Chapter 3 Newtonian Mechanics II

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 3 3-1

Chapter 3 Newtonian Mechanics II

! What are human cannon balls and how do their lives depend on physics knowledge?

! Would an adult or a small child win a race down a water slide?

! How does friction help us walk?

Make sure you know how to

1. Use trigonometric functions

2. Use motion diagrams and mathematical equations to describe motion. (Sections 1.3, 1.6, and

1.7)

3. Identify a system, construct a force diagram for it, and use the force diagram to apply

Newton’s second law. (Sections 2.1, 2.6 and 2.8)

In 1998, David “Cannonball” Smith set the record for the distance traveled by a human cannon

ball—reported as 56.64 m. He is the father of a cannonball family that includes his son David Jr.

and two daughters Stephanie and Jennifer.

Human Cannonball

It is a dangerous business. In 2006 Stephanie overshot her landing airbag at the Royal Adelaide in

Australia. She survived with minor injuries. In 1974 another cannonball woman, 21-year-old Mary

Connors, went for a record-breaking shot in England. Her cannon shot across the River Avon was

short, and she landed in the river. The rescue boat capsized, leaving her and two rescuers

floundering until a frogman pulled them to safety. How could these adventurers use physics to

improve their chances for a safe landing?

Lead In the last chapter, we learned that a system object’s acceleration equals the sum of the forces

that other objects exert on it divided by the mass of the object. We considered situations in which

the forces exerted on an object were mainly along one axis—the axis along which motion

occurred. An example is a person in a moving elevator. The forces exerted on the person by Earth

and by the elevator’s floor point parallel to the vertical axis that the elevator moves.


Most situations that occur in everyday life are more complex. In these cases, several

objects simultaneously exert forces on the system object of interest, and the forces are not all

directed only along the axis of motion. For example, when you push a lawn mower (Fig. 3.1), the

force that the handle exerts on the mower points partially toward the ground, even though the

mower moves parallel to the ground. The gravitational force exerted by Earth points down. The

force exerted by the grass can be represented by the normal force perpendicular to the surface and

the friction force parallel to the surface - opposite the mower’s direction of motion. To understand

how the mower’s acceleration is related to all these forces, we will need to learn how to apply

Newton’s second law for this and other more complex situations. In this chapter you will use sins

and cos’s often, so make sure you feel comfortable with these trigonometric functions before you

start.

Figure 3.1 Forces exerted on mower

3.1 Force Components

Vector quantities such as force, acceleration, and velocity have magnitudes and directions.

Vectors are very different mathematical entities than ordinary numbers (called scalars). Because of

this, we need new techniques to work mathematically with them. In particular, to apply Newton’s

second law in situations when the force vectors do not point along the coordinate axes, we need to

learn how to break forces into their components. In this and the next section, you’ll learn how to

do this. Let’s start by considering a different vector quantity—displacement.

Suppose you want to take a trip and can reach your final destination in two different ways

(see Fig. 3.2). Route 1 is a direct path that is represented by a displacement vector C

!. Its tail

represents your initial location and its head represents your final destination. Route 2 goes along

two roads represented by displacement vectors A

! and B

! that are perpendicular to each other. You

first travel along A

!, then along B

!, and end at the same final destination.


Figure 3.2

Graphical Vector Addition and Components of a Vector

The latter two-part trip just described is an example of how to graphically add

displacement vectors. You saw an example of vector addition in Chapters 1 and 2; we will repeat it

again here. You place the tail of A

! at your initial position; then, place the tail of B

! at head of A

!.

In this example, the head of B

! will be at your final position. Because the initial and final positions

are the same for either route, we say that C A B" #! ! !

. Any displacement vector (for example C!

)

can be replaced by two perpendicular displacements (for example A

! and B

!) that graphically add

to equal C!

(as illustrated in Fig. 3.2).

We can apply the same approach to forces. Often it is useful to replace a force F

! by two

perpendicular forces and x y

F F

! !that graphically add to equal the original force. Suppose for

example that we place a small box on a very smooth surface. The box is stationary. Then we attach

strings and spring scales on opposite sides of the box and pull exerting a 5-N horizontal force on

each string (Fig. 3.3a). We choose the coordinate system so that these string forces point at 037

angles relative to the positive and the negative x-axis. These two forces balance, and the box

remains stationary.

Figure 3.3(a) Graphically adding force components


Now, let’s replace the string and spring scale pulling at 037 relative to the positive x -axis

with two strings and two spring scales, one pulling along the x -axis and the other along the y -

axis (Fig. 3.3b). We find that if the x -axis scale pulls with a 4-N force and the y -axis scale pulls

with a 3-N force, the box again remains stationary. Thus, these two forces have the same effect on

the block as the single 5-N string force exerted on the box at a 037 above the positive x -direction.

