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me274_e2_fa2008.pdf-ME 274 – Fall 2008 SOLUTION
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me274 e2 fa2008.pdf-ME 274 – Fall 2...
me274_e2_fa2008.pdf-ME 274 – Fall 2008 SOLUTION
##### Page 1
ME 274 – Fall 2008
SOLUTION
Examination No. 2
PROBLEM NO. 1
Given
:
A vehicle having a mass of
500 kg
is traveling on a banked track on a path
R = 1000 meters
. At the instant showing, the vehicle
is traveling with a speed of
v = 50 m/sec
with this speed
decreasing
at a rate
of
4 m/sec
2
(due to braking).
Find
:
For this instant, determine
the magnitude of the TOTAL friction force
acting
on the vehicle by the roadway. (Here the total friction force is that due to both
braking and turning.) Use the figures provided below for two views of the
P
θ
= 36.87°
R
k
e
n
O
R
O
e
n
e
t
P
v
TOP VIEW
P
k
e
n
FBD’s
e
n
e
t
P
TOP VIEW
f
turn
N
mg
f
brake

##### Page 2
ME 274 – Fall 2008
SOLUTION
Examination No. 1
FBD:
previous page
Newton:
(1)
F
t
!
=
"
f
brake
=
ma
t
(2)
F
n
!
=
"
f
turn
cos
#
+
N
sin
#
=
ma
n
(3)
F
z
!
=
f
turn
sin
#
+
N
cos
#
"
mg
=
ma
z
\$
f
turn
sin
#
+
N
cos
#
=
mg
+
a
z
(
)
Multiply (2) by cos
θ
, (3) by sin
θ
and subtract:
f
turn
cos
2
!
+
sin
2
!
(
)
=
"
ma
n
cos
!
+
mg
+
a
z
(
)
sin
!
#
(4)
f
turn
=
m
"
a
n
cos
!
+
g
+
a
z
(
)
sin
!
\$
%
&
Kinematics:
(5)
a
t
=
dv
dt
=
!
4
m
/
sec
2
(6)
a
n
=
v
2
R
(7)
a
z
=
0
Solve:
Combining (1), (4)-(7):
f
brake
=
!
ma
t
=
!
500
(
)
!
4
(
)
=
2000
N
f
turn
=
m
!
v
2
R
cos
"
+
g
sin
"
#
\$
%
%
&
(
(
=
500
(
)
!
50
2
1000
0.8
(
)
+
9.806 0.6
(
)
#
\$
%
%
&
(
(
=
1941
N
Therefore,
f
total
=
f
brake
2
+
f
turn
2
=
2000
2
+
1941
2
=
2787
N

##### Page 3
ME 274 – Fall 2008
SOLUTION
Examination No. 2
PROBLEM NO. 2
Given
:
Particles A and B (having masses of
m
A
= m
B
= 10 kg
)
are interconnected by
the cable-pulley system shown in the figure. Both particles are constrained to
vertical motion with particle A able to slide on a smooth vertical rod. The
system is released at
s
A
= 0
with
A traveling downward with a speed of 5
m/sec
. Assume the pulleys to be small, massless and frictionless.
Find
:
You are asked here to find the
speed of particle A when A has reached the
position of
s
A
= 2 m
. In your solution clearly indicate the following of the
four-step solution method:
1.
FBD
: Complete the free body diagram of the system of A, B and the
cable shown below right. Identify any nonconservative forces that do
work on the system.
2.
Work-energy equation:
Clearly indicate the gravitational datum
line(s) used.
3.
Kinematics:
Here you need to relate
s
B
to
s
A
as well as v
B
to v
A
for
the two positions.
4.
Solve
B
A
O
O
y
O
x
m
B
g
m
A
g
N
B
A
smooth
rod
g
s
A
s
B
1.5 m
O
datum for B
datum for A

##### Page 4
ME 274 – Fall 2008
SOLUTION
Examination No. 2
FBD
: shown on previous page. All forces doing work are conservative and will be
included in the potential energy.
Work-energy
:
T
1
=
1
2
mv
A
1
2
+
1
2
mv
B
1
2
V
1
=
0
T
1
=
1
2
mv
A
2
2
+
1
2
mv
B
2
2
V
2
=
!
mgs
A
+
mgh
B
U
1
"
2
nc
(
)
=
0
Therefore,
T
1
+
V
1
+
U
1
!
2
nc
(
)
=
T
2
+
V
2
"
v
A
1
2
+
v
B
1
2
=
v
A
2
2
+
v
B
2
2
+
2
g
#
s
A
+
h
B
(
)
Kinematics
2
s
B
+
s
A
2
+
1.5
2
=
const
.
!
2
!
s
B
+
s
A
!
s
A
s
A
2
+
1.5
2
=
0
!
v
B
=
1
2
s
A
s
A
2
+
1.5
2
v
A
Also,
!
s
B
=
1
2
1.5
2
+
s
A
2
"
1.5
#
\$
%
&
(
=
h
B
At position 1:
v
B
1
=
1
2
0
1.5
v
A
1
=
0
At position 2:
v
B
2
=
1
2
2
2.5
v
A
2
=
0.4
v
A
2
and
h
B
=
1
2
2.5
!
1.5
(
)
=
0.5
m
Solve
:
v
A
2
=
v
A
1
2
+
2
gs
A
!
h
B
(
)
1
+
0.4
2
=
5
()
2
+
2
()
9.806
(
)
2
!
0.5
(
)
1
+
0.16
=
6.85
m
/
sec

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