Download Examples to Illustrate Newton`s Third Law of Motion

Examples to Illustrate Newton’s Third Law of Motion
Somnath Datta
656, ”Snehalata”, 13th Main, 4th Stage, T K Layout, Mysore 570009, India
Email :[email protected]
April 6, 2010
Abstract
This article, a follow up of an earlier article on Newton’s Third law of Motion, discusses
some examples that illustrate this law. The examples are drawn from (a) motion along an
inclined plane, (b) rider on the vertical wall of a cylinder (Wall of Death), (c) man stuck
to the wall of a rotating drum, (d) traction of a locomotive, and (f) oscillation of a simply
supported beam following the impact of a boy jumping on it.
All these examples illuminate the meaning of action and reaction and show Newton’s
third law of motion in a better perspective.
1 Introduction
This is a follow up of of my earlier article “Newton’s Third Law of Motion”, to be referred
to as N3L, which was published in this journal.1 N3L was written to eradicate some popular
misconceptions that undermine one’s faith and foundation in classical mechanics. A number of
illustrative drawings, especially freebody diagrams (FBDs), contained in the article should have
given a beginner a better appreciation of the concepts surrounding Newton’s laws of motion,
especially the third law, and avoid the misconceptions. N3L however didn’t contain numerical
problems which could have achieved our objective better.
With this objective this follow up article has been written. The worked out problems in this
article will illuminate the meaning of action and reaction and show Newton’s third law of motion
in a better perspective. The best example of action-reaction comes when I push a hard surface and
the hard surface pushes me back equally. We have therefore selected five examples based on this
experience. The first four examples will also highlight the constructive role played by friction in
the motion of vehicles and other objects, especially how friction provides the necessary traction
force which is the driving force behind all locomotions.
2 Example 1 : Inclined Plane
Ex.1. As our first example we take two rectangular blocks connected to each other by a weightless string, and moving along an inclined plane. One of them, m, is moving up the plane, while
the other, M, is moving vertically down (Fig. 1). We have to analyze the motion and find the
acceleration.
Solution The string provides a very good example of Newton’s 3rd Law of motion. If we mentally divide a string AB into two parts A and B across a plane C-C (see Fig. b), and imagine what
is happening at the molecular level, then we see that the molecules on the left side A, adjacent
1
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
2
C
B
B
m
A
M
TAB = T = T
BA
m
Ν
Z
tion
f mo
no
ctio
dire
T
T
m
P
direction of motion
N
X
M
F
θ
θ
mg
(c)
(b)
A
(a)
F
TBA
TAB
θ
R
C
(d)
Mg
Z
(e)
Figure 1: Mass connected by a string moving up an inclined plane. (b) FBD of mass m, (c) FBD of mass
M.
to the cut section, are attracting the molecules on the right side B with a force TBA . Similarly,
the molecules in B are attracting those in A with an equal and opposite force TAB . In the true
meaning of the 3rd law these two forces are equal and opposite. Since they are equal in magnitude we do not make any distinction between them and write either of them as T . We call T the
tension in the string.
Let us examine some other action-reaction forces. Concentrate on the block m. This block is
pressing against plane with a force P and the plane is answering back with an equal and opposite
reaction force R (see Fig. c). Our interest is in the forces acting on this block. We resolve the
reaction force R into a component N which is perpendicular to the plane, and another component
F, which is tangential to the plane. We call N the normal reaction force, and F the friction force.
When the block is moving this tangential component equals µk times the normal reaction, i.e.,
F = µk N, where µk is called the coefficient of kinetic friction (see Chapter 5.)
What are the other forces acting on m? The ubiquitous gravity force mg of course, pulling
it down? What pulls it up then? None other that the tension force T.
We have shown an FBD of the block m in Fig. d. In the battle of the forces, the tension T
wins and the block moves up.
Fig. e is an FBD of the block M. In the battle involving only two forces, namely, the gravity
force M g and the tension T, the former wins and the block moves down.
