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