308i7ZRoTV24_1210016284_jwk572 - homework 09 KIM JI Due...

homework 09 – KIM, JI – Due: Oct 31 2007, 4:00 am 1
Question 1, chap 14, sect 1.
part 1 of 1 10 points
A string provides a horizontal force which
acts on a rectangular block of weight 490 N at
top right-hand corner as shown in the figure
below.
1 m
0 . 7 m
F
490 N
If the block slides with constant speed, find
the tension in the string required to start to
tip the block over.
Correct answer: 171 . 5 N (tolerance ± 1 %).
Explanation:
Let : F = 171 . 5 N ,
W = 490 N ,
h = 1 m , and
w = 0 . 7 m .
h
w
F
W
Locate the origin at the bottom left corner
of the block.
In equilibrium (at constant velocity; i.e. , no
acceleration), we have
summationdisplay
vector
F = 0, so
summationdisplay
F
x
= F - μ N = 0 (1)
summationdisplay
F
y
= N - W = 0 , (2)
The block is trying to tip over its lower
right-hand corner (the fulcrum). The string
tension T pulls to the right (clockwise) at a
distance h from that corner. The weight W
acts down (counter-clockwise) at a distance
w
2
from the corner. From the torques
summationdisplay
vector τ = 0 ,
so
summationdisplay
vector τ =
1
2
W w - F h = 0 . (3)
F h =
W w
2
F =
W w
2 h
=
(490 N) (0 . 7 m)
2 (1 m)
= 171 . 5 N .
Question 2, chap 14, sect 1.
part 1 of 1 10 points
A student of mass 54 kg wants to walk be-
yond the edge of a cliff on a heavy beam of
mass 300 kg and length 5 . 5 m. The beam lies
on the horizontal surface of the clifftop (with-
out any attachment), with one end sticking
out beyond the cliff’s edge:
d
The student wants to position the beam so
it sticks out as far as possible beyond the edge,
but he also wants to make sure he can walk to
the beam’s end without falling down.
How far from the edge of the ledge can the
beam extend?
Correct answer: 2 . 33051 m (tolerance ± 1
%).
Explanation:

homework 09 – KIM, JI – Due: Oct 31 2007, 4:00 am 2
Let : m = 54 kg ,
M = 300 kg , and
ℓ = 5 . 5 m ,
The stability of the beam (with the student
standing on it) requires that the center of
gravity of the beam + student system must
lie above the cliff itself.
Let the cliff’s edge be the origin of our co-
ordinate system. The beam extends from
X
1
= d - L < 0 (on the cliff) to X
2
= d > 0
(off the cliff), so assuming the beam is uni-
form, its center of mass is located at
X
beam
c . m .
=
X
1
+ X
2
2
= d -
L
2
.
When the student walks all the way to the
beam’s end, his own center of mass is located
at
X
student
c . m .
≈ d ,
where the approximation neglects the size of
the student compared to the beam’s length.
The overall center of mass of the beam +
student system is therefore located at
X
overall
c . m .
=
m
m + M
X
student
c . m .
+
M
m + M
X
beam
c . m .
= d -
M
m + M
L
2
.
The stability condition is that this overall
center of mass should stay on the cliff, so
X
overall
c . m .
< 0
and
d < d
max
=
M
m + M
L
2
=
300 kg
54 kg + 300 kg
5 . 5 m
2
= 2 . 33051 m .
Question 3, chap 14, sect 2.
part 1 of 2 10 points
A 22 . 1 kg person climbs up a uniform 118 N
ladder. The upper and lower ends of the lad-
der rest on frictionless surfaces. The bottom
of the ladder is fastened to the wall by a
horizontal rope that can support a maximum
tension of 81 . 6 N . The angle between the hor-
izontal and the ladder is 64
◦
.