homework 28 – HETTIGER, CHRISTOF – Due: Apr 8 2008, 4:00 am
1
Question 1, chap 14, sect 2.
part 1 of 3
10 points
Consider a uniform ladder leaning against
a smooth wall and resting on a smooth floor
at point
P.
There is a rope stretched horizon
tally, with one end tied to the bottom of the
ladder essentially at
P
and the other end to
the wall. The top of the ladder is at a height
is
h
up the wall and the base of the ladder is
at a distance
b
from the wall.
The weight of the ladder is
W
1
.
Jill, with a
weight
W
2
, is onefourth the way
parenleftbigg
d
=
ℓ
4
parenrightbigg
up
the ladder. The force which the wall exerts on
the ladder is
F.
P
W
1
W
2
ℓ
d
T
θ
h
b
F
Note:
Figure is not to scale.
The torque equation about
P
is given by
1.
b
4
W
2
+
b
2
W
1
=
Fh
correct
2.
h
4
W
2
+
h
2
W
1
=
Fb
3.
h
2
W
2
+
hW
1
=
Fb
4.
(
W
1
+
W
2
)
h
2
=
Fb
5.
b
2
W
2
+
bW
1
=
Fh
6.
(
W
1
+
W
2
)
b
2
=
Fh
Explanation:
Pivot
F
T
N
f
W
2
W
1
θ
summationdisplay
F
x
:
T
−
F
= 0
,
(1)
summationdisplay
F
y
:
N
f
−
W
2
−
W
1
= 0
,
and (2)
summationdisplay
τ
P
:
W
2
d
cos
θ
+
W
1
ℓ
2
cos
θ
(3)
−
Fℓ
sin
θ
= 0
,
where
d
is the distance of the person from the
bottom of the ladder. Therefore
2
Fℓ
sin
θ
= 2
W
2
d
cos
θ
+
W
1
ℓ
cos
θ.
Since sin
θ
=
h
ℓ
and cos
θ
=
b
ℓ
,
the torque
equation about
P
is given by
b
4
W
2
+
b
2
W
1
=
Fh
.
(4)
Question 2, chap 14, sect 2.
part 2 of 3
10 points
Given:
W
2
= 3
W
1
=
W
,
h
=
b
.
Determine the force
F
the wall exerts on
the ladder.
1.
F
=
3
4
W
2.
F
=
5
12
W
correct
3.
F
=
5
6
W
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homework 28 – HETTIGER, CHRISTOF – Due: Apr 8 2008, 4:00 am
2
4.
F
=
2
3
W
5.
F
=
1
3
W
6.
F
=
1
12
W
7.
F
=
1
4
W
8.
F
=
1
2
W
9.
F
=
1
6
W
10.
F
=
7
12
W
Explanation:
For
W
2
= 3
W
1
=
W
and
h
=
b
, Eq. 4 gives
b
4
W
+
b
2
W
3
=
Fb,
so that
F
=
W
4
+
W
6
=
(3 + 2)
12
W
=
5
12
W
.
Question 3, chap 14, sect 2.
part 3 of 3
10 points
Given:
W
2
= 3
W
1
=
W
,
h
=
b.
When Jill has climbed up the ladder such
that the rope tension reaches
T
=
W
2
deter
mine Jill’s height
y
from the floor.
1.
y
=
1
4
b
2.
y
=
5
6
b
3.
y
=
5
12
b
4.
y
=
2
3
b
5.
y
=
1
12
b
6.
y
=
1
3
b
correct
7.
y
=
1
6
b
8.
y
=
3
4
b
9.
y
=
1
2
b
10.
y
=
7
12
b
Explanation:
For rope tension
T
=
W
2
, and by applying
Newton’s 2nd law in the horizontal direction
(giving
T
−
F
= 0), we have
F
=
T
=
W
2
.
With the origin at
P
, denote Jill’s coordinates
on the ladder by (
x,y
).
Then the torque equation gives
xW
2
+
b
2
W
1
=
Fh
=
bW
2
.
(5)
For
W
2
= 3
W
1
=
W
, Eq. 5 gives
xW
+
b
2
W
3
=
bW
2
.
Finally, since
h
=
b
, Jill is at height
y
, where
y
=
x
=
1
2
b
−
1
6
b
=
1
3
b
.
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 Spring '10
 Turner
 Physics, Mass, Work, Sin, Correct Answer, Jill

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