Version 130 – Midterm 1 – sai – (58015)
1
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001
10.0 points
The distance between two telephone poles is
58 m.
When a 1
.
36 kg bird lands on the
telephone wire midway between the poles, the
wire sags 0
.
111 m.
The acceleration oF gravity is 9
.
8m
/
s
2
.
How much tension in the wire does the bird
produce? Ignore the weight oF the wire.
1. 1308.58
2. 925.697
3. 1242.66
4. 959.181
5. 549.983
6. 1741.06
7. 978.017
8. 889.274
9. 1123.86
10. 1197.95
Correct answer: 1741
.
06 N.
Explanation:
0.11 m
58 m
θ
Given :
m
=1
.
36 kg
,
Δ
x
= 58 m
,
and
Δ
y
=0
.
111 m
.
T
T
mg
tan
θ
=
Δ
y
1
2
Δ
x
=
2 Δ
y
Δ
x
θ
= tan

1
±
2 Δ
y
Δ
x
²
= tan

1
±
2
×
0
.
111 m
58 m
²
=0
.
219303
◦
.
³
F
y
=2
T
y

mg
=2
T
sin
θ

mg
=0
since it is in equilibrium, so
T
=
mg
2 sin
θ
=
(1
.
36 kg)(9
.
8m
/
s
2
)
2(sin0
.
219303
◦
)
=
1741
.
06 N
.
002
(part 1 oF 2) 10.0 points
A pulley is massless and Frictionless. 1 kg,
3 kg, and 8 kg masses are suspended as in the
fgure.
3
.
1m
20
.
6m
ω
3 kg
1 kg
8 kg
T
2
T
1
T
3
What is the tension
T
1
in the string be
tween the two blocks on the leFthand side
oF the pulley? The acceleration oF gravity is
9
.
8m
/
s
2
.
1.
T
1
=
±
17
12
kg
²
(9
.
8m
/
s
2
)
2.
T
1
=
±
7
6
kg
²
(9
.
8m
/
s
2
)
3.
T
1
=
±
7
4
kg
²
(9
.
8m
/
s
2
)
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2
4.
T
1
=
±
13
12
kg
²
(9
.
8m
/
s
2
)
5.
T
1
=
±
11
6
kg
²
(9
.
8m
/
s
2
)
6.
T
1
=
±
19
12
kg
²
(9
.
8m
/
s
2
)
7.
T
1
=
±
5
3
kg
²
(9
.
8m
/
s
2
)
8.
T
1
=
±
4
3
kg
²
(9
.
8m
/
s
2
)
correct
9.
T
1
=
±
5
4
kg
²
(9
.
8m
/
s
2
)
10.
T
1
=
±
3
2
kg
²
(9
.
8m
/
s
2
)
Explanation:
Let :
R
= 20
.
6m
,
m
1
= 1 kg
,
m
2
= 3 kg
,
m
3
= 8 kg
,
and
h
=3
.
1m
.
Consider the free body diagrams
1 kg
3 kg
8 kg
T
1
2
3
m
g
a
For each mass in the system
"
F
net
=
m"a.
Since the string changes direction around
the pulley, the forces due to the tensions
T
2
and
T
3
are in the same direction (up). The
acceleration of the system will be down to
the right (
m
3
>m
1
+
m
2
), and each mass in
the system accelerates at the same rate (the
string does not stretch). Let this acceleration
rate be
a
and the tension over the pulley be
T
≡
T
2
=
T
3
.
For the lower lefthand mass
m
1
the accel
eration is up and
T
1

m
1
g
=
m
1
a.
(1)
For the upper lefthand mass
m
2
the acceler
ation is up and
T

T
1

m
2
g
=
m
2
a.
(2)
For the righthand mass
m
3
the acceleration
is down and

T
+
m
3
g
=
m
3
a.
(3)
Adding Eqs. 1, 2, and 3, we have
(
m
3

m
1

m
2
)
g
=(
m
1
+
m
2
+
m
3
)
a
(4)
a
=
m
3

m
1

m
2
m
1
+
m
2
+
m
3
g
(5)
=
8 kg

1 kg

3 kg
1 kg + 3 kg + 8 kg
g
=
4 kg
12 kg
(9
.
8m
/
s
2
)
=
1
3
(9
.
8m
/
s
2
)
.
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 Spring '08
 Kaplunovsky
 Physics

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