UNIT-07

Newton’s Three Laws of Motion 1. Learning Objectives:

1. Understand the three laws of motion, their proper areas of applicability and especially the

difference between the statements of the first and the third laws

2. Know the difference between inertial and non-inertial frames of reference.

3. Be able to draw free-body diagrams for a single or a system of several objects.

4. Be able to use F = Ma (with item #3 above) to solve problems for a system of one or

several objects.

5. Be familiar with concepts of static and kinetic friction.

6. Understand the concepts of tangential velocity and centripetal acceleration and be able to

solve problems involving circular motion

2. Uniform Motion

One of the central concepts in Newtonian dynamics is that of uniform motion. In Figure 1a-e are shown the positions of a particle at regular intervals of time. The motion of the particle can be described as follows:

[a] The velocity V = constant = 0 , the particle is at rest.

[b] The velocity V = Vi, where V is constant .

[c] The speed is increasing, the direction remains unchanged. Thus, velocity is not constant and the particle has an acceleration.

[d] The speed is constant, the direction of motion is changing. Thus, velocity is not constant and the particle has an acceleration.

[e] The speed and the direction of motion are changing. Thus, velocity is not constant and the particle has an acceleration.

Points to note:

[i] Motion with constant velocity implies motion in a straight line with constant speed. When a particle moves with constant velocity it is said to be in a state of uniform motion. Figure1a,b are examples of uniform motion.

X[b]

X[a]

[e]

X[c]

[d]

FIGURE 1

[ii] A particle not moving in a straight line is necessarily accelerating (Fig.1d,e).

3. Newton’s Laws

The first two laws are about change in the state of motion of a mass. The First Law: A mass will continue in a state of uniform motion unless an external agent exerts an influence on the mass. The Second Law: F = ma = m dV/dt …[1] Comments: [a] The “external agent” mentioned in the first law is now identified as the force F. [b] If F = 0, then dV = 0 ,which implies uniform motion. Thus the statement of Newton’s first law is contained in the second law. [c] F in eq.[1] is the net force on m, F = F1+ F2+ F3…… , where F1, F2, F3, etc. are the individual forces acting on m. To emphasize this point, sometimes the second law (eq.[1]) is written as: !F =!F1! +"F2 +

"F3 + .. = m

"a …[2]

[d] Remember  F  =  ma  is  a  vector  equation  which  means  it  expresses  three  

relations:  

ΣFx  =  max          ;        ΣFy  =  may                    ;      ΣFz  =  maz  

These  equations  imply  that  the  acceleration  of  an  object  in  a  given  direction  is  

governed  only  by  the  net  force  in  that  direction.  

When  ΣFx  =  ΣFy  =  ΣFz  =  0,  the  system  is  said  to  be  in  equilibrium  and  the  

equation,    

ΣFx  =  ΣFy  =  ΣFz  =  0  is  known  as  the  first  condition  of  equilibrium.  

[e] Newton had actually stated the second law as: F = dp/dt = d(mV)/dt = m dV/dt + Vdm/dt …[2] Here p = mV is called the linear momentum. We will discuss the linear momentum in a few weeks. If the mass of the system doesn’t change during motion, dm/dt = 0 and eq.[2] reduces to eq.[1].

 

 

4. The Free-body Force Diagram.

To apply F = ma to an object you must first find out the net force acting on it. This is

done by drawing a free-body force diagram or simply free-body diagram (FBD) for

the object. Review all the FBD examples discussed in the lectures.

The Third Law

The third law deals with interaction between two objects (call them 1 and 2) and

states that the force exerted by 1 on 2 is equal in magnitude and opposite in direction

to the force exerted by 2 on (BE CAREFUL :Every time you come across a pair of

forces that are equal and opposite, don’t confuse it with the action and reaction forces

of the third Law.)

Try the following.

1. You are pulling a crate by applying a force on it. The crate exerts an equal and

opposite force on you. If these forces add up to zero according to the third Law

then what makes the crate move?

(Ans. The forces act on two different objects – you and the crate – and cannot be

added to yield zero force. The motion of the crate is determined by the force, F,

acting on the crate. Since there is a net force exerted on it (by you), it moves

according to F = ma.

2. A book of mass m is resting on the top of a table which in turn is resting on the

floor (Fig.a). The FBD for the book is shown in Fig.(b) below. The force exerted by

the Earth

on the book, Fbe = mg is equal and opposite to the normal reaction force N exerted by

the table on the book. Is the fact that Fbe – N = 0 and therefore the book is in

equilibrium an instance of Newton’s third law or Newton’s first Law?

