Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 1
We start with a projectile problem.
A golf ball is hit from the ground at 35 m/s at an angle of 55º. The ground is level.
1. How long is the ball in the air?
2. What is the maximum height of the ball?
3. How far from the launching point does the ball hit the ground?
4. What is the ball’s position after 2.0 seconds? When does it reach this height again?
5. When is the ball 20 m above the ground?
In the plot below, all units are in meters.
Solution:
1. When the ball hits the ground, y = 0.
)(0
)(
21
2
21
tavt
tatvy
yiy
yiy
Since the only way a product can equal zero is when one of the factors equals zero,
0)(or0 21 tavt yiy
0
5
10
15
20
25
30
35
40
45
0 20 40 60 80 100 120
iyv
ixv
iv
= 55o
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 2
The first condition tells us that the golf ball starts from the ground. The second gives us
the time of flight, tf
y
iy
f
iyy
yiy
a
vt
vta
tav
2
0)(
21
21
This is more useful if we use ay = −g and viy = vi sin (see the diagram below the
trajectory plot).
s85.5
m/s8.9
55sin)m/s35(2
sin2
2
2
g
v
a
vt
i
y
iy
f
What angle maximizes the time of flight? The angle that maximizes sin . The largest
sine can be is 1 and that occurs at 90º. Hit the ball straight up!
2. At the maximum height, vfy = 0.
g
vt
tgv
tavv
ih
hi
yiyfy
sin
sin0
But this is ½ the time of flight. When the ball is shot over level ground half of the time
the ball is going up, the other half of the time it is going down. It takes half the total time
to reach the highest point.
The height at this time is
2
21
2
21
2
21
sinsinsin
)(sin
)(
g
vg
g
vv
tgtvh
tatvy
ii
i
hhi
yiy
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 3
m9.41
)m/s8.9(2
55sin)m/s35(
2
sin
sin
2
1sin
2
22
22
2
2222
g
v
g
vg
g
vh
i
ii
What angle maximizes the height? We need to find the maximum of sin2 . The
maximum of sin2 occurs at the maximum of sin , which again is 90º. Hit it straight up.
3. The ball hits the ground when t = tf. The horizontal displacement from the launch
point to where the ball hits the ground is called the range (R).
g
v
g
vv
tvR
tvx
i
ii
fi
ix
cossin2
sin2cos
cos
2
This can be rewritten using the identity sin 2 = 2sin cos ,
m117
m/s8.9
)]55(2sin[)m/s35(
2sin
2
2
2
g
vR i
What angle maximizes the range? This time we want to find the maximum of sin 2.
The maximum of sine occurs at 90º. This time 2 = 90º or = 45º. The angle required
for maximum range over level ground is 45º. Also the range is symmetric about 45º. For
some angle ,
2cos)290sin()]45(2sin[
and
2cos)2cos()290sin()]45(2sin[
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 4
The range for 1 = 45º + is the same for 2 = 45º − . Another way is to say if
90)45()45(21
the ranges are the same!
4. Where is the ball at 2.0 seconds? Its horizontal position is found using
m2.40
s255cos)m/s35(
cos
fi
ix
tv
tvx
Its vertical position is
m7.37
)s0.2)(m/s8.9()s0.2(55sin)m/s35(
)(sin
)(
22
21
2
21
2
21
tgtv
tatvy
i
yiy
When is it at this height again? There are many ways to find this. First, note that the
time of flight of the ball is 5.85 s from part 1. Since the trajectory is symmetric, if it
reaches this height 2.0 s after launch, it will reach it again 2.0 s before it lands,
s85.3s00.2s85.5 t
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 5
Another way is to use the symmetry of the y-component of the velocity. The y-
components of the velocities at the same heights have the same magnitudes but opposite
signs. At 2.0 s
m/s07.9
)s2)(m/s8.9(55sin)m/s35(
sin
2
tgvv
tavv
ify
yiyfy
When is the speed −9.07 m/s?
s85.3
m/s8.9
55sinm/s)35(m/s07.9
sin
sin
2
g
vvt
tgvv
tavv
ify
ify
yiyfy
Finally, the worst way is to just solve for t using the quadratic formula,
s86.3,s00.2
8.9
11.967.28
)9.4(2
)7.37)(9.4(4)67.28()67.28(
2
4
0m7.37)m/s67.28())(m/s9.4(
0m7.3755sin)m/s35())(m/s8.9(
))(m/s8.9(55sin)m/s35(m7.37
)(sin
)(
2
2
22
22
21
22
21
2
21
2
21
a
acbbt
tt
tt
tt
tgtv
tatvy
i
yiy
The first answer is when it reaches 37.7 m while ascending (we knew it would be 2.0 s),
the second is when it reaches 37.7 m while descending. (I quit writing units in the
quadratic because it makes the equation even more unwieldy.)
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 6
5. To find when the ball reaches 20 m we use the quadratic equation again,
s04.5,s81.0
8.9
74.2067.28
)9.4(2
)20)(9.4(4)67.28()67.28(
2
4
0m20)m/s67.28())(m/s9.4(
0m2055sin)m/s35())(m/s8.9(
))(m/s8.9(55sin)m/s35(m20
)(sin
)(
2
2
22
22
21
22
21
2
21
2
21
a
acbbt
tt
tt
tt
tgtv
tatvy
i
yiy
Notice that the sum of these times is 5.85 s, the time of flight.
