This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A student oF mass 58 kg wants to walk beyond
the edge oF a cli± on a heavy beam oF mass
190 kg and length 6
.
4 m.
The beam is not
attached to the cli± in any way, it simply lays
on the horizontal surFace oF the cli±top, with
one end sticking out beyond the cli±’s edge:
d
The students want 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?
1. 5.04425
2. 2.6775
3. 2.28928
4. 3.54768
5. 3.44697
6. 4.54128
7. 2.64988
8. 3.14286
9. 2.45161
10. 3.76736
Correct answer: 2
.
45161 m.
Explanation:
Let :
m
= 58 kg
,
M
= 190 kg
,
and
L
= 6
.
4 m
.
An unsecured body supported From below
remains stable iF and only iF its center oF grav
ity projects down inside the area supporting
the body. IF the body in turn supports some
thing else above it, then the overall center oF
gravity (oF the body plus everything on top oF
it) must project down inside the supporting
area.
²or the problem in question, the stability
oF the beam – with the student standing on
it – requires that the center oF gravity oF the
beam+student system must lay above the cli±
itselF. IF this overall center oF gravity were to
move beyond the cli±’s edge, the beam would
tilt down and Fall o± the cli± – and the student
would Fall down too.
Let the cli±’s edge be the origin oF our co
ordinate system.
The beam extends From
X
1
=
d

L <
0 (on the cli±) to
X
2
=
d >
0
(o± the cli±), 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 is neglecting 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 cli±, thus
X
overall
c
.
m
.
<
0
and hence
d < d
max
=
M
m
+
M
×
L
2
= 2
.
45161 m
.
002
10.0 points
midterm 03 – Turner – (58185)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentVersion 051/AADAD – midterm 03 – Turner – (58185)
2
A block of mass 3 kg and one of mass 7 kg are
connected by a massless string over a pulley
that is in the shape of a disk having a radius
of 0
.
18 m, and a mass of 4 kg. In addition, the
blocks are allowed to move on a Fxed block
wedge of angle 44
◦
, as shown. The coe±cient
of kinetic friction is 0
.
26 for both blocks.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Kinetic Energy, Mass, Momentum, kg, Isquare

Click to edit the document details