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Engineering Mechanics Statics and Dynamics 14th Edition Hibbeler
SOLUTIONS MANUAL People also search: https://testbanklive.com/do https://testbankliv e.com/download/engineering-m wnload/engineering-mechanics-statics-and-dynam echanics-statics-and-dynamicsics14th-edition-hibbeler-solutions-manual/ 22 – 1. 1.
A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibr ium position and given a downward downward velocity of 1.50 m > s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.
SO LUT ION + T Σ F = ma ; y
mg
k ( y y + y ) =
-
y
st
$
k
y +
Hence
p =
=
k
$
my
m
= mg st
y = 0
Where k =
Bm
where ky
8(9.81) = 448.46 N > m 0.175
448.46 = 7.487 B 8 $
$
y + (7.487)2 y = 0
6
y + 56.1 y = 0
Ans.
The solution of the above differen tial equatio equation n is of the form: y = A sin pt + B cos pt #
v = y = Ap cos pt
(1)
- Bp
sin pt
(2)
At t = 0, y = 0.1 m and v = v0 = 1.50 m > s From Eq. (1)
0 .1 =
A
sin 0 + B cos 0
v0 1.50 = = 0.2003 m p 7.487
From Eq. (2)
v0 = Ap cos 0
Hence
y = 0.2003 sin 7.487t + 0.1 cos 7.487t
At t = 0.22 s,
y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]
= 0.192 m
-
0
B = 0.1 m A =
Ans.
Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m 1190
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22 – 2. 2.
A spring has a stiffness of 800 N > m. If a 2-kg block is br ium attached to the spring, pushed 50 mm above its equili br position, and released from rest, determine the equation that describes the block ’s motion. Assume that positive displacement is downward.
SO LUT ION k
p =
Am
=
800 = 20 A 2
x = A sin pt + B cos pt x =
- 0.05
- 0.05
m
when t = 0,
= 0 + B;
v = Ap cos pt
- Bp
B =
- 0.05
sin pt
v = 0 when t = 0, 0 = A(20)
-
0;
A
= 0
Thus, x =
- 0.05
cos (20t )
Ans.
Ans: x = - 0.05 cos (20t )
1191
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22 – 3. 3.
A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its it s equili br br ium position
and given a downward
velocity of 0.75 m> s,
determine the equation which describes the motion. What is the phase angle? Assume that positive displacement is downward.
SOLUTION k =
F
y
=
15(9.81) = 735.75 N> m 0.2
k vn =
Am
=
735.75 = 7.00 A 15
y = A sin v n t + B cos v n t y = 0.1 m when t = 0, 0.1 = 0 + B; v = A v n cos vn t
B = 0.1 -
Bvn sin vn t
v = 0.75 m> s when t = 0,
0.75 = A(7.00) A = 0.107 y = 0.107 sin (7.00t) + 0.100 ccos os (7.00 t) f
= tan - 1 a
B 0.100 b = tan - 1 a b = 43.0° A
Ans. Ans.
0.107
Ans: y = 0.107 sin (7.00 (7.00 t ) + 0.100 cos (7.00 t ) f
= 43.0°
1192
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*22 – 4. 4.
When a 20-lb weight weight is suspended from a spring, the spring spr ing is stretched a distance of 4 in. Determine the natural frequency and the period of vibrati vibration on for a 10-lb weigh t attached to the same spring.
S O LUT IO N k =
20 12
vn =
t
=
4
= 60 lb> ft
60 k = 13.90 rad > s = 10 A 32.2 Am
Ans.
2p = 0.452 s vn
Ans.
Ans: vn = 13.90 rad > s t = 0.452 s
1193
22 – 5. 5.
When a 3-kg block is suspended from a spring, the spring is of 60 mm. Determine the natural stretched a distance frequency and the period of vibratio n for a 0.2-kg block attached to the same spring.
S O LUT IO N k =
vn =
f
=
t
=
3(9.81) F = = 490.5 N> m ¢x 0.060
490.5 k = 49.52 = 49.5 rad > s = m A A 0.2
Ans.
vn 49.52 = 7.88 Hz = 2p 2p 1 f
=
1 = 0.127 s 7.88
Ans.
