G4 – Newton’s Laws of Motion Sections 1-8
G4
Q1,2,9,15 ( 3,5,19,26 emergency on 9/19/2008)
P4,5,7,10,11, 19,23,31, 37,49,45
Sections
4-6
7
8
CHAPTER 4: Dynamics: Newton’s Laws of Motion
Questions
1. Why does a child in a wagon seem to
fall backward when you give the wagon a
sharp pull forward?
1.
The child tends to remain at rest (Newton’s 1st Law), unless a force acts on her. The
force is applied to the wagon, not the child, and so the wagon accelerates out from
under the child, making it look like the child falls backwards relative to the wagon.
If the child is standing in the wagon, the force of friction between the child and the
bottom of the wagon will produce an acceleration of the feet, pulling the feet out
from under the child, also making the child fall backwards.
2.
A box rests on the (frictionless) bed of a truck. The truck driver starts the truck and accelerates
forward. The box immediately starts to slide toward the rear of the truck bed. Discuss the motion of
the box, in terms of Newton’s laws, as seen (a) by Mary standing on the ground beside the truck, and
(b) by Chris who is riding on the truck (Fig. 4–35).
2.
(a) Mary sees the box stay stationary with respect to the ground. There is no
horizontal force on the box since the truck bed is smooth, and so the box cannot
accelerate. Thus Mary would describe the motion of the box in terms of
Newton’s 1st law — there is no force on the box, so it does not accelerate.
(b) Chris, from his non-inertial reference frame, would say something about the
box being “thrown” backwards in the truck, and perhaps use Newton’s 2nd law
to describe the effects of that force. But the source of that force would be
impossible to specify.
9.
A stone hangs by a fine thread from the ceiling, and a section of the same thread dangles from the
bottom of the stone (Fig. 4–36). If a person gives a sharp pull on the dangling thread, where is the
thread likely to break: below the stone or above it? What if the person gives a slow and steady pull?
Explain your answers.
9.
When giving a sharp pull, the key is the suddenness of the application of the force. When a large,
sudden force is applied to the bottom string, the bottom string will have a large tension in it. Because
of the stone’s inertia, the upper string does not immediately experience the large force. The bottom
string must have more tension in it, and will break first.
If a slow and steady pull is applied, the tension in the bottom string
increases. We approximate that condition as considering the stone to
be in equilibrium until the string breaks. The free-body diagram for the
stone would look like this diagram. While the stone is in equilibrium,
Newton’s 2nd law states that Fup  Fdown  mg. Thus the tension in the
upper string is going to be larger than the tension in the lower string
because of the weight of the stone, and so the upper string will break
first.
15.
According to Newton’s third law,
each team in a tug of war (Fig. 4–37)
pulls with equal force on the other team.
What, then, determines which team will
win?
15.
In a tug of war, the team that pushes
hardest against the ground wins. It is true
that both teams have the same force on
them due to the tension in the rope. But
the winning team pushes harder against
the ground and thus the ground pushes
harder on the winning team, making a net
unbalanced force. The free body diagram
below illustrates this. The forces are

FT1G , the force on team 1 from the

FT2G ,
ground,
the force on team 2 from

the ground, and FTR , the force on each
team from the rope.
Thus
the
net force on the winning team


F

 FTR is in the winning direction.
T1G
Problems
Sections
4–4 to 4–6
P4,5,7,10,11, 19,23,31, 37,49,45
4-6
7
8
Newton’s Laws, Gravitational Force, Normal Force
3. (I) How much tension must a rope
withstand if it is used to accelerate a 960kg car horizontally along a frictionless
surface at 1.20 m s ?
2
4. (I) What is the weight of a 76-kg
astronaut (a) on Earth, (b) on the Moon


g  1.7 m s , (c) on Mars g  3.7 m s , (d) in


outer
2
space
velocity?
2
traveling
with
constant
4.
5.
In all cases, W  mg, where g changes with location.


