homework 28 – BAUTISTA, ALDO – Due: Apr 10 2006, 4:00 am
1
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Question 1
Part 1 of 1.
10 points.
Two weights attached to a uniform beam
of mass 19 kg are supported in a horizontal
position by a pin and cable as shown in the
figure.
The acceleration of gravity is 9
.
8 m
/
s
2
.
25 kg
13 kg
3
.
2 m
6
.
6 m
44
◦
19 kg
What is the tension in the cable which sup
ports the beam?
Correct answer: 0
.
488424 kN (tolerance
±
1
%).
Explanation:
Let :
m
= 19 kg
,
M
1
= 25 kg
,
M
2
= 13 kg
,
L
1
= 3
.
2 m
,
L
2
= 6
.
6 m
,
and
θ
= 44
◦
.
Basic Concepts:
X
~
F
= 0
X
~
τ
= 0
.
Solution:
The sum of the torques about the
pivot is
T L
2
sin
θ

mg
L
2
2

M
1
gL
1

M
2
gL
2
= 0
Solving for the tension
T
T
=
m g
L
2
2
+
M
1
g L
1
+
M
2
g L
2
L
2
sin
θ
=
(19 kg)(9
.
8 m
/
s
2
)
6
.
6 m
2
(6
.
6 m) sin
θ
+
(25 kg)(9
.
8 m
/
s
2
)(3
.
2 m)
(6
.
6 m) sin
θ
+
(13 kg)(9
.
8 m
/
s
2
)(6
.
6 m)
(6
.
6 m) sin
θ
= 0
.
488424 kN
.
Question 2
Part 1 of 3.
10 points.
A uniform rod pivoted at one end “point
O
” is free to swing in a vertical plane in a
gravitational field.
However, it is held in
equilibrium by a force
F
at its other end.
x
y
‘
F
F
x
F
y
W
R
x
R
R
y
θ
O
Force vectors are drawn to scale.
What
is
the
condition
for
translational
equilibrium along the horizontal
x
direction?
1.
F
x
= 0
2.
R
x

F
x
sin
θ
= 0
3.
R
x

F
x
cos
θ
= 0
4.

R
x
+
F
x
= 0
correct
5.
F
x
cos
θ

R
x
sin
θ
= 0
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homework 28 – BAUTISTA, ALDO – Due: Apr 10 2006, 4:00 am
2
Explanation:
Using the distances, angles and forces de
picted in the figure, the condition
X
F
x
= 0
for translational equilibrium in the
x
direction
is

R
x
+
F
x
= 0
.
Question 3
Part 2 of 3.
10 points.
What
is
the
condition
for
translational
equilibrium along the vertical
y
direction?
1.
W

R
y
+
F
y
= 0
2.
R
y
sin
θ
+
F
y
sin
θ

W
cos
θ
= 0
3.
R
y
+
F
y
= 0
4.
R
y

F
y
+
W
= 0
5.
R
y
+
F
y

W
= 0
correct
Explanation:
For the
equilibrium
in
the
y
direction,
X
F
y
= 0 turns out to be
R
y
+
F
y

W
= 0
.
Question 4
Part 3 of 3.
10 points.
Taking the origin (point
O
) as the pivot
point, what is the condition for rotational
equilibrium?
1.
2
F
y
‘
sin
θ

W ‘
cos
θ

2
F
x
‘
sin
θ
= 0
2.
F
y
‘
sin
θ

W
‘
2
sin
θ

F
x
‘
sin
θ
= 0
3.
F
y
‘
cos
θ

W
‘
2
cos
θ

F
x
‘
sin
θ
= 0
correct
4.
W
‘
2

F
x
‘
cos
θ

F
y
‘
cos
θ
= 0
5.
F
y
‘
cos
θ

W ‘
sin
θ
+
F
x
‘
sin
θ
= 0
Explanation:
For rotational equilibrium about point O,
the net torque on the system about that point
must vanish. The angle
θ
appears as follows
x
y
‘
F
x
F
y
W
W
cos
θ
F
y
cos
θ
F
x
sin
θ
θ
θ
θ
θ
O
and we see that the forces
R
x
and
R
y
exert
no torque on the point O. From the figure we
have, counterclockwise
‘ F
y
cos
θ

‘
2
W
cos
θ

‘ F
x
sin
θ
= 0
.
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 Spring '08
 Turner
 Mass, Work, Sin, Trigraph, Correct Answer, kg, BAUTISTA

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