Dynamics 14th edition by r c hibbeler chapter 05

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5–10.
Determine the components of the support reactions at the
fixed support A on the cantilevered beam.

6 kN

30Њ

SOLUTION

30Њ

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig. a,
Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about
point A.
+ ©F = 0;
:
x

1.5 m
1.5 m

1.5 m

4 kN

4 cos 30° - A x = 0
Ans.

A x = 3.46 kN
+ c ©Fy = 0;

A

A y - 6 - 4 sin 30° = 0
Ans.

A y = 8 kN

a+ ©MA = 0;MA - 6(1.5) - 4 cos 30° (1.5 sin 30°) - 4 sin 30°(3 + 1.5 cos 30°) = 0
MA = 20.2 kN # m

Ans.

Ans:
Ax = 3.46 kN
Ay = 8 kN
MA = 20.2 kN # m
397


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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.

5–11.
Determine the reactions at the supports.

400 N/m

5

3
4

B
A
3m

3m

Solution
Equations of Equilibrium. NA and By can be determined directly by writing the
moment equations of equilibrium about points B and A, respectively, by referring to
the beam’s FBD shown in Fig. a.
4
1
a+ ΣMB = 0;   (400)(6)(3) - NA a b(6) = 0
2
5
NA = 750 N

a+ ΣMA = 0;  By(6) -

Ans.

1
(400)(6)(3) = 0
2
Ans.

By = 600 N

Using the result of NA to write the force equation of equilibrium along the x axis,
+ ΣFx = 0;  750 a 3 b - Bx = 0
S
5

Ans.

Bx = 450 N

Ans:
NA = 750 N
By = 600 N
Bx = 450 N
398


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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.

*5–12.
4 kN

Determine the horizontal and vertical components of
reaction at the pin A and the reaction of the rocker B on
the beam.

B

A

30Њ

SOLUTION
Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can
be obtained by writing the moment equation of equilibrium about point A.
a
+ ©MA = 0;

6m

2m

NB cos 30°(8) - 4(6) = 0
Ans.

NB = 3.464 kN = 3.46 kN

Using this result and writing the force equations of equilibrium along the x and
y axes, we have
+ ©F = 0;
:
x

A x - 3.464 sin 30° = 0
Ans.

A x = 1.73 kN

+ c ©Fy = 0;

A y + 3.464 cos 30° - 4 = 0
Ans.

A y = 1.00 kN

Ans:
NB = 3.46 kN
Ax = 1.73 kN
Ay = 1.00 kN
399


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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.

5–13.
Determine the reactions at the supports.

900 N/m
600 N/m

B

A
3m

3m

Solution
Equations of Equilibrium. NA and By can be determined directly by writing the
moment equations of equilibrium about points B and A, respectively, by referring to
the FBD of the beam shown in Fig. a.
a+ ΣMB = 0;  600(6)(3) +

1
(300)(3)(5) - NA(6) = 0
2
Ans.

NA = 2175 N = 2.175 kN
a+ ΣMA = 0;  By(6) -

1
(300)(3)(1) - 600(6)(3) = 0
2
Ans.

By = 1875 N = 1.875 kN

Also, Bx can be determined directly by writing the force equation of equilibrium
along the x axis.
+ ΣFx = 0;      Bx = 0
S

Ans.

Ans:
NA = 2.175 kN

By = 1.875 kN
Bx = 0
400


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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.

5–14.
Determine the reactions at the supports.
800 N/m
A

3m

B
1m

3m

Solution
Equations of Equilibrium. NA can be determined directly by writing the moment
equation of equilibrium about point B by referring to the FBD of the beam shown
in Fig. a.
a+ ΣMB = 0;  800(5)(2.5) - NA(3) = 0
Ans.

NA = 3333.33 N = 3.33 kN

Using this result to write the force equations of equilibrium along the x and y axes,

+ ΣFx = 0;  Bx - 800(5) a 3 b = 0
S
5

Ans.

Bx = 2400 N = 2.40 kN

4
+ c ΣFy = 0;  3333.33 - 800 (5)a b - By = 0
5

Ans.

By = 133.33 N = 133 N

Ans:
NA = 3.33 kN
Bx = 2.40 kN
By = 133 N
401


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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.

5–15.
Determine the reactions at the supports.