Ans: vn = 49.5 rad > s t = 0.127 s
1194
22 – 6. 6.
An 8-kg block is suspen suspended ded from a spring having h aving a stiffness k = 80 N> m. If the block is given an upward velocity of 0.4 m> s when when it is 90 mm above its equilibrium position, motion and the determine the equation which describes the motion maximum upward displacement of the block measured from the equilibriu m position. Assume that positive displacement is measured downward.
S O LUT IO N vn =
80 k = A 8 = 3.162 rad > s Am 0.4 m> s,
x
y =
- 0.09
m at t
= = 0 -
Ans: vn = 19.7 rad > s C = 1 in. y = (0.0833 cos 19.7t ) f t
*22 – 8. 8.
A 6-lb weight weight is suspended from a spring having a stiffness b> in. If the weight is given an upward k = 3 l b upward velocity of 20 ft> s when it is 2 in. above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.
S O LUT IO N k = 3(12) = 36 lb> ft 36 k vn = = 13.90 rad > s = 6 A Am 32.2
t = 0,
y =
- 20
ft> s,
y = -
1 ft 6
From Eq. 22 – 3, 3, -
1 = 0 + B 6
B =
- 0.167
From Eq. 22 – 4, 4, - 20
= A(13.90) + 0
A =
- 1.44
Thus,
y = [ - 1.44 sin (13.9t)
-
0.167 cos (13.9t)] ft
Ans.
From Eq. 22 – 10, 10, C =
2
2A
+ B2 =
2
2 (1.44)
+ ( - 0.167)2 = 1.45 ft
Ans.
Ans: y = [ - 1.44 sin (13.9t ) C = 1.45 f t 1197
-
0.167 cos (13.9 t )] )] f t
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22 – 9. 9.
A 3-kg block is suspend ed from a spring having a stiffn ess of k = 200 N> m. If the block is pushed 50 mm upward from its equilibriu m position posit ion and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward.
SOLUTIO N vn =
k
200 = 8.16 rad >s A 3
=
Am
Ans.
x = A sin v n t + B cos vn t x =
- 0.05
m when t = 0,
- 0.05
= 0 + B; v = Ap cos vn t
B = - 0.05 - Bv n sin vn t
v = 0 when t = 0,
0 = A(8.165)
-
0;
A = 0
Hence,
x = C =
2
2A
+ B2 =
- 0.05 2
2 (0)
cos (8.16 t)
Ans.
+ ( - 0.05) = 0.05 m = 50 mm
Ans.
Ans: vn = 8.16 rad > s x = - 0.05 cos (8.16t ) C = 50 mm 1198
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22 – 10. 10.
The uniform rod of mass m is supported by a pin at A and a spring at B. If B is given a small sideward displacement and released, determine the natural period of vibr ation. ation.
A
L
Solution Equation of Motio n. The mass moment of inertia of the rod about A is I A =
a + Σ M A
However;
= I a ;
- mg
A
a
sin u b
-
mL 2.
3
Referring to the FBD. of the rod, Fig. a, L
1
B
1
k
(kx cos u)( L L) = a mL 2 b a
2
3
x = L sin u. Then - mg L
sin u
2
-
1 2 mL a 3
kL 2 sin u cos u =
Using the trigonometry identity sin 2u = 2 sin u cos u, 2
- mg L
sin u
2
1
KL -
2
sin 2u =
3
2
mL a
$ Here since u is small sin u equation becomes
u and sin 2u
mg L 1 2 $ mL u + a + 3 2
2u. Also a = u . Then the above
2
kL b u
= 0
$ L 3mg + 6k u + u = 0 2m L Comparing to that of the Standard form, vn =
t
=
2p vn
= 2 p
L 3mg + 6k . Then 2m L A
2m L A 3mg + 6kL
Ans.
