x
= {6i - 1 j - 14k} N M RO
= ©MO ;
M RO
= 0.15
3
i
6
j 0 -3
k 0.3 - 10
3 3
Ans.
i
+ 0
j
- 0.25
0
2
k 0.3 -4
3
= 0.9i + 3.30 j - 0.450k + 0.4i
= {1.30i + 3.30 j - 0.450k} N # m Note that FRz (MRO)x
=
= - 14 N pushes the drill bit down into the stock.
#
Ans.
= 3.30 N # m cause the drill bit to bend. # - 0.450 N m causes the drill case and the spinning drill bit to rotate
1.30 N m and (MRO)y
(MRO)z = 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
30 deg
Solution:
Fv
sin
F cos
0
Fv
a MO
b 0
187.5
MO
Ans.
0
Fv
324.76 N
0 0
100.481
N m
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
30 deg
Solution:
Fv
sin
F cos
0
Fv
MP
a c MP
bd 0
187.5
324.76 N
Fv
0
0
0
736.538
N m
Ans.
Ans.
4–115
.
Determine the magnitude and direction 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
103 N
Given:
F 1 5 kN
a 3m
F 2 3 kN
b 4m
F R 12 kN
c 6m
M 50 kN m
e 7
f 24
Solution:
F 1 kN
Initial guesses:
Given
e 2
e f
2
f 2
e f
2
f 2
e f
2
30 deg
d 2m
F F cos 0 1 F 1 F sin F2 F R F a F sin ( a b d) F ( a b) M 1 2
F
d
Find F d
F 4.427 kN
71.565 deg
d 3.524 m
Ans.
4–116
.
Determine the magnitude and direction 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: kN
103 N
Given:
F 1 5 kN
a 3m
F 2 3 kN
b 4m
F R 10 kN
c 6m
M 45 kN m e 7 f 24 Solution:
F 1 kN
Initial guesses:
Given
e
30 deg
d 1m
e f
f F 1 F sin F 2 F R 2 2 e f
2
f 2
e f
2
2
F F cos 0 1
F a F sin ( a b d) F ( a b) M 1 2
F
d
Find F d
F 2.608 kN
57.529 deg
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
300 N
30
60 B A
2m
m 4
SOLUTION
+
:
FRx
= © Fx ;
+ c FRy = © Fy ; F u
=
2( - 125)2 +
= tan-1
FRx
= 450 cos 60° - 700 sin 30° = - 125 N = 125 N
FRy
= - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N T
( - 1296)2
a 1296 b = 84.5° 125
c + M RA = © MA ;
= 1302 N d
1296(x) x
;
Ans. Ans.
= 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 3000
= 8.51 m
Ans.
m 3
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
300 N
30
60 B A
2m
m 4
SOLUTION
+
= © Fx ;
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 u
FRx
=
2( - 125)2 +
= tan-1
( - 1296)2
a b = 84.5° 1296 125
c + M RB = © MB ;
= 1302 N
x
Ans. Ans.
d
1296(x)
;
= - 450 sin 60°(4) + 700 cos 30°(3) + 3000
= 2.52 m (to the right)
Ans.
m 3
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 m1
A
.
.
m1
m1
500 N m1
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
3m
500 N
+ © Fx =
:
FRx ;
a b - 500(cos 60°) = - 450N = 450 N
4 - 250 5
=
FRx
;
A
+ c © Fy = © Fy ; FR u
=
2( - 450)2 +
= tan-1
FRy
= - 300 -
( - 883.0127)2
a 883.0127 b = 63.0° 450
a + MRA = © MA ;
450y
y
=
3 250 5
a b - 500 sin 60° = - 883.0127 N = 883.0127 N T
= 991 N
d
= 400 + (500 cos 60°)(3) + 250
800 450
= 1.78 m
Ans. Ans.
a 45 b (5) - 300(2) - 250 a 35 b (5) 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
+ © Fx =
:
FRx ;
= - 250
FRx
a 45 b - 500(cos 60°) = - 450 N = 450 N
3m
500 N ;
A
+ c © Fy = © Fy ; FR u
=
2( - 450)
= tan-1
a
2
FRy
= - 300 -
3 250 5
a b - 500 sin 60° = - 883.0127 N = 883.0127 N T
+ ( - 883.0127) = 991 N
883.0127 450
c + MRA = © MC ;
2
b = 63.0°
d
883.0127x
x
=
Ans.