5 kN

2m
B

A
6 kN
2m

2m

8 kN
2m

Solution
Equations of Equilibrium. Ay and NB can be determined by writing the moment
equations of equilibrium about points B and A, respectively, by referring to the FBD
of the truss shown in Fig. a.
a+ ΣMB = 0;  8(2) + 6(4) - 5(2) - Ay(6) = 0
Ans.

Ay = 5.00 kN
a+ ΣMA = 0;  NB(6) - 8(4) - 6(2) - 5(2) = 0

Ans.

NB = 9.00 kN

Also, Ax can be determined directly by writing the force equation of equilibrium
along x axis.
+ ΣFx = 0;  5 - Ax = 0  Ax = 5.00 kN

S

Ans.

Ans:
Ay = 5.00 kN
NB = 9.00 kN
Ax = 5.00 kN
402


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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.

*5–16.
Determine the tension in the cable and the horizontal and
vertical components of reaction of the pin A. The pulley at
D is frictionless and the cylinder weighs 80 lb.

D
2
1

A

SOLUTION

B

5 ft

C
5 ft

3 ft

Equations of Equilibrium: The tension force developed in the cable is the same
throughout the whole cable. The force in the cable can be obtained directly by
summing moments about point A.
a + ©MA = 0;

+ ©F = 0;
:
x

T152 + T ¢

≤ 1102 - 801132 = 0
25
T = 74.583 lb = 74.6 lb
2

Ax - 74.583 ¢

1
25

Ans.

≤ = 0

Ans.

Ax = 33.4 lb
+ c ©Fy = 0;

74.583 + 74.583

2
25

- 80 - By = 0
Ans.

Ay = 61.3 lb

Ans:
T = 74.6 lb
Ax = 33.4 lb
Ay = 61.3 lb
403


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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.

5–17.
The man attempts to support the load of boards having a
weight W and a center of gravity at G. If he is standing on a
smooth floor, determine the smallest angle u at which he can
hold them up in the position shown. Neglect his weight.

4 ft

u

SOLUTION
a + ©MB = 0;

G

4 ft

- NA (3.5) + W(3 - 4 cos u) = 0

As u becomes smaller, NA : 0 so that,

A

W(3 - 4 cos u) = 0

0.5 ft

B
3 ft

Ans.

u = 41.4°

Ans:

u = 41.4°
404


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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.

5–18.
Determine the components of reaction at the supports A and
B on the rod.

P
L
––
2

SOLUTION
Equations of Equilibrium: Since the roller at A offers no resistance to vertical
movement, the vertical component of reaction at support A is equal to zero. From
the free-body diagram, Ax, By, and MA can be obtained by writing the force
equations of equilibrium along the x and y axes and the moment equation of
equilibrium about point B, respectively.
+ ©F = 0;
:
x

Ax = 0

+ c ©Fy = 0;

By - P = 0

Pa

B

Ans.

Ans.

By = P
a + ©MB = 0;

A

L
––
2

L
b - MA = 0
2

MA =

PL
2

Ans.

Ans:
Ax = 0
By = P
PL
MA =
2
405


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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.

5–19.
The man has a weight W and stands at the center of the
plank. If the planes at A and B are smooth, determine the
tension in the cord in terms of W and u.

B

SOLUTION
a + ©MB = 0;

L
W a cos f b - NA(L cos f ) = 0
2

+ ©F = 0;
:
x

T cos u -NB sin u = 0

+ c ©Fy = 0;

T sin u +NB cos u +

NA

f

W
=
2

L

u

A

(1)

W
- W= 0
2

(2)

Solving Eqs. (1) and (2) yields:
T=

NB =

W
sin u
2

Ans.

W
cos u
2

Ans:
W
T=
sin u
2
406


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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.

*5–20.
A uniform glass rod having a length L is placed in the smooth
hemispherical bowl having a radius r. Determine the angle of
inclination u for equilibrium.

u
B

r
A

SOLUTION
By observation f = u.
Equilibrium:
a + ©MA = 0;

NB (2r cos u) - W a

L
cos ub = 0
2

+Q ©Fx = 0;

NA cos u - W sin u = 0

+a©Fy = 0;

(W tan u) sin u +

WL
4r

NA = W tan u

WL
- W cos u = 0
4r

sin2 u - cos2 u +

L
cos u = 0
4r

(1 - cos2 u) - cos2 u +
2 cos2 u cos u =

NB =

L
cos u = 0
4r

L
cos u - 1 = 0
4r

L ; 2L2 + 128r2
16r

Take the positive root
cos u =

L + 2L2 + 128r2
16r

u = cos - 1 ¢

L + 2L2 + 128r2
≤
16r

Ans.