Answer: Both forces are acting on the same object. Therefore this is an instance of

Newton’s first Law (and the second Law), but not of the third Law. The reaction to

Fbe is the force exerted by the book on the Earth Feb (see Fig.(c), it is NOT an FBD).

The fact that Feb = - Fbe is an instance of the third Law.

 

 

book

table

floor (earth)

Fbe

book

EARTH

Feb

FbeN

book

(a) (b)

(C)

 

   

 

4.1  Using  the  FBD  –  some  examples.    

(We  may  not  discuss  this  topic  due  to  time  constraints)  

Problem  1:A  block  of  mass  m  =  50.0kg  is  released  from  rest  on  a  frictionless  

inclined  plane  of  angle  370.    What  is  the  magnitude  of  the  acceleration  of  the  block?  

370

Α

             

 

 

Step1.  Draw  the  FBD  for  each  part  of  the  system.  Make  sure  you  label  each  force  (or  

force  component)  and  any  angles  involved.  Remember  you  must  draw  all  the  forces  

acting  ON  the  object  of  your  interest.  Do  not  include  in  the  FBD  any  forces  that  the  

object  exerts  on  other  parts  of  the  system  or  some  external  system.  Improper  or  

incomplete  labeling  can  lead  to  an  incorrect  application  of  the  second  law.  

Step  2.  Choose  appropriate  set  of  axes  and  decompose  all  forces  along  the  two  axes.  

In  the  example  above  we  have  chosen  the  x-­‐axis  along  the  incline  pointing  upward  

and  the  y-­‐axis  normal  to  the  incline.  

Step-­‐3.  Apply  the  first  condition  of  equilibrium:   ΣFx  =  max    ;          ΣFy  =  may    to  each  

FBD.  

In  the  present  case  we  get  the  following  equations  for  mass  m:  

N  –  400  =  0          -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[1]  

500sin37o=  300    =  50a    -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[2]  

Or  a  =  6.0m/s2  

37ο

500N500cos37o

500sin37ο

A

N

=400N=300N

Problem  2  :  A  block  of  mass  m  =  50.0kg  on  a  frictionless  inclined  plane  of  angle  370  

is  connected  by  a  cord  over  a  massless  and  frictionless  pulley  to  a  second  mass  M=  

100.0  kg.    

[a]  What  is  the  magnitude  of  the  acceleration  of  each  block  ?  

[b]  What  is  the  tension  T  in  the  cord?  

(Note:  In  this  problem  we  have  simply  attached  another  block  B  to  the  existing  block  

A..  Notice  how  the  FBD  of  block  A  and  the  resulting  F  =  ma  equation  changes.)    

 

A B

37o                      

37ο

500N500cos37o

500sin37ο

A

NT

1000N

T

B

=400N=300N

 

 

 

Step1.  Draw  the  FBD  for  each  part  of  the  system.  Make  sure  you  label  each  force  (or  

force  component)  and  any  angles  involved.  Remember  you  must  draw  all  the  forces  

acting  ON  the  object  of  your  interest.  Do  not  include  in  the  FBD  any  forces  that  the  

object  exerts  on  other  parts  of  the  system  or  some  external  system.  Improper  or  

incomplete  labeling  can  lead  to  an  incorrect  application  of  the  second  law.  

Step  2.  Choose  appropriate  set  of  axes  and  decompose  all  forces  along  the  two  axes.  

In  the  example  above  we  have  chosen  the  x-­‐axis  along  the  incline  pointing  upward  

and  the  y-­‐axis  normal  to  the  incline.  

Step-­‐3.  Apply  the  first  condition  of  equilibrium:   ΣFx  =  max    ;          ΣFy  =  may    to  each  

FBD.  

In  the  present  case  we  get  the  following  equations  for  mass  m:  

N  –  400  =  0          -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[1]  

T  –  300    =  50a    -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐[2]  

(Note:  we  are  assuming  the  mass  would  move  up  the  incline.  If  our  assumption  is  

right  the  value  of  a  would  turn  out  to  be  positive,  if  we  made  the  wrong  assumption  

the  value  of  a  would  be  negative.  The  negative  value  of  a  does  not  render  our  

solution  wrong.  The  physical  interpretation  would  then  tell  us  that  the  mass  would  

be  moving  downward.)          

From  the  FBD  of  M,  we  get  

1000  –T  =  100a      -­‐-­‐-­‐-­‐-­‐-­‐-­‐[3]    

Step-­‐4.  Solve  the  equations  obtained  in  Step-­‐3.  Thus  we  get,  

N  =  400  N  

Adding  [3]  and  [4]  gives,  700  =  150a    or  a  =  4.7m/s2.  

From  [2]  T  =  300  +50a  =  533.3  N.  