= - 400 + 300(3) + 250
2333 883.0127
= 2.64 m
a 35 b (6) + 500 cos 60°(2) + (500 sin 60°)(1) 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
60deg
a 3 m
f 12
b 3 m
g 5
c 4 m
F1 8kN
d 4 m
F2 6kN
e 6m Solution:
g
F1
FR
2
f g
2
f
0
cos (
)
F 2 sin ( )
0
0.077 FR
2.188
kN
0.000 0 c
MO
0
e
M 0
MO
0.00
6.92
F1 2
f g
Ans.
g 2
f
0
cos (
)
d F2 sin ( )
0
0
0
kN m
0.00
FR 2.19 kN
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
60deg
a 3 m
f 12
b 3 m
g 5
c 4 m
F1 8kN
d 4 m
F2 6kN
e 6m Solution:
g
F1
FR
2
f g
2
f
0
cos (
)
F 2 sin ( )
0
0.077 FR
2.188
F R 2.19 kN
kN
0.000
0 c b MP
0
M
MP
0.000
0
g
F1 2
2
f g
f
0
b
cos (
)
d F 2 sin ( )
0
0
0.000
e
Ans.
0.358
kN m
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, 4
+
:
(FR)x
= © Fx;
= 250 5 - 500cos 30° - 300 = - 533.01 N = 533.01 N
(FR)x
+ c (FR)y = © Fy;
ab
(FR)y
= 500 sin 30° - 250
1m A ;
a 35 b = 100 N c
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 R= (FR)y
(FR)x
tan - 1
100 c 533.01 d = 10.63° = 10.6°
b
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;
533.01(d) d
= 500 cos 30°(2) - 500 sin 30°(0.2) - 250
= 0.8274 mm = 827 mm
a 35 b (0.5) - 250 a 45 b (3) + 300(1) 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
500 N
5
3
1m
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 componen ts algebraically along the x and y axes,
+ © (FR)x = © Fx;
= 250
(FR)x
:
+ c (FR)y = © Fy;
(FR)y
4
- 500cos 30° - 300 = - 533.01N = 533.01 N
a5b
= 500 sin 30° - 250
1m A ;
a 35 b = 100 N c
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 R= (FR)y
(FR)x
tan - 1
100 c 533.01 d = 10.63° = 10.6°
b
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;
- 533.01(d) = - 500 cos 30°(1) - 500 sin 30°(0.2) - 250 d
= 2.17 m
a 35 b (0.5) - 300(2)
Ans.
250 N
4–126. Replace the force and couple moment system acting on the overhang bea m by a resultant forc e, and specify its location along AB measured from point A.
30 kN
26 kN
30 13
12
0.3 m A
5
45 kNm
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.
z
FD
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)
© My = 0;
500(0.2)
+ 600(0.4) -
FC(0.4)
- 500(0.4) = 0
FC - FD = 100 + 600(0.2) - FC(0.2) FC
+
FD
(1) FD(0.2)
= 1100
=0 (2)
Solving Eqs. (1) and (2) yields: FC
= 600 N
FD
= 500 N
Ans.
500 N
O
C
400 mm
x
200 mm
z B
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: kN
103 N
Given: 8
6 kN
F
8
20 M
70 kN m
20
a 3m b 3m
e 5m
c 4m
f 6m
d 6m
g 5m
Solution:
f FR
F
MR
M
e
d g
46
8 F
FR
6 kN
8
MR
66
56
kN m
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:
103 N
kN Given:
8
6 kN
F
8 20
M
70 kN m
20
a 3m b 3m
e 5m
c 4m
f 6m
d 6m
g 5m
Solution:
FR
F
MR
M
0 e F g
FR
8 8
6 kN
MR
10 30 kN m 20
Ans.
4–130.
The building slab is subjected to four parallel column loadings. Determine the equival ent resultant force and specify its location (x, y) on the slab. Take F1 = 30 kN, F2 = 40 kN.
z
20 kN
F1
50 kN
F2
SOLUTION
x
+ c FR = © Fz; (MR)x
FR
= - 20 - 50 - 30 - 40 = - 140kN = 140 kN T
= © Mx;
= © My;
8m
Ans.
- 140y = - 50(3) - 30(11) - 40(13) y
(MR)y
= 7.14 m
140x x
Ans.
= 50(4) + 20(10) + 40(10)
= 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
+ T FR = © Fz;
x
FR
MR y
= 20 + 50 + 20 + 50 = 140 kN
= © My;
140(x) x
MR x
= © Mx;
8m
Ans.