Ans:
u = cos - 1a
407

L + 2L2 + 12r 2
b
16r


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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.

5–21.
The uniform rod AB has a mass of 40 kg. Determine the
force in the cable when the rod is in the position shown.
There is a smooth collar at A.

A

3m

60Њ
C

Solution

B

Equations of Equilibrium. TBC can be determined by writing the moment equation
of equilibrium about point O by referring to the FBD of the rod shown in Fig. a.
a+ ΣMO = 0;  40(9.81)(1.5 cos 600°) - TBC (3 sin 60°) = 0
Ans.

TBC = 113.28 N = 113 N

Ans:
TBC = 113 N
408


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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.

5–22.
If the intensity of the distributed load acting on the beam
is w = 3 kN>m, determine the reactions at the roller A and
pin B.

A
w
30Њ

B

3m
4m

Solution
Equations of Equilibrium. NA can be determined directly by writing the moment
equation of equilibrium about point B by referring to the FBD of the beam shown
in Fig. a.
a+ ΣMB = 0;  3(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0
Ans.

NA = 3.713 kN = 3.71 kN

Using this result to write the force equation of equilibrium along the x and y axes,
+ ΣFx = 0;  3.713 sin 30° - Bx = 0
S
Ans.

Bx = 1.856 kN = 1.86 kN
+ c ΣFy = 0;  By + 3.713 cos 30° - 3(4) = 0

Ans.

By = 8.7846 kN = 8.78 kN

Ans:
NA = 3.71 kN
Bx = 1.86 kN
By = 8.78 kN
409

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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.

5–23.
If the roller at A and the pin at B can support a load up
to 4 kN and 8 kN, respectively, determine the maximum
intensity of the distributed load w, measured in kN>m, so
that failure of the supports does not occur.

A
w
30Њ

B

3m
4m

Solution
Equations of Equilibrium. NA can be determined directly by writing the moment
equation of equilibrium about point B by referring to the FBD of the beam shown
in Fig. a.
a+ ΣMB = 0;  w(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0
NA = 1.2376 w
Using this result to write the force equation of equilibrium along x and y axes,
+ ΣFx = 0;  1.2376 w sin 30° - Bx = 0
S

Bx = 0.6188 w

+ c ΣFy = 0;  By + 1.2376 w cos 30° - w(4) = 0

By = 2.9282 w

Thus,
FB = 2Bx2 + By2 = 2(0.6188 w)2 + (2.9282 w)2 = 2.9929 w

It is required that
FB 6 8 kN;

2.9929 w 6 8

w 6 2.673 kN>m

1.2376 w 6 4

w 6 3.232 kN>m

And
NA 6 4 kN;

Thus, the maximum intensity of the distributed load is
Ans.

w = 2.673 kN>m = 2.67 kN>m

Ans:
w = 2.67 kN>m

410


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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.

*5–24.
The relay regulates voltage and current. Determine the force
in the spring CD, which has a stiffness of k 120 N m, so
that it will allow the armature to make contact at A in figure
(a) with a vertical force of 0.4 N. Also, determine the force
in the spring when the coil is energized and attracts the
armature to E, figure (b), thereby breaking contact at A.

50 mm 50 mm 30 mm
10°

A

B

A

C

E

B

C

k

k

D

D

SOLUTION
From Fig. (a):
a + ©MB = 0;

0.4(100 cos 10°) - Fs (30 cos 10°) = 0

(a)

Ans.

Fs = 1.333 N = 1.33 N
Fs = kx;

(b)

1.333 = 120 x
x = 0.01111 m = 11.11 mm

From Fig (b), energizing the coil requires the spring to be stretched an additional
amount
¢x = 30 sin 10° = 5.209 mm.
Thus

x¿ = 11.11 + 5.209 = 16.32 mm
Ans.

Fs = 120 (0.01632) = 1.96 N

Ans:
Fs = 1.33 N
Fs = 1.96 N
411


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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.

5–25.
Determine the reactions on the bent rod which is supported
by a smooth surface at B and by a collar at A, which is fixed
to the rod and is free to slide over the fixed inclined rod.

