practice 05 – GADHIA, TEJAS – Due: Feb 15 2007, 2:00 pm
1
Question 1
part 1 of 5
10 points
A car travels at a speed of 20 m
/
s around a
curve of radius 40 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
9 Mg
μ
21
◦
What is the
net
centripetal force needed to
keep the car from skidding sideways?
Correct answer: 29000 N (tolerance
±
1 %).
Explanation:
Let :
m
= 2900 kg
,
v
= 20 m
/
s
,
r
= 40 m
,
θ
= 21
◦
,
and
μ
= 0
.
457386
Part5
.
The
centripetal
acceleration
of
the
car
rounding a curve is
a
c
=
v
2
R
and the
net
centripetal force needed to provide such ac
celeration is
F
c
=
m a
c
=
m v
2
R
= 29000 N
.
Question 2
part 2 of 5
10 points
Were there no friction between the car’s
tires and the road,
what centripetal force
could be provided just by the banking of the
road?
Correct answer:
10909
.
4
N (tolerance
±
1
%).
Explanation:
In the absence of friction, there are only two
forces acting on the car, namely its weight
mvectorg
and the normal force
vector
N
provided by the road.
The weight is directed vertically down while
the normal force is directed perpendicular to
the banked road surface and thus at the angle
θ
from the vertical.
The centripetal force
due to banking comes from the horizontal
component of the normal force,
F
banking
c
=
N
sin
θ
.
The free body diagram in the vertical di
rection gives
N
cos
θ
=
m g
or
N
=
m g
cos
θ
and
horizontally gives
F
banking
c
=
N
sin
θ
=
m g
cos
θ
sin
θ
=
m g
tan
θ
= 10909
.
4 N
.
Question 3
part 3 of 5
10 points
Now suppose the friction force is sufficient
to keep the car from skidding. Calculate the
magnitude of the normal force exerted on the
car by the road’s surface.
Hint:
Check the
correctness of your answer to the first question
before proceeding with this and the following
questions.
Correct answer: 36925 N (tolerance
±
1 %).
Explanation:
Consider the free body diagram for the car
m g
sin
θ
N
=
m g
cos
θ
μ N

W
bardbl

m a
bardbl

m g
W
bardbl
=
m g
sin
θ
a
bardbl
=
v
r
2
cos
θ
Now in addition to the car’s weight
mvectorg
and
the normal force
vector
N
there is also the friction
force
vector
F
f
whose direction is across the road
but along the road’s surface. In other words,
vector
F
f
lies in the same vertical plane as the net
centripetal force
vector
F
c
that makes the car follow
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
practice 05 – GADHIA, TEJAS – Due: Feb 15 2007, 2:00 pm
2
the curving road and at the angle
θ
below the
horizontal direction in that plane.
The net force
vector
F
net
=
mvectorg
+
vector
N
+
vector
F
f
must be equal to the centripetal force
vector
F
c
since
otherwise the car would skid.
Writing this
vector equation in terms of the horizontal and
the vertical components, we have
F
c
=
N
sin
θ
+
F
f
cos
θ
(1)
0 =
N
cos
θ

F
f
sin
θ

m g .
(2)
Solve Eqs. 1 & 2 equations for
N
, by isolating
F
f
we obtain
F
f
sin
θ
=
N
cos
θ

m g
F
f
cos
θ
=
F
c

N
sin
θ ,
dividing the two equation we have
sin
θ
cos
θ
=
N
cos
θ

m g
F
c

N
sin
θ
,
by cross multiplication we have
N
cos
2
θ

m g
cos
θ
=
F
c
sin
θ

N
sin
2
θ ,
and
N
(sin
2
θ
+ cos
2
θ
) =
F
c
sin
θ
+
m g
cos
θ ,
thus
N
=
F
c
sin
θ
+
m g
cos
θ
= 36925 N
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Acceleration, Force, Friction, Gravity, Correct Answer, centripetal force Fc

Click to edit the document details