practice 11 – ALIBHAI, ZAHID – Due: Apr 8 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
Two weights attached to a uniform beam
of mass 38 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
.
23 kg
20 kg
2
.
9 m
8 m
37
◦
38 kg
What is the tension in the cable which sup
ports the beam?
Correct answer: 0
.
770847 kN (tolerance
±
1
%).
Explanation:
Let :
m
= 38 kg
,
M
1
= 23 kg
,
M
2
= 20 kg
,
L
1
= 2
.
9 m
,
L
2
= 8 m
,
and
θ
= 37
◦
.
Under static equilibrium
summationdisplay
vector
F
= 0 and
summationdisplay
vector
τ
= 0
.
The sum of the torques about the pivot is
T L
2
sin
θ

m g
L
2
2

M
1
g L
1

M
2
g L
2
= 0
T
=
m g
L
2
2
+
M
1
g L
1
+
M
2
g L
2
L
2
sin
θ
=
m g
2 sin
θ
+
M
1
g L
1
L
2
sin
θ
+
M
2
g
sin
θ
=
(38 kg)(9
.
8 m
/
s
2
)
2 sin 37
◦
+
(23 kg) (9
.
8 m
/
s
2
) (2
.
9 m)
(8 m) sin 37
◦
+
(20 kg) (9
.
8 m
/
s
2
)
sin 37
◦
= 770
.
847 N
=
0
.
770847 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
ℓ
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
= 0
correct
3.
R
x

F
x
cos
θ
= 0
4.
R
x

F
x
sin
θ
= 0
5.
F
x
cos
θ

R
x
sin
θ
= 0
Explanation:
Using the distances, angles and forces de
picted in the figure, the condition
summationdisplay
F
x
= 0
for translational equilibrium in the
x
direction
gives

R
x
+
F
x
= 0
.
Question 3
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practice 11 – ALIBHAI, ZAHID – Due: Apr 8 2007, 4:00 am
2
part 2 of 3
10 points
What is the condition for translational
equilibrium along the vertical
y
direction?
1.
R
y
+
F
y

W
= 0
correct
2.
R
y

F
y
+
W
= 0
3.
R
y
+
F
y
= 0
4.
R
y
sin
θ
+
F
y
sin
θ

W
cos
θ
= 0
5.
W

R
y
+
F
y
= 0
Explanation:
For the equilibrium in the
y
direction,
summationdisplay
F
y
= 0 gives
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.
F
y
ℓ
sin
θ

W
ℓ
2
sin
θ

F
x
ℓ
sin
θ
= 0
2.
F
y
ℓ
cos
θ

W
ℓ
2
cos
θ

F
x
ℓ
sin
θ
= 0
correct
3.
2
F
y
ℓ
sin
θ

W ℓ
cos
θ

2
F
x
ℓ
sin
θ
= 0
4.
W
ℓ
2

F
x
ℓ
cos
θ

F
y
ℓ
cos
θ
= 0
5.
F
y
ℓ
cos
θ

W ℓ
sin
θ
+
F
x
ℓ
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
ℓ
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
ℓ F
y
cos
θ

ℓ
2
W
cos
θ

ℓ F
x
sin
θ
= 0
.
Question 5
part 1 of 1
10 points
A light string has its ends tied to two walls
separated by a distance equal to fourfifth the
length
L
of the string as shown in the figure.
A 25 kg mass is suspended from the center of
the string, applying a tension in the string.
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 Acceleration, Force, Gravity, Mass, Work, Sin, Correct Answer

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