100 lb
3 ft

3 ft

200 lb и ft

A

2 ft

3
5

4

B

13
12

5

SOLUTION
12
5
b (6) - NB a b (2) = 0
13
13

a + ©MA = 0;

MA - 100 (3) - 200 + NB a

+ ©F = 0;
:
x

4
5
NA a b - NB a b = 0

5
13

+ c ©Fy = 0;

3
12
NA a b + NB a b - 100 = 0
5
13

Solving,
NA = 39.7 lb

Ans.

NB = 82.5 lb

Ans.

MA = 106 lb # ft

Ans.

Ans:
NA = 39.7 lb
NB = 82.5 lb
MA = 106 lb # ft
412

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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.

5–26.
The mobile crane is symmetrically supported by two
outriggers at A and two at B in order to relieve the
suspension of the truck upon which it rests and to provide
greater stability. If the crane and truck have a mass of
18 Mg and center of mass at G1, and the boom has a mass
of 1.8 Mg and a center of mass at G2, determine the vertical
reactions at each of the four outriggers as a function of the
boom angle u when the boom is supporting a load having a
mass of 1.2 Mg. Plot the results measured from u = 0° to
the critical angle where tipping starts to occur.

6.25 m

G2
6m

SOLUTION
+ ©MB = 0;

θ

- NA (4) + 18 A 103 B (9.81)(1) + 1.8 A 103 B (9.81) (2 - 6 sin u)

G1

+ 1.2 A 103 B (9.81) (2 - 12.25 sin u) = 0
NA = 58 860 - 62 539 sin u

A
2m

Tipping occurs when NA = 0, or

1m 1m

Ans.

u = 70.3°
+ c ©Fy = 0;

B

NB + 58 860 - 62 539 sin u - (18 + 1.8 + 1.2) A 103 B (9.81) = 0
NB = 147 150 + 62 539 sin u

Since there are two outriggers on each side of the crane,
NA
¿
= (29.4 - 31.3 sin u) kN
=
NA
2
NB¿ =

Ans.

NB
= (73.6 + 31.3 sin u) kN
2

Ans.

Ans:
u = 70.3°
=
NA
= (29.4 - 31.3 sin u) kN
NB= = (73.6 + 31.3 sin u) kN
413


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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.

5–27.
Determine the reactions acting on the smooth uniform bar,
which has a mass of 20 kg.

B

4m

A

30º

60º

Solution
Equations of Equilibrium. NB can be determined directly by writing the moment
equation of equilibrium about point A by referring to the FBD of the bar shown in
Fig. a.
a+ ΣMA = 0;  NB cos 30°(4) - 20(9.81) cos 30°(2) = 0


Ans.

NB = 98.1 N

Using this result to write the force equation of equilibrium along the x and y axes,
+ ΣFx = 0;  Ax - 98.1 sin 60° = 0
S

Ax = 84.96 N = 85.0 N

Ans.

+ c ΣFy = 0;  Ay + 98.1 cos 60° - 20(9.81) = 0
Ans.

Ay = 147.15 N = 147 N

Ans:
NB = 98.1 N
Ax = 85.0 N

Ay = 147 N
414


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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.

*5–28.
A linear torsional spring deforms such that an applied couple
moment M is related to the spring’s rotation u in radians by
the equation M = (20 u) N # m. If such a spring is attached to
the end of a pin-connected uniform 10-kg rod, determine
the angle u for equilibrium. The spring is undeformed
when u = 0°.

A

u

M ϭ (20 u) N и m
0.5 m

Solution
a+ ΣMA = 0;   - 98.1 (0.25 cos u) + 20(u) = 0
Solving for u,
Ans.

u = 47.5°

Ans:

u = 47.5°
415


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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.

5–29.
Determine the force P needed to pull the 50-kg roller over
the smooth step. Take u = 30°.

P

u
A

50 mm

300 mm

B

Solution
Equations of Equilibrium. P can be determined directly by writing the moment equation of
Equilibrium about point B, by referring to the FBD of the roller shown in Fig. a.
a+ ΣMB = 0;  P cos 30°(0.25) + P sin 30° ( 20.32 - 0.252 2 - 50(9.81) 20.32 - 0.252 = 0
Ans.

P = 271.66 N = 272 N

Ans:
P = 272 N
416


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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.

5–30.
Determine the magnitude and direction u of the minimum
force P needed to pull the 50-kg roller over the smooth step.