= (50)(4) + 20(10) + 50(10)
= 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 = 35kN , determine the magnitude of the resultant force and specify the location of its point of application (x, y) on the slab.
z
30kN 0.75 m
90kN
FB 2.5 m
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, Fig. b,
+ c F = ©F ; R
z
- F = - 30 - 20 - 90 - 35 - 40 R
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
(MR)y
= © My;
= 3.68 m
215(x) x
Ans.
= 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75)
= 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
30kN 0.75 m
90 kN
FB 2.5 m
20 kN
2.5 m 0.75 m FA
0.75 m
x
3m
SOLUTION
3m 0.75 m
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 FR
= 140 +
FA
+
FA
-
FB
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) FR
(MR)y
= © My;
= 0.2FB + 1.8FA + 132
FR(3.25) FR
= 30(0.75) + 20(0.75) + 90(3.25) +
= 1.769FA + 1.769FB + 101.54
FA(5.75)
(3)
Solving Eqs.(1) through (3) yields FA
= 30kN
FB
= 20 kN
FR
FA(6.75)
(2)
= 190kN
Ans.
+
FB(5.75)
y
4–134.
Replace the two wrenches and the force, acting on the pipe assembly, by an equivalent resultant force and cou ple moment at point O.
100 N · m
300 N
z
C
SOLUTION
O 0.5 m
Force And Moment Vectors:
F2
5 6 = 5100 6 N = 2005cos 45° - sin 45° 6 N = 5141.42 - 141.42 6 N = 5100 6 N # m = 1805cos 45° - sin 45° 6 N # m = 5127.28 - 127.28 6 N # m F3
i
M2
k
i
FR
k
k
= F1 + F2 + F3
1
2
= 141.42i + 100.0 j + 300 - 141.42 k
5
6
= 141i + 100 j + 159k N The position vectors are r1 M RO
= ©MO ;
Ans.
5 6
5 6
= 0.5j m and r2 = 1.1 j m.
M RO
= r1 * F1 + r2 * F2 + M1 + M2 i
= 0
3
0
+
j 0.5 0 i 0 141.42
k 0 300
3
j 1.1 0
k 0 - 141.42
+ 100k + 127.28i - 127.28k = 122i - 183k N # m
200 N
180 N · m
Equivalent Force and Couple Moment At Point O:
= © F;
0. 8m
°
k
i
FR
0. 6m
45
j
k
i
M1
B
x
= 300k N
F1
A
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 0.6 m b 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
M C 1 N m
Given
M0 C 0 F MC
P( c0 h) 00 ba 0Q 0 F ( e f) 0 0
Find F MC
MC 30N 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
{800k} N
{500i} N
FA A
B y
P
x y
x
SOLUTION
C
FR
= {500i + 300 j + 800k} N
FR
=
uFR
6m
4m
2
(500)2
+ (300)2 + (800)2 = 990 N
FC
Ans.
= {0.5051i + 0.3030 j + 0.8081k}
MRx¿
= © Mx¿;
MRy¿
= © My¿;
MRy¿
= 800x
MRz¿
= © Mz¿;
MRz¿
= 500y + 300(6 - x)
MRx¿
= 800(4 - y)
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) = 3.07 kN # m
Ans.
x
= 1.16 m
Ans.
y
= 2.06 m
Ans.
MR
{300j} N
–
.
The loading on the bookshelf is dis tributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O.
30
50
1
1.5
0.5
1
1
=
= 2 2
(
)
70
0.107
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
m3
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 1
+ T FR = © F;
FR
1
= 2 (3)(3) + 2 (3)(1.5) = 6.75 kN T
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 2 2.5 m
1 2
- 6.75(x) = - (3)(3)(2) - (3)(1.5)(3.5) x
=
Ans.
m 1.5
4–140.
Replace the loading by an equivalent force and couple moment acting at point O.
200 N/m
O
m4
SOLUTION Equivalent Force and Couple Moment At Point O:
+ c FR = © Fy ;
FR = - 800 - 300
= - 1100 N = 1.10 kN T a+ MRO = © MO ;
12
Ans.
12
MRO = - 800 2 - 300 5
= - 3100 N # m
= 3.10 kN # m (Clockwise)
Ans.
m3
4–141
.
Replace the loading by an equivalent resultant force and couple moment acting at point A . Units Used: kN
103 N
Given:
w1 600
N m
w2 600 N m a 2.5 m b 2.5 m Solution:
F R w1 a w2 b
a b 2
MRA w1 a
FR 0 N
Ans.
