COSMOS: Complete Online Solutions Manual Organization System
Chapter 13, Solution 4.
Given: Mass of stone,
m = 2 kg
Velocity of stone,
v = 24 m/s
g m = 1.62 m/s 2
Acceleration of gravity on the moon,
Find:
(a) Kinetic energy, T
Height h, from which the stone was dropped
(b) T and h on the Moon
(a) On the Earth
T =
1 2 1
2
mv = ( 2 kg )( 24 m/s ) = 576 N ⋅ m
2
2
(
T = 576 J )
W = mg = ( 2 kg ) 9.81 m/s 2 = 19.62 N
T1 = U1− 2 = T2
Wh = T2
T1 = 0
h=
U1− 2 = Wh
T2 = 576 J
( 576 N ⋅ m ) = 29.36 m
T2
=
W
(19.62 N )
h = 29.4 m (b) On the Moon
m = 2 kg
Mass is unchanged.
T = 576 J Thus T is unchanged.
(
Wm = mg m = ( 2 kg ) 1.62 m/s 2
Weight on the moon is,
)
Wm = 3.24 N
hm =
( 576 N ⋅ m ) = 177.8 m
T
=
3.24 N
Wm
hm = 177.8 m Vector Mechanics for Engineers: Statics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, George H. Staab, © 2004 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 13, Solution 14.
Given: Crane moves at velocity, v = 10 ft/s and stops suddenly.
Find: Maximum horizontal distance, d moved by the bucket.
Refer to Problem 13.13 free body.
T1 =
1
1W
W
mv 2 =
(10 ft )2 = 50
g
2
2 g
T2 = 0
U1− 2 = −Wh
T1 + U1− 2 = T2
h=
50
W
− Wh = 0
g
50
50
=
= 1.55279 ft
g
32.2
AB = ( 30 ) = d 2 + y 2 = d 2 + ( 30 − 1.55279 )
2
2
d 2 = 90.7562
2
d = 9.5266
d = 9.53 ft Vector Mechanics for Engineers: Statics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, George H. Staab, © 2004 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 13, Solution 19.
Given: System at rest when 500 N force is applied to collar A. No
friction. Ignore pulleys mass.
Find: (a) Velocity, v A of A just before it hits C.
(b) v A If counter weight B is replaced by a 98.1 N
downward force.
Kinematics
X B = 2X A
vB = 2v A
(a) Blocks A and B
1
1
mBvB2 + mAv A2
2
2
1
1
2
T2 = (10 kg )( 2v A ) + ( 20 kg ) v A2
2
2
T1 = 0
T2 =
( )
T2 = ( 30 kg )( v A )
2
U1− 2 = ( 500 ) X A + (WA )( X A ) − (WB )( X B )
(
)
U1− 2 = ( 500 N )( 0.6 m ) + 20 kg × 9.81 m/s 2 ( 0.6 m )
(
)
− 10 kg × 9.81 m/s 2 (1.2 m )
U1− 2 = 300 + 117.72 − 117.72 = 300 J
T1 + U1− 2 = T2
0 + 300 J = ( 30 kg ) v A2
v A2 = 10
v A = 3.16 m/s (b) Since the 10 kg mass at B is replaced by a 98.1 N force, kinetic
energy at 2 is,
1
1
T2 = mAv A2 = ( 20 kg ) v A2
T1 = 0
2
2
The work done is the same as in part (a)
U1− 2 = 300 J
T1 + U1− 2 = T2
0 + 300 J = (10 kg ) v A2
v A2 = 30
v A = 5.48 m/s Vector Mechanics for Engineers: Statics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, George H. Staab, © 2004 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 13, Solution 28.
3 N/mm = 3000 N/m
(a) Spring constants
2 N/mm = 2000 N/m
Max deflection at 2 when velocity of C = 0
v1 = 0, T1 = 0, v2 = 0, T2 = 0
U1 − 2 = U e + U g
0.05
U1− 2 = ∫0
U1 − 2 =
( Fe )1 dx − ∫0ym ( Fe )2 dx + WC ( 0.3 +
( 3000 N/m )
2
( 0.05 m )2 −
(
( 2000 N/m )
2
ym )
( ym ) 2
)
+ ( 3 kg ) 9.81 m/s 2 ( 0.3 + ym )
= 3.750 − 1000 ( ym ) + 8.829 + 29.43 ( ym )
2
T1 + U1− 2 = T2 : 0 − 1000 ( ym ) + 29.43 ( ym ) + 12.579 = 0
2
ym = 0.12783 m
ym = 127.8 mm (b) Maximum velocity occurs as the lower spring is compressed a
distance y′
T1 = 0;
T1 + U1− 2 = T2 ;
Substitute
dv 2
=0
dy′
T2 =
1
1
mC v 2 = ( 3 kg ) v 2 = (1.5 ) v 2
2
2
0 − (1000 )( y′ ) + 29.43 ( y′ ) + 12.579 = (1.5 ) v 2
2
y′ = 0.014715 m
− 2000 ( y′ ) + 29.43 = 0;
y′ = 0.014715 m
− 0.0002165 + 0.43306 + 12.579 = 1.5v 2
v 2 = 2.9452 m/s
Vector Mechanics for Engineers: Statics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, George H. Staab, © 2004 The McGraw-Hill Companies.
v = 2.95 m/s COSMOS: Complete Online Solutions Manual Organization System
Chapter 13, Solution 46.
N =0
From Problem 13.45, block leaves the surface when
g cosθ =
vC2
h
h = 8 ft, θ = 40°
vC2 = hg cosθ = 8 ( 32.2 )( cos 40° ) = 197.33
Work-energy principle
TB =
TC =
1 2
mv
2
1 2
1
mvC = m (197.33) = 98.667m
2
2
U B − C = mgh (1 − cosθ )
TB + U B − C = TC
1 2
1
mv + mgh (1 − cosθ ) = mvC2
2
2
vC2 = 197.33 − 2 gh (1 − cos θ )
= 197.33 − 2 ( 32.2 )( 8 )(1 − cos 40° ) = 76.796
vC = 8.76 ft/s Vector Mechanics for Engineers: Statics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, George H. Staab, © 2004 The McGraw-Hill Companies.