lec 32 (planar kinetic equation of motion) 051
TRANSCRIPT
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King Fahd University of Petroleum & Minerals
Mechanical Engineering Dynamics ME 201
BY Dr. Meyassar N. Al-Haddad
Lecture # 32
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I M am F
am F
GG
yG y
xG x
)(
)(
General Application of the Equations of Motion
P P M )( k
Summation of moment in FBD = summation of the kinetic moment in K.D
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Rectilinear Translation
0
)()(
G
yG y
xG x
M
am F am F
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Curvilinear Translation
0
)(
)( 22
G
t Gt
nGn
M
mam F
mmam F
])([])([)( nGt G B B amhame M k
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slipping Vs. Tipping
x
N
If x > 1.5 ft tipping
If x < 1.5 ft slipping
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Wheely Vs. Non -Wheely
NB=0
Wheely : lift the front wheel off the ground
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?
250
2
a
slipping wheelsback
.
Mg m
k
)3.0(2000
)25.1)(81.9(2000)2()(
G
B A A
a
N M k
Example
G B xG x a N am F 200025.0;)(
0)81.9(2000;)( B A yG y N N am F
2/59.1,7.12,88.6 smakN N kN N G B A
0)3.0(25.0)75.0()25.1(0 B B AG N N N M
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GC xG x akg N N am F )50(2.0600;)(
05.490;)( N N am F C yG y
0)5.0(2.0)()3.0(600;0 C C G N x N M
2/0.10,467.0,490 sma x N N GC
sliding mm x 5.0467.0 If x > 0.5 m tipping will take place
Example
mk=0.2a= ?Flip Or Tipp?
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Example
G B xG x akg kg F am F )12575(;)(
025.122675.735;)( N N N am F B yG y
)6.0)(125()9.0()75(
)8.0)(25.1226()4.0)(75.735(;)(
makg makg
m N m N M
GG
B B k
N F
N N
sma
B
B
G
1790
1962
/94.8 2
912.019621790
min)( B
B s N
F m
m m=125 kgm R= 75 kga= ?N
B=?
FB=?ms=? Minimum toLift front wheel
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kN T T sma D Bt G 32.1/9.4)( 2
Examplem BD=100kgm AB=m CD= Neglectq = 30 o
= 6 rad/s
T A= ?TD= ?a G=?
222 /18)5.0()6()( smr a nG )18(10030cos981)( D BnGn T T am F
t Gt Gt aam F )(10030sin981)(
0)4.0(30cos)4.0(30cos0 D BG T T M
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Problem
m= 80 kgmB=0.8N A=?
NB=?When rear wheel locks for breaka =? Deceleration
N A NB G B xG x a N am F 808.0;)(
mBNB
0)81.9(80;)( A B yG y N N am F
)2.1(80)55.0)(81.9(80)95.0(;)( G B A A a N M k
N N
N N
sma
A
B
G
559
226
/26.2 2
a G
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Problem
m= 80 kgmB= ?N A=?
NB=?a =?When traveling at constant velocityand no break was applied
0)81.9(80;)( A B yG y N N am F
0)55.0)(81.9(80)95.0(;0 B A N M
N A NB
mBNB
a = 0
N N
N N
sma
B
A
G
454
330
/0 2
0;0 Bk x N F m
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Problem
m= 80 kgN A=?NB=?
a =?mk=? minimumWhen rider applies the front breakand back wheel start to lift off the ground
G Ak xG x a N am F 80;)( m
0)81.9(80;)( A yG y N am F
)2.1(80)55.0)(81.9(80;)( G A A a M k
458.0
785
/5.4 2
k
A
G
N N
sma
m
0 B N
N A NB=0
mkN A
a
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