practice 15 – – Due: May 8 2007, noon
1
Question 1, chap 18, sect 3.
part 1 of 1
5 points
The
pressure
exerted on the ground by a
man is greatest when:
1.
He stands with both feet flat on the
ground.
2.
He stands on the toes of one foot.
cor
rect
3.
The pressure on the ground is the same,
however he stands, sits or lies.
4.
He sits on his butt.
5.
He stands flat on one foot.
6.
He lies down on his back.
Explanation:
P
=
F
A
The forces exerted on the earth in different
cases are the same while when the man stands
on the toes of one foot, the area is the smallest
and the pressure is the greatest.
Question 2, chap 18, sect 4.
part 1 of 1
5 points
The four tires of an automobile are inflated
to a gauge pressure of 88100 Pa.
Each tire
has an area of 0
.
0212 m
2
in contact with the
ground.
Determine the weight of the automobile.
Correct answer:
7470
.
88
N (tolerance
±
1
%).
Explanation:
Let W be its weight.
Then each tire sup
ports
W
4
,
so that
P
=
F
A
=
W
4
A
,
yielding:
W
= 4
A P
= 4 (0
.
0212 m
2
)(88100 Pa)
= 7470
.
88 N
.
Question 3, chap 18, sect 3.
part 1 of 2
5 points
In 1657, Otto von Guericke, inventor of
the air pump, evacuated a sphere made of
two brass hemispheres.
Two teams of eight
horses each could pull the hemispheres apart
only on some trials, and then only “with the
greatest of difficulty.” (See figure for details.)
F
F
P
R
P
a
If
P
a
is the atmospheric pressure, P is the
pressure inside the hemispheres and R is the
radius of the hemispheres, then the force F
required to pull the hemispheres apart is given
by
1.
F
=
π R
2
(
P
a

P
)
correct
2.
F
=
2
π R
3
3
(
P
a

P
)
3.
F
=
π R
2
2
(
P
a

P
)
4.
F
= 4
π R
2
(
P
a

P
)
5.
F
=
4
π R
3
3
(
P
a

P
)
6.
F
= 2
π R
2
(
P
a

P
)
Explanation:
Basic Concepts
P
≡
F
A
Solution:
Force = (Difference in pressure on
2 sides)
×
(Area).
We must choose the area
carefully. Each team of horses is pulling in the
z
direction with a force
F
. The hemispheres
will come apart only when
F
is
≥
the
z
com
ponent of the net force on each hemisphere
due to the pressure difference (see figure). We
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practice 15 – – Due: May 8 2007, noon
2
must therefore pick the effective area which is
perpendicular to the
z
direction. If you stand
far away on the
z
axis and look at the hemi
sphere, you see a circle of area
πR
2
. Hence,
F
=
π R
2
(
P
a

P
)
.
θ
φ
x
y
dA
F
z
Hemisphere of radius R
dF
If we wanted to justify this more rigorously,
we would have to examine the amount of force
dF
caused by the pressure difference acting on
a small area of the hemisphere
dA
(see figure).
We would then take the zcomponent of this
and integrate this amount over the surface
of the hemisphere (a double integral).
The
result is the same as our “intuitive” argument
above.
Question 4, chap 18, sect 3.
part 2 of 2
5 points
Determine the force, if
P
= 0
.
0534 atm and
R
= 0
.
225 m.
Assume:
P
a
= 101300 Pa.
Correct answer: 15
.
2507
kN (tolerance
±
1
%).
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 Spring '08
 Turner
 Force, Lift

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