Dynamics 14th edition by r c hibbeler chapter 04

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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

j

Ay
Dy

k
Az 3
Dz

= (A * B) + (A * D)

(QED)

228


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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)

229


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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.

230


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*4–4.
Determine the moment about point A of each of the three
forces acting on the beam.

F2 = 500 lb

F1 = 375 lb
5

A

4
3

8 ft

6 ft

SOLUTION

B

0.5 ft

5 ft
30˚
F3 = 160 lb

a + 1MF12A = - 375182

= - 3000 lb # ft = 3.00 kip # ft (Clockwise)

Ans.

4
a + 1MF22A = - 500 a b 1142
5

= -5600 lb # ft = 5.60 kip # ft (Clockwise)

Ans.

a + 1MF32A = - 1601cos 30°21192 + 160 sin 30°10.52

= - 2593 lb # ft = 2.59 kip # ft (Clockwise)

Ans.

Ans:

( MF1 ) A = 3.00 kip # ft (Clockwise)
( MF2 ) A = 5.60 kip # ft (Clockwise)
( MF3 ) A = 2.59 kip # ft (Clockwise)

231


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4–5.
Determine the moment about point B of each of the three
forces acting on the beam.

F2 = 500 lb

F1 = 375 lb
5

A

8 ft

SOLUTION

4
3

6 ft

B

0.5 ft

5 ft
30˚
F3 = 160 lb

a + 1MF12B = 3751112

= 4125 lb # ft = 4.125 kip # ft (Counterclockwise)

Ans.

4
a + 1MF22B = 500 a b 152
5

= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)

Ans.

a + 1MF32B = 160 sin 30°10.52 - 160 cos 30°102
= 40.0 lb # ft (Counterclockwise)

Ans.

Ans:
( MF1 ) B = 4.125 kip # ftd
( MF2 ) B = 2.00 kip # ftd
( MF3 ) B = 40.0 lb # ftd
232


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4–6.
The crowbar is subjected to a vertical force of P = 25 lb at the
grip, whereas it takes a force of F = 155 lb at the claw to pull
the nail out. Find the moment of each force about point A and
determine if P is sufficient to pull out the nail. The crowbar
contacts the board at point A.

60Њ

F

O
20Њ
3 in.
P
14 in.

A
1.5 in.

Solution

a + MP = 25 ( 14 cos 20° + 1.5 sin 20° ) = 341 in # lb (Counterclockwise)
c + MF = 155 sin 60°(3) = 403 in # lb (Clockwise)

Since MF 7 MP,  P = 25 lb is not sufficient to pull out the nail.

Ans.

Ans:
MP = 341 in. # lbd
MF = 403 in. # lbb
Not sufficient
233


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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

4

5
3

d2 = 5 sin 60° = 4.330 m

F3

500 N

d3 = 2 sin 53.13° = 1.60 m
Using each force where MA = Fd, we have
a + 1MF12A = - 25011.7322

= - 433 N # m = 433 N # m (Clockwise)

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)

Ans.

Ans:
( MF1 ) A = 433 N # mb
( MF2 ) A = 1.30 kN # mb
( MF3 ) A = 800 N # mb

234


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*4–8.
Determine the moment of each of the three forces about point B.

F1

F2

250 N 30

300 N

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
Ans.

MB = 0

Ans:

MB = 150 N # md
MB = 600 N # md
MB = 0
235


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4–9.
Determine the moment of each force about the bolt located
at A. Take FB = 40 lb, FC = 50 lb.

0.75 ft
B

2.5 ft

30 FC

20

A

C

FB

25

SOLUTION
a +MB = 40 cos 25°(2.5) = 90.6 lb # ft d

Ans.

a +MC = 50 cos 30°(3.25) = 141 lb # ftd

Ans.

Ans:
MB = 90.6 lb # ftb
MC = 141 lb # ftd
236


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4–10.
If FB = 30 lb and FC = 45 lb, determine the resultant
moment about the bolt located at A.

0.75 ft
B

2.5 ft

A

C

30 FC

20
FB

25

SOLUTION
a + MA = 30 cos 25°(2.5) + 45 cos 30°(3.25)
= 195 lb # ft d

Ans:
MA = 195 lb # ftd
237


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4–11.
The towline exerts a force of P = 6 kN at the end of the
8-m-long crane boom. If u = 30°, determine the placement
x of the hook at B so that this force creates a maximum
moment about point O. What is this moment?

A
P ϭ 6 kN
8m
u

O

1m

B
x

Solution
In order to produce the maximum moment about point O, P must act perpendicular
to the boom’s axis OA as shown in Fig. a. Thus


a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a,


x = x' + x" =

8
+ tan 30° = 9.814 m = 9.81 m
cos 30°

Ans.

Ans:
(MO)max = 48.0 kN # m d

x = 9.81 m
238


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*4–12.
A

The towline exerts a force of P = 6 kN at the end of the
8-m-long crane boom. If x = 10 m, determine the position u
of the boom so that this force creates a maximum moment
about point O. What is this moment?

