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Version 097/ABCAB – midterm 03 – Turner – (58220)
1
This printout should have 19 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A17
.
5kgpersonclimbsupauniForm142N
ladder. The upper end oF the ladder rests on
aFr
ict
ion
lesswa
l
l
. Thebottomo
Ftheladder
rests on a ±oor with a rough surFace. The
angle between the horizontal and the ladder
is 43
◦
.
The person can climb 35% up the
ladder beFore the base oF the ladder slips on
the ±oor.
17
.
5kg
142 N
d
2m
43
◦
μ
s
>
0
μ
=0
Note:
²igure is not to scale.
What is the coe³cient oF static Friction
μ
s
between the base oF the ladder and the ±oor?
The acceleration oF gravity is 9
.
8m
/
s
2
.
1. 0.611755
2. 0.228951
3. 0.33071
4. 0.259405
5. 0.211536
6. 0.270395
7. 0.448189
8. 0.516792
9. 0.279139
10. 0.299018
Correct answer: 0
.
448189.
Explanation:
Pivot
N
w
f
N
f
mg
W
θ
Let :
g
=9
.
/
s
2
,
θ
=43
◦
,
L
=2m
,
d
.
35
L,
W
=142N
,
and
m
=17
.
.
Applying translational equilibrium hori
zontally,
±
F
x
:
f

N
w
f
=
N
w
±
τ
◦
:
mgd
cos
θ
+
W
L
2
cos
θ

N
w
L
sin
θ
where
d
is the distance oF the person From the
bottom oF the ladder
2
fL
sin
θ
=2
cos
θ
+
W
L
cos
θ
f
=
1
2
W
tan
θ
+
d
L
tan
θ
The ladder may slip when
f
=
f
max
=
N
w
,
so
f
max
≡
μ
(
W
+
)
≥
1
2
W
tan
θ
+
d
L
tan
θ
μ
≥
1
(
W
+
)tan
θ
²
1
2
W
+
d
L
³
≥
1
[142 N + (17
.
5kg)(9
.
/
s
2
)] tan43
◦
×
´
1
2
(142 N)
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2
+(0
.
35)(17
.
5kg)(9
.
8m
/
s
2
)
±
≥
0
.
448189
.
002
10.0 points
Amobileconsistingoffourweightshanging
on three rods of negligible mass.
24 N
W
1
W
2
W
3
5m
9m
7m
4m
Find the weight of
W
3
.
1. 155.048
2. 19.5556
3. 45.3968
4. 29.3333
5. 60.8395
6. 62.3333
7. 23.8333
8. 34.0476
9. 63.0667
10. 116.111
Correct answer: 29
.
3333 N.
Explanation:
Let :
W
=24N
,
)
r
1
=9m
,
)
l
1
=5m
,
)
r
2
=4m
,
)
l
2
=7m
,
)
r
3
=8m
,
and
)
l
3
.
We can apply the balance condition
²
*τ
=
0successively
,startingwiththelowestpartof
the mobile, to ±nd the value of each unknown
weight.
Apply
²
=0aboutanax
isthroughthe
point of suspension of the lowest part of the
mobile, we obtain
)
l
1
W
)
r
1
W
1
=0
W
1
=
)
l
1
W
)
r
1
=
(5 m) (24 N)
=13
.
3333 N
.
Apply
²
point of suspension of the middle part of the
mobile, we obtain
)
l
w
W
2

)
r
2
[
W
+
W
1
]=0
W
2
=
)
r
2
[
W
+
W
1
]
)
l
w
=
(4 m) (24 N + 13
.
3333 N)
=21
.
3333 N
.
Apply
²
point of suspension of the top part of the
mobile, we obtain
)
l
e
[
W
+
W
1
+
W
2
]

)
r
3
W
3
W
3
=
)
l
3
[
W
+
W
1
+
W
2
]
)
r
3
=
(24 N + 13
.
3333 N
+21
.
3333 N)
=
29
.
3333 N
.
keywords:
003
10.0 points
Version 097/ABCAB – midterm 03 – Turner – (58220)
3
The puck in the fgure has a mass oF 0
.
145 kg.
Its original distance From the center oF rota
tion is 51
.
9cm
,andthepuckismov
ingw
ith
aspe
edo
F75
.
7cm
/
s. The string is pulled
downward 15
.
1cmth
roughth
eho
l
einth
e
Frictionless table.
51
.
75
.
/
s
0
.
145 kg
What is the magnitude oF the work was
done on the puck? Treat the hockey puck as
apointmass.
1. 0.0460518
2. 0.0410899
3. 0.0650961
4. 0.0963719
5. 0.0580828
6. 0.055173
7. 0.0345982
8. 0.127202
9. 0.0349704
10. 0.039917
Correct answer: 0
.
0410899 J.
Explanation:
Given :
m
p
=0
.
145 kg
,
r
i
=51
.
9cm=0
.
519 m
,
v
i
=75
.
/
s=0
.
757 m
/
s
,
and
Δ
r
=15
.
1cm=0
.
151 m
.
r
v
F
c
m
The kinetic energy is
KE
=
1
2
Iω
2
=
1
2
(
mr
2
)
±
v
r
²
2
=
1
2
mv
2
Angular momentum is conserved, so
I
i
ω
i
=
I
f
ω
f
(
i
2
)
v
i
r
i
=
(
f
2
)
v
f
r
f
v
f
=
v
i
r
i
r
f
Thus the work done on the puck is
W
=Δ
=
1
2
2
f

1
2
2
i
=
1
2
m
³
v
i
r
i
r
f
´
2

1
2
2
i
=
1
2
m
µ
r
2
i
r
2
f

1
¶
v
2
i
=
1
2
(0
.
145 kg)
³
(0
.
519 m)
2
(0
.
368 m)
2

1
´
×
(0
.
757 m
/
s)
2
.
0410899 N
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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