Figure 3.3(b)

This experimental result is consistent with graphical vector addition, where we find that a

4-N force in the positive x -direction plus a 3-N force in the positive y -direction add to produce a

5-N force at 037 relative to the positive x -axis (Fig. 3.3c). The vectors form a “3-4-5 right

triangle.”

In summary, we can always replace any force F

! with two perpendicular forces

and x y

F F

! !, as long as the perpendicular forces graphically add to F

! (Fig. 3.3c). In this case, the

perpendicular forces are along the perpendicular x- and y-axes and are called the x- and y-vector

components of the original force F

!. Since the vector components are also vectors, they are

written with a vector symbol above them. Note that x

F

!+

yF

! = F

!, just as displacement C

! was

identical to the displacement A

! + B

! in Fig. 3.2.

Figure 3.3(c)


Scalar components of a vector

When working with well-defined x - and y -axes, we do not need to use the vector

components of F

!, but can instead provide the same information about the vector by specifying

what are called the scalar components x

F and y

F of the force F

!. For example, in Fig. 3.4, we see

that F

! is 5 N and points

037 above the negative x -axis. The x vector component of force F

! is

4 N long and points in the negative x direction. Thus, the scalar x component of F

! is

–4 Nx

F " . Similarly, the y vector component of F

! is 3 N in the positive y direction. Thus, we

can specify the y effect of F

! as 3 N

yF " # . Thus, the x and y components –4 N

xF " and

3 Ny

F " # tell us everything we need to know about the force F

!. Note that since the scalar

components of a vector are scalars, they are not written with a vector symbol above them.

Figure 3.4 Components of a vector

Finding the magnitude and direction of a force from its scalar components

The scalar components of a vector contain all the information needed to reconstruct the

original vector. To do this, first place the x -vector component of the vector so that its tail is at the

origin (Fig. 3.5). The x -scalar component indicates the direction and magnitude of the x -vector

component along the x -axis. Then place the y -vector component so it its tail is at the head of the

x -vector component. We also know the direction and magnitude of the y -vector component

relative to the y -axis from the sign and magnitude of the y -scalar component. The force vector

points from the tail of the x -vector component to the head of the y -vector component. Together,

the original vector and its two vector components form a right triangle, and the lengths of the sides

of the triangle are the magnitudes of the respective vectors. Since the magnitudes of the vector

components are equal to the magnitudes of the scalar components, we can use the Pythagorean

theorem to determine the magnitude F of the force (how strong the force is, in newtons):

2 2 = x y

F F F# . (3.1)


We can also use this triangle, along with some trigonometry to determine the angle $ !!!( 090 or less)

that the force makes relative to the positive or negative x -axis:

tan = y

x

F

F$ ,

or

$ = tan–1

y

x

F

F

. (3.2)

Thus, if we know the components and x y

F F , we can determine the magnitude and direction of

the force F

!.

Figure 3.5 Use scalar components Fx and F y to find magnitude and direction of a force

Finding the scalar components of a force from its magnitude and direction

What about the opposite operation? Suppose that we know the magnitude of the force F

and the angle $ ! that the force makes above or below the positive or negative x -axis. Then, how

do we find the x and y scalar components of the force? The following skill box summarizes how

we can calculate the scalar components of a force.


Tip! Note that we do not use vector arrows over the expressions for the scalar components. The

sign (+ or –) of a scalar component indicates the orientation of the corresponding vector

component relative to the axis. If the vector component points in the positive direction of the axis,

the scalar component is positive. If the vector component points in the negative direction, the

scalar component is negative.

Notice that the expressions for components contain sin and cos. Every time you use sin and

cos, make sure that you visualize the angle and check that the result makes sense. For example,

when you have a force pointed 450 above horizontal, its x component should be the same

magnitude as the y ; if you have a force that points 300 above horizontal, its x component should

be larger than the y component, and so forth.

Quantitative Exercise 3.1 Components of forces exerted on knot Three ropes pull on the knot

shown in Fig. 3.6a. A force diagram for the knot is shown in b (the diagram is on a grid that allows

you to see the components of the three forces that the ropes exert on the knot). Rope 2 exerts a

force 2 on KT

! of magnitude 500 N pointing

037 above the positive x -axis. By visual inspection

using the diagram and grid, determine the x and y -scalar components of the force 2 on KT

! that

rope 2 exerts on the knot. Then use the mathematical method described in the above skill box to

calculate the components of this force. The two methods should produce consistent results.