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Ans:
2m L t
= 2 p
L A 3mg + 6k
1199
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22 – 11. 11. While standing in an elevator, the man holds a pen pendu dulum which consists of an 18-in. cord cord and a 0.5-lb bob. bob. If the elevator is descending with an an acceleration a = 4 f t> s2, determine the natural period of vibration for small amplitudes of swing. a
4 f t/s2
S O LUT IO N Since the acceleration of the pendulum is (32.2
-
4) = 28.2 ft> s2
Using the result of Example 22 – 1, 1, We have vn =
t
=
g Al 2p vn
=
=
28.2 = 4.336 rad > s A 18 > 12 2p = 1.45 s 4.336
Ans.
Ans: t = 1.45 s 1200
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*22 – 12. 12. Determine the natural period of vibration of the uniform bar of mass m when it is displaced downward slightly and released.
O
k L —
L —
2
2
Solution Equation of Motion . The mass moment of inertia of the bar about O is I 0 =
1 mL 2. 12
Referring to the FBD of the rod, Fig. a, a + Σ M 0
= I a ;
- ky
L
1
2
12
cos u a
0
However, y =
L
2
b = a
sin u. Then L
-k
mL 2 b a
a
2
1
L
sin u b cos u a
2
b =
12
2
mL a
Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain 1 mL2a + 12
2
L k
8
sin 2u = 0
Here since u is small, sin 2u 12 12
2u. Also, a = u . Then the above equ ation becomes
$kL 2
mL u +
4
u = 0
$ 3k u + u = 0 m Comparing to that of the Standard form, vn =
t
=
2p vn
= 2 p
m A 3k
3k . Then Am Ans.
Ans: t
= 2 p
m A 3 k
1201
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22 – 13. 13.
The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of k G. If it is displaced a slight amount u from its equilibriu m position and released, determine the natural period of vibration.
O
d u G
SOLUTION a +
©M O = IO a;
- mgd
$ u +
$ 2 2 + md Du sin u = C mk G gd 2
2
k
G
sin u = 0
+ d
However, for small rotation sin u L u. Hence $ gd u + 2 u = 0 2 k G + d gd
From the above differential equation, vn = t
=
2p vn
=
2p gd
B k 2G + d 2
= 2 p
.
2 k G + d 2 C gd
Ans.
A k G2 + d 2
Ans: t
= 2 p C
k 2G + d 2 gd
1202
22 – 14. 14.
The 20-lb rectangul ar plate has a natural period of vibration t = 0.3 s, as it oscillates around the axis of rod AB. ad, Determine the torsional stiffness k , measured in lb ft > r ad, of the rod. Neglect the mass of the rod.
A
k B
Solution T =
k u
1 Σ M z
= I za ;
- k u
=
20
2
4 ft
$
a b (2) u 12 32.2
$ u + k (4.83)u (4.83)u = 0 2 ft t
=
2p 2k (4.83 )
= 0.3
k = 90.8 lb # ft > rad
Ans.
Ans: k = 90.8 lb # ft > rad
Engineering Mechanics Statics and Dynamics 14th Edition Hibbeler
SOLUTIONS MANUAL Full download: https://testbanklive.com/download/engineering-m https://testbanklive.com/d ownload/engineering-mechanics-statics-andechanics-statics-anddynamics14th-edition-hibbeler-solutions-manual/ People also search: engineering mechanics statics & dynamics (14th edition) pdf engineering mechanics statics and dynamics 14th edition solutions statics and dynamics 13th edition statics and dynamics 14th edition chegg engineering mechanics: statics & dynamics pdf statics and dynamics hibbeler pdf engineering mechanics statics and dynamics hibbeler pdf engineering mechanics statics and dynamics 14th edition pdf free download
SOLUTIONS MANUAL People also search: https://testbanklive.com/do https://testbankliv e.com/download/engineering-m wnload/engineering-mechanics-statics-and-dynam echanics-statics-and-dynamicsics14th-edition-hibbeler-solutions-manual/ 22 – 1. 1.
A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibr ium position and given a downward downward velocity of 1.50 m > s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.
SO LUT ION + T Σ F = ma ; y
mg
k ( y y + y ) =
-
y
st
$
k
y +
Hence
p =
=
k
$
my
m
= mg st
y = 0
Where k =
Bm
where ky
8(9.81) = 448.46 N > m 0.175
448.46 = 7.487 B 8 $
$
y + (7.487)2 y = 0
6
y + 56.1 y = 0
Ans.
