Ch 9 Applications Newton - Physics2000

Chapter 9

Applications of

Newton¡¯s Second

Law

CHAPTER 9

APPLICATIONS OF

NEWTON¡¯S SECOND LAW

In the last chapter our focus was on the motion of

planets and satellites, the study of which historically

lead to the discovery of Newton¡¯s law of motion and

gravity. In this chapter we will discuss various applications of Newton¡¯s laws as applied to objects we

encounter here on earth in our daily lives. This chapter

contains many of the examples and exercises that are

more traditionally associated with an introductory

physics course.

9-2

Applications of Newton¡¯s Second Law

ADDITION OF FORCES

The main new concepts discussed in this chapter are

how to deal with a situation in which several forces are

acting at the same time on an object. We had a clue for

how to deal with this situation in our discussion of

projectile motion with air resistance, where in Figure

(1) reproduced here, we saw that the acceleration a of

the Styrofoam projectile was the vector sum of the

acceleration g produced by gravity and the acceleration a air produced by the air resistance

a = g + a air

(1)

If we multiply Equation 1 through by m, the mass of the

ball, we get

ma = mg + ma air

(2)

We know that mg is the gravitational force acting on

the ball, and it seems fairly clear that we should identify

ma air as the force Fair that the air is exerting on the ball.

Thus Equation 2 can be written

ma = Fg + Fair

3

In other words the vector ma , the ball¡¯s mass times its

acceleration, is equal to the vector sum of the forces

acting upon it. More formally we can write this

statement in the form

ma =

¦² Fi =

i

the vector sum of

the forces acting

on the object

more general

form of Newton's

second law

(4)

Equation 4 forms the basis of this chapter. The basic

rule is that, to predict the acceleration of an object, you

first identify all the forces acting on the object. You

then take the vector sum of these forces, and the result

is the object¡¯s mass m times its acceleration a .

When we begin to apply Equation 4 in the laboratory,

we will be somewhat limited in the number of different

forces that we can identify. In fact there is only one

force for which we have an explicit and accurate

formula, and that is the gravitational force mg that acts

on a mass m. Our first step will be to identify other

forces such as the force exerted by a stretched spring,

so that we can study situations in which more than one

force is acting.

d"

in

"w

a3

g

a air

Fs

v3

m

a 3 = g + a air

mg

Figure 1

Vector addition of accelerations.

Figure 2

Spring force balanced by the gravitational force.

9-3

SPRING FORCES

The simplest way to study spring forces is to suspend

a spring from one end and hang a mass on the other as

shown in Figure (2). If you wait until the mass m has

come to rest, the acceleration of the mass is zero and

you then know that the vector sum of the forces on m

is zero. In this simple case the only forces acting on m

are the downward gravitational force mg and the

upward spring force Fs . We thus have by Newton¡¯s

second law

¦² Fi =

Fs + mg = ma = 0

(5)

i

and we immediately get that the magnitude Fs of the

spring force is equal to the magnitude mg of the

gravitational force.

As we add more mass to the end of the spring, the spring

stretches. The fact that the more we stretch the spring,

the more mass it supports, means that the more we

stretch the spring the harder it pulls back, the greater Fs

becomes.

To measure the spring force, we started with a spring

suspended from a nail and hung 50 gm masses on the

end, as shown in Figure (3). With only one 50 gram

mass, the length S of the spring, from the nail to the

hook on the mass, was 45.4 cm. When we added

another 50 gm mass, the spring stretched to a length of

54.8 centimeters. We added up to five 50 gram masses

and plotted the results shown in Figure (4).

Looking at the plot in Figure (4) we see that the points

lie along a straight line. This means that the spring force

is linearly proportional to the distance the spring has

been stretched.

To find the formula for the spring force, we first draw

a line through the experimental points and note that the

line crosses the zero force axis at a length of 35.9 cm.

We will call this distance the unstretched length So .

Thus the distance the spring has been stretched is

S ¨C So , and the spring force should be linearly proportional to this distance. Writing the spring force

formula in the form

Fs = k S ¨C So

(6)

all we have left is determine the spring constant k.

Mass (in Grams)

300

250

(200 gm, 73.7 cm)

200

150

Fs = K(S ¨C S0)

S

100

50

Length of spring S

50 gm

20

40

60

80

S0 = 35.9cm

Figure 3

50 gm

Calibrating the spring force.

50 gm

Figure 4

Plot of the length of the spring as

a function of the force it exerts.

100

120 cm

9-4

Applications of Newton¡¯s Second Law

The easy way to find the value of k is to solve Equation

6 for k and plug in a numerical value that lies on the

straight line we drew through the experimental points.