P ϭ 6 kN
8m
u

O

1m

B
x

Solution
In order to produce the maximum moment about point O, P must act perpendicular
to the boom’s axis OA as shown in Fig. a. Thus,

a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a,
10 =

8
+ tan u
cos u



10 =

8
sin u
+
cos u
cos u



10 cos u - sin u = 8



x = x' + x";

10
1
8
cos u sin u =

1101
1101
1101



(1)

From the geometry shown in Fig. b,



a = tan-1 a
sin a =

Then Eq (1) becomes

1
b = 5.711°
10

1
1101

cos a =

10
1101

cos u cos 5.711° - sin u sin 5.711° =

8
1101

Referring that cos (u + 5.711°) = cos u cos 5.711° - sin u sin 5.711°


cos (u + 5.711°) =

8
1101

u + 5.711° = 37.247°
Ans.

u = 31.54° = 31.5°

Ans:
(MO)max = 48.0 kN # m (counterclockwise)
u = 31.5°
239


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4–13.
z

The 20-N horizontal force acts on the handle of the socket
wrench. What is the moment of this force about point B.
Specify the coordinate direction angles a, b, g of the
moment axis.

20 N
B

200 mm
A

60Њ
10 mm

50 mm
O

y

x

Solution
Force Vector And Position Vector. Referring to Fig. a,
F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rBA = { - 0.01i + 0.2j} m

Moment of Force F about point B.
MB = rBA * F
 




i
= † - 0.01
17.32

j
0.2
- 10

k
0†
0

= { - 3.3641 k} N # m
= { - 3.36 k} N # m

Ans.

Here the unit vector for MB is u = - k. Thus, the coordinate direction
angles of MB are

a = cos-1 0 = 90°

Ans.



b = cos-1 0 = 90°

Ans.



-1

Ans.

g = cos

( - 1) = 108°

Ans:
MB = { - 3.36 k} N # m
a = 90°
b = 90°
g = 180°
240


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4–14.
z

The 20-N horizontal force acts on the handle of the socket
wrench. Determine the moment of this force about point O.
Specify the coordinate direction angles a, b, g of the
moment axis.

20 N
B

200 mm
A

60Њ
10 mm

50 mm
O

y

x

Solution
Force Vector And Position Vector. Referring to Fig. a,
F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rOA = { - 0.01i + 0.2j + 0.05k} m

Moment of F About point O.





MO = rOA * F
i
= † - 0.01
17.32

j
0.2
- 10

k
0.05 †
0

= {0.5i + 0.8660j - 3.3641k} N # m
= {0.5i + 0.866j - 3.36k} N # m

Ans.

The magnitude of MO is
MO = 2(MO)2x + (MO)2y + (MO)2z = 20.52 + 0.86602 + ( -3.3641)2
= 3.5096 N # m

Thus, the coordinate direction angles of MO are


a = cos-1 c



b = cos-1 c



g = cos-1 c

(MO)x
MO
(MO)y
MO
(MO)z
MO

d = cos-1 a
d = cos-1 a
d = cos-1 a

0.5
b = 81.81° = 81.8°
3.5096
0.8660
b = 75.71° = 75.7°
3.5096

-3.3641
b = 163.45° = 163°
3.5096

Ans.
Ans.
Ans.

Ans:
MO = {0.5i + 0.866j - 3.36k} N # m
a = 81.8°
b = 75.7°
g = 163°
241


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4–15.
Two men exert forces of F = 80 lb and P = 50 lb on the
ropes. Determine the moment of each force about A. Which
way will the pole rotate, clockwise or counterclockwise?

6 ft

P

F

45
B

3

12 ft

5
4

C

SOLUTION

A

4
c + (MA)C = 80 a b (12) = 768 lb # ftb
5

Ans.

a + (MA)B = 50 (cos 45°)(18) = 636 lb # ftd

Ans.

Since (MA)C 7 (MA)B
Ans.

Clockwise

Ans:
(MA)C = 768 lb # ftb
(MA)B = 636 lb # ftd
Clockwise
242


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*4–16.
If the man at B exerts a force of P = 30 lb on his rope,
determine the magnitude of the force F the man at C must
exert to prevent the pole from rotating, i.e., so the resultant
moment about A of both forces is zero.

6 ft

P

F

45
B

SOLUTION
a+

3

12 ft

5
4

C

4
30 (cos 45°)(18) = Fa b(12) = 0
5

A

Ans.

F = 39.8 lb

Ans:
F = 39.8 lb
243


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4–17.
The torque wrench ABC is used to measure the moment or
torque applied to a bolt when the bolt is located at A and a

force is applied to the handle at C. The mechanic reads the
torque on the scale at B. If an extension AO of length d is
used on the wrench, determine the required scale reading if
the desired torque on the bolt at O is to be M.

F
M
O

A
d

B

l

C

Solution
Moment at A = m = Fl
Moment at O = M = (d + l)F


M = (d + l)



m = a

m

l

l
bM
d + l

Ans.