Figure 3.6(a) Knot pulled by three ropes

Represent Mathematically Using the force diagram shown in Fig. 3.6b, we see that the x -scalar

component of this force is +400 N (4 grid units in the positive x direction) and the y -scalar

component of this force is +300 N (3 grid units in the positive y direction). In symbols,

2 on K 400 Nx

T " # and 2 on K 300 Ny

T " # .

Figure 3.6(b)

Solve and Evaluate Now use the mathematical method outlined in the previous Skill Box to

calculate the components of the force that rope 2 exerts on the knot:

0

2 on K (500 N)cos 37 400 Nx

T " # " #

0

2 on K (500 N)sin 37 300 Ny

T " # " # .

The diagrammatical and the mathematical methods are consistent.

Try It Yourself: Use the same two methods to determine the components of the other two forces

whose magnitudes and directions are given in Fig. 3.6b.

Answer: 3 on K –400 Nx

T " and 3 on K 100 N

yT " # ; and 1 on K 0

xT " and

1 on K 400 Ny

T " % .

Note that in calculating the components, we had to consider three things:


! What is the sign of the scalar component—does the vector component point in the + or –

direction relative to the axis?

! What is the magnitude of the force (always a positive number)?

! What is the angle that the force makes relative to the positive or negative x -axis?

Tip! Be careful when applying Eqs. (3.3 x ) and (3.3 y ). The angles that appear in those equations

must be measured with respect to the positive or negative x axis; if a different angle is given in

the problem, you need to use geometry to find the desired angle or alternatively use the given

angle and write your own expressions for the components.

Review Question 3.1

When does a vector have a positive scalar component? When does a vector have a negative scalar

component?

3.2 Newton’s second law in component form

Now that we have learned how to work with vectors and their components, we can start to

use Newton’s second law to analyze more complex situations in which one or more of the forces

exerted on the system do not point along one of the coordinate axes. The forces that the three ropes

exert on the knot shown in Fig. 3.6 is such an example. We know that the knot is not accelerating.

Thus, the sum of the forces exerted on the knot is zero:

on K 1 on K 2 on K 3 on K 0F T T T" # # "&! ! ! !

.

(The subscript K refers to the knot). It is difficult to use this vector equation to analyze the

situation further—for example to determine one of the forces that is unknown if you know the

other two forces. However, we can do such tasks if the above situation is represented in

component form.

Notice in the previous example (see the force diagram in Fig. 3.6b) that the y -scalar

components of the forces that the three ropes exert on the knot add to zero. Ropes 2 and 3 exert on

the knot forces whose y -scalar components are 300 N + 100 N = 400 N and rope 1 exerts a

force whose y -scalar component is –400 N. Consequently, the sum of the y -scalar components

of the forces exerted on the knot is zero, and consequently the y -component of the acceleration of

the knot is zero. Mathematically:

1 on K 2 on K 3 on K

K

K

K

400 N 300 N 100 N= 0

y y y

y

T T Ta

m

m

# #"

% # #"


Similarly, the x -scalar components of the forces exerted on the knot also add to zero. The -

x -scalar component of the force exerted by rope 2 on the knot is +400 N. The x -scalar

component of the force exerted by rope 3 on the knot is -400 N. The x -scalar component of the

force exerted by rope 1 on the knot is zero; it does not pull the knot either left or right. Since the

x -scalar components of the forces exerted on the knot add to zero, the x -component of the

acceleration of the knot is zero. Mathematically:

1 on K 2 on K 3 on K K

K

0 400 N 400 N0

x x x

x

K

T T Ta

m

m

# #"

# # %" "

Thus we can infer that when the x - and y -components of the sum of the forces exerted on the

system is zero, it does not accelerate:

on K 0 if 0x x

a F" ' "

on K y0 if 0y

a F" ' "

Newton’s second law in component form

We learned in Chapter 2 that when the sum of the forces exerted on the system is not zero,

the system accelerates. Newton’s 2nd law gives us a way of predicting the system’s acceleration.

Suppose that objects 1, 2, 3 and so forth exert forces 1 on O 2 on O 3 on O, , F F F

! ! !… on an object of

interest—the system and that the forces point in arbitrary directions. If the system has a mass Sm ,

it will has acceleration Sa!

as a consequence of these forces (Newton’s second law):

1 on S 2 on S 3 on SS

S

+ + +...=

F F Fa

m

! ! !!

.

Because this equation involves vectors, we can’t work with it directly. However, if we

split it into its and x y -component forms, we can use Newton’s 2nd law in component form to

determine the system’s acceleration.