Summary: Derived equations for a projectile launched from level ground with initial
velocity vi at an angle above the ground:
Time of flight g
vt i
f
sin2
Time to maximum height fi
h tg
vt 2
1sin
Maximum height g
vh i
2
sin22
Range g
vR i 2sin
2
If the ground is not level, for example throwing a ball from the top of a building, these
equations will not apply (unless the initial and final heights are the same)!
The projectile travels in a parabolic path as long as we neglect air resistance. The motion is
symmetric about the maximum height (the vertex of the parabola).
Relative velocity is a great example of adding vectors.
Have you ever had this happen to you? While sitting in your car at a red traffic light, the car
beside you slowly drifts forward. You mash on the brake to stop your car from rolling
backwards, but your car is not moving.
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 7
Within your environment, there is no way to distinguish between your car moving backwards
and the car besides you moving forward. The velocity is relative. We need a reference frame
(the traffic light, for example) to define who is moving.
The train moves at 10 m/s and Wanda can walk at 1 m/s. How fast will Greg see Wanda walk?
Wanda’s velocity relative to Greg is the sum of the velocity of the Wanda relative to the train
plus the velocity of train relative to Greg.
TGWTWG vvv
Notice the order of the subscripts. We have the Ts cancelling from the two terms on the right.
This equation will always hold, but how do we use it? What is our rule about vectors?
WE DO NOT DEAL WITH VECTORS. WE DEAL WITH THEIR COMPONENTS.
Take the x-component:
m/s11
m/s)10(m/s)1(
TGxWTxWGx vvv
Greg sees Wanda walking to the right at 11 m/s. What happens when she walks back to her seat?
m/s9
m/s)10(m/s)1(
TGxWTxWGx vvv
According to Greg, Wanda is walking at 9 m/s to the right.
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 8
Hopefully, this is pretty easy. But
what about this?
From Example 3.10. Jack wants to
row directly across the river from the
east shore to a point on the west
shore. The current 0.61 m/s and Jack
can row at 1.16 m/s. What direction
must he point the boat and what is his
velocity across the river?
The velocity of the rowboat relative
to the shore is equal to the velocity of
the rowboat relative to the water plus
the velocity of the water relative to
the shore.
WSRWRS vvv
The rowboat is to head directly to the west.
Take components.
WSyRWyRSyWSxRWxRSx vvvvvv and
The diagram is the key to solving relative velocity problems. For the x-component,
cos
0cos
RW
RWRS
WSxRWxRSx
v
vv
vvv
The y-component,
sin
sin0
RWWS
WSRW
WSyRWyRSy
vv
vv
vvv
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 9
Our unknowns are and vRS. From the y-component equation,
7.31
526.0m/s16.1
m/s61.0sin
sin
RW
WS
RWWS
v
v
vv
From the x-component equation.
m/s987.0
7.31cosm/s)16.1(
cos
RWRS vv
The boat must point 31.7º N of W upstream. Its speed across the water is 0.99 m/s.
Chapter 4 Force and Newton’s Laws of Motion
We can describe motion, but why do things move?
Forces: Objects interact through forces. “A force is a push or pull.” Forces can be long range
(gravity, electric, magnetic, etc.) or contact (normal force, tension, etc.).
Fig. 04.01
Obviously, forces are vector quantities since their effect depends on the directions of the forces.
The net force is the vector sum of all forces acting on an object.
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 10
nFFFFF
21net
A free-body diagram (FBD) is an essential tool for finding the net force acting on an object.
(See page 96.)
Draw the object in a simplified way
Identify all the forces that are exerted on the object.
Draw vector arrows representing all the forces on the object.
Examples
1. Freely falling object.
2. Object hanging from a rope.
3. Object sitting on a horizontal table.
4. Object sitting on a horizontal table being pulled by a rope.
Drawing the free-body diagram is the key to solving problems.
Newton’s First Law (law of inertia): An object’s velocity vector v
remains constant if and only
if the net force acting on the object is zero (page 97).
An object moving at constant velocity has no net force! A revolutionary idea.
An object moving at constant velocity is said to be in translational equilibrium. That velocity
could be zero.
T
N N T
W W W
W
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 11
Inertia is the resistance to changes in velocity.
Newton’s Second law: The rate of change of an object’s velocity is proportional to the net force
acting on it and inversely proportional to its mass (page 101).
aF
m
Recall our rule: we never deal with vectors, we deal with their components. A far more useful
form of Newton’s second law will be
xx maF yy maF
The left hand side is supplied by the free-body diagram. The right hand side is supplied by our
knowledge of the motion.
The SI unit of force is the newton. 1 N = 1 kg·m/s2.
What is mass? Mass is a measure of inertia. Mass is not the same as weight. When
an object is dropped it is pulled down by its weight and its acceleration is g
downward. Applying Newton’s second law gives
W
Lesson 4: Relative motion, Forces, Newton’s laws (sections 3.6-4.4)
Lesson 4, page 12
gmW
maF yy
and the relationship between weight and mass is W = mg.
Newton’s Third Law: In an interaction between two objects, each object exerts a force on the
other. These forces are equal in magnitude and opposite in direction (page 103).
If two objects A and B are exerting forces on each other,
BAAB FF
The forces are equal in magnitude and opposite in direction.
Newton’s Laws of Motion (pages 97-103)
1. An object’s velocity vector v
remains constant if and only if the net force acting on the
object is zero.
2. When a nonzero net force acts on an object, the object’s velocity changes. The object’s
acceleration if proportional to the net force acting on it and inversely proportional to its
mass.
xx maF yy maF
3. In an interaction between two objects, each object exerts a force on the other. These
forces are equal in magnitude and opposite in direction.
BAAB FF
BAF
ABF
A B