MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall 2010
Problem Set 2: One and Two Dimensional Kinematics Solutions
W03D1 Reading Question Monday/Tuesday Class September 20/21.
The static friction force
!
f
s
can have a magnitude
!
f
s
!
f
s
"
μ
s
N
(Eq. 5.6).
Suppose you
have a block with a rope attached on opposite sides (right and left sides). (a) Describe the
direction and magnitude of the static friction force as you increase your pull on the right
side until the block just slips. (b) Describe the direction and magnitude of the static
friction force as you increase your pull on the left side until the block just slips. (c).
Describe the direction and magnitude of the static friction force when you pull both sides
with the same magnitude of force.
Answer. (a) As you increase your pull on the right, the static friction force increases in
magnitude and is directed oppose your pull (to the left). When the block just slips the
magnitude of the static friction force reaches its maximum value
(
f
s
)
max
=
μ
s
N
. (b)
When you pull on the left, the direction of the static friction force is now pointing to the
right. The magnitude of the static friction force increases until the block just slips when
the static friction force reaches its maximum value
(
f
s
)
max
=
μ
s
N
. (c) When you pull both
sides equally the static friction force is zero.
W03D2 Reading Question Wednesday/Thursday Class September 22/23.
The magnitude of gravitational force on an object of mass
m
at the surface of the earth
due to the interaction of the object and the earth is given by
!
F
grav
=
mg
E
(Eq. 4.10). The
magnitude of this force is also given by Eq. 12.2,
!
F
grav
=
Gm
E
m
/
R
E
2
. If you placed the
same object on the surface of a planet with the same mass as the earth,
m
P
=
m
E
but only
one fifth the mass density,
!
P
=
(1/ 5)
!
E
, what is the ratio of the magnitudes of the
gravitational accelerations at the surface of the planet and earth, i.e. what is
g
P
/
g
E
?
Answer.
The gravitational constant for the earth is found by setting the two different expressions
for the gravitational force at the surface of the earth equal,
mg
E
=
Gm
E
m
/
R
E
2
which
becomes
g
E
=
Gm
E
/
R
E
2
. Thus the ratio of the magnitudes of the gravitational
accelerations at the surface of the planet and earth is
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g
P
g
E
=
Gm
p
/
R
P
2
Gm
E
/
R
E
2
=
m
p
/
R
P
2
m
E
/
R
E
2
.
Because we are told that the masses of the earth and the planet are the same,
m
P
=
m
E
,
the ratio of the magnitudes of the gravitational accelerations is
g
P
g
E
=
R
E
2
R
P
2
.
The mass, density and volume of a spherically symmetric planet of uniform mass density
is given by
m
P
=
!
P
(4 / 3)
"
R
P
3
. We can express the condition that the masses of the earth
and the planet are equal as
!
P
(4 / 3)
"
R
P
3
=
!
E
(4 / 3)
"
R
E
3
.
Therefore we can solve for the ratio of the radii
R
E
R
P
=
!
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 Fall '09
 guth
 Physics, Force, Friction, Mass, General Relativity, Velocity, Car, static friction force

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