The solution of the above differen tial equatio equation n is of the form: y = A sin pt + B cos pt #
v = y = Ap cos pt
(1)
- Bp
sin pt
(2)
At t = 0, y = 0.1 m and v = v0 = 1.50 m > s From Eq. (1)
0 .1 =
A
sin 0 + B cos 0
v0 1.50 = = 0.2003 m p 7.487
From Eq. (2)
v0 = Ap cos 0
Hence
y = 0.2003 sin 7.487t + 0.1 cos 7.487t
At t = 0.22 s,
y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]
= 0.192 m
-
0
B = 0.1 m A =
Ans.
Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m 1190
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22 – 2. 2.
A spring has a stiffness of 800 N > m. If a 2-kg block is br ium attached to the spring, pushed 50 mm above its equili br position, and released from rest, determine the equation that describes the block ’s motion. Assume that positive displacement is downward.
SO LUT ION k
p =
Am
=
800 = 20 A 2
x = A sin pt + B cos pt x =
- 0.05
- 0.05
m
when t = 0,
= 0 + B;
v = Ap cos pt
- Bp
B =
- 0.05
sin pt
v = 0 when t = 0, 0 = A(20)
-
0;
A
= 0
Thus, x =
- 0.05
cos (20t )
Ans.
Ans: x = - 0.05 cos (20t )
1191
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22 – 3. 3.
A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its it s equili br br ium position
and given a downward
velocity of 0.75 m> s,
determine the equation which describes the motion. What is the phase angle? Assume that positive displacement is downward.
SOLUTION k =
F
y
=
15(9.81) = 735.75 N> m 0.2
k vn =
Am
=
735.75 = 7.00 A 15
y = A sin v n t + B cos v n t y = 0.1 m when t = 0, 0.1 = 0 + B; v = A v n cos vn t
B = 0.1 -
Bvn sin vn t
v = 0.75 m> s when t = 0,
0.75 = A(7.00) A = 0.107 y = 0.107 sin (7.00t) + 0.100 ccos os (7.00 t) f
= tan - 1 a
B 0.100 b = tan - 1 a b = 43.0° A
Ans. Ans.
0.107
Ans: y = 0.107 sin (7.00 (7.00 t ) + 0.100 cos (7.00 t ) f
= 43.0°
1192
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*22 – 4. 4.
When a 20-lb weight weight is suspended from a spring, the spring spr ing is stretched a distance of 4 in. Determine the natural frequency and the period of vibrati vibration on for a 10-lb weigh t attached to the same spring.
S O LUT IO N k =
20 12
vn =
t
=
4
= 60 lb> ft
60 k = 13.90 rad > s = 10 A 32.2 Am
Ans.
2p = 0.452 s vn
Ans.
Ans: vn = 13.90 rad > s t = 0.452 s
1193
22 – 5. 5.
When a 3-kg block is suspended from a spring, the spring is of 60 mm. Determine the natural stretched a distance frequency and the period of vibratio n for a 0.2-kg block attached to the same spring.
S O LUT IO N k =
vn =
f
=
t
=
3(9.81) F = = 490.5 N> m ¢x 0.060
490.5 k = 49.52 = 49.5 rad > s = m A A 0.2
Ans.
vn 49.52 = 7.88 Hz = 2p 2p 1 f
=
1 = 0.127 s 7.88
Ans.
Ans: vn = 49.5 rad > s t = 0.127 s
1194
22 – 6. 6.
An 8-kg block is suspen suspended ded from a spring having h aving a stiffness k = 80 N> m. If the block is given an upward velocity of 0.4 m> s when when it is 90 mm above its equilibrium position, motion and the determine the equation which describes the motion maximum upward displacement of the block measured from the equilibriu m position. Assume that positive displacement is measured downward.
S O LUT IO N vn =
80 k = A 8 = 3.162 rad > s Am 0.4 m> s,
x
y =
- 0.09
m at t
= = 0 -
Ans: vn = 19.7 rad > s C = 1 in. y = (0.0833 cos 19.7t ) f t
*22 – 8. 8.