 76 kg 1.7 m s 
(a)
WEarth  mgEarth  76 kg  9.8 m s 2  7.4  10 2 N
(b)
WMoon  mgMoon
2
1.3  10 2 N
(II) A 20.0-kg box rests on a table. (a) What is the weight of
the box and the normal force acting on it? (b) A 10.0-kg box is
placed on top of the 20.0-kg box, as shown in Fig. 4–38.
Determine the normal force that the table exerts on the 20.0-kg
box and the normal force that the 20.0-kg box exerts on the
10.0-kg box.
5. (a) The 20.0 kg box resting on
the table has the free-body
diagram shown. It’s
weight is


mg  20.0 kg  9.80 m s 2  196 N .
Since the box is at rest, the net force on
the box must be 0, and so the normal
force must also be 196 N .
(b) Free-body diagrams are shown for

both boxes. F12 is the force on box
1 (the top box) due to box 2 (the bottom
box). That is the normal
force on box 1. F is the
force on box 2 due to box
1, and has the same
magnitude as F by
Newton’s 3rd law. F is the
force of the table on box
2. That is the normal force
on box 2. Since both
boxes are at rest, the net force on each
box must be 0. Write Newton’s 2nd law
in the vertical direction for each box,
taking the upward direction to be
positive.
21
12
N2
F  F
1
N1
 m1g  0


FN1  m1g  10.0 kg  9.80 m s 2  98.0 N  F12  F21
F
2
 FN 2  F21  m2 g  0


FN2  F21  m2 g  98.0 N  20.0 kg  9.80 m s 2  294 N
7. (II) What average force is needed to
accelerate a 7.00-gram pellet from rest to
125 m s over a distance of 0.800 m along
the barrel of a rifle?
7.
The average force on the pellet is its mass times its average acceleration. The
average acceleration is found from Eq. 2-11c. For the pellet, v0  0, v  125 m s , and
x  x0  0.800 m.
125 m s   0

v 2  v02


 9770 m s2
2 x  x0 
2 0.800 m 
2
aavg



Favg  maavg  7.00  10 3 kg 9770 m s2  68.4 N
10.
(II) How much tension must a rope
withstand if it is used to accelerate a
1200-kg
car
2
0.80 m s ?
vertically
upward
at
Choose up to be the positive
direction. Write Newton’s 2nd law
for the vertical
direction, and solve for the tension
force.
 F  FT  mg  ma  FT  m g  a 
10.


FT  1200 kg  9.80 m s2  0.80 m s2  1.3  10 4 N
11.
(II) A particular race car can cover a
quarter-mile track (402 m) in 6.40 s
starting from a standstill. Assuming the
acceleration is constant, how many “g’s”
does the driver experience? If the
combined mass of the driver and race car
is 485 kg, what horizontal force must the
road exert on the tires?
11. Use Eq. 2-11b with
acceleration.
x  x0  v0t  12 at 2

a
v0  0
to find the
2 x  x0   402 m 
1 "g" 
2


19.6
m
s
 9.80 m s2   2.00 g's
t2
6.40 s2
The accelerating force is found by
Newton’s 2nd law.


F  ma  485 kg  19.6 m s2  9.5110 3N
4–7 Newton’s Laws and Vectors
19. (I) A box weighing 77.0 N rests on a
table. A rope tied to the box runs
vertically upward over a pulley and a
weight is hung from the other end (Fig. 4–
40). Determine the force that the table
exerts on the box if the weight hanging on
the other side of the pulley weighs (a)
30.0 N, (b) 60.0 N, and (c) 90.0 N.
19. Free body diagrams for the box and the weight are shown
below. The tension exerts the same magnitude of force on
both objects.
(a) If the weight of the hanging weight is less than the
weight of the box, the objects will not move, and the
tension will be the same as the weight of the hanging
weight. The acceleration of the box will also be zero,
and so the sum of the forces on it will be zero. For the
box,
FN  FT  m1 g  0  FN  m1 g  FT  m1 g  m2 g  77.0 N  30.0 N  47.0 N
(b) The same analysis as for part (a) applies here.
FN  m1 g  m2 g  77.0 N  60.0 N  17.0 N
(c) Since the hanging weight has more weight than the box on the table, the box on
the table will be lifted up off the table, and normal force of the table on the box
will be 0 N .
23. (II) Arlene is to walk across a “high wire” strung horizontally between two buildings 10.0 m apart.
The sag in the rope when she is at the midpoint is 10.0º as shown in Fig. 4–42. If her mass is 50.0 kg,
what is the tension in the rope at this point?
23. Consider the point in the rope directly below Arlene.
That point can be analyzed as having three forces on
it – Arlene’s weight, the tension in the rope towards
the right point of connection, and the tension in the
rope towards the left point of connection. Assuming
the rope is massless, those two tensions will be of the same magnitude. Since the
point is not accelerating the sum of the forces must be zero. In particular, consider
the sum of the vertical forces on that point, with UP as the positive direction.
F  F
T
FT 
26.
sin 10.0º  FT sin 10.0º  mg  0



