217
Chapter 6
1. The greatest deceleration (of magnitude
a
) is provided by the maximum friction force
(Eq. 61, with
F
N
=
mg
in this case).
Using Newton’s second law, we find
a
=
f
s,max
/
m
=
s
g
.
Eq. 216 then gives the shortest distance to stop: 
x
 =
v
2
/2
a
= 36 m.
In this calculation,
it is important to first convert
v
to 13 m/s.
2. Applying Newton’s second law to the horizontal motion, we have
F
k
m g = ma
,
where we have used Eq. 62, assuming that
F
N
=
mg
(which is equivalent to assuming
that the vertical force from the broom is negligible). Eq. 216 relates the distance traveled
and the final speed to the acceleration:
v
2
= 2
a
x
.
This gives
a
= 1.3 m/s
2
. Returning to
the force equation, we find (with
F =
25 N and
m =
3.5 kg) that
k
= 0.60.
3. The freebody diagram for the bureau is
shown to the right. We do not consider the
possibility that the bureau might tip, and treat
this as a purely horizontal motion problem (with
the person’s push
F
in the +
x
direction).
Applying Newton’s second law to the
x
and
y
axes, we obtain
, max
0
s
N
F
f
ma
F
mg
respectively. The second equation yields the
normal force
F
N
=
mg
, whereupon the maximum
static friction is found to be (from Eq. 61)
f
mg
s
s
,max
.
Thus, the first equation becomes
F
mg
ma
s
0
where we have set
a
= 0 to be consistent with the fact that the static friction is still (just
barely) able to prevent the bureau from moving.
(a) With
s
0 45
.
and
m
= 45 kg, the equation above leads to
F
= 198 N.
To bring the bureau into a state of motion, the person should push with any force greater
than this value. Rounding to two significant figures, we can therefore say the minimum
required push is
F
= 2.0
10
2
N.
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CHAPTER 6
218
(b) Replacing
m
= 45 kg with
m
= 28 kg, the reasoning above leads to roughly
2
1.2
10
N
F
.
Note: The values found above represent the minimum force required to overcome the
friction. Applying a force greater than
s
mg
results in a net force in the +
x
direction,
and hence, nonzero acceleration.
4. We first analyze the forces on the pig of mass
m
. The incline angle is
.
The +
x
direction is “downhill.’’ Application of Newton’s second law to the
x
 and
y
axes
leads to
sin
cos
0.
k
N
mg
f
ma
F
mg
Solving these along with Eq. 62 (
f
k
=
k
F
N
) produces the following result for the pig’s
downhill acceleration:
sin
cos
.
k
a
g
To compute the time to slide from rest through a downhill distance
, we use Eq. 215:
v t
at
t
a
0
2
1
2
2
.
We denote the frictionless (
k
= 0) case with a prime and set up a ratio:
t
t
a
a
a
a
2
2
/
/
which leads us to conclude that if
t
/
t'
= 2 then
a'
= 4
a
. Putting in what we found out
above about the accelerations, we have
sin
4
sin
cos
.
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 Spring '11
 Force, Friction

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