Bell, Evan – Homework 8 – Due: Mar 20 2005, 11:00 pm – Inst: Furlan
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The due time is Central
time.
001
(part 1 of 1) 10 points
Two objects, of masses 30 kg and 54 kg, are
hung from the ends of a stick that is 70 cm
long and has marks every 10 cm, as shown.
30 kg
54 kg
A
B
C
D
E
F
G
10
20
30
40
50
60
If the mass of the stick is negligible, at
which of the points indicated should a cord be
attached if the stick is to remain horizontal
when suspended from the cord?
1.
E
2.
D
3.
F
correct
4.
C
5.
G
6.
A
7.
B
Explanation:
Let :
‘
= 70 cm
,
m
1
= 30 kg
,
m
2
= 54 kg
.
For static equilibrium,
τ
net
= 0.
Denote
x
the distance from the left end
point of the stick to the point where the cord
is attached.
m
1
g x

m
2
g
[
‘

x
] =
τ
= 0
(
m
1

m
2
)
x
=
m
2
‘
x
=
m
2
‘
m
1

m
2
=
(54 kg) (70 cm)
30 kg

54 kg
=
45 cm
.
Therefore the point should be point
F
.
002
(part 1 of 2) 10 points
A uniform rod pivoted at one end “point
O
”
is free to swing in a vertical plane in a gravita
tional field. However, it is held in equilibrium
by a force
F
at its other end.
The rod makes an angle 36
◦
with the hori
zontal. The length of the rod is 7 m. The force
makes and angle 58
◦
with the horizontal. The
weight of the rod is 7
.
4 N.
x
y
7 m
F
58
◦
36
◦
O
What is the magnitude of the force
F
?
Correct answer: 7
.
99068 N.
Explanation:
Let :
‘
= 7 m
,
W
= 7
.
4 N
,
θ
= 36
◦
,
and
φ
= 58
◦
.
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Bell, Evan – Homework 8 – Due: Mar 20 2005, 11:00 pm – Inst: Furlan
2
x
y
ψ
‘
W
F
φ
θ
O
R
Basic Concepts:
For static equilibrium,
we have
X
i
F
x i
= 0
,
(1)
X
i
F
y i
= 0
,
and
(2)
X
i
τ
i
= 0
.
(3)
Solution:
Using
the
torque
equation,
about point
O
, (Eq. 3 with counter clock
wise positive), we have
‘
2
W
cos
θ

‘ F
cos
θ
sin
φ
+
‘ F
sin
θ
cos
φ
= 0
W
2

F
sin
φ
+
F
tan
θ
cos
φ
= 0
,
so
F
=
W
2 [sin
φ

cos
φ
·
tan
θ
]
(4)
=
(7
.
4 N)
2 [sin 58
◦

cos 58
◦
·
tan 36
◦
]
=
7
.
99068 N
.
Alternate Solution:
The angle between
the force
F
and the rod is
β
=
φ

θ
= (58
◦
)

(36
◦
) = 22
◦
.
Using the torque equation,
about point
O
, (Eq. 3 with counter clockwise
positive), we have
‘
2
W
cos
θ
=
‘ F
sin
β ,
so
F
=
W
2
cos
θ
sin
β
(5)
=
(7
.
4 N)
2
cos 36
◦
sin 22
◦
=
7
.
99068 N
.
003
(part 2 of 2) 10 points
What is the magnitude of the force the pivot
exerts on the rod at point
O
?
Correct answer: 4
.
28008 N.
Explanation:
Using the translational equations (Eqs.
1
and 2), where
R
x
and
R
y
are the horizontal
and vertical components of the force
R
at the
pivot, we have
F
x
+
R
x
= 0
,
so
(1)
R
x
=

F
x
=

F
cos
φ
=

(7
.
99068 N) cos 58
◦
=

4
.
23442 N
,
and
F
y
+
R
y
=
W ,
so
(2)
R
y
=
W

F
y
=
W

F
sin
φ
= (7
.
4 N)

(7
.
99068 N) sin 58
◦
= 0
.
623516 N
,
and
k
~
R
k
=
q
R
2
x
+
R
2
y
=
q
(

4
.
23442 N)
2
+ (0
.
623516 N)
2
=
4
.
28008 N
.
The angle
ψ
from the positive
x
axis to the
vector
R
is
ψ
= arctan
R
y
R
x
¶
= arctan
0
.
623516 N

4
.
23442 N
¶
= 171
.
623
◦
.
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 Spring '07
 GALLIS,MICHAELRO
 Equilibrium, Force, Mass, Work

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