50.0 kg  9.80 m s2
mg

 1.41  10 3 N
2 sin 10.0º
2 sin 10.0º
(II) A person pushes a 14.0-kg lawn mower at constant speed with
a force of F  88.0 N directed along the handle, which is at an
angle of 45.0º to the horizontal (Fig. 4–45). (a) Draw the freebody diagram showing all forces acting on the mower. Calculate
(b) the horizontal friction force on the mower, then (c) the normal
force exerted vertically upward on the mower by the ground. (d)
What force must the person exert on the lawn mower to accelerate it from rest to
1.5 m s in 2.5 seconds, assuming the same friction force?
31. (II) Figure 4–49 shows a block mass m1  on a smooth horizontal surface, connected by a thin cord
that passes over a pulley to a second block m2 , which hangs vertically. (a) Draw a free-body
diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord,
and any normal force. (b) Apply Newton’s second law to find formulas for the acceleration of the
system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.
31. (a) See the free-body diagrams included.
(b) For block 1, since there is no motion in the vertical
direction., we have FN1  m1 g. We write Newton’s
2nd law for the x direction:  F1x  FT  m1 a1x . For
block 2, we only need to consider vertical forces:
 F2 y  m2 g  FT  m2 a 2 y . Since the two blocks are
connected, the magnitudes of their accelerations
will be the same, and so let a1x  a 2 y  a. Combine
the two force equations from above, and solve for a by substitution.
FT  m1 a m 2 g  FT  m 2 a  m 2 g  m1 a  m 2 a
m1 a  m 2 a  m 2 g

ag
m2
m1  m 2

FT  m1 a  g
m1 m 2
m1  m 2
4–8 Newton’s Laws with Friction; Inclines
37.
(I) A force of 48.0 N is required to start a 5.0-kg box moving across a horizontal
concrete floor. (a) What is the coefficient of static friction between the box and the
floor? (b) If the 48.0-N force continues, the box accelerates at 0.70 m s 2 . What is the
coefficient of kinetic friction?
37. A free-body diagram for the box is shown. Since the box does not
accelerate vertically, FN  mg
(a) To start the box moving, the pulling force must just
overcome the force of static friction, and that means the force
of static friction will reach its maximum value of Ffr   s FN .
Thus we have for the starting motion,
Fp  Ffr   s FN  s mg

s 
Fp
mg

48.0 N
5.0 kg 9.8
m s2

 0.98
(b) The same force diagram applies, but now the friction is kinetic friction, and the
pulling force is NOT equal to the frictional force, since the box is accelerating
to the right.
F  F
p
k 
 Ffr  ma
Fp  ma

 Fp  k FN  ma

48.0 N  5.0 kg  0.70 m s
 Fp   k mg  ma
2


0.91
5.0 kg 9.8 m s2 
40. (II) The coefficient of static friction between hard rubber and normal street
pavement is about 0.8. On how steep a hill (maximum angle) can you leave a car
parked?
40. See the included free-body diagram. To find the maximum
angle, assume that the car is just ready to slide, so that the
force of static friction is a maximum. Write Newton’s 2nd
law for both directions. Note that for the both directions, the
net force must be zero since the car is not accelerating.
mg
F
F
s 
y
 FN  mg cos   0  FN  mg cos
x
 mg sin   Ffr  0  mg sin   Ffr   s FN   s mg cos 
mg sin 
 tan   0.8    tan 1 0.8  39 º  40 º
mg cos 
1 sig fig 
45. (II) The coefficient
of kinetic friction for a
22-kg bobsled on a track
is 0.10. What force is
required to push it down a 6.0º incline and
achieve a speed of 60 km h at the end of 75
m?
45. A free-body
diagram for the
bobsled is shown.
The acceleration of
the sled is found
from Eq. 2-11c. The
final velocity also needs to be converted
to m/s.
 1ms 
v0  60 km h 
 16.667 m s.

3.6
km
h


v 2  v02  2a x  x0  
16.667 m s  1.852 m s2
v 2  v02
ax 

2 x  x0 
2 75 m 
2