Newtons Second Law

1. Newton’s Second Law
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.
Statement- When particle is subjected to force and accelerates in the
direction of force. Then the resultant force is proportional to the
magnitude of the mass and acceleration.
. . .
F m a
i e When no of forces are acting then
F m a



2. Newton’s Second Law
• When more than one force acting on a particle, the resultant
force is determined by taking vector summation of all forces.
 1
(2)
In - Direction
In - Direction
x x
y y
x
F m a
y
F m a
      
      


• In the equation ma is called as Inertia Force
• For Dynamic equilibrium 0
0
and
x x
y y
F m a
F m a
 
 



3. • Newton’s Second Law can be applied for curvilinear
motion in Normal and tangential direction.
Newton’s Second Law
2
(2)
,
In Tangential Direction
where,
In Normal Direction
t t
t
n n
n
F m a
dv
a
dt
F m a
v
a where Radious of curvature



      
 


𝜌

4. Newton’s Second Law
Example 1- The 50 Kg crate is traveled along the floor with an
initial velocity 7m/sec at x=0. The Coe. of kinetic friction is 0.40.
Calculate the time required for the crate to come to rest and the
corresponding distance x traveled.
Given: Mass of the Crate M= 50Kg,
Weight W=Mg, Hence , W=50 X 9.81=490.5 N
Initial Velocity = 7 m/sec (at x = 0)
Crate come to rest, i.e. final velocity = 0
Coe. of kinetic friction = 0.40k

5. Newton’s Second Law
Example 1 Continue …….
As crate moves in x- direction i.e. no no
displacement in Y-direction Hence, = 0
Using Second law in y-direction
See Fig add force in y-direction
ya
(1)y yF m a      
490.5
50 , 0
. (1),
490.5 50 0 Re 490.5
y
y
y
F W N
F N
m Kg a
Put in Eq
N X We will get Normal action N N
  
  
 
   



6. Example 1 Continue….
Newton’s Second Law
Now, Use Newton’s Second law in x -direction
(2)x xF m a      
2
2 2
2
0.4 490.5
50 , 0.4 ,
.(2)
0.4 490.5 50
3.924 / tan
2
0 (7) 2 ( 3.924) 0 7 ( 3.924)
6.24 1.724
To find
x k
k x
x x
x
x
F Frictional Force N X
m Kg a
Put in Eq F m a
X X a
a m s Also find dis ce x and time t
v u a x v u at
x and t
x m t Sec


   
  

 

   
     
 



7. Newton’s Second Law
Example 2- A Man moves a crate by pushing horizontally against
until slides on the floor. If , with what
acceleration does the crate begin to move? Assume that the force
exerted by the man at impending motion is maintained when
sliding begins.
Given: Mass of the Crate M, Weight W = Mg,
Coe. of Kinetic friction
Initial velocity of the crate = 0 (rest)
Let P be the force required to push the crate on the floor
0.5 0.4k sand  
0.5k 

8. Example 2- Continue…
(1)y yF m a 
Using Second law in y-direction
,
0,
. (1),
0
y
y
F W N
Accl in y direction a
Put in Eq
mg N we get N mg
  
 
   

For Impending Motion,
0
0
0.5
0.5
x
s
F
P N
P N as N mg
P mg


 
 



9. Using Second law in X-direction i.e. Eq. (2)
Example 2- Continue…
2
(2)
0.5 0.4
0.1
0.981 / s Acceleration of the Crate
x x
k x
x
x
x
F m a
P N m a
put P and N
mg mg m a
Taking m common and cancaling
a g
a m

      
 
 




10. Example -3
Application of Second law to Lift Problem
Prob- A lift carries a weight of 3600N is moving with a uniform
acceleration of 3.5 m /sec2. Determine the tension in the supporting
cable. When 1) Lift is moving upward 2) Lift is moving downward.
Solution: Given Wt. of Lift = 3600N
Acceleration in y-direction a = 3.5 m/s2
Case-I When lift moving Up
Use Newton’s second Law in Y-direction
Put W=mg and m = W / g,
(1)y yF m a 
1 (1)yF T W put in Eq 

11. Problem 3 Continue
1
1
1
1
(1)
3600
3600 3.5
9.81
4884.404
y y
y
y
F m a
W
T W a
g
W
T W a
g
T X
T N
      
 
   
 
 
   
 
 
   
 


When lift moving Up
For upward motion, Tension in cable should be more than W

12. Problem 3 Continue
1
2
2
2
sin (2)
3600
3600 3.5
9.81
2315.60
nd
y y
y
y
U g II law F m a
W
W T a
g
W
T W a
g
T X
T N
  
 
   
 
 
   
 
 
   
 


Case-II When lift moving Down
To have motion downward W should more than Tension in cable

13. Example -4
Application of Second law for Curvilinear Motion
Prob- A bob of 1m pendulum describes an arc of a circle in a vertical
plane. When the angle of the cord is with the vertical, the tension in
the cord is two times the weight of the bob. Find the velocity and
acceleration of the in this position.
Solution: Given Weight of the Bob= W
Tension in the Cord T = 2W
Radius of the curvature, = 1m
Let v be the velocity of the bob,
a is the acceleration of the bob
0
30


14. Example -4 continue…
Draw F.B.D. at position shown
Using Second law in Normal direction
2
2
,
3.335 / sec
,
(1)
s in
30, / , 2
/ , 1 .(1)
cos30
2 0.866
1
x x
n n
n
W get Cancels
v m
F m a here
F m a
Force Normal direction
F T W Cos m w g Given T W
an v m Put in Eq
T W m an
W v
W W X
g
 


      
   
 
 
 




15. Example -4 continue…
2 2 2
2
2 2
. .
/ 3.335 /1 11.125 /
. .
sin .
(2)
sin 30
4.905 /
( )
To Find
To Find
n
n
t
t t
t
t
n t
a Normal Comp of Accl
a v m s
a Tangeential Comp of Accl
U g Newtons Second law in Tangential dir
F m a
W
W a Canceling W
g
a m s
Total Accleration a
a a a


  

      


 

2 2 2
11.125 4.905 12.16 /
n tput Values of a and a
a a m s  

16. Thank You