Dynamics 14th edition by r c hibbeler chapter 03

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3–1.
The members of a truss are pin connected at joint O.
Determine the magnitudes of F1 and F2 for equilibrium.
Set u = 60°.

y

5 kN

F2

70Њ
30Њ

x
O

SOLUTION
+ ©F = 0;
:
x

5
4

4
F2 sin 70° + F1 cos 60° - 5 cos 30° - (7) = 0
5

7 kN

u

3

F1

0.9397F2 + 0.5F1 = 9.930
+ c ©Fy = 0;

F2 cos 70° + 5 sin 30° - F1 sin 60° -

3
(7) = 0
5

0.3420F2 - 0.8660F1 = 1.7
Solving:
F2 = 9.60 kN

Ans.

F1 = 1.83 kN

Ans.

Ans:
 F2 = 9.60 kN

F1 = 1.83 kN
161


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3–2.
The members of a truss are pin connected at joint O.
Determine the magnitude of F1 and its angle u for
equilibrium. Set F2 = 6 kN.

y

5 kN

F2

70Њ
30Њ

x
O

SOLUTION
+ ©F = 0;
:
x

5

4

4
6 sin 70° + F1 cos u - 5 cos 30° - (7) = 0
5

7 kN

u

3

F1

F1 cos u = 4.2920
+ c ©Fy = 0;

6 cos 70° + 5 sin 30° - F1 sin u -

3
(7) = 0
5

F1 sin u = 0.3521
Solving:
u = 4.69°

Ans.

F1 = 4.31 kN

Ans.

Ans:
u = 4.69°
F1 = 4.31 kN
162


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3–3.
Determine the magnitude and direction u of F so that the
particle is in equilibrium.

y
8 kN

30Њ

x
5 kN

60Њ

4 kN

Solution

u

Equations of Equilibrium. Referring to the FBD shown in Fig. a,
+ ΣFx = 0;   F sin u + 5 - 4 cos 60° - 8 cos 30° = 0
   S


(1)

F sin u = 3.9282

F

   +cΣFy = 0;      8 sin 30° - 4 sin 60° - F cos u = 0


(2)

F cos u = 0.5359

Divide Eq (1) by (2),
sin u
= 7.3301
cos u
sin u
Realizing that tan u =
, then
cos u
tan u = 7.3301

Ans.

u = 82.23° = 82.2°

Substitute this result into Eq. (1),
F sin 82.23° = 3.9282


Ans.

F = 3.9646 kN = 3.96 kN

Ans:
 u = 82.2°
F = 3.96 kN
163


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*3–4.

The bearing consists of rollers, symmetrically confined
within the housing. The bottom one is subjected to a 125-N
force at its contact A due to the load on the shaft.
Determine the normal reactions NB and NC on the bearing
at its contact points B and C for equilibrium.

40°

SOLUTION
+ c ©Fy = 0;

NB

125 - NC cos 40° = 0
Ans.

NC = 163.176 = 163 N
+ ©F = 0;
:
x

NC

C

B
A
125 N

NB - 163.176 sin 40° = 0
Ans.

NB = 105 N

Ans:
NC = 163 N

NB = 105 N
164


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3–5.
The members of a truss are connected to the gusset plate. If
the forces are concurrent at point O, determine the
magnitudes of F and T for equilibrium. Take u = 90°.

y

9 kN
F
A
5 3 B
4

SOLUTION
3
f = 90° - tan - 1 a b = 53.13°
4

O

+ ©F = 0;
:
x

4
T cos 53.13° - F a b = 0
5

+ c ©Fy = 0;

3
9 - T sin 53.13° - Fa b = 0
5

x

u

C
T

Solving,
T = 7.20 kN

Ans.

F = 5.40 kN

Ans.

Ans:
T = 7.20 kN
F = 5.40 kN

165


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3–6.
The gusset plate is subjected to the forces of three members.
Determine the tension force in member C and its angle u for
equilibrium. The forces are concurrent at point O. Take
F = 8 kN.

y

9 kN
F
A
5 3 B
4

SOLUTION
+ ©F = 0;
:
x

4
T cos f - 8a b = 0
5

(1)

+ c ©Fy = 0;

3
9 - 8 a b - T sin f = 0
5

(2)

Rearrange then divide Eq. (1) into Eq. (2):

O

x

u

C
T

tan f = 0.656, f = 33.27°
T = 7.66 kN

Ans.