Newton’s second law rewritten in component form becomes:

1 on S 2 on S 3 on S S

S

...x x x

x

F F Fa

m

# # #" (3.4 x )

1 on S y 2 on S y 3 on S y

S

S

...y

F F Fa

m

# # #" (3.4 y )

In practice we usually choose the axes so that the object accelerates along only one of these axes

and has zero acceleration along the other axis. The process of analyzing a situation using Newton’s

second law in component form is illustrated in the next skill box.


Skill: Using Newton’s second law in component form

The sequence of steps below illustrates how we use Newton’s second law in component form

to analyze a situation (subscript notations for agent of force and system are omitted for

simplicity).

1. Draw a force diagram (the force arrows represent the forces that

other objects exert on the system).

2. Determine the sum of the forces in the x direction by adding the x-

scalar components of all the forces exerted on the system.

3. Write Newton’s second law in component form:

1 2 3 4 1 3cos 0 0

x

x x x xxF F F F F FF

am m m

$# # # # # % #'" " "

Notice that:

1, 1 cosx

F F $" #

0

2, 2 cos90 0x

F F" # "

0

3, 3 3cos 0x

F F F" % " %

0

4, 4 cos90 0x

F F" # "

4. Determine the sum of the forces in the y direction by adding the y-scalar

components of all the forces exerted on the system.

5. Write Newton’s second law in component form:

1 2 3 4 1 42sin 0

y

y y y yyF F F F F F FF

am m m

$# # # # # # %'" " "

Notice that:

1, 1 siny

F F $" #

0

2, 2 2sin 90y

F F F" # " #

0

3, 3 sin 0 0y

F F" % "


Using components to solve a problem

Now, let’s use this force component method to analyze a process. In the first example

below, we analyze the process diagrammatically to get a better feel for what is happening. Then, in

the example that follows, we analyze the same process again, this time mathematically by applying

the component form of Newton’s second law.

Conceptual Exercise 3.2 Book slides down a wall Alice exerts a 21.0-N force on a 2.0-kg book

as it slides down a slippery vertical wall. She pushes 450 above the horizontal. Determine

diagrammatically (without doing any mathematics) the magnitude of the force that the wall exerts

on the book in the direction perpendicular to the wall’s surface.

Sketch and Translate The process is sketched in Fig. 3.7a and includes the known information.

The book is chosen as the system object. Note that Earth exerts a

(2.0 kg)(9.8 N/kg) 19.6 Nmg " " downward force on the book.

Figure 3.7(a) Person lowers book down wall

Simplify and Diagram Model the book as a point like object and assume that the book–wall

interface is smooth enough so that the friction force the wall exerts on the book is small and can be

ignored. A partially completed force diagram for the book is shown in Fig. 3.7b. The known force

arrows have been carefully drawn to scale (the distance between grid lines is 5 newtons). The

force diagram includes three forces: the 21.0-N contact force that Alice exerts on the book A on BF

!,

the long-range non-contact downward 19.6-N gravitational force that Earth exerts on the book

E on BF

!, and the unknown contact normal force

W on BN

!that the wall exerts on the book (its

length is not yet known—no arrow on its end yet). Notice that the normal force that the wall exerts

on the book is perpendicular outward from the wall’s surface. This is a general feature of normal

forces.


Figure 3.7(b)

The book does not leave contact with the wall (it does not accelerate horizontally—only

vertically); therefore, the x -component of its acceleration is zero. Hence, the x -component of the

sum of the force exerted on the book must also be zero. The x -component of the force exerted by

Earth on the book is zero since it points vertically down. The woman exerts a 21.0-N pushing

force on the book. By inspection of the diagram, the x -scalar component is A on B 15.0 Nx

F " #

(three grid lines). Thus, the normal force exerted by the wall on the book has to balance the x -

component of the pushing force. This means its x -scalar component must be W on B –15.0 Nx

N "

(three grid lines to the left).

Try It Yourself: Suppose the woman pushed in the same direction as in this example, only now

exerting a 14.0-N force on the book instead of a 21.0-N force. Now what is the normal force that

the wall exerts on the book?

Answer: The normal force would have to balance the x-component of the force that the woman

exerted on the book. Thus, the normal force would have a magnitude of 10.0 N and would point

toward the left (two grid lines).

In the last conceptual exercise, we used the idea that the x -components of the forces

exerted on the book had to add to zero in order to determine the magnitude of the normal force that

the wall exerts on the book. It is very helpful to visualize the components when you do vector

addition when analyzing processes and evaluating your solutions. Let’s repeat this analysis, but

quantitatively this time by applying Newton’s second law in component form.