MRA 3.75 kN m
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 ;
1 FR = 1(0.3) + (2.5 - 1)(0.3) = 0.525 kN c 2
Ans.
Location of Equivalent Resultant Force:
1 2
a + MR
O
= © MO ;
1 2 = 0.30010.152 + 0.22510.22
0.525 d
d = 0.171 m
Ans.
.
.
4 –145
.
Replace the distributed loading by an e quivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used:
kN
103N
Given:
N
8000 m a 5m w
b
5m 30deg
Solution :
FR
w a
F R x = w a
x
w a a 2
w b
FR
2 a
2
w b
2
a
w2b a b3 FR
60 kN
Ans.
b
3
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,
+ T FR = © F;
FR =
a b + 12 a L2 b = 12
1 L w0 2 2
w0
w0L T
Ans.
If we equate the moments of FR, Fig. b, to the sum of the moment of the fo rces in Fig. a about point A,
a + (MR)A = © MA;
-
1 w L(x) 2 0
= -
x =
a b a L6 b - 12 a L2 b a 23 L b
1 L w 2 0 2
5 L 12
w0
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:
kN
103 N
Given:
w1 1.5
kN
a 3m
m
w2 1 kN m w3 2.5
b 3m
kN
c 1.5 m
m
Solution:
F R w1 a w2 b w3 c M A w1 a
a 2
w2 b a
MA 45.563kN m
b 2
w3 ca b
d
MA FR
F R 11.25 kN
Ans.
d 4.05 m
Ans.
c 2
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.
80 kN 60 kN m
1
m 2.5 m
1 3.5 m
50 kN
SOLUTION 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 w2
8 3
60
>
8 3
1
80 3.5
-
8 3
≤- ¢ - ≤= 50 7
>
= 17.1875 kN m = 17.2 kN m
8 3
0
Ans.
Using the result of w2 and writing the force equation of equilibrium along the y axis, we obtain
+ c © Fy = 0;
1 (w - 17.1875)8 + 17.1875(8) - 60 2 1 w1 = 30.3125 kN m = 30.3 kN m
>
>
- 80 - 50 = 0 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: kN
103 N
Given:
w1 4
kN
w2 2.5
m
M 8 kN m
kN m
c 9m
Solution:
a 1m
Initial Guesses:
1
Given
2 1 2
a b
w1 b
w1 b a
Find ( a b)
1 2
w2 c 0
2b 3
b 1m
1
2c
2
3
w2 c
M
a 1.539 b 5.625
m
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 pointO.
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 1 kN/m lb/ft B
1.2 4 ftm
61.8 ft
m
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
F1 =
1 2
FRx = 2.338 kN
(1.8) (1) = 0.9 kN
F2 = (1.8) (1) = 1.8 kN F3 = (1.2) (1) = 1.2 kN +
→ FRx = ΣFx;
FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN
+↓ FRy = ΣFy; FR=
FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN 2
(2.338) + (2.55)
2
2.55 = 47.5° 2.338
= tan–1
+ MRA = ΣMA;
= 3.460 kN
Ans. Ans.
MRA = 0.9 (0.6) + 1.8 (0.9) + 1.2 (1.8 cos 60° + 0.6) = 3.96 kN · m 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 2 kN/m lb/ft 60 A
F1 =
1 2
(1.8) (1) = 0.9 kN
F2 = (1.8) (1) = 1.8 kN F3 = (1.2) (1) = 1.2 kN +
→ FRx = ΣFx;
FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN
+↓ FRy = ΣFy; FR=
FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN 2
(2.338) + (2.55)
2
= 3.460 kN
Ans.
2.55 = 47.5° 2.338
= tan–1
+ MRB = ΣMB;
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
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 ;
;
+ T FRy = © Fy ; FR =
2(1000)2 +
FRx = 1000 N FRy = 900 N (900)2
= 1345 N
FR = 1.35 kN u
=
Ans.
c d = 42.0°
900 tan-1 1000
a + M RA = © MA ;
200 N /m
Ans.
d
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 ;
;
+ T FRy = © Fy ; FR =
2(1000)2 +
FRx = 1000 N FRy = 900 N (900)2
= 1345 N
FR = 1.35 kN u
=
Ans.
c d = 42.0°
900 tan-1 1000
a + M RC = © MC ;
200 N/m
d
900x
Ans.
= 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 w w0 sin (2–– x) L
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
=
p
w0 sin
x dx
2L
¢
≤
Integrating dFR over the entire length of the beam gives the resultant force FR .
