219
wagon, results in F
T
=F
W
: we simply support the
wagon and it glides up the slope like a chair-lift on a
ski slope. In the case of
.
=180
°
-
.
, F
T
becomes
infinite. Physically this is because we are pulling
directly into the ground, so no amount of force will
suffice.
11. (a) If there was no friction, the angle of repose
would be zero, so the coefficient of static friction,
µ
s
,
will definitely matter. We also make up symbols
.
, m
and g for the angle of the slope, the mass of the
object, and the acceleration of gravity. The forces
form a triangle just like the one in section 8.3, but
instead of a force applied by an external object, we
have static friction, which is less than
µ
s
F
N
. As in that
example, F
s
=mg sin
.
, and F
s
<
µ
s
F
N
, so
mg sin
.
<
µ
s
F
N
.
From the same triangle, we have F
N
=mg cos
.
, so
mg sin
.
<
µ
s
mg cos
.
.
Rearranging,
.
< tan
–1
µ
s
.
(b) Both m and g canceled out, so the angle of
repose would be the same on an asteroid.
Chapter 9
5. Each cyclist has a radial acceleration of v
2
/r=5 m/
s
2
. The tangential accelerations of cyclists A and B
are 375 N/75 kg=5 m/s
2
.
A
B
C
scale:
5 m/s
2
6. (a) The inward normal force must be sufficient to
produce circular motion, so
F
N
= mv
2
/ r .
We are searching for the minimum speed, which is
the speed at which the static friction force is just
barely able to cancel out the downward gravitational
force. The maximum force of static friction is
|F
s
|=
µ
s
F
N
,
and this cancels the gravitational force, so
|F
s
|= mg .
Solving these three equations for v gives
v=
gr
µ
s
.
(b) Greater by a factor of
3
.
7. The inward force must be supplied by the inward
component of the normal force,
F
N
sin
.
= mv
2
/ r .
The upward component of the normal force must
cancel the downward force of gravity,
F
N
cos
.
= mg .
Eliminating F
N
and solving for
.
, we find
.
=tan
–1
v
2
gr
.
Chapter 10
10. Newton’s law of gravity tells us that her weight
will be 6000 times smaller because of the asteroid’s
smaller mass, but 13
2
=169 times greater because of
its smaller radius. Putting these two factors together
gives a reduction in weight by a factor of 6000/169,
so her weight will be (400 N)(169)/(6000)=11 N.
11. Newton’s law of gravity says F=Gm
1
m
2
/r
2
, and
Newton’s second law says F=m
2
a, so Gm
1
m
2
/r
2
=m
2
a.
Since m
2
cancels, a is independent of m
2
.
12. Newton’s second law gives
F = m
D
a
D
,
where F is Ida’s force on Dactyl. Using Newton’s
universal law of gravity, F = Gm
I
m
D
/r
2
,and the
equation a = v
2
/ r for circular motion, we find
Gm
I
m
D
/ r
2
= m
D
v
2
/ r .
Dactyl’s mass cancels out, giving
Gm
I
/ r
2
= v
2
/ r .
Dactyl’s velocity equals the circumference of its orbit
divided by the time for one orbit: v=2
p
r/T. Inserting
this in the above equation and solving for m
I
, we find
m
I
=
4
p
2
r
3
GT
2
,
so Ida’s density is
.
= m
I
/ V
=
4
p
2
r
3
GVT
2
.