Quantitative Exercise 3.3 Book Sliding Down Wall Alice exerts a 21.0-N force on a 2.0-kg

book as it slides down a slippery smooth vertical wall (Fig. 3.7a). She pushes 045 above the


horizontal. Determine the magnitude of the normal force that the wall exerts on the book and the

acceleration of the book. Use the force diagram from the previous example.

Represent Mathematically Apply the x -scalar component form of Newton’s second law to the

force diagram shown in Fig. 3.7b to determine the magnitude of the normal force so that the book

stays on the surface. Since B 0x

a " , the sum of the x -components of the forces exerted on the

book by other objects must be zero:

A on B W on B E on B B

B B

10x x x

x x

F N Fa F

m m

# #" " "&

Now, insert expressions for the x -scalar components into the above equation:

0 0 0

W on B(21.0 N)cos 45 ( cos 0 ) (19.6 N)cos90 0N# # % # "

Since 0cos90 0" and

0cos0 1.0" , we find that:

0

W on B(21.0 N)cos 45 0 0N% # " . (3.5)

This equation can be solved to determine the magnitude of the normal force that the wall exerts on

the book (done in the Solve and Evaluate part of solution).

Apply the y -scalar component form of Newton’s second law to the force diagram in Fig.

3.7b in order to determine the y -scalar component of the acceleration of the box:

A on B W on B E on B

B

B B

+ + y y y y

y

F F N Fa

m m" "&

Substitute into this equation expressions for the y -scalar components of the forces:

0 0 0

W on BB

+(21.0 N)sin 45 + sin 0 +[ (19.6 N)sin 90 ]

2.0 kgy

Na

%" (3.6)

Solve and Evaluate Solve Eq. (3.5) for the magnitude of the normal force:

0

W on B (21.0 ) cos 45 15.0 NN N" " .

Then solve Eq. (3.6) for the acceleration of the box in the y-direction, noting that sin 00 = 0 and

sin 090 = 1.0:

0

W on B

+ (21.0 N)sin 45 + 0 ( –19.6 N 1)

2.0 kgy

Na

( # ("

2 15.0 N + 0 – 19.6 N–2.3 N/kg = –2.3 m/s .

2.0 kg" "

The units are correct, and the magnitude is reasonable. Consider some limiting cases. If the mass

of the book were zero, the gravitational force that Earth exerts on the book would also be zero, and

the book would accelerate upward at 215.0 N

m s0 kg

" ) . What if the woman were not pushing


on the book at all? In that case, the acceleration of the book would be 2–19.6 N

= –9.8 m/s2.0 kg

,

which is the acceleration of a freely falling object. That also makes sense.

Try It Yourself: What would the y -scalar component of the force that Alice needs to exert on the

book be so that the book moves downward at constant velocity?

Answer: +19.6 N to balance the downward –19.6 N gravitational force that Earth exerts on the

book. Remember that when an object’s acceleration is zero, its velocity is constant.

Let’s summarize the steps we used in the previous example to analyze the process. We:

! Sketched the situation and chose an object of interest—the system,

! Constructed a force diagram for the system,

! Used the force diagram to help apply the x – and y -scalar component forms of Newton’s

second law, and

! Used these component equations to determine the desired physical quantities.

This is an abbreviated form of a general strategy that will be formalized in the next section and

then used to analyze a wider range of processes.

Tip! Perhaps the part of this procedure that causes the most difficulty is the construction of the

force diagram. Be sure to include in the diagram only forces exerted on the system by external

objects (outside the system). Do not include forces that the system exerts on external objects

(objects that are not included in the system).

Review Question 3.2

Write Newton’s second law in component form for the force diagram and process sketched in Fig.

3.8 (both x and y axes).

Figure 3.8


3.3 Problem-solving strategies for analyzing dynamics processes

In analyzing future dynamics processes, we will use the same steps that we used in the last

example in Section 3.2, and in the examples in Chapter 2. Dynamics processes involve forces

exerted on an object and using the component form of Newton’s second law to determine the

effect of these forces on the velocity of the object. In the examples in this chapter, the forces that

other objects exert on an object of interest can be in any direction in the x y% plane. An example

is a skier sliding down a hill or a wagon pulled by a rope oriented at an angle above the horizontal.

The general procedure for analyzing such dynamics process is described on the left side of the next

example and illustrated on the right side.

Example 3.4 Pulling a Sled You pull with a rope a sled with two children (a total mass of 60.0

kg) on a hard snow. The rope exerts a 100-N force on the sled and is oriented 37o above the

horizontal. If the sled starts at rest, how fast will it be moving after being pulled for 10.0 m?

Skills for analyzing dynamics

processes

Example

Sketch and translate

! Make a sketch of the process.

! Choose a system.

