Niu (qn269) – HW04 – Tsoi – (58020)
1
This printout should have 27 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 5) 10.0 points
A car travels at a speed oF 22 m
/
s around a
curve oF radius 31 m.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
2
.
9 Mg
μ
30
◦
What is the
net
centripetal Force needed to
keep the car From skidding sideways?
Correct answer: 45277
.
4 N.
Explanation:
Let :
m
= 2900 kg
,
v
= 22 m
/
s
,
r
= 31 m
,
θ
= 30
◦
,
and
μ
= 0
.
529117 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
=
ma
c
=
mv
2
R
= 45277
.
4 N
.
002
(part 2 oF 5) 10.0 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: 16408
.
3 N.
Explanation:
In the absence oF Friction, there are only two
Forces acting on the car, namely its weight
mVg
and the normal Force
V
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
=
mg
or
N
=
mg
cos
θ
and
horizontally gives
F
banking
c
=
N
sin
θ
=
mg
cos
θ
sin
θ
=
mg
tan
θ
= 16408
.
3 N
.
003
(part 3 oF 5) 10.0 points
Now suppose the Friction Force is su±cient 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 frst question
beFore proceeding with this and the Following
questions.
Correct answer: 47251
.
2 N.
Explanation:
Consider the Free body diagram For the car
mg
sin
θ
N
=
cos
μN

W
b
−
ma
mg
W
b
=
mg
sin
θ
a
b
=
v
r
2
cos
θ
Now in addition to the car’s weight
mVg
and
the normal Force
V
N
there is also the Friction
Force
V
F
f
whose direction is across the road
but along the road’s surFace. In other words,
V
F
f
lies in the same vertical plane as the net
centripetal Force
V
F
c
that makes the car Follow
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2
the curving road and at the angle
θ
below the
horizontal direction in that plane.
The net force
v
F
net
=
mvg
+
v
N
+
v
F
f
must be equal to the centripetal force
v
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
−
mg .
(2)
N
, by isolating
F
f
we obtain
F
f
sin
θ
=
N
cos
θ
−
mg
F
f
cos
θ
=
F
c
−
N
sin
θ ,
dividing the two equation we have
sin
θ
cos
θ
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 Spring '09
 Staff
 Physics, Force, Friction, Mass, Correct Answer

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