A 6-lb weight weight is suspended from a spring having a stiffness b> in. If the weight is given an upward k = 3 l b upward velocity of 20 ft> s when it is 2 in. above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.
S O LUT IO N k = 3(12) = 36 lb> ft 36 k vn = = 13.90 rad > s = 6 A Am 32.2
t = 0,
y =
- 20
ft> s,
y = -
1 ft 6
From Eq. 22 – 3, 3, -
1 = 0 + B 6
B =
- 0.167
From Eq. 22 – 4, 4, - 20
= A(13.90) + 0
A =
- 1.44
Thus,
y = [ - 1.44 sin (13.9t)
-
0.167 cos (13.9t)] ft
Ans.
From Eq. 22 – 10, 10, C =
2
2A
+ B2 =
2
2 (1.44)
+ ( - 0.167)2 = 1.45 ft
Ans.
Ans: y = [ - 1.44 sin (13.9t ) C = 1.45 f t 1197
-
0.167 cos (13.9 t )] )] f t
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22 – 9. 9.
A 3-kg block is suspend ed from a spring having a stiffn ess of k = 200 N> m. If the block is pushed 50 mm upward from its equilibriu m position posit ion and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward.
SOLUTIO N vn =
k
200 = 8.16 rad >s A 3
=
Am
Ans.
x = A sin v n t + B cos vn t x =
- 0.05
m when t = 0,
- 0.05
= 0 + B; v = Ap cos vn t
B = - 0.05 - Bv n sin vn t
v = 0 when t = 0,
0 = A(8.165)
-
0;
A = 0
Hence,
x = C =
2
2A
+ B2 =
- 0.05 2
2 (0)
cos (8.16 t)
Ans.
+ ( - 0.05) = 0.05 m = 50 mm
Ans.
Ans: vn = 8.16 rad > s x = - 0.05 cos (8.16t ) C = 50 mm 1198
currently currently © 2016 © 2016 Pearson Pearson Inc., Inc., Upper Upper S addle Saddle Saddle River, River, NJ.NJ. AllAll rights rights reserved. reserved. ThiThis This s material material i s protected is is protected under under all all copyright copyright laws laws as they as they Education, Education, reproduced, reproduced, publisher. publish publisher. publish er. er. exist. exist. No No portion portion of this of this material material maymay be be in any in any form form form or by or any by any means, means, without with out without permission permission permis sion in writing i n writing in from from thethe
22 – 10. 10.
The uniform rod of mass m is supported by a pin at A and a spring at B. If B is given a small sideward displacement and released, determine the natural period of vibr ation. ation.
A
L
Solution Equation of Motio n. The mass moment of inertia of the rod about A is I A =
a + Σ M A
However;
= I a ;
- mg
A
a
sin u b
-
mL 2.
3
Referring to the FBD. of the rod, Fig. a, L
1
B
1
k
(kx cos u)( L L) = a mL 2 b a
2
3
x = L sin u. Then - mg L
sin u
2
-
1 2 mL a 3
kL 2 sin u cos u =
Using the trigonometry identity sin 2u = 2 sin u cos u, 2
- mg L
sin u
2
1
KL -
2
sin 2u =
3
2
mL a
$ Here since u is small sin u equation becomes
u and sin 2u
mg L 1 2 $ mL u + a + 3 2
2u. Also a = u . Then the above
2
kL b u
= 0
$ L 3mg + 6k u + u = 0 2m L Comparing to that of the Standard form, vn =
t
=
2p vn
= 2 p
L 3mg + 6k . Then 2m L A
2m L A 3mg + 6kL
Ans.