P

u
A

50 mm

300 mm

B

Solution
Equations of Equilibrium. P will be minimum if its orientation produces the greatest
moment about point B. This happens when it acts perpendicular to AB as shown in
Fig. a. Thus
u = f = cos-1 a

0.25

b = 33.56° = 33.6°
0.3

Ans.

Pmin can be determined by writing the moment equation of equilibrium about point
B by referring to the FBD of the roller shown in Fig. b.
a+ ΣMB = 0;  Pmin (0.3) - 50(9.81)(0.3 sin 33.56°) = 0
Ans.

Pmin = 271.13 N = 271 N

Ans:
Pmin = 271 N
417


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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.

5–31.
The operation of the fuel pump for an automobile depends
on the reciprocating action of the rocker arm ABC, which
is pinned at B and is spring loaded at A and D. When the
smooth cam C is in the position shown, determine the
horizontal and vertical components of force at the pin and
the force along the spring DF for equilibrium. The vertical
force acting on the rocker arm at A is FA = 60 N, and at C
it is FC = 125 N.

E

30°
F
FC = 125 N

FA = 60 N

B
A

C

D

SOLUTION
a + ©MB = 0;

- 60(50) - FB cos 30°(10) + 125(30) = 0

50 mm

- Bx + 86.6025 sin 30° = 0
Ans.

Bx = 43.3 N
+ c ©Fy = 0;

20 mm

Ans.

FB = 86.6025 = 86.6 N
+ ©F = 0;
:
x

10 mm

60 - By - 86.6025 cos 30° + 125 = 0
Ans.

By = 110 N

Ans:
FB = 86.6 N
Bx = 43.3 N
By = 110 N
418


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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.

*5–32.
Determine the magnitude of force at the pin A and in the
cable BC needed to support the 500-lb load. Neglect the
weight of the boom AB.

B

8 ft
C

22Њ

A

35Њ

SOLUTION
Equations of Equilibrium: The force in cable BC can be obtained directly by
summing moments about point A.
a + ©MA = 0;

FBC sin 13°(8) - 500 cos 35°(8) = 0
Ans.

FBC = 1820.7 lb = 1.82 kip
+
Q ©Fx = 0;

A x - 1820.7 cos 13° - 500 sin 35° = 0
A x = 2060.9 lb

a + ©Fy = 0;

A y + 1820.7 sin 13° - 500 cos 35° = 0
Ay = 0

Thus,

Ans.

FA = A x = 2060.9 lb = 2.06 kip

Ans:
FBC = 1.82 kip
FA = 2.06 kip
419


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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.

5–33.
The dimensions of a jib crane, which is manufactured by the
Basick Co., are given in the figure. If the crane has a mass of
800 kg and a center of mass at G, and the maximum rated
force at its end is F 15 kN, determine the reactions at its
bearings. The bearing at A is a journal bearing and supports
only a horizontal force, whereas the bearing at B is a thrust
bearing that supports both horizontal and vertical components.

3m
A
0.75 m
2m

G

F

SOLUTION
a + ©MB = 0;

Ans.

Ax = 25.4 kN
+ c ©Fy = 0;

By - 800 (9.81) - 15 000 = 0
Ans.

By = 22.8 kN
+ ©F = 0;
:
x

B

Ax (2) - 800 (9.81) (0.75) - 15 000(3) = 0

Bx - 25.4 = 0
Ans.

Bx = 25.4 kN

Ans:
Ax = 25.4 kN
By = 22.8 kN

Bx = 25.4 kN
420


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5–34.
The dimensions of a jib crane, which is manufactured by the
Basick Co., are given in the figure. The crane has a mass of
800 kg and a center of mass at G. The bearing at A is a journal
bearing and can support a horizontal force, whereas the
bearing at B is a thrust bearing that supports both horizontal
and vertical components. Determine the maximum load F that
can be suspended from its end if the selected bearings at A
and B can sustain a maximum resultant load of 24 kN and
34 kN, respectively.

3m
A
0.75 m
2m

F

SOLUTION
a + ©MB = 0;

G

B

Ax (2) - 800 (9.81) (0.75) - F (3) = 0

+ c ©Fy = 0;

By - 800 (9.81) - F = 0

+ ©F = 0;
:
x

Bx - Ax = 0

Assume Ax = 24 000 N.
Solving,
Bx = 24 kN
By = 21.9 kN
Ans.

F = 14.0 kN
FB =

(24)2 + (21.9)2 = 32.5 kN 6 34 kN

OK

Ans:
F = 14.0 kN
421