oldhomewk 28 – VALENCIA, DANIEL – Due: Apr 6 2008, 4:00 am
1
Question 1, chap 14, sect 3.
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
= 0
2.
T
= (
m
1
+
m
2
)
g
sin
θ
3.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
correct
4.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
5.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
6.
T
= (
m
1
+
m
2
)
g
cos
θ
7.
T
= (
m
1
+
m
2
)
parenleftbigg
L
x
cord
parenrightbigg
parenleftBig
g
2
parenrightBig
8.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
9.
T
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
10.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
Explanation:
Under static equilibrium,
summationdisplay
F
= 0 and
summationdisplay
τ
= 0
.
Since we know little about the reaction
force between the bar and the wall, it is easi
est to begin by examining the torques around
the connection between the bar and the wall,
because the reaction forces will produce no
torques around that point.
summationdisplay
τ
=

L m
2
g
+
x
cord
T

L
2
m
1
g
= 0
.
T
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
.
Question 2, chap 14, sect 3.
part 2 of 2
10 points
Find the the horizontal component of the
force exerted on the bar by the wall.
(Take
right to be the positive direction.)
1.
F
x
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
2.
F
x
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
cos
θ
3.
F
x
= 0
correct
4.
F
x
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
5.
F
x
= (
m
1
+
m
2
)
g
sin
θ
6.
F
x
= (
m
1
+
m
2
)
parenleftbigg
L
x
cord
parenrightbigg
parenleftBig
g
2
parenrightBig
7.
F
x
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
sin
θ
8.
F
x
=
parenleftbigg
m
1
+
1
2
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
9.
F
x
=
parenleftbigg
1
2
m
1
+
m
2
parenrightbigg parenleftbigg
L
x
cord
parenrightbigg
g
10.
F
x
= (
m
1
+
m
2
)
g
cos
θ
Explanation:
In general, there will be a horizontal reac
tion force
R
x
at the connection between an
object and a wall. However, none of the other
forces in this situation act in the horizontal
direction, so
summationdisplay
F
x
=
R
x
= 0
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
oldhomewk 28 – VALENCIA, DANIEL – Due: Apr 6 2008, 4:00 am
2
Question 3, chap 14, sect 1.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Force, Mass, Correct Answer, Daniel, maximum overhang

Click to edit the document details