currently currently © 2016 © 2016 Pearson Pearson Inc., Inc., Upper Upper S addle Saddle Saddle River, River, NJ.NJ. AllAll rights rights reserved. reserved. ThiThis This s material material i s protected is is protected under under all all copyright copyright laws laws as they as they Education, Education, reproduced, reproduced, publisher. publish publisher. publish er. er. exist. exist. No No portion portion of this of this material material maymay be be in any in any form form form or by or any by any means, means, without with out without permission permission permis sion in writing i n writing in from from thethe
Ans:
2m L t
= 2 p
L A 3mg + 6k
1199
© 2016 Pearson Education, Inc., Upper S Saddle addle River, NJ. All rights reserved. This Thi s material is i s protected under all copyright laws as they currently publisher. er. exist. No portion of this material may be reproduced, in any form form or by any means, without permission in writing from the publish
22 – 11. 11. While standing in an elevator, the man holds a pen pendu dulum which consists of an 18-in. cord cord and a 0.5-lb bob. bob. If the elevator is descending with an an acceleration a = 4 f t> s2, determine the natural period of vibration for small amplitudes of swing. a
4 f t/s2
S O LUT IO N Since the acceleration of the pendulum is (32.2
-
4) = 28.2 ft> s2
Using the result of Example 22 – 1, 1, We have vn =
t
=
g Al 2p vn
=
=
28.2 = 4.336 rad > s A 18 > 12 2p = 1.45 s 4.336
Ans.
Ans: t = 1.45 s 1200
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently publisher. er. exist. No portion of this material may be reproduced, in any form or by any means, without with out permis permission sion in i n writing writing from the publish
*22 – 12. 12. Determine the natural period of vibration of the uniform bar of mass m when it is displaced downward slightly and released.
O
k L —
L —
2
2
Solution Equation of Motion . The mass moment of inertia of the bar about O is I 0 =
1 mL 2. 12
Referring to the FBD of the rod, Fig. a, a + Σ M 0
= I a ;
- ky
L
1
2
12
cos u a
0
However, y =
L
2
b = a
sin u. Then L
-k
mL 2 b a
a
2
1
L
sin u b cos u a
2
b =
12
2
mL a
Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain 1 mL2a + 12
2
L k
8
sin 2u = 0
Here since u is small, sin 2u 12 12
2u. Also, a = u . Then the above equ ation becomes
$kL 2
mL u +
4
u = 0
$ 3k u + u = 0 m Comparing to that of the Standard form, vn =
t
=
2p vn
= 2 p
m A 3k
3k . Then Am Ans.
Ans: t
= 2 p
m A 3 k
1201
© 2016 Pearson Education, Inc., Upper S Saddle addle River, NJ. All rights reserved. This Thi s material is i s protected under all copyright laws as they currently publisher. er. exist. No portion of this material may be reproduced, in any form form or by any means, without permission in writing from the publish
22 – 13. 13.
The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of k G. If it is displaced a slight amount u from its equilibriu m position and released, determine the natural period of vibration.
O
d u G
SOLUTION a +
©M O = IO a;
- mgd
$ u +
$ 2 2 + md Du sin u = C mk G gd 2
2
k
G
sin u = 0
+ d
However, for small rotation sin u L u. Hence $ gd u + 2 u = 0 2 k G + d gd
From the above differential equation, vn = t
=
2p vn
=
2p gd
B k 2G + d 2
= 2 p
.
2 k G + d 2 C gd
Ans.
A k G2 + d 2
Ans: t
= 2 p C
k 2G + d 2 gd
1202
22 – 14. 14.
The 20-lb rectangul ar plate has a natural period of vibration t = 0.3 s, as it oscillates around the axis of rod AB. ad, Determine the torsional stiffness k , measured in lb ft > r ad, of the rod. Neglect the mass of the rod.
A
k B
Solution T =
k u
1 Σ M z
= I za ;
- k u
=
20
2
4 ft
$
a b (2) u 12 32.2
$ u + k (4.83)u (4.83)u = 0 2 ft t
=
2p 2k (4.83 )
= 0.3
k = 90.8 lb # ft > rad
Ans.
Ans: k = 90.8 lb # ft > rad
Engineering Mechanics Statics and Dynamics 14th Edition Hibbeler
SOLUTIONS MANUAL Full download: https://testbanklive.com/download/engineering-m https://testbanklive.com/d ownload/engineering-mechanics-statics-andechanics-statics-anddynamics14th-edition-hibbeler-solutions-manual/ People also search: engineering mechanics statics & dynamics (14th edition) pdf engineering mechanics statics and dynamics 14th edition solutions statics and dynamics 13th edition statics and dynamics 14th edition chegg engineering mechanics: statics & dynamics pdf statics and dynamics hibbeler pdf engineering mechanics statics and dynamics hibbeler pdf engineering mechanics statics and dynamics 14th edition pdf free download
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Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ 2–1. If
60° and
450 N, determine the magnitude of the y
resultant force and its direction, measured counterclockwise from the positive x axis.
