homework 29 – BAUTISTA, ALDO – Due: Apr 12 2006, 4:00 am
1
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Question 1
Part 1 of 2.
10 points.
A sphere of weight
W
is held against a
wall by a string being pulled at an angle
θ
as
shown.
θ
R
O
F
The torque equation about point
O
, the
center of the sphere, leads to :
(
f
is the
magnitude of the frictional force, and weight
W
=
Mg
)
1.
F
sin
θ
cos
θ
=
f
2.
F
=
f
correct
3.
W
=
f
4.
F
cos
2
θ
=
f
5.
F
+
W
=
f
6.
F
sin
θ
=
f
Explanation:
For static Equilibrium,
X
i
F
i
= 0
,
X
i
~
τ
i
= 0
.
From
X
i
~
τ
i
= 0, we get
τ
c
=
τ
cc
about
O
, the
center of the sphere. That is,
F R
=
f R,
where
τ
c
=
F R
and
τ
cc
=
f R
. Or
F
=
f.
Question 2
Part 2 of 2.
10 points.
The vertical component of the force equa
tion leads to :
1.
F
cos
θ
+
W
=
f
2.
F
sin
θ
=
W
3.
F
sin
θ
=
f
+
W
4.
F
sin
θ
=
f
5.
F
sin
θ
+
f
=
W
correct
Explanation:
From
X
i
(
F
i
)
y
= 0, we get
F
sin
θ
+
f
=
W.
Question 3
Part 1 of 1.
10 points.
In
the
figure
the
radius
of
the
rod
is
0
.
221 cm, the length of aluminum part is
1
.
3 m, the copper part is 2
.
79 m.
For alu
minum, Young’s modulus is 70000000000 Pa,
for
copper,
110000000000
Pa.
Aluminum
Copper
r
L
a
L
b
Determine the elongation of the rod if it is
under a tension of 7790 N
Correct answer: 2
.
23056
cm (tolerance
±
1
%).
Explanation:
The crosssectional area of the wire
A
=
π
(0
.
221 cm)
2
= 1
.
53439
×
10

5
m
2
m
2
,
and
the tension = 7790 N N throughout both
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homework 29 – BAUTISTA, ALDO – Due: Apr 12 2006, 4:00 am
2
pieces. Let us compute the elongation of each
part separately: For aluminum:
∆
L
al
=
L
0
F
Y A
=
(1
.
3 m)(7790 N)
(70000000000 Pa)(1
.
53439
×
10

5
m
2
)
= 0
.
00942862 m
.
Similarly, for the copper part:
∆
L
cu
=
L
0
F
Y A
=
(2
.
79 m)(7790 N)
(110000000000 Pa)(1
.
53439
×
10

5
m
2
)
= 0
.
012877 m
.
So, the total elongation = 0
.
0223056 m =
2
.
23056 cm
.
Question 4
Part 1 of 1.
10 points.
A stainlesssteel orthodontic wire is applied
to a tooth that is out of line by 40
◦
.
The
wire has an unstretched length of 4
.
5 cm and
a diameter of 0
.
22 mm.
40
◦
40
◦
If the wire is stretched 0
.
11 mm, find the
magnitude of the force on the tooth. (Disre
gard the width of the tooth). Young’s modu
lus for stainless steel is 180000000000 Pa.
Correct answer:
21
.
5023
N (tolerance
±
1
%).
Explanation:
Given :
L
0
= 4
.
5 cm = 0
.
045 m
,
θ
= 40
◦
,
r
= 0
.
11 cm = 0
.
00011 m
,
∆
L
= 0
.
11 mm = 0
.
00011 m
,
and
Y
= 180000000000 Pa
.
40
◦
40
◦
F
F
We need the tension required to stretch the
wire by 0
.
11 mm:
Y
=
F L
0
A
∆
L
F
=
Y A
∆
L
L
0
=
Y
(
π r
2
)
∆
L
L
0
The resultant force exerted on the tooth has
components
X
F
x
=
F
cos
θ

F
cos
θ
= 0
and
X
F
y
=

F
sin
θ

F
sin
θ
so
F
net
=

2
F
sin
θ
=

2
π Y r
2
∆L sin
θ
L
0
=

2
π
(180000000000 Pa) (0
.
00011 m)
2
·
(0
.
00011 m) sin 40
◦
0
.
045 m
=

21
.
5023 N
.
which has a magnitude of
21
.
5023 N
,
and is
directed down the page.
Question 5
Part 1 of 1.
10 points.
A small piece of wood is dropped from a
height 4
.
1 m
m above water. The wood has
density density 790 kg
/
m
3
while the water’s
density is 1000 kg
/
m
3
.
How deep would the wooden piece sink into
the water before floating back to the surface
if there were no viscosity or any other kind of
water resistance to its motion?
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 Spring '08
 Turner
 Conservation Of Energy, Energy, Force, Simple Harmonic Motion, Work, Correct Answer, BAUTISTA

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