3
u = f + tan - 1 a b = 70.1°
4

Ans.

Ans:
T = 7.66 kN
u = 70.1°
166


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3–7.
C

The man attempts to pull down the tree using the cable and
small pulley arrangement shown. If the tension in AB is
60 lb, determine the tension in cable CAD and the angle u
which the cable makes at the pulley.

D

θ
20°

A
B
30°

SOLUTION
+ R©Fx¿ = 0;

60 cos 10° - T - T cos u = 0

+Q©Fy¿ = 0;

T sin u - 60 sin 10° = 0

Thus,
T(1 + cos u) = 60 cos 10°
u
T(2cos2 ) = 60 cos 10°
2
2T sin

(1)

u
u
cos = 60 sin 10°
2
2

(2)

Divide Eq.(2) by Eq.(1)
tan

u
= tan 10°
2
Ans.

u = 20°
T

Ans.

30.5 lb

Ans:
u = 20°
T = 30.5 lb
167


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*3–8.
The cords ABC and BD can each support a maximum load
of 100 lb. Determine the maximum weight of the crate, and
the angle u for equilibrium.

D

u
B

A

Equations of Equilibrium. Assume that for equilibrium, the tension along the

length of rope ABC is constant. Assuming that the tension in cable BD reaches the
limit first. Then, TBD = 100 lb. Referring to the FBD shown in Fig. a,


5

C

+ ΣFx = 0;   W a 5 b - 100 cos u = 0
S
13




13

12

Solution

100 cos u =

5W

13

+cΣFy = 0;      100 sin u - W - W a

100 sin u =

25
W
13

(1)
12
b = 0
13
(2)

Divide Eq. (2) by (1),
   

sin u
= 5
cos u

Realizing that tan u =

sin u
,
cos u

       tan u = 5
Ans.

     u = 78.69° = 78.7°

Substitute this result into Eq. (1),
   100 cos 78.69° =

5
W
13

     W = 50.99 lb = 51.0 lb 6 100 lb (O.K)

Ans.

Ans:
u = 78.7°
W = 51.0 lb
168


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3–9.
Determine the maximum force F that can be supported in
the position shown if each chain can support a maximum
tension of 600 lb before it fails.

B
3
4

5

30Њ
A

C

F

Solution
Equations of Equilibrium. Referring to the FBD shown in Fig. a,
4
   +cΣFy = 0;   TAB a b - F sin 30° = 0          TAB = 0.625 F
5

+ ΣFx = 0;        TAC + 0.625 F a 3 b - F cos 30° = 0   TAC = 0.4910 F
   S
5

Since chain AB is subjected to a higher tension, its tension will reach the limit first.
Thus,


TAB = 600;  0.625 F = 600
Ans.

F = 960 lb

Ans:
F = 960 lb
169

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3–10.
The block has a weight of 20 lb and is being hoisted at
uniform velocity. Determine the angle u for equilibrium and
the force in cord AB.

B
20Њ

A

u
C

Solution

D

F

Equations of Equilibrium. Assume that for equilibrium, the tension along the
length of cord CAD is constant. Thus, F = 20 lb. Referring to the FBD shown in
Fig. a,


+ ΣFx = 0;   20 sin u - TAB sin 20° = 0

S
TAB =



20 sin u

sin 20°

(1)

+cΣFy = 0;    TAB cos 20° - 20 cos u - 20 = 0

(2)

Substitute Eq (1) into (2),


20 sin u
cos 20° - 20 cos u = 20
sin 20°



sin u cos 20° - cos u sin 20° = sin 20°

Realizing that sin (u - 20°) = sin u cos 20° - cos u sin 20°, then
sin (u - 20°) = sin 20°
u - 20° = 20°

Ans.

u = 40°

Substitute this result into Eq (1)


TAB =

20 sin 40°
= 37.59 lb = 37.6 lb
sin 20°

Ans.