Step-­‐5.  SMILE  

 

5.  Concept  Questions  

 

1.  Which  of  the  following  statements  is  most  correct?  (a)  It  is  possible  for  an  object  to  have  motion  in  the  absence  of  forces  on  the  object.  (b)  It  is  possible  to  have  forces  on  an  object  in  the  absence  of  motion  of  the  object.  (c)  Neither  (a)  nor  (b)  is  correct.  (d)  Both  (a)  and  (b)  are  correct.  

 

Answer:  (d).  Choice  (a)  is  true.  Newton’s  first  law  tells  us  that  motion  requires  no  force:  An  object  in  motion  continues  to  move  at  constant  velocity  in  the  absence  of  

external  forces.  Choice  (b)  is  also  true:  A  stationary  object  can  have  several  forces  acting  on  it,  but  if  the  vector  sum  of  all  these  external  forces  is  zero,  there  is  no  net  force  and  the  object  remains  stationary.  

 

2.  You  push  an  object,  initially  at  rest,  across  a  frictionless  floor  with  a  constant  force  for  a  time  interval  Δt,  resulting  in  a  final  speed  of  v  for  the  object.  You  repeat  the  experiment,  but  with  a  force  that  is  twice  as  large.  What  time  interval  is  now  required  to  reach  the  same  final  speed  v?  (a)  4  Δt  (b)  2  Δt  (c)  Δt  (d)  Δt/2  (e)  Δt/4  

 

Answer:  (d).  With  twice  the  force,  the  object  will  experience  twice  the  acceleration.  Because  the  force  is  constant,  the  acceleration  is  constant,  and  the  speed  of  the  object,  starting  from  rest,  is  given  by  v  =  at.  With  twice  the  acceleration,  the  object  will  arrive  at  speed  v  at  half  the  time.  

 

 

3.  If  a  fly  collides  with  the  windshield  of  a  fast-­‐moving  bus,  which  experiences  an  impact  force  with  a  larger  magnitude?  (a)  The  fly  does.  (b)  The  bus  does.  (c)  The  same  force  is  experienced  by  both.  Which  experiences  the  greater  acceleration?  (d)  The  fly  does.  (e)  The  bus  does.  (f)  The  same  acceleration  is  experienced  by  both.  

 

Answer:  (c),  (d).  In  accordance  with  Newton’s  third  law,  the  fly  and  the  bus  experience  forces  that  are  equal  in  magnitude  but  opposite  in  direction.  Because  the  fly  has  such  a  small  mass,  Newton’s  second  law  tells  us  that  it  undergoes  a  very  large  acceleration.  The  huge  mass  of  the  bus  means  that  it  more  effectively  resists  any  change  in  its  motion  and  exhibits  a  small  acceleration.  

 

4.  Which  of  the  following  is  the  reaction  force  to  the  gravitational  force  acting  on  your  body  as  you  sit  in  your  desk  chair?  (a)  the  normal  force  from  the  chair  (b)  the  force  you  apply  downward  on  the  seat  of  the  chair  (c)  neither  of  these  forces  

 

Answer:  (c).  The  reaction  force  to  your  weight  is  an  upward  gravitational  force  on  the  Earth  caused  by  you.  

 

PROBLEMS  

 

1.   A  force  F  applied  to  an  object  of  mass  m1  produces  an  acceleration  of  3.00  

m/s2.  The  same  force  applied  to  a  second  object  of  mass  m2  produces  an  acceleration  of  1.00  m/s2.  (a)  What  is  the  value  of  the  ratio  m1/m2?  (b)  If  m1  and  m2  are  combined,  find  their  acceleration  under  the  action  of  the  force  F

.    

 

Solution:  

  For  the  same  force  F,  acting  on  different  masses  F m a= 1 1    

  and  F m a= 2 2    

  so  m a m a1 1 2 2= .  

 

(a) mm

aa

1

2

2

1

13

= =  

m m2 13=

(b) F = m1 + m2( )a = 4m1a = m1 3.00 m s2( )  

a= 0750.   m s2  

2.   (a)  A  car  with  a  mass  of  1000.0  kg  is  moving  to  the  right  with  a  constant  speed  of  1.8  m/s.  What  is  the  total  force  on  the  car?  (b)  What  is  the  total  force  on  the  car  if  it  is  moving  to  the  left?    

Solution:  

  Since  the  car  is  moving  with  constant  speed  and  in  a  straight  line,  the  resultant  force  on  it  must  be   zero  regardless  of  whether  it  is  moving    

(a) toward the right or (b) the left.