For analyzing the motion we have taken the axes X − Z for m and Z ′ for M differently, as
shown in Figures (d) and (e) respectively.
We shall now write the equations of motion. We shall denote the masses of the two blocks
as m and M respectively.
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
3
Consider first the force components in the Z direction. The block m is not moving in the
Z direction. Hence the normal reaction N balances the component of the gravity force on m,
namely mg cos θ. That means, N = mg cos θ. The friction force that the block experiences in
the negative X direction is F = µk N = µk mg cos θ. Both the blocks are moving with the same
acceleration (in magnitude) a. The remaining equations of motion, written in the form “external
force= mass × acceleration” are now
Eqn. of motion for M (Z ′ direction):
M g − T = M a (a)
Eqn. of motion for m (X direction): T − mg sin θ − µk mg cos θ = ma (b)
(1)
We multiply Eq. (1a) with m, and (b) with M , subtract one from the other, and then divide each
side with (M + m). We get the following.
M mg
(1 + sin θ + µk cos θ) .
(a)
M +m
mg
a = g−
(1 + sin θ + µk cos θ)
M +m
g
=
[M − m(sin θ + µk cos θ)] . (b)
M +m
T
Hence, from (1a)
=
(2)
We shall give a numerical edge to the example, by assuming that θ = 300 , M = m = 10kg and
that the blocks are moving with constant velocity, so that a = 0. Let us now find the value of
µk . Setting these values in Eq. (2 b ) we get
1 = sin 300 + µk cos 300 = 0.5 + 0.87µk . ⇒ µk = 0.57.
3 Example 2 : Motorcyclist on Wall of Death
Ex.2. Fig. 2 gives a diagrammatic view of a motorcycle rider going around a “wall-of-death”,2 .3
The mass of the rider and bike, taken together, is 150 kg. The radius of the drum is 4.5 m.
(a) Show FBDs of the rider and the drum and write the equation of motion. (b) Find expressions for the angle θ that the rider makes with the horizontal, and the thrust force P that
he exerts on the wall when going at speed v. (c) Given that the coefficient of friction between
rubber and wood is µ = 0.6, find the minimum speed at which he must drive in order not to slip,
and the angle θ at this speed. (d) Find the maximum speed below which the rider must ride so
that the wall does not collapse and the angle θ at this speed. The maximum side thrust (i.e., the
horizontal component of P) that the wall can withstand is 20,000 N.
Solution.
(a) Fig. (b) is a “partial” FBD of the drum wall, which shows only the thrust force P with which
the rider-bike system (abbreviated as RB) presses against the wall of the drum (we have suppressed the other forces, viz., the gravity force and the forces from the ground, acting on the
drum.)
Fig. (c) is a complete FBD of the rider-bike system. The only two forces that are acting on RB
are the force of gravity M g acting downward through the centre of mass C of this system, and
the reaction force R. RB gets this reaction force by pressing against it with a push force P, and
hence, by Newton’s 3rd Law, R = −P.
Fig. (d) is a modified FBD of RB. We have resolved the reaction force R into a vertical component F and a normal component N. In this case F is the friction force between the tyre and the
wall, and it is this force which exactly balances the force of gravity with which the earth pulls
RB. On the other hand N provides the the centripetal acceleration of the rider.
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
4
P
Wall
Z
Drum
θ
r
(b)
R
θ
C
Path
of th
e rid
Mg
(c)
er
F
X
θ
C
N
(d)
(a)
Mg
Figure 2: Rider on “Wall of Death”. (a) General view. (b) Partial FBD of the drum. (b) FBD of the
rider-bike showing the only two forces acting on him. (d) Modified FBD of the man-bike after resolving
the reaction force into horizontal and vertical components.
The equations of motion are now written as follows. At the instant of time when we are
viewing the rider, the Z axis is taken upward and the X axis in the direction from the point of
contact (between the tyre and the wall) to the axis of the drum.