Ans:
u = 40°
TAB = 37.6 lb
170


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–11.
Determine the maximum weight W of the block that can be
suspended in the position shown if cords AB and CAD can
each support a maximum tension of 80 lb. Also, what is the
angle u for equilibrium?

B
20Њ

A

u
C

Solution

D

F

Equations of Equilibrium. Assume that for equilibrium, the tension along the
length of cord CAD is constant. Thus, F = W. Assuming that the tension in cord
AB reaches the limit first, then TAB = 80 lb. Referring to the FBD shown in Fig. a,


+ ΣFx = 0;   W sin u - 80 sin 20° = 0
S
W =



80 sin 20°

sin u

(1)

+cΣFy = 0;      80 cos 20° - W - W cos u = 0
W =

80 cos 20°

1 + cos u

(2)

Equating Eqs (1) and (2),
80 sin 20°
80 cos 20°
=
sin u
1 + cos u
sin u cos 20° - cos u sin 20° = sin 20°
Realizing then sin (u - 20°) = sin u cos 20° - cos u sin 20°, then
sin (u - 20°) = sin 20°
u - 20° = 20°


Ans.

u = 40°

Substitute this result into Eq (1)
  W =
      

  

80 sin 20°
= 42.56 lb = 42.6 lb 6 80 lb (O.K)
sin 40°

Ans.

Ans:
u = 40°
W = 42.6 lb
171


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*3–12.
The lift sling is used to hoist a container having a mass of
500 kg. Determine the force in each of the cables AB and
AC as a function of u. If the maximum tension allowed in
each cable is 5 kN, determine the shortest lengths of cables
AB and AC that can be used for the lift. The center of
gravity of the container is located at G.

F
A

SOLUTION

B

Free-Body Diagram: By observation, the force F1 has to support the entire weight
of the container. Thus, F1 = 50019.812 = 4905 N.

θ

θ

1.5 m

C

1.5 m

Equations of Equilibrium:
+ ©F = 0;
:
x

FAC cos u - FAB cos u = 0

+ c ©Fy = 0;

4905 - 2F sin u = 0

FAC = FAB = F
G

F = 52452.5 cos u6 N

Thus,
FAC = FAB = F = 52.45 cos u6 kN

Ans.

If the maximum allowable tension in the cable is 5 kN, then
2452.5 cos u = 5000
u = 29.37°
From the geometry, l =

1.5
and u = 29.37°. Therefore
cos u
l =

1.5
= 1.72 m
cos 29.37°

Ans.

Ans:
FAC = {2.45 cos u} kN
l = 1.72 m
172


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–13.
A nuclear-reactor vessel has a weight of 500 ( 103 ) lb.
Determine the horizontal compressive force that the
spreader bar AB exerts on point A and the force that each
cable segment CA and AD exert on this point while the
vessel is hoisted upward at constant velocity.

30Њ

C 30Њ

A

B

D

E

Solution
At point C :
+ ΣFx = 0;   FCB cos 30° - FCA cos 30° = 0  
S
FCB = FCA

+cΣFy = 0;      500 ( 103 ) - FCA sin 30° - FCB sin 30° = 0


500 ( 103 ) - 2FCA sin 30° = 0

FCA = 500 ( 103 ) lb Ans.


At point A :

+ ΣFx = 0;          500 ( 103 ) cos 30° - FAB = 0
S
FAB = 433 ( 103 ) lb


+cΣFy = 0;      500 ( 10

3



Ans.

) sin 30° - FAD = 0

FAD = 500 ( 103 ) sin 30°

FAD = 250 ( 103 ) lb

Ans.

Ans:
FCA = 500 ( 103 ) lb
FAB = 433 ( 103 ) lb
FAD = 250 ( 103 ) lb

173


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–14.
Determine the stretch in each spring for equlibrium of the
2-kg block. The springs are shown in the equilibrium
position.

3m

4m

C

3m

B
kAC ϭ 20 N/m
kAB ϭ 30 N/m

SOLUTION
FAD = 2(9.81) = xAD(40)
+ ©F = 0;
:
x
+ c ©Fy = 0;

xAD = 0.4905 m

A

Ans.

