HRW Hints for exercise. Chapter 06.pdf

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C
HAPTER
6
H
INT FOR
E
XERCISE
8
The climber pushes against the rock with her back and the rock pushes on her back with a
force of equal magnitude. The rock at her back provides a frictional force that helps hold her
against the downward force of gravity. Since she is one the verge of slipping the magnitude
of the force of friction on her back is given by
f
back
=
µ
back
N
back
, where
N
back
is the normal
force of the rock on her back. A similar statement can be made about the forces at her feet:
f
shoes
=
µ
shoes
N
shoes
. Solve Newton’s second law, with the acceleration equal to zero, for
N
back
and
N
shoes
.
(a) Draw a free-body diagram for the climber. Indicate and label the two normal forces
N
back
and
N
shoes
, the two forces of friction
f
back
and
f
shoes
, and the force of gravity
mg.
Take
the
x
axis to be horizontal and the
y
axis to be vertical.
(b) Write Newton’s second law in component form:
N
shoes
−
N
back
= 0
and
f
shoes
+
f
back
−
mg
= 0
.
Substitute
f
shoes
=
µ
shoes
N
shoes
and
f
back
=
µ
back
N
back
, then solve simultaneously for
N
shoes
and
N
back
. According to Newton’s third law the magnitude of the push of the feet against
the rock is
N
shoes
and the magnitude of the push of the back against the rock is
N
back
.
(c) You want to calculate
f
shoes
/mg.
Use
f
shoes
=
µ
shoes
N
shoes
.
C
HAPTER
6
H
INT FOR
P
ROBLEM
14
Take the
x
axis to be parallel to the slide and positive down the slide. Place the origin at
the point where the pig starts (from rest). If
x
is the length of the slide and
t
is the time
the pig takes to slide down it, then
x
=
1
at
2
, where
a
is the acceleration of the pig. Now let
2
a
f
be the acceleration when friction is present and let
a
nf
be acceleration for a frictionless
slide. Let
t
be the time of the slide when the slide is frictionless. Then
x
=
1
a
nf
t
2
and
2
1
2
x
=
2
a
f
(2t) . This means that
a
nf
= 4a
f
.
Use Newton’s second law to find expressions for
a
f
and
a
nf
. Draw a free-body diagram for
the pig sliding on a slide with friction. Three forces act on the pig: the force of gravity
mg,
down, the normal force
N
of the slide, perpendicular to the slide, and the force of kinetic
friction
µ
k
N
, parallel to the slide and up the slide. Here
m
is the mass of the pig and
µ
k
is
the coefficient of kinetic friction between the slide and the pig.
The
x
component of Newton’s second law is
mg
sin
θ
−
µ
k
N
=
ma
f
,
where
θ
is the angle the slide makes with the horizontal.
Take the
y
axis to be in the direction of the normal force. The
y
component of Newton’s
second law is
N
−
mg
cos
θ
= 0
.
Use this equation to find
N
=
mg
cos
θ,
then replace
N
in the
x
component equation with
mg
cos
θ.
Solve the resulting equation for
a
f
.
You can find an expression for
a
nf
by setting
µ
k
equal to zero in the expression you found
for
a
f
. Finally, set
a
nf
equal to 4a
f
and solve for
µ
k
.
C
HAPTER
6
H
INT FOR
P
ROBLEM
18
(a) When the applied force is a minimum, the force of friction is up the plane (helping
the applied force) and has its maximum value
µ
s
N
. Use Newton’s second law, with the
acceleration equal to zero, to solve for
F
.
Draw a free-body diagram for the block. Indicate and label the force of friction
f
, the applied
force
F
, the normal force
N
, and the weight
mg.
Take the
x
axis to be up the plane and
the
y
axis to be normal to the plane. Newton’s second law yields
f
+
F
−
mg
sin
θ
= 0
and
N
−
mg
cos
θ
= 0
.
Replace
f
with
µ
s
N
in the first equation. Solve the second equation for
N
and use the
resulting expression to substitute for
N
. Solve for
F
.
(b) As
F
is increases the force of friction diminishes to zero, then it increases in the other
direction (down the plane) until it reaches its greatest possible magnitude,
µ
s
N
. Then the
block starts to slide. Repeat the calculation above, but with the force of friction down the
plane.
(c) The Newton’s second law equations are exactly like those for part (b) (the frictional force
is down the plane) but now
f
=
µ
k
N
.
C
HAPTER
6
H
INT FOR
P
ROBLEM
22
Block A is pulled down the incline by the force of gravity and is restrained by the force of
friction and the tension force of the rope. The components of these forces parallel to the
incline must sum to zero since the velocity of the block is constant. Similarly block
B
is
pulled upward by the tension force of the rope and is restrained by the force of gravity.
These forces must also sum to zero. The tension force depends on the mass of block
B:
the
greater the mass the greater the tension force. Thus, the mass of B must be just right to
bring about the proper tension force to just balance all other forces on A.
Draw a free-body diagram for block B. Show the force of gravity (m
B
g,
down) and the
tension force of the rope (T , up). Take the axis to be positive in the upward direction and
write the Newton’s second law equation:
T
−
m
B
g
= 0
.
Draw a free-body diagram for block A. Show the force of gravity (m
A
g,
down), the normal
force (N , perpendicular to the plane), the force of friction (f , parallel to the plane and up
the plane), and the tension force of the rope (T , parallel to the plane and up the plane).
Take the
x
axis to be parallel to the plane and positive down the plane; take the
y
axis to
be perpendicular to the plane. Write the second law equations:
m
A
g
sin
θ
−
T
−
f
= 0
and
N
−
m
A
g
cos
θ
= 0
,
where
θ
is the angle of the plane above the horizontal.
Replace
f
with
µ
k
N
and solve the three equations for
m
B
. From the first equation
T
=
m
B
g.
Substitute this expression into the other two equations. You should now have
N
−
m
A
g
cos
θ
= 0
and
m
A
g
sin
θ
−
m
B
g
−
µ
k
m
A
g
cos
θ
= 0
.
Substitute
N
=
m
A
g
cos
θ,
from the first equation, into the second equation and solve for
m
B
. You should get
m
B
= (sin
θ
−
µ
k
cos
θ)m
A
.
Evaluate this expression.
C
HAPTER
6
H
INT FOR
P
ROBLEM
25
A large force
F
clearly holds the smaller block against the larger block. The force of static
friction between the blocks balances the force of gravity on the smaller block. If
F
is reduced,
the force of the smaller block on the larger block is reduced, the reaction force of the larger
block on the smaller block is reduced, and the maximum allowable force of static friction is
reduced. When it becomes less than the force of gravity on the smaller block, that block
falls. If the minimum force is applied, the force of friction is given by
f
=
µ
s
N
, where
N
is
the (horizontal) normal force of the larger block on the smaller block.
Draw a free-body diagram for the smaller block. The forces acting on this block are the
force of gravity
mg
(down), the force of friction
f
(up), the applied force
F
(right), and the
normal force
N
exerted by the larger block (left). Take the
x
axis to be horizontal and the
y
axis to be vertical. Write the Newton’s second law equations:
F
−
N
=
ma
and
f
−
mg
= 0
.
Since
f
=
mg
and
f
=
µ
s
N
, you should obtain
N
=
mg/µ
s
and
F
−
mg/µ
s
=
ma.
Now, consider the two blocks as a single object and write Newton’s second law as
F
= (m +
M
)a
.
Solve this equation and
F
−mg/µ
s
=
ma
simultaneously for
F
. The other unknown quantity
is the acceleration. It must be eliminated.
ans:
490 N

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