2
X component:
N = M vr . (a)
Z component: F − M g = 0.
(b)
(b) Since tan θ =
F
N,
(3)
it follows from Eqs. (3a and b) that
tan θ =
F
gr
= 2.
N
v
(4)
(c) It is seen from Eq. (4) that the angle θ increases as v decreases, until a value is reached when
the friction force F is unable to equal the gravity force M g. At this moment the bike starts
slipping and there is accident. This criticality is reached when F = µN . That means
tan θmax =
F
= µ = 0.6. ⇒ θmax = 310 .
N
(5)
It is seen from (4) that for this speed
gr
µ = 2 , ⇒ vmin =
v
r
gr
=
µ
r
9.8 × 4.5
= 8.75 m/s = 30.86 km/hr.
0.6
(d) At this other end of criticality N =12,000 N. Therefore from (3a).
(6)
Examples to illustrate Newton’s 3rd Law of Motion -
12, 000 =
2
150 vmax
4.5
⇒ vmax =
Now from (4) tan θmin =
gr
2
vmax
q
( April 6, 2010)
12000×4.5
=
150
9.8×4.5
= 18.97×18.97
5
18.97m/s = 68.3km/hr.
(7)
= 0.1225. ⇒ θmin = 70 .
(8)
4 Example 3 : Stuck Against the Wall of a Rotating Drum
Ex.3. I am reminded of a strange and somewhat scary experience I had in some amusement park
in the USA about 15 years ago. I was led into a cylindrical chamber of which the top was open
and, along with the other visitors, I was standing against the wall as the chamber started rotating.
Soon I was horrified to see that there was no floor under my feet. It had been removed after the
chamber started rotating. I had no other feeling except that I felt pressed against the wall without
any support from below. I didn’t feel at all that I was in rotation, except when I looked up and
saw that the people standing outside surrounding the cylindrical wall were revolving around us.
How fast were they revolving? Not too fast, nor too slow, as far as I remember. But I can find
out.
Solution We have explained the situation in Fig. 3(a), along with an FBD of the man in (b). This
is one more example of interaction between the wall of a cylindrical drum and a man. The man
presses the wall with a force P and the wall presses him back with an equal and opposite reaction
force R. The vertical component of R is the the friction force F which just counterbalances the
gravity force W, same as mg. The horizontal component N provides the centripetal acceleration
of the man. In the limiting case (i.e., the lowest rotational velocity below which the man won’t
stick to the wall) the friction force equals µs N . The equations of motion are same as in Eq. (3).
Let us take the diameter of the chamber (roughly what I recall) to be 12 feet, or about 4
m, so that r = 2 m and the coefficient of static friction to be µs = 0.6. (this coefficient
mv 2
mv 2
= N ; mg = µs N = µs
.
applies to friction between leather against steel.) Then
r
r
Z
ω
F
N
W
X
(a)
(b)
Figure 3: Man stuck against the wall of a rotating drum. (a) General view. (b) FBD of the man
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
6
q
gr
9.8×2
=
0.6 = 5.7 m/s. This corresponds to a rotational angular frequency
µs
ω
ω = vr = 5.7
2 = 2.86 rad/s. The frequency in r.p.s is then f = 2π = 0.46 revolutions per second.
The actual rotation speed must have been somewhat larger that this minimum value.
Note: This problem can also be found in some standard textbooks.4
Hence, v =
r
5 Example 4 : Locomotive Pulling Train
Ex.4. Refer to the example of “locomotive pulling train” in Ref. 1. We have redrawn Fig. 5
of that article by changing the number of coaches from 2 to 20. The new drawing is shown in
Fig. 4.
A locomotive engine weighing 90 tonnes is pulling 20 coaches on a level terrain. Each coach
weighs 70 tonnes. Coefficients of friction between the steel wheel and the steel rail are (i) µs =
coefficient of static friction = 0.58; (ii) µr = coefficient of rolling friction = 0.001. For simplicity
assume (i) the entire weight of the engine is resting on the driving wheels; (ii) friction in the axles
of the engine and the coaches can be ignored; (iii) drag forces due to wind etc can be ignored.