4
1
FAB a b - FAC a
b = 0
5
22
FAC a

kAD ϭ 40 N/m

1

3
b + FAB a b - 2(9.81) = 0
5
22

D

FAC = 15.86 N
xAC =

15.86
= 0.793 m

20

Ans.

FAB = 14.01 N
xAB =

14.01
= 0.467 m
30

Ans.

Ans:
xAD = 0.4905 m
xAC = 0.793 m
xAB = 0.467 m
174


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3–15.
The unstretched length of spring AB is 3 m. If the block is
held in the equilibrium position shown, determine the mass
of the block at D.

3m

4m

C

3m

B
20 N/m

kAC

kAB

SOLUTION

A

F = kx = 30(5 - 3) = 60 N
+ ©F = 0;
:
x

4
Tcos 45° - 60 a b = 0
5
T = 67.88 N

+ c ©Fy = 0;

30 N/m

D

3
-W + 67.88 sin 45° + 60 a b = 0
5
W = 84 N
m =

84
= 8.56 kg
9.81

Ans.

Ans:
m = 8.56 kg
175


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*3–16.
Determine the mass of each of the two cylinders if they
cause a sag of s = 0.5 m when suspended from the rings at
A and B. Note that s = 0 when the cylinders are removed.

2m

1.5 m

s

1m

2m
D

C

k

SOLUTION

100 N/m

k
A

100 N/m

B

TAC = 100 N>m (2.828 - 2.5) = 32.84 N
+ c ©Fy = 0;

32.84 sin 45° - m(9.81) = 0
Ans.

m = 2.37 kg

Ans:
m = 2.37 kg
176


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–17.

Unstretched
position

Determine the stiffness kT of the single spring such that the
force F will stretch it by the same amount s as the force F
stretches the two springs. Express kT in terms of stiffness k1
and k2 of the two springs.

kT
F

s
k1

k2
F

s

Solution
F = ks
s = s1 + s2
s =


F
F
F
=
+
kT
k1
k2

1
1
1
=
+
kT
k1
k2

Ans.

Ans:
1
1

1
=
+
kT
k1
k2
177


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3–18.
If the spring DB has an unstretched length of 2 m, determine
the stiffness of the spring to hold the 40-kg crate in the
position shown.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,




+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0
S
113
12
+cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2
1
b + TCDa
b - 40(9.81) = 0
113
12

(2)

TBD = 282.96 N   TCD = 332.96 N

The stretched length of the spring is

Then, x = l - l0 =

l = 232 + 22 = 213 m

( 113 - 2 ) m. Thus,
Fsp = kx;



282.96 = k ( 113 - 2 )

Ans.

k = 176.24 N>m = 176 N>m

Ans:
k = 176 N>m
178


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3–19.
Determine the unstretched length of DB to hold the 40-kg
crate in the position shown. Take k = 180 N>m.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,




+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0
S
113
12
+cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2
1
b + TCDa
b - 40(9.81) = 0
113
12

(2)

TBD = 282.96 N   TCD = 332.96 N
The stretched length of the spring is
l = 232 + 22 = 213 m

Then, x = l - l0 = 113 - l0. Thus
Fsp = kx;



282.96 = 180 ( 113 - l0 )

Ans.

l0 = 2.034 m = 2.03 m

Ans:
l0 = 2.03 m
179

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*3–20.
A vertical force P = 10 lb is applied to the ends of the 2-ft
cord AB and spring AC. If the spring has an unstretched
length of 2 ft, determine the angle u for equilibrium. Take
k = 15 lb>ft.

