homework 12 – BAUTISTA, ALDO – Due: Feb 20 2006, 4:00 am
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Question 1
Part 1 of 1.
10 points.
Given:
g
= 9
.
8 m
/
s
2
.
Consider a boxcar accelerating
up
a 15
.
9
◦
slope.
Inside the boxcar, an object of un
known mass
m
hangs on a string attached to
the boxcar’s ceiling.
When the car accelerates uphill at a
steady
rate
a
, the string hangs at a constant an
gle
θ
= 30
.
4
◦
from the perpendicular to the
boxcar’s ceiling and ±oor.
a
m
30
.
4
◦
15
9
Given the angles
θ
= 30
.
4
◦
> φ
= 15
.
9
◦
,
calculate the boxcar’s acceleration
a
.
Correct answer: 2
.
84485 m
/
s
2
(tolerance
±
1
%).
Explanation:
Basic Concept:
In a noninertial frame such
as an accelerating boxcar, the inertial force

m~a
combines with the true gravitational
force
m~g
into a single
apparent weight
force
~
W
app
=
m
(
~g

~a
frame
)
.
(1)
In fact, the Equivalence Principle says that
there is no observable di²erence between the
true gravity and the inertial forces, so in a
noninertial frame there is a
net efective grav
ity
~g
eF
=
~g

~a
frame
.
(2)
Solution:
Consider the hanging object in the
noninertial frame of the accelerating boxcar.
In the car’s frame, the objects hangs without
motion so its
apparent
weight
(1) must be
balanced by the string’s tension.
Hence, the
direction of the
efective gravity
(2) must be
opposite to the string’s pull on the object,
which is 15
.
9
◦
from the perpendicular to the
boxcar’s ±oor and ceiling and 30
.
4
◦

15
.
9
◦
=
14
.
5
◦
from the true vertical.
At this point, the problem reduces to ge
ometry: Given the directions of vectors
~g
,
~a
and
~g

~a
and the magnitude
g
= 9
.
8 m
/
s
2
,
³nd the magnitude
a
. We can solve this ques
tion using the sine theorem, but it is just as
easy to solve in Carthesian coordinates.
Let
the
x
axis run uphull along the boxcars’s ±oor
while the
y
axis is perpendicular the the ±oor:
In these coordinates,
a
x
=
a,
a
y
= 0
,
(3)
g
x
=

g
sin
φ,
g
y
=

g
cos
φ
(4)
where
φ
= 15
.
9
◦
is the hill’s slope.
At the
same time, the string’s direction indicates
g
eF
x
=

g
eF
sin
θ,
g
eF
y
=

g
eF
cos
θ
(5)
where
θ
= 30
.
4
◦
is the angle between the
string and the
y
axis. Consequently,
tan
θ
=
g
eF
x
g
eF
y
=
g
x

a
x
g
y

a
y
=

g
sin
φ

a

g
cos
φ
= tan
φ
+
a
g
cos
φ
.
(6)
and therefore
a
= (tan
θ

tan
φ
)
g
cos
φ
(7)
=
2
.
84485 m
/
s
2
.
Question 2
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 Spring '08
 Turner
 Force, Mass, Potential Energy, Work, Correct Answer

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