Version 108/ABCDA – final – Demkov – (59910)
1
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printout
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have
12
questions.
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before answering.
001
10.0 points
A uniform bar of length
L
and weight
W
is
attached to a wall with a hinge that exerts on
the bar a horizontal force
H
x
and a vertical
force
H
y
. The bar is held by a cord that makes
a 90
◦
angle with respect to bar and angle
θ
with respect to wall.
The acceleration of gravity
g
= 9
.
8 m
/
s
2
.
L
90
◦
W
θ
What is the magnitude of the horizontal
force
H
x
on the pivot?
1.
H
x
=
1
2
W
tan
θ
2.
H
x
=
1
2
W
cos
θ
3.
H
x
=
1
2
W
sin
θ
cos
θ
correct
4.
H
x
=
1
2
W
sin
θ
5.
H
x
=
1
2
W
cos
2
θ
6.
H
x
=
1
2
W
sin
2
θ
Explanation:
Analyzing the torques on the bar, with the
hinge at the axis of rotation, we have
summationdisplay
τ
=
L T
−
parenleftbigg
L
2
sin
θ
parenrightbigg
W
= 0
,
so,
T
=
1
2
W
sin
θ .
Analyzing the force on the
bar, we have
summationdisplay
F
x
=
H
x
−
T
cos
θ
= 0
.
Put
T
into this equation and get
H
x
=
parenleftbigg
1
2
W
sin
θ
parenrightbigg
cos
θ
.
002
10.0 points
A triangular wedge 7 m high, 14 m base
length, and with a 13 kg mass is placed on
a frictionless table. A small block with a 8 kg
mass (and negligible size) is placed on top of
the wedge as shown in the figure below.
13 kg
14 m
7 m
8 kg
Δ
X
wedge
13 kg
14 m
7 m
8 kg
All surfaces are frictionless, so the block
slides down the wedge while the wedge slides
sidewise on the table. By the time the block
slides all the way down to the bottom of the
wedge, how far Δ
X
wedge
does the wedge slide
to the right?
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Version 108/ABCDA – final – Demkov – (59910)
2
1. 4.72727
2. 4.66667
3. 5.71429
4. 4.95238
5. 4.58824
6. 3.94737
7. 3.61111
8. 5.89474
9. 5.33333
10. 4.11765
Correct answer: 5
.
33333 m.
Explanation:
Let :
M
= 13 kg
,
m
= 8 kg
,
L
= 14 m
,
and
H
= 7 m
.
Consider the wedge and the block as a two
body system.
The
external forces
acting on
this system — the weight of the wedge, the
weight of the block and the normal force from
the table — are all vertical, hence the net
hor
izontal momentum
of the system is conserved,
P
wedge
x
+
P
block
x
= constant
.
Furthermore, we start from rest =
⇒
center
ofmass is not moving, and therefore the
X
coordinate of the centerofmass will remain
constant while the wedge slides to the right
and the block slides down and to the left,
X
cm
=
m X
block
+
M X
wedge
m
+
M
= constant
.
Note:
Only the
X
coordinate of the center
ofmass is a constant of motion;
i.e.
, the
Y
cm
accelerates downward because the
P
y
compo
nent of the net momentum is not conserved.
Constant
X
cm
means Δ
X
cm
= 0 and there
fore
m
Δ
X
block
+
M
Δ
X
wedge
= 0
.
Note that this formula does not depend on
where the wedge has its own centerofmass;
as long as the wedge is rigid, its overall dis
placement Δ
X
wedge
is all we need to know.
Finally, consider the geometry of the prob
lem: By the time the block slides all the way
down, its displacement
relative to the wedge
is equal to the wedge length
L
, or rather
−
L
because the block moves to the left of the
wedge. In terms of displacements relative to
the inertial frame of the table, this means
Δ
X
block
−
Δ
X
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 Spring '07
 Swinney
 mechanics, Energy, Force, Kinetic Energy, Mass, Potential Energy, Version 108/ABCDA

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