2 ft

B

2 ft

C

u

k
A

SOLUTION
+ ©F = 0;
:
x

Fs cos f - T cos u = 0

(1)

+ c ©Fy = 0;

T sin u + Fs sin f - 10 = 0

(2)
P

s = 2(4)2 + (2)2 - 2(4)(2) cos u - 2 = 2 25 - 4 cos u - 2
Fs = ks = 2k( 25 - 4 cos u - 1)
From Eq. (1): T = Fs a

cos f
b
cos u

T = 2k A 25 - 4 cos u - 1 B ¢

2 - cos u
25 - 4 cos u

≤a

1
b
cos u

From Eq. (2):
2k a 25 - 4 cos - 1 b(2 - cos u)

2k a 25 - 4 cos u - 1 b2 sin u
tan u +

25 - 4 cos u
a 25 - 4 cos u - 1 b
25 - 4 cos u

(2 tan u - sin u + sin u) =

tan u a 25 - 4 cos u - 1 b
25 - 4 cos u

=

= 10

225 - 4 cos u

10
2k

10
4k

Set k = 15 lb>ft
Solving for u by trial and error,
Ans.

u = 35.0°

Ans:
u = 35.0°
180


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3–21.
Determine the unstretched length of spring AC if a force
P = 80 lb causes the angle u = 60° for equilibrium. Cord
AB is 2 ft long. Take k = 50 lb>ft.

2 ft

B

2 ft

C

u

k
A

SOLUTION
l = 242 + 22 - 2(2)(4) cos 60°
l = 212

P

2
212
=
sin 60°
sin f
f = sin - 1 ¢

2 sin 60°
212

≤ = 30°

+ c ©Fy = 0;

T sin 60° + Fs sin 30° - 80 = 0

+ ©F = 0;
:
x

-T cos 60° + Fs cos 30° = 0

Solving for Fs,
Fs = 40 lb
Fs = kx
40 = 50(212 - l¿)

l = 212 -

40
= 2.66 ft
50

Ans.

Ans:
l = 2.66 ft
181


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3–22.
The springs BA and BC each have a stiffness of 500 N>m and an
unstretched length of 3 m. Determine the horizontal force F
applied to the cord which is attached to the small ring B so
that the displacement of the ring from the wall is d = 1.5 m.

A
k

500 N/m
B

6m

F

SOLUTION
+ ©F = 0;
:
x

k

1.5
211.25

500 N/m

(T)(2) - F = 0

C
d

T = ks = 500(232 + (1.5)2 - 3) = 177.05 N
Ans.

F = 158 N

Ans:
F = 158 N
182


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3–23.
The springs BA and BC each have a stiffness of 500 N>m and an
unstretched length of 3 m. Determine the displacement d of the
cord from the wall when a force F = 175 N is applied to the cord.

A
k

500 N/m
B

6m

F

SOLUTION
+ ©F = 0;
:
x

k

500 N/m

175 = 2T sin u

C

T sin u = 87.5
TC

d
2

23 + d 2

d

S = 87.5

T = ks = 500( 232 + d 2 - 3)
d a1 -

3
29 + d2

b = 0.175

By trial and error:
Ans.

d = 1.56 m

Ans:
d = 1.56 m
183

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*3–24.
Determine the distances x and y for equilibrium if F1 = 800 N
and F2 = 1000 N.

F1

D

C
y

B

F2

2m

Solution
Equations of Equilibrium. The tension throughout rope ABCD is constant, that is
F1 = 800 N. Referring to the FBD shown in Fig. a,


A
x

+cΣFy = 0;   800 sin f - 800 sin u = 0   f = 0

+ ΣFx = 0;     
 
1000 - 2[800 cos u] = 0     
u = 51.32°
S
Referring to the geometry shown in Fig. b,


y = 2 m

Ans.

2
= tan 51.32°;  x = 1.601 m = 1.60 m
x

Ans.

and


Ans:
y = 2m
x = 1.60 m
184


© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–25.
Determine the magnitude of F1 and the distance y if x = 1.5 m
and F2 = 1000 N.

F1

D

C
y

B

F2

2m

Solution
Equations of Equilibrium. The tension throughout rope ABCD is constant,
that is F1. Referring to the FBD shown in Fig. a,


+cΣFy = 0;      F1a

y
2

2

2y + 1.5

y

2




b - F1a
2

2y + 1.5

=

A
x

2
b = 0
2.5

2
2.5

Ans.

y = 2 m

+ ΣFx = 0;         1000 - 2c F1a 1.5 b d = 0
S

2.5

Ans.

F1 = 833.33 N = 833 N
 

Ans:
y = 2m
F1 = 833 N
185