Chapter 6 – Student Solutions Manual
1. 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
G
F
in the +
x
direction). Applying
Newton’s second law to the
x
and
y
axes, we obtain
, max
0
s
N
F
f
m
F
mg
a
−
=
−
=
respectively. The second equation yields the normal force
F
N
=
mg
, whereupon the
maximum static friction is found to be (from Eq. 6-1)
f
s
s
,max
mg
=
μ
. Thus, the first
equation becomes
F
mg
ma
s
−
=
=
μ
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
μ
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.
(b) Replacing
m
= 45 kg with
m
= 28 kg, the reasoning above leads to roughly
.
2
1.2
10
N
F
=
×
3. We denote
as the horizontal force of the person exerted on the crate (in the +
x
direction),
G
is the force of kinetic friction (in the –
x
direction),
is the vertical
normal force exerted by the floor (in the +
y
direction), and
G
F
f
k
N
F
mg
G
is the force of gravity.
The magnitude of the force of friction is given by
f
k
=
μ
k
F
N
(Eq. 6-2). Applying Newton’s
second law to the
x
and
y
axes, we obtain
0
k
N
F
f
ma
F
mg
−
=
−
=
respectively.
(a) The second equation yields the normal force
F
N
=
mg
, so that the friction is
(
)(
)
2
2
0.35
55 kg (9.8 m/s )
1.9
10
N .
k
k
f
mg
μ
=
=
=
×
(b) The first equation becomes
F
mg
m
k
a
−
=
μ

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