homework 11 – FRENNEA, KYLE – Due: Apr 13 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
A weight (with a mass of 66 kg) is sus
pended from a point near the righthand end
of a uniform boom with a mass of 73 kg
.
To
support the uniform boom a cable runs from
this same point to a wall (the lefthand verti
cal coordinate in the figure) and by a pivot on
the same wall at an elevation of 5 m
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
Vertical Height (m)
Horizontal Distance (m)
Boom and Weight
73 kg
T
66 kg
Figure:
Drawn to scale.
Calculate the tension in the cable.
Correct answer:
2729
.
23
N (tolerance
±
1
%).
Explanation:
Let :
M
b
= 73 kg
,
M
w
= 66 kg
,
ℓ
= 10 m boom length
,
x
b
=
ℓ
2
= 5 m
,
x
= 8 m
,
y
= 5 m
,
and
h
= 8
.
5 m
.
The static equilibrium conditions are
summationdisplay
T
x
= 0
,
(1)
summationdisplay
T
y
= 0
,
and
(2)
summationdisplay
τ
= 0
,
or
(3)
W
b
x
b
+
W
w
x
=
T
y
x .
(4)
From the figure,
θ
= arctan
parenleftbigg
h

y
x
parenrightbigg
(5)
= arctan
parenleftbigg
8
.
5 m

5 m
8 m
parenrightbigg
= 23
.
6294
◦
.
0
1
2
3
4
5
6
7
8
9
10
2
1
0
1
2
3
4
5
6
7
8
9
10
Vertical Height (m)
Horizontal Distance (m)
Force Scale is
312
.
55 N
/
m
(0
, h
)
(
x, y
)
(0
, y
)
W
w
T
x
T
T
y
W
b
θ
Let the cable make an angle
θ
with the
horizontal boom.
Then using Eq. 5 and the
figure, we have
sin
θ
=
T
y
T
=
h

y
radicalbig
(
h

y
)
2
+
x
2
,
T
y
=
T
(
h

y
)
radicalbig
(
h

y
)
2
+
x
2
,
(6)
Solving Eq. 4 for
T
y
, we have
T
y
=
(
W
b
x
b
+
W
w
x
)
x
,
and substituting
T
y
from Eq. 6, we have
T
(
h

y
)
radicalbig
(
h

y
)
2
+
x
2
=
W
b
x
b
+
W
w
x
x
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homework 11 – FRENNEA, KYLE – Due: Apr 13 2007, 4:00 am
2
T
= (
W
b
x
b
+
W
w
x
)
radicalbig
(
h

y
)
2
+
x
2
x
(
h

y
)
= [(73 kg) (5 m) + (66 kg) (8 m)]
×
(9
.
8 m
/
s
2
)
×
radicalbig
(8
.
5 m

5 m)
2
+ (8 m)
2
(8 m) (8
.
5 m

5 m)
=
2729
.
23 N
.
Question 2
part 1 of 2
10 points
A solid bar of length
L
has a mass
m
1
. The
bar is fastened by a pivot at one end to a
wall which is at an angle
θ
with respect to the
horizontal. The bar is held horizontally by a
vertical cord that is fastened to the bar at a
distance
x
cord
from the wall.
A mass
m
2
is
suspended from the free end of the bar.
T
m
2
m
1
θ
x
cord
L
Find the tension
T
in the cord.
1.
T
= (
m
1
+
m
2
)
g
cos
θ
2.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
3.
T
= 0
4.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
5.
T
= (
m
1
+
m
2
)
parenleftbigg
L
x
cord
parenrightbigg
parenleftBig
g
2
parenrightBig
6.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
7.
T
= (
m
1
+
m
2
)
g
sin
θ
8.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
correct
9.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
10.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbiggparenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
Explanation:
Under static equilibrium,
summationdisplay
F
= 0 and
summationdisplay
τ
= 0
.
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 Spring '08
 Turner
 Force, Mass, Work, Correct Answer, kg, KYLE

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