+T
FR
=
L
L¢ L
dFR
=
w0 sin
0
L
≤ = ¢-
p x dx 2L
2w0L p cos x p 2L
≤`
L
0
=
2w0L p
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 given by
= L L
x
L
L ¢ = L
xcdFR
L
dFR
x w0 sin
0
≤
p x dx 2L
2w0L p
4w0L2
=
p2 2w0L p
=
2L 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
w
10 kN/m
1 (x2 4x 60) kN/m –– 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 6
- 4x + 60)dx
Integrating dFR over the entire length of the beam gives the resultant force FR .
+T
FR
=
L
=
dFR
L 0
L
6m
1 ( - x2 6
- 4x + 60)dx =
1 6
B-
x3 3
- 2x2 + 60x
= 36 kN T
x
=
L L
L B = 6m
L
dFR
x
0
6m 0
Ans.
Location:The location of dFR on the beam is xc the location x of FR measured from point A is
xcdFR
R`
1 (- x 2 6
- 4x + 60)
36
R
= x , measured from point A. Thus
dx
=
L 0
6m
1 ( - x3 6
- 4x2 + 60x)dx 36
=
L
= 2.17 m
Ans.
1 6
¢-
x4 4
-
4x3 3
+ 30x2
36
≤`
6m 0
x
4 –158
.
Determine the equivalent resultant force and couple moment at point O. Units Used: kN
2
3
10 N
Given:
a 3m wO 3 kN m w ( x) wO
x a
2
Solution:
a 0
F R w ( x) dx
a 0
MO w ( x) ( a x) dx
F R 3 kN
Ans.
MO 2.25 kN m
Ans.
4–159.
Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of th is 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:
+ FR = © Fx;
- FR = - LdA = -
:
L
h
z
8 kPa
dz
w
0
4m
L a bA B
FR =
1
20z2
103 dz
0
A B
= 106.67 103 N = 107 kN
z
;
Ans.
Location of Equivalent Resultant Force:
= L L
z
zdA
z
= L L cA = L L A cA = L L A =
zwdz
0
A
z
dA
A
wdz
0
4m
B d B
1
z 20z2 (103) dz
0
4m
1
20z2 (103)dz
0
4m
3
B d B
20z2 (10 3) dz
0
4m
20z12 (103)dz
0
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 ;
FR = FR = -
L
L
dA = -
x
wdx
0
A
L A B 9m
1
200x2 dx
0
= - 3600 N = 3.60 kN T
Ans.
x
a + MRO = © MO ;
L = L A B = L A B
MRO = -
xwdx
0
9m
1
x 200x2 dx
0
9m
3
200x2 dx
0
= - 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 locati on, measured from point A. Evaluate the integral using Simpson’s rule.
w w
5x
(16
x
2 1/2
)
kN/m 5.07 kN/m
2 kN/m x A
m 3
SOLUTION FR =
L
4
wdx =
L4
5x
1
+ (16 + x2)2 dx
0
FR = 14.9 kN
L
4
x dF =
0
L
Ans.
4
(x) 45x
1 2
+ (16x + x2 ) dx
0
= 33.74 kN # m x =
B
33.74 14.9
= 2.27 m
Ans.
m 1
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: M C1
600 mm
1 2
= rAB * 400i i
= 0.6 cos 45°
3
400
j
k
- 0.5
- 0.6 sin 45°
0
B
3
0
= - 169.7 j + 200k
400 mm C
150 N
For the 150-N forces: M C2
1 2
= rOB * 150 j =
3
i 0.6c os 45° 0
j 0 150
k
- 0.6 sin 45° 0
3
= 63.6i + 63.6k M CR
= MC1 + MC2
M CR
= 63.6i - 170 j + 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
SOLUTION 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 = 30(sin 45°i - cos 45° j) N = [21.213i - 21.213 j] N
Moment of ForceF About z Axis:The unit vector along the Eq. 4–11, we have
z
axis is k. Applying
#
Mz = k (rBA * F)
3
0
= - 0.01 21.213
0 0.2 - 21.213
1 0 0
3
= 0 - 0 + 1[( - 0.01)( - 21.213) - 21.213(0.2)] = - 4.03 N # m
Ans.
Or
#
Mz = k (rOA * F)
3
0
= - 0.01 21.213
0 0.2 - 21.213
1 0.05 0
3
= 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
F
30 N
200 mm
B
y
4–165.