! Choose coordinate axes with one

axis in the direction of

acceleration, and the other axis

perpendicular to that direction.

! Indicate in the sketch everything

you know about the process

relative to these axes. Identify

the unknown quantity of interest.

The process is sketched. Choose the sled

plus kids as the system.

The process starts with the sled at rest ( 0 0x

v " ) and ends when it has

traveled 1 0 10.0 mx x% " and has an unknown final speed 1 xv .

Simplify and diagram

! Simplify the process. For

example, decide if it is

appropriate to model the system

as a point-like object, or if

friction can be ignored.

! Represent the process

diagrammatically with a motion

diagram and/or a force diagram.

! Check for consistency of the

diagrams—for example, is the

net force in the direction of the

acceleration.

Consider the system as a point-like object. Since we have no

information about friction, we’ll assume its effects on the sled are

minor.

A motion diagram for the sled is shown below. The sled moves at

increasing speed toward the right.

Draw a force diagram for the system. Earth exerts a downward

gravitational force on the system E on SLF

!, the rope pulls on the sled

R on SLT

! at a 370 angle above the horizontal, and the snow exerts

normal force on the sled perpendicular to the surface (in this case

upward) S on SLN

!. Are the diagrams consistent? The sum of the forces

is towards right in the direction of the acceleration.

ALG

3.4.1-

3.4.4


Represent mathematically

! Convert these qualitative

representations into quantitative

mathematical descriptions of the

process using Newton’s second

law and kinematics equations.

The horizontal x-scalar component form of Newton’s second law is:

x xma F" &

SL R on SL S on SL E on SL x x x xm a T N F" # #

Substitute expressions for the x-scalar components of these three

forces: 0 0 0

SL R on SL S on SL E on SLcos 37 cos 90 cos 90

xm a T N F" # #

Noting that 0

cos 90 0" , and dividing both sides by the mass of the

sled with the children, we find that: 0

R on SL

SL

cos 37 0 0x

Ta

m

# #"

We could at this point use the y-scalar component equation to find the

magnitude of the normal force that the surface exerts on the sled, but

we don’t need to do that to answer the question of interest.

The above equation can be used to determine the sled’s acceleration.

We are then left with a kinematics problem to determine the speed

1 xv of the sled after pulling it for 1 0

10.0 mx x% " . We can use

kinematics Eq. (2.6): 2 2

1 0 1 02 ( )

x x xv v a x x" # %

Solve and evaluate

! Substitute the given values into

the mathematical expressions

and solve for the unknowns.

! Decide whether the assumptions

that you made were reasonable.

! Finally, evaluate your work to

see if it is reasonable (check

units, limiting cases, and whether

the answer has a reasonable

magnitude).

! Make sure the answer is

consistent with other

representations.

The acceleration is 0 0

2R on W

SL

cos 37 (100 N)(cos 37 )1.33 m/s

60.0 kgx

Ta

m

# #" " " # .

The speed of the sled after being pulled for 10 m will be: 2 2 2 2

1 0 1 02 ( ) 0 2(1.33 m/s )(10.0 m – 0)

x x xv v a x x" # % " #

or

1 5.1 m/s

xv " .

This is fast but not unreasonable. In real life it is probably smaller as

there is definitely some friction between the sled and the snow. The

units are correct. The speeding up of the sled is consistent with the

motion and force diagrams.

Try It Yourself: (a) Determine the magnitude of the gravitational force that Earth exerts on the sled

plus children. (b) Then determine the y-scalar component of the forces that the rope and Earth

exert on the sled plus children. (c) Finally, use the y-scalar component form of Newton’s second

law to determine the magnitude of the normal force that the snow exerts on the sled (which is in

the positive y-direction).


Answers: (a) E on SL SL = = 588 NF m g ; (b) 0

R on SL = + sin 37 60 Ny

T T " # and

0

E on SL W = – sin 90 = –588 Ny

F m g ; (c) S on SL 528 NN " . Look at the force diagram to see if

these numbers make sense. Think about why the magnitude of the normal force is less than SLm g .

The next example involves a more difficult to analyze process. Following the strategy

we just developed should help. Construct a sketch of the situation, choose an object of interest

for a force diagram, construct the force diagram, use it to apply the component forms of

Newton’s second law, and …. You will start seeing a way to solve the problem.

Example 3.5 Acceleration of a BART train You carry a pendulum and a protractor with you

onto a Bay Area Rapid Transit (BART) train ride. As the train starts moving, the pendulum

string swings back to an angle of 08.0 with respect to vertical. Determine the acceleration of

the train while this is happening.