(iv) Rolling friction in the “other” wheels of the engine can be ignored.
Determine (a) the maximum tractive effort, i.e., the maximum pulling force Pmax that the
engine can exert on the train; (b) maximum acceleration amax of the train; (c) the tensions
T1 , T20 in the chains that connect the engine to the coach # 1, and coach #19 to coach #20. (d)
What will be velocity of the train if it is dragged for 1 min with the maximum traction force?
Solution (a) Pmax is same as the maximum force that comes from the rails on the engine wheels,
which is equal and opposite to the force with which the engine pushes the rails backward without
slipping, which equals µs N where N is the normal reaction that comes from the rails on the
driving wheels. Since there is no motion (hence no acceleration) in the vertical direction, N −
M g = 0, where M is the mass of the engine, so that N = M g. Hence, Pmax = µs N = µs M g.
Using the values given: Pmax = 0.58 × 90 × 103 × 9.8 = 51.16 × 104 N.
(b) Ignoring wind resistance, friction in the wheels, etc., the only drag force is the resistance
coming from rolling friction. This resistance from each coach equals Fr = µr mg where m is
#1
#2
#20
(a)
#1
#2
#20
T1
(b)
#20
T 20
Fr
Fr
Fr
(c)
Fr
Figure 4: Locomotive pulling 20 coaches (a) General view. (b) FBD of the 20 coaches. (c) FBD of the
last coach. (gravity suppressed)
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
7
the mass of each coach. With the engine hauling n coaches the equation of motion will be
(M + nm)a = P − nFr .
(9)
The acceleration is maximum when P is maximum. Using the values given and just calculated,
Fr = .001×70×103 ×9.8 = 686 N; Hence, (90+20×70)×103 amax = 51.16×104 −20×686 =
498 × 103 . Or, amax = 0.33 m/s2 .
(c) The force that drives all the 20 coaches is T1 − 20Fr (See Fig. 4b ). The force that drives
the last coach is T20 − Fr (See Fig. 4c ).. Equating these expressions to the mass of the driven
coaches times the acceleration amax , we get the maximum values of tension.
T20,max − 686 = 70 × 103 × 0.33; Hence, T20,max = 23.8 × 103 N.
Similarly, T1,max − 20 × 686 = 20 × 70 × 103 × 0.33; Hence, T1,max = 462 × 103 N.
(d) v = .33 × 60 = 20m/s = 72km/hr.
6 Example 5 : Jumping Upon a Wooden Plank
Ex.5. A man jumps onto a wooden plank which is simply supported on two pillars at its ends,
setting it vibrating (Fig.5a). Let the deflection of the centre point of the plank be δ(t) (measured
positive downward) at time t. (a) Draw a freebody diagram of the boy and the plank, showing all
the forces acting on each of them at t. (b) What is the net force acting on each of them at t. (c)
Write the equation of motion of the centre of mass C of the man, and hence, an expression for the
frequency of its oscillation. (d) The plank is 12 feet long and its cross sectional dimensions are
shown in Fig. (e). Mass of the man is 100 kg. The modulus of elasticity of the wood is E = 10
GPa. Neglecting mass of the plank find the frequency of vertical oscillation of the Centre of
Mass C of the man.
Solution. In section 5 (Newton’s Second Law of Motion) we had treated the system of our
example as a spring-mass system with a spring constant5
k=
48EI
,
ℓ3
(10)
where ℓ is the length of the beam, E is the modulus of elasticity of the material of the plank, I
is the area moment of inertia of the cross section of the beam. In Fig. (d) we have shown the
equivalent spring-mass system which is also subjected to the force of gravity.
(a) We have shown the freebody diagrams in Figs. (b) and (c).