 

3.   A  3.00-­‐‑kg  object  undergoes  an  acceleration  given  by   !a = (2.00 i + 5.00 j)  

m/s2.  Find  the  resultant  force  acting  on  it  and  the  magnitude  of  the  resultant  force.    

Solution:  

 

 

m = 3.00 kg!a = 2.00 i + 5.00 j( ) m s2

!F = m!a = 6.00 i +15.0 j( ) N!!F! = 6.00( )2 + 15.0( )2 N = 16.2 N

 

4.   Two  forces   1F  and   2F

 act  on  a  5.00-­‐‑kg  object.  If  F1  =  20.0  N  and  F2  =  15.0  N,  

find  the  accelerations  in  (a)  and  (b)  of  Figure  4.      

Solution:  

 

(a) !F! =!F1 +!F2 = 20.0 i +15.0 j( ) N

!F! = m!a : 20.0 i +15.0 j = 5.00!a

!a = 4.00 i + 3.00 j( ) m s2

or

a= = °500 369. .   m s  at  2 θ

(b)

F2x = 15.0cos60.0° = 7.50 NF2y = 15.0sin60.0° = 13.0 N!F2 = 7.50 i +13.0 j( ) N!F! =!F1 +!F2 = 27.5 i +13.0 j( ) N = m!a = 5.00!a

!a = 5.50 i + 2.60 j( ) m s2 = 6.08 m s2 at 25.3°

F2

F1

F1

F2

 

 

FIG.  4a(top)  and  b  

 

 

5.   Three  forces,  given  by   !F1 = (!2.00 i + 2.00 j)  N,  

!F2 = (5.00 i ! 3.00 j)  N,  and  

!F3 = (!45.0 i)  N  act  on  an  object  to  give  it  an  acceleration  of  magnitude  3.75  m/s2.  (a)  What  is  the  direction  of  the  acceleration?  (b)  What  is  the  mass  of  the  object?  (c)  If  the  object  is  initially  at  rest,  what  is  its  speed  after  10.0  s?  (d)  What  are  the  velocity  components  of  the  object  after  10.0  s?    

Solution:  

  F a∑ =m  reads  

 

  !2.00 i + 2.00 j + 5.00 i ! 3.00 j ! 45.0 i( ) N = m 3.75 m s2( ) a    

  where   a  represents  the  direction  of  a  

 

    !42.0 i !1.00 j( ) N = m 3.75 m s2( ) a  

    !F! = 42.0( )2 + 1.00( )2 N  at   tan!1 1.00

42.0"#$

%&'  below    

the  –x-­‐axis  

    !F! = 42.0 N at 181° = m 3.75 m s2( ) a .  

For  the  vectors  to  be  equal,  their  magnitudes  and  their  directions  must  be  equal.  

 

(a) ! a is at 181° counterclockwise from the x-axis

(b) m= =420 112. .  N

3.75  m s  kg2

(d)

!v f =!v i +!at = 0 + 3.75 m s2 at 181°( )10.0 s so

!v f = 37.5 m s at 181°

!v f = 37.5m s cos181° i + 37.5m s sin181° j so

!v f = !37.5 i ! 0.893 j( ) m s

(c)  

v f = + =375 0893 3752 2. . .   m s   m s  

 

6.   If  a  man  weighs  900  N  on  the  Earth,  what  would  he  weigh  on  Jupiter,  where  the  free-­‐‑fall  acceleration  is  25.9  m/s2?    

Solution:  

Fg = mg = 900 N ,  m =900 N

9.80 m s2 = 91.8 kg  

 

  Fg( )on Jupiter= 91.8 kg 25.9 m s2( ) = 2.38 kN  

 

7.   An  electron  of  mass  9.11  ×  10–31  kg  has  an  initial  speed  of  3.00  ×  105  m/s.  It  travels  in  a  straight  line,  and  its  speed  increases  to  7.00  ×  105  m/s  in  a  distance  of  5.00  cm.  Assuming  that  its  acceleration  is  constant,  (a)  determine  the  net  force  exerted  on  the  electron  and  (b)  compare  this  force  with  the  weight  of  the  electron.    

Solution:

(a) F ma∑ = and v v axf i f2 2 2= + or a

v vx

f i

f=

−2 2

2.

Therefore,

F! = mvf

2 " vi2( )

2x f

F! = 9.11#10"31 kg7.00 #105 m s2( )2

" 3.00 #105 m s2( )2$%

&'

2 0.050 0 m( ) = 3.64 #10"18 N .

(b) The weight of the electron is

Fg = mg = 9.11!10"31 kg( ) 9.80 m s2( ) = 8.93!10"30 N

The accelerating force is 4.08 !1011 times the weight of the electron.