Sketch and Translate A sketch of the protractor and pendulum is shown in Fig. 3.9a. Choose

the pendulum bob attached to the bottom of the string as the system of interest. The pendulum

bob is accelerating, thus the objects interacting with it should exert forces whose sum points in

the direction of acceleration. There are two objects interacting with the bob – the string that

touches it and Earth. As Earth exerts a vertical force, it must be the string that exerts the force

that has a horizontal component. That is why the string tilts from the vertical orientation.

Figure 3.9(a)

Simplify and Diagram Assume that the pendulum bob is a point-like object. Next, construct a

force diagram for the pendulum bob (Fig. 3.9b). The string exerts a force S on BT

! that is oriented

at an 08.0 angle relative to the vertical or

082.0 relative to the horizontal. Earth exerts a

downward gravitational force on the bob E on BF

!. Choose our object of reference to be the train

station. The pendulum bob is accelerating horizontally, so we choose the x -axis to point in the

horizontal direction and the y -axis to point perpendicular in the vertical direction.


Figure 3.9(b)

Represent Mathematically Use the force diagram to apply Newton’s second law in the

- and -x y scalar component forms:

x -component form of Newton’s 2nd

law: B B S on B E on B x x xm a T F" #

0 0

B B S on B E on Bcos82 cos90x

m a T F" # #

y -component form of Newton’s 2nd

law: B B S on B E on B +y y y

m a T F"

0 0

B B y S on B E on Bsin 82 (– sin 90 )m a T F" # #

Solve and Evaluate Note that the bob’s velocity is not changing in the vertical direction; thus

B, 0y

a " . The bob’s velocity is changing in the horizontal direction (the bob accelerates with

the train), so the x -component of acceleration is not zero. Also, recall that the gravitational

force that Earth exerts on the bob has magnitude E on B BF m g" . Finally, note that

0cos90 0" and 0sin 90 1.0" . With these substitutions and a bit of algebra, the above

- and -x y scalar component equations become:

0

B B, S on B

0

S on B B

= cos 82

0 sin 82 – .

xm a T

T m g"

We want to determine B xa but do not know S on BT and Bm . In this particular case a

bit of mathematical creativity will help us. Move the Bm g in the second equation to the left

side, and divide the left side of the first equation by the left second of the second equation; and

divide the right side of the first equation by the right side of the second equation. This is a

legitimate operation as we are dividing each side of the first equation by terms that are equal

and non-zero. We get:

0

B B S on B

0

B S on B

cos 82.

sin 82

xm a T

m g T"

Now cancel the TS on B from the numerator and denominator on the right side, and the

pendulum bob mass mB from the numerator and denominator on the left side. We now have a

single equation with the desired single unknown quantity—the acceleration of the bob, which

is the acceleration of the train:


00 2 2

B, 0

cos82= cot 82 (9.8 m/s ) 1.4 m/s

sin82x

a g " " .

The units are correct, and the magnitude is reasonable. As a limiting case analysis, we

find from the above equation that if the acceleration of the train is zero, then the string hangs

straight down at a 900 angle relative to the horizontal (

0

0

cos900

sin 90" as

0cos90 0" and

0sin 90 1" ). This is what we expect if the train is parked at the train station. What if the train

is moving with constant velocity? See the next “Try It Yourself” question.

Try It Yourself: Suppose the train is moving forward at a constant velocity of magnitude 18

m/s. At what angle will the pendulum bob string hang?

Answer: The train has constant velocity and hence has zero acceleration. Thus, the bob has

zero acceleration and B, 0.x

a " If the left side of the final equation in the solution is zero, then

the angle of the string on the right side must be 090 as

0cot 90 0" . The bob should hang

straight down. Notice that the behavior of the pendulum on a train moving at constant velocity

is the same as on a stationary train.

Processes involving inclines

Processes involving inclines are common in everyday life. For example, a skier moves

along an inclined ski slope or a wagon is pulled uphill. How do we use the component form of

Newton’s second law and kinematics to help analyze such processes?

Example 3.6 Who wins a water slide race? A child of mass m and an adult of mass 4m start

simultaneously at the top of a water slide that has a downward slope of 030 relative to the

horizontal. Which person reaches the bottom first?

Sketch and Translate Sketch of the process and choose either the child or the adult as the system

object. The sketch shows the person at one moment during the downhill slide and includes the

known information (Fig. 3.10a). The person who has a greater acceleration along the slide will win

the race, so we will construct an expression for the person’s acceleration and see how that

acceleration depends on the person’s mass.

Figure 3.19(a) Person going down waterslide


Simplify and Diagram The person is modeled as a point-like object. Since a water slide is very

smooth and slippery, we assume that the friction force exerted by the slide on the person can be

ignored.