(b) The plank is acting like a simply supported beam, and we shall refer to it as beam. It follows
from Eq. (10) that the force on this beam is F = −kδ k. Here we are neglecting the weight of
the beam (i.e., we are assuming that the force Fbeam to be large compared to the gravity force
mplank g.) The reaction force on the man is R = −F = kδ k. The gravity force on him is his
weight W = −M g k. Hence the total force on him is Fman = (kδ − M g)k.
(c) Since the forces are one dimensional. we shall avoid vector symbols. The equation of motion
is
M z̈ = Fman = kδ(t) − M g.
(11)
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
8
Z
C
z
h
1111
0000
0000
1111
X
A
B
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
δ
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
2
0000
1111
0000
1111
0000000000000000000000000000000000000000000000000000
1111111111111111111111111111111111111111111111111111
1111111111111111111111111111111111111111111111111111
0000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000
1111111111111111111111111111111111111111111111111111
0000000000000000000000000000000000000000000000000000
1111111111111111111111111111111111111111111111111111
0000000000000000000000000000000000000000000000000000
1111111111111111111111111111111111111111111111111111
(a)
Z
35 cm
k
10 cm
Y
(e) cross section of the plank
C
(d)
Mg
Equivalent spring−mass system
subjected to gravitational force
Mg
−F
(b)
FBD of the man
A
R
B
F
R
(c) F.B.D. of the plank
Figure 5: Man on plank. (a) General view. (b) FBD of the man. (c) FBD of the beam.
Let us assume that the height h of the CM of the man above the centre of the plank does not
change appreciably during a few oscillations of the beam, so that we can take h to be constant.
Since z = h − δ, z̈ = −δ̈. Therefore we can rewrite the equation of motion as
M δ̈ = M g − kδ
(12)
This is the equation of motion of a spring mass system which is also subjected to the force of
gravity, as shown in Fig. (d). We had treated this problem in Chapter 5. The force of gravity has
no effect of the frequency of oscillation, which is now given as
r
r
k
48EI
1
1
=
.
(13)
f=
2π M
2π M ℓ3
Examples to illustrate Newton’s 3rd Law of Motion -
( April 6, 2010)
(d) We set the numerical values as follows. E = 1010 N/m2 ; I =
10−4 m4 ; ℓ = 12 × .3048 = 3.66 m; M = 100 kg.
Inserting these values in Eq.(11) we get f = 8.4 c.p.s..
bd3
12
9
=
.35×.13
12
= 0.29 ×
References
[1] S. Datta, “Newton’s Third Law of Motion”, Physics Education, Vol 26, No.3 (2009)
[2] Wikipedia gives the following description. The Wall of Death or motordrome is a carnival
sideshow featuring drum- or barrel-shaped wooden cylinder, ranging from 20 to 36 feet in
diameter, in which stunt motorcyclists ride and carry out tricks.
... These motordromes
with perfectly straight walls were soon dubbed the “Wall of Death.” ... The audience
views from the top of the drum, looking down. The riders start start at the bottom of the
drum, in the centre, and ascend an initial ramped section until they gain enough velocity to
drive horizontally to the floor.
[3] A similar problem has been given in
A.P.French, Newtonian Mechanics, W.W,Norton and Co., New York (1971.) See Prob. 7-21
on p. 239.
[4] D. Halliday, R.Resnick and J.Walker, Fundamentals of Physics, 4th Edn, Asian Books Pvt
Ltd, New Delhi (1994) See “Sample Problem” 6-12 on p. 146.
[5] The formula can be found in any standard engineering book discussing deflection of beams
and vibration of mechanical systems. Here are some standard references.
W.T.Thomson, Vibration Theory and Applications, Prentice-Hall, N.J. (1965). See formula
given as answer to Prob. 31 in Ch. 1, p.32.
S. Timoshenko and D.H.Young, Vibration Problems in Engineering, 3rd Edn., Affiliated
East West Press Pvt Ltd, New Delhi (1955). Set c = ℓ/2 in Prob. 1 at the end of Ch. 1, p.
6.