Ans:
m = a
244

l
bM
d + l


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4–18.
The tongs are used to grip the ends of the drilling pipe P.
Determine the torque (moment) MP that the applied force
F = 150 lb exerts on the pipe about point P as a function of
u. Plot this moment MP versus u for 0 … u … 90°.

F
u

P

SOLUTION

MP

MP = 150 cos u(43) + 150 sin u(6)

43 in.

= (6450 cos u + 900 sin u) lb # in.
= (537.5 cos u + 75 sin u) lb # ft
dMP
= -537.5 sin u + 75 cos u = 0
du

tan u =

75
537.5

6 in.

Ans.
u = 7.943°

At u = 7.943° , MP is maximum.
(MP)max = 538 cos 7.943° + 75 sin 7.943° = 543 lb # ft
Also (MP)max = 150 lb ¢ a

1

43 2
6 2 2
b + a b ≤ = 543 lb # ft
12
12

Ans:
MP = (537.5 cos u + 75 sin u) lb # ft
245


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4–19.
The tongs are used to grip the ends of the drilling pipe P. If
a torque (moment) of MP = 800 lb # ft is needed at P to
turn the pipe, determine the cable force F that must be
applied to the tongs. Set u = 30°.

SOLUTION
MP = F cos 30°(43) + F sin 30°(6)

F
u

P

Set MP = 800(12) lb # in.

6 in.

MP
43 in.

800(12) = F cos 30°(43) + F sin 30°(6)
Ans.

F = 239 lb

Ans:
F = 239 lb
246


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*4–20.
The handle of the hammer is subjected to the force of
F = 20 lb. Determine the moment of this force about the
point A.

F
30

5 in.
18 in.

SOLUTION
Resolving the 20-lb force into components parallel and perpendicular to the
hammer, Fig. a, and applying the principle of moments,

A
B

a +MA = - 20 cos 30°(18) - 20 sin 30°(5)
= -361.77 lb # in = 362 lb # in (Clockwise)

Ans.

Ans:
MA = 362 lb # in (Clockwise)
247


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4–21.
In order to pull out the nail at B, the force F exerted on the
handle of the hammer must produce a clockwise moment of
500 lb # in. about point A. Determine the required magnitude
of force F.

F
30

5 in.

18 in.

SOLUTION
Resolving force F into components parallel and perpendicular to the hammer, Fig. a,
and applying the principle of moments,

A
B

a + MA = - 500 = -F cos 30°(18) - F sin 30°(5)
Ans.

F = 27.6 lb

Ans:
F = 27.6 lb
248


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4–22.
y

Old clocks were constructed using a fusee B to drive the
gears and watch hands. The purpose of the fusee is to
increase the leverage developed by the mainspring A
as it uncoils and thereby loses some of its tension. The
mainspring can develop a torque (moment) Ts = ku,

where k = 0.015 N # m>rad is the torsional stiffness and u
is the angle of twist of the spring in radians. If the torque
Tf developed by the fusee is to remain constant as the
mainspring winds down, and x = 10 mm when u = 4 rad,
determine the required radius of the fusee when u = 3 rad.

x

A

B
y

t

x
12 mm
Ts

Tf

Solution
When u = 4 rad, r = 10 mm
Ts = 0.015(4) = 0.06 N # m
F =

0.06
= 5N
0.012

Tf = 5(0.010) = 0.05 N # m (constant)
When u = 3 rad,

Ts = 0.015(3) = 0.045 N # m
F =

0.045
= 3.75 N
0.012

For the fusee require
0.05 = 3.75 r
Ans.

r = 0.0133 m = 13.3 mm

Ans:
r = 13.3 mm
249


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4–23.
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

B

C
G3

9.5m

7.5 m

D
12.5 m

G1

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;

A

(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.

Ans:
(MR)A = (MR)B = 76.0 kN # md
250


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*4–24.
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
C
G3

9.5m
B

7.5 m

D
12.5 m

G1

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)
Ans.

MC = 4966.67 kg = 4.97 Mg

Ans:
MC = 4.97 Mg
251


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4–25.
If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb
man have centers of gravity located at points G1 , G2 and G3 ,

respectively, determine the resultant moment produced by
each weight about point A.

G3

B

D

G2
C
2.5 ft 1.75 ft

20 ft

G1

SOLUTION

10 ft
75

Moment of the weight of boom AB about point A:

A

a + (MAB)A = -1500(10 cos 75°) = -3882.29 lb # ft

= 3.88 kip # ft (Clockwise)

Ans.

Moment of the weight of cage BCD about point A:
a + (MBCD)A = -200(30 cos 75° + 2.5) = - 2052.91 lb # ft

= 2.05 kip # ft (Clockwise)

Ans.

Moment of the weight of the man about point A:
a + (Mman)A = -175(30 cos 75° + 4.25) = - 2102.55 lb # ft

= 2.10 kip # ft (Clockwise)

Ans.

Ans:
( MAB ) A = 3.88 kip # ftb
( MBCD ) A = 2.05 kip # ftb
( Mman ) A = 2.10 kip # ftb
252