A motion diagram for the person while going down the slide is shown in Fig. 3.10b. The

velocity arrows are parallel to the slide, and the person’s velocity is increasing so the velocity

change arrow v*!

(and the acceleration a!

) points down the slide.

Figure 3.10(b)

A force diagram for the person (Fig. 3.10c) shows that Earth exerts a downward

gravitational force E on PF

! on the person and the slide exerts a normal force S on PN

! on the person

perpendicular to the surface. Choose the x -axis) parallel to the inclined surface and the y -axis

perpendicular to the inclined surface. With this choice, the acceleration points in the positive x -

direction.

Figure 3.10(c)

Represent Mathematically The motion of the person is along the x -axis. Use the force diagram to

apply the x -scalar component form of Newton’s second law:

P E on P S on P x x xm a F N" #

The slide is inclined at 300 relative to the horizontal. This means that the gravitational

force makes a 600 angle relative to the slide (and therefore relative to the x -axis as in Fig. 3.10c).

We can calculate the x -scalar component of the gravitational force in two ways:

0

E on P P cos 60x

F m g" or 0

E on P P sin 30x

F m g" . The normal force exerted by the slide on the

person is perpendicular to the slide (and therefore to the x-axis). The above component equation

becomes:

0 0

P P S on P ( ) cos 60 cos 90x

m a m g N" # #


Solve and Evaluate Substituting 0cos90 0" , we get:

0

P P ( ) cos 60 0x

m a m g" # # .

Divide by the Pm on each side of the equation. The

result is 0cos 60

xa g" . The person’s acceleration is

independent of her/his mass! Thus, the acceleration of

the child and the adult is the same. If they start at same

time at same position and both initially at rest and have

the same acceleration, they arrive at bottom at same

time. The race is a tie (see the photo).

Try It Yourself: Suppose the slide above is 20 m long. How fast will the person be moving when

they reach the bottom of the slide?

Answer: 14 m/s..

Somehow, the result of the Example 3.6 does not seem reasonable. The component

gravitational force that Earth exerts on the adult parallel to the slide is four times that exerted by

Earth on the child—why doesn’t the adult win? Remember that the acceleration of an object

depends on the forces exerted on it and also on the mass of that object. It’s true that Earth exerts a

four times greater force on the adult, but since the adult’s mass is four times greater than the

child’s, the accelerations of the adult and of the child are equal. This result is similar to what we

learned in Chapters 1 and 2 where objects of all different masses have the same free-fall

acceleration. Earth exerts a force on an object proportional to its mass which when divided by its

mass to find the acceleration leads to the same free-fall acceleration g independent of the mass of

the object.

Tip! Notice in the last example that the inclined surface and the x -axis were at a 030 angle

relative to the horizontal. The gravitational force exerted by Earth on the system then makes a 060

angle relative to the x-axis (060 is the complement of

030 ). In general, the angle $ of the inclined

surface relative to the horizontal is the complement 090 –$ of the angle that the gravitational

force makes with respect to the inclined surface (see Fig. 3.11).


Figure3.11

Two objects linked together

Another common situation involves two or more objects connected together by a cable or

rope, such as a van pulling a rope connected to a wagon that is pulling a rope connected to a

second wagon. Perhaps the first “scientific” quantitative application of Newton’s second law to a

situation with two objects connected together by a string or rope involved two blocks of different

mass at the ends of a string that passed over a pulley (Fig. 3.12). George

Atwood invented this experiment and used it to determine the value of g .

In 1784 he published a description of the apparatus, which became

affectionately known as the Atwood machine. Atwood was one of the early

scientists who described experimental results mathematically. He was

famous for his lectures, particularly his lecture demonstrations. William

Pitt, an English prime minister who attended Atwood’s lectures, enlisted

him to help apply mathematical models to economic issues in England.

How did Atwood use it to determine free-fall acceleration g ? Figure 3.12 Atwood Machine

Remember, this was 1784. There were no motion sensors, no precision stopwatches;

nothing that would allow the accurate measurement of the motion of a rapidly accelerating object

(29.8 m s would have been considered a rapid acceleration in those days). Atwood’s machine

was significant in that it allowed the determination of g despite these difficulties. Observations of

Atwood’s machine show that when the blocks are of different mass, and the system is released

from rest, the heavier block always accelerates downward, and the lighter block always accelerates

upward. For the blocks in Fig. 3.12, no matter how different their masses are, both always have

accelerations of the same magnitude (since they are attached to one another by a string), and the

magnitude of this acceleration was always less than g . If the blocks are nearly the same mass, the

magnitude of their acceleration is small and measurements of the time interval it took them to

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Chapter 3 Newtonian Mechanics II