The horizontal 30-N f orce acts on the handle of the wrench. Determine the moment of this f orce about point O. Specify the coordinate direction angles a, b , g of the moment axis.
z
O
SOLUTION x
Position Vector And Force Vectors:
F
=
{( - 0.01
-
0)i
=
{ - 0.01i
+
0.2 j
=
30(sin 45°i
-
=
{21.213i
21.213 j} N
-
(0.2
+ +
0) j
-
+
(0.05
-
0)k} m
0.05k} m
cos 45° j) N
Moment of ForceF About Point O:Applying Eq. 4–7, we have MO
=
rOA
=
3
F
*
i - 0.01 21.213
=
{1.061i
=
{1.06i
j 0.2 - 21.213 +
+
1.061 j
1.06 j
-
k 0.05 0
3
4.031k} N # m
-
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
b
=
cos - 1
g
=
cos - 1
a 1.061 b 4.301 a 1.061 b 4.301
=
75.7°
Ans.
=
75.7°
Ans.
4.301 a 4.301 b -
=
160°
Ans.
A
45 45
10 mm
50 mm
rOA
30 N
200 mm
B
y
4–166.
The forces and couple moments that are exerted on the t oe an d hee l plat es of a s now s ki ar e Ft = 5- 50i + 80 j - 158k6 N, M t = 5- 6i + 4 j + 2k6 N # m, and Fh = 5- 20i + 60 j - 250k6 N, M h = 5- 20i + 8 j + 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
Mt
800 mm
120 mm y x
SOLUTION FR
=
Ft
MRP
=
MRP
=
+
3
Fh
=
i 0.8 - 20
(200 j
{ - 70i
j 0 60 +
+
140 j
k 0 - 250
48k)
+
MRP
=
{ - 26i
+
357.36 j
MRP
=
{ - 26i
+
357 j
+
33 +
-
i 0.92 - 50
(145.36 j +
408k}
+
Ans.
N
j 0 80
73.6k)
3
+
( - 6i
+
( - 6i
+
4j
2k)
k 0 - 158 +
+
4j
+
+
2k)
+
( - 20i
( - 20i
+
8j
+
+
8j
+
3k)
3k)
126.6k} N # m
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: rAB
=
{[ - 0.5
( - 0.5)]i
-
+
[0
-
[ - 0.5 FC
=
=
250
{0
§B
-
{0
[159.33i
+
-
[ - (1
30
( - 1)] j -
+
(0
( - 2.5)]i
-
0)k} m
=
a
+
+
1.5 cos 30°)]} j + (0 - 1.5 sin 30°)k [ -0.5 - (-2.5)]2 + [ - (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°)2
183.15 j
-
59.75k]N
¥
=
=
=
rAB
3
*
x
F
i 0 159.33 -
59.7i
j 1 183.15 -
k 0 - 59.75
159k N # m
B
1m
0.5 m
N
Mo me nt of F or ceFc Ab ou t P oi nt A: Applying Eq. 4–7, we have MA
A
{1 j} m
3 Ans.
y
4–171.
Determine the magnitude of the moment of the force about the hinged axis aa of the door.
Fc
z
C
1.5 m
2.5 m FC
250 N
a
SOLUTION Position Vector And Force Vectors: rAB
=
{[ - 0.5
( - 0.5)]i
-
[0
+
-
{ - 0.5 FC
=
=
250
{0
§B
-
{0
[159.33i
-
[
[
-
+
-
+
(0
( - 2.5)]i
-
0)k} m
=
a
+
+
183.15 j
-
¥
59.75k] N
Mo me nt of For ceFc Ab ou t a - a Ax isThe : unit vector along the Applying Eq. 4–11, we have Ma - a
=
=
=
i # (rAB
3
*
1 0 159.33
1[1( - 59.75)
= -
a
–
x
a
axis is i.
FC)
0 1 183.15 -
0 0 - 59.75
3
(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
A
{1 j} m
1.5 cos 30°)]} j + (0 - 1.5 sin 30°)k N [ - 0.5 - ( - 2.5)]2 + (1 + 1.5 cos 30°)]}2 + (0 - 1.5 sin 30°)2
(1 -
30
( - 1)] j
Ans.
B
1m
0.5 m
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:
F 1 600 N F 2 250 N a 1m
40 deg
M 400 N m Solution: Initial Guess
Given
F1
F 1N
a F a F a M 2 2cos 2cos
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
F {60i 20j 15k} N
0.25 m
0.4 m
. 30
x
y