Using the value Fs = 200 gm ¡Á 980 cm /sec

= 19.6 ¡Á 10 4 dynes when the spring is stretched to a

distance S = 73.7 cm gives

k =

Fs

S ¨C So

=

= 5.18 ¡Á 10 3

19.6 ¡Á 10 4 dynes

73.7 ¨C 35.9 cm

dynes

cm

Equation 6, the statement that the force exerted by a

spring is linearly proportional to the distance the spring

is stretched, is known as Hooke¡¯s law. Hooke was a

contemporary of Isaac Newton, and was one of the first

to suspect that gravitational forces decreased as 1/r2.

There was a dispute between Hooke and Newton as to

who understood this relationship first. It may be more

of a consolation award that the empirical spring force

¡°law¡± was named after Hooke, while Newton gets

credit for the basic gravitational force law.

Hooke¡¯s law, by the way, only applies to springs if you

do not stretch them too far. If you exceed the ¡°elastic

limit¡±, i.e., stretch them so far that they do not return to

the original length, you have effectively changed the

spring constant k.

The Spring Pendulum

The spring pendulum experiment is one that nicely

demonstrates that an object¡¯s acceleration is proportional to the vector sum of the forces acting on it . In this

experiment, shown in Figure (5), we attach one end of

a spring to a nail, hang a ball on the other end, pull the

ball back off to one side, and let go. The ball loops

around as seen in the strobe photograph of Figure (6).

The orbit of the ball is improved, i.e., made more open

and easier to analyze, if we insert a short section of

string between the end of the spring and the nail, as

indicated in Figure (5).

This experiment does not appear in conventional textbooks because it cannot be analyzed using calculus¡ª

there is no analytic solution for this motion. But the

analysis is quite simple using graphical methods, and a

computer can easily predict this motion. The graphical

analysis most clearly illustrates the point we want to

make with this experiment, namely that the ball¡¯s

acceleration is proportional to the vector sum of the

forces acting on the ball.

In this experiment, there are two forces simultaneously

acting on the ball. They are the downward force of

gravity Fg = mg , and the spring force Fs . The spring

force Fs always points back toward the nail from which

the spring is suspended, and the magnitude of the

nail

string

spring

ball

Figure 5

Figure 6

Experimental setup.

Strobe photograph of a spring pendulum.

9-5

spring force is given by Hooke¡¯s law Fs = k S ¨C So .

Since we can calibrate the spring before the experiment

to determine k and So , and since we can measure the

distance S from a strobe photograph of the motion, we

can determine the spring force at each position of the

ball in the photograph.

In Figure (7) we have transferred the information about

the positions of the ball from the strobe photograph to

graph paper and labeled the first 17 positions of the ball

from ¨C 1 to 15. Consider the forces acting on the ball

when it is located at the position labeled 0. The spring

force Fs points from the ball up to the nail which in this

photograph is located at a coordinate (50, 130). The

distance S from the hook on the ball to the nail, the

distance we have called the stretched length of the

Experimental Coordinates

10

0

-1) ( 91.1, 63.1)

0) ( 88.2, 42.8)

9

1) ( 80.2, 24.4)

2) ( 68.0, 12.0)

3) ( 52.9, 8.6) 90

4) ( 37.4, 14.7)

8

5) ( 24.0, 28.8)

6) ( 14.2, 47.5) 80

7) ( 9.0, 67.0)

8) ( 8.2 , 83.9)

9) ( 11.1, 95.0) 70

10) ( 16.7, 98.8)

7

11) ( 23.9, 94.1)

12) ( 32.2, 81.5)

60

13) ( 41.9, 62.1)

14) ( 52.1, 39.9)

15) ( 62.2, 19.4)

16) ( 70.3, 6.0) 50

17) ( 75.3, 2.8)

6

10 20

30

spring, is 93.0 cm. You can check this for yourself by

marking off the distance from the edge of the ball to the

nail on a piece of paper, and then measuring the

separation of the marks using the graph paper (as we

did back in Figure (1) of Chapter 3). We measure to the

edge of the ball and not the center, because that is where

the spring ends, and in calibrating the spring we measured the distance S to the end of the spring. (If we

measured to the center of the ball, that would introduce

Nail (50,130)

93.0 cm

40

50

60

70

80

90

11

90

12

80

70

13

¨C1

0

14

1

15

20

Figure 7

4

Spring pendulum 10

data transferred

to graph paper.

20

30

20

10

2

3

10

40

30

5

0

60

50

40

30

100

40

50

60

70

80

90

100

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