F 15 700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N
4502
2(700)(450) cos 45°
497 N
Ans.
This yields sin 700
sin 45° 497.01
Thus, the direction of angle positive axis, is
60°
of F
95.19°
95.19°
measured counterclockwise from the
60°
155°
Ans.
x
Ans: FR = 497 N f = 155 22
2–2. y
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
F u 15
x
700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N
Ans.
Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78 u = 45.2°
Ans.
Ans: F = 960 N u = 45.2 23
2–3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis.
y F1
250 lb
30
SOLUTION
x
FR = 2(250)2 + (375)2 - 2(250)(375) cos 75° = 393.2 = 393 lb 393.2
Ans.
45
250 =
sin 75° sin u u = 37.89° f = 360° - 45° + 37.89° = 353°
F2
375 lb
Ans.
Ans: FR = 393 lb f = 353 24
*2–4. The vertical force F acts downward at on the two-membered frame. Determine the magnitudes of the two components of 500 N. F directed along the axes of and . Set
B
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have
sin 60°
448 N
sin 45°
F
500 sin 75°
C
Ans.
500 sin 75°
366 N
Ans.
Ans: FAB = 448 N FAC = 366 N
25
2–5. Solve Prob. 2-4 with F = 350 lb. B
45
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have FAB
F
350
30 C
=
sin 60° sin 75° FAB = 314 lb FAC
Ans.
350
= sin 45° sin 75° FAC = 256 lb
Ans.
Ans: FAB = 314 lb FAC = 256 lb 26
2–6. v
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.
30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b, FR = 242 + 6 2 - 2(4)(6) cos 105 = 8.026 kN = 8.03 kN
Ans.
Using this result to apply Law of sines, Fig. b, sin 105 sin u ; = 8.026 6
u = 46.22
Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22 - 45 = 1.22
Ans.
Ans: f = 1.22 27
2–7. v
Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b. (F1)v 4 = ; sin 45 sin 105
(F1)v = 2.928 kN = 2.93 kN
Ans.
(F1)u 4 = ; sin 30 sin 105
(F1)u = 2.071 kN = 2.07 kN
Ans.
Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28
*2–8. v
Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b, (F2)u 6 = ; sin 75 sin 75
(F2)u = 6.00 kN
Ans.
(F2)v 6 = ; sin 30 sin 75
(F2)v = 3.106 kN = 3.11 kN
Ans.
Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29
2–9. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.
F A u B 60
900 lb
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30 = 615.94 lb = 616 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 900
sin 30 =
615.94
;
u = 46.94 = 46.9
Ans.
Ans: F = 616 lb u = 46.9 30
2–10. y
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
800 lb 40
x 35
Solution 500 lb
Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, FR = 28002 + 5002 - 2(800)(500) cos 95 = 979.66 lb = 980 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 500
sin 95 =
979.66
;
u = 30.56
Thus, the direction f of FR measured counterclockwise from the positive x axis is f = 50 - 30.56 = 19.44 = 19.4
Ans.
Ans: FR = 980 lb f = 19.4 31
2–11. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° 40
= 10.80 kN = 10.8 kN
Ans. B
The angle u can be determined using law of sines (Fig. b).
FB
6 kN
sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°
Ans.
Ans: FR = 10.8 kN f = 3.16 32
*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8
40 B
sin (90° - u) = 0.5745
FB
u = 54.93° = 54.9°
6 kN
Ans.
From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN
Ans.
Ans: u = 54.9 FR = 10.4 kN 33
2–13.
Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ engineering mechanics statics 14th edition free pdf statics hibbeler 14th edition solutions engineering mechanics statics 14th edition ebook engineering mechanics statics and dynamics 14th edition engineering mechanics statics hibbeler 14th edition pdf download engineering mechanics statics rc hibbeler 14th edition solution manual pdf engineering mechanics statics 13th edition pdf engineering mechanics statics pdf
60° and
450 N, determine the magnitude of the y
resultant force and its direction, measured counterclockwise from the positive x axis.
F 15 700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N
4502
2(700)(450) cos 45°
497 N
Ans.
This yields sin 700
sin 45° 497.01
Thus, the direction of angle positive axis, is
60°
of F
95.19°
95.19°
measured counterclockwise from the
60°
155°
Ans.
x
Ans: FR = 497 N f = 155 22
2–2. y
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
F u 15
x
700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N
Ans.
Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78 u = 45.2°
Ans.
Ans: F = 960 N u = 45.2 23
2–3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis.
y F1
250 lb
30
SOLUTION
x
FR = 2(250)2 + (375)2 - 2(250)(375) cos 75° = 393.2 = 393 lb 393.2
Ans.
45
250 =
sin 75° sin u u = 37.89° f = 360° - 45° + 37.89° = 353°
F2
375 lb
Ans.
Ans: FR = 393 lb f = 353 24
*2–4. The vertical force F acts downward at on the two-membered frame. Determine the magnitudes of the two components of 500 N. F directed along the axes of and . Set
B
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have
sin 60°
448 N
sin 45°
F
500 sin 75°
C
Ans.
500 sin 75°
366 N
Ans.
Ans: FAB = 448 N FAC = 366 N
25
2–5. Solve Prob. 2-4 with F = 350 lb. B
45
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have FAB
F
350
30 C
=
sin 60° sin 75° FAB = 314 lb FAC
Ans.
350
= sin 45° sin 75° FAC = 256 lb
Ans.
Ans: FAB = 314 lb FAC = 256 lb 26
2–6. v
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.
30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b, FR = 242 + 6 2 - 2(4)(6) cos 105 = 8.026 kN = 8.03 kN
Ans.
Using this result to apply Law of sines, Fig. b, sin 105 sin u ; = 8.026 6
u = 46.22
Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22 - 45 = 1.22
Ans.
Ans: f = 1.22 27
2–7. v
Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b. (F1)v 4 = ; sin 45 sin 105
(F1)v = 2.928 kN = 2.93 kN
Ans.
(F1)u 4 = ; sin 30 sin 105
(F1)u = 2.071 kN = 2.07 kN
Ans.
Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28
*2–8. v
Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b, (F2)u 6 = ; sin 75 sin 75
(F2)u = 6.00 kN
Ans.
(F2)v 6 = ; sin 30 sin 75
(F2)v = 3.106 kN = 3.11 kN
Ans.
Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29
2–9. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.
F A u B 60
900 lb
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30 = 615.94 lb = 616 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 900
sin 30 =
615.94
;
u = 46.94 = 46.9
Ans.
Ans: F = 616 lb u = 46.9 30
2–10. y
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
800 lb 40
x 35
Solution 500 lb
Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, FR = 28002 + 5002 - 2(800)(500) cos 95 = 979.66 lb = 980 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 500
sin 95 =
979.66
;
u = 30.56
Thus, the direction f of FR measured counterclockwise from the positive x axis is f = 50 - 30.56 = 19.44 = 19.4
Ans.
Ans: FR = 980 lb f = 19.4 31
2–11. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° 40
= 10.80 kN = 10.8 kN
Ans. B
The angle u can be determined using law of sines (Fig. b).
FB
6 kN
sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°
Ans.
Ans: FR = 10.8 kN f = 3.16 32
*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8
40 B
sin (90° - u) = 0.5745
FB
u = 54.93° = 54.9°
6 kN
Ans.
From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN
Ans.
Ans: u = 54.9 FR = 10.4 kN 33
2–13.
Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ engineering mechanics statics 14th edition free pdf statics hibbeler 14th edition solutions engineering mechanics statics 14th edition ebook engineering mechanics statics and dynamics 14th edition engineering mechanics statics hibbeler 14th edition pdf download engineering mechanics statics rc hibbeler 14th edition solution manual pdf engineering mechanics statics 13th edition pdf engineering mechanics statics pdf