This document contains a presentation on Newton's second law of motion. The presentation topics include the relation between force, mass and acceleration, applications of Newton's second law, equations of motion, and an introduction to kinetics of particles. The document provides definitions and explanations of key concepts such as force, mass, acceleration, momentum, impulse, and kinetics. It also includes sample problems demonstrating applications of Newton's second law and equations of motion, along with step-by-step solutions. The presentation was made by Danyal Haider and Kamran Shah and covers fundamental principles of classical mechanics.

1. Presentation
Made by :
• Danyal Haider (01-133132-014)
• Kamran Shah (01-133132-027)

3. • PRESENTATION TOPICS
• Relation between force , mass and
acceleration .
• Applications of Newton second law of motion.
• Equations of motions .
• Introduction to kinetics of particles .

4. Relation between force , mass and acceleration :
Relation between force , mass and acceleration is
refered to Newton second law of motion .
FORCE :• Force is the action of one body on another body .
A force tends to move a body in the direction of
its action . The action of the force is charaterized
by its magnitude , by the direction of its action ,
and its point of application .
OR

5. Force is a vector quantity having both magnitude and direction .
The magnitude of the force can be measured using a spring
scale.
Unit of force is Newton .

6. MASS :
Mass is a measure of Inertia of a body .
Mass is a scalar quantity .
Unit of mass is Kilogram (kg) .
ACCELERATION :
• The rate of change of velocity is called acceleration .
• Acceleration is a vector quantity so it can be changed in two ways:
1. changing its magnitude
2. changing its direction
• Acceleration can be positive or negative. If an object’s speed
increases with time, it will have positive acceleration. If an
object’s speed decreases with time, it will have negative
acceleration that’s also known as deceleration. Si unit of
acceleration is meters /second squared .

7. “NEWTON”
Sir Isaac Newton (4 January 1643 – 31 March 1727) was
an English physicist and mathematician. He is famous for his work on
the laws ofmotion, optics, gravity, and calculus. In 1687, Newton
published a book called the Principia Mathematica in which he
presents his theory of universal gravitation and three laws of motion.

8. Newton’s second law:
• A particle will have an acceleration proportional to
the magnitude of the resultant force acting on it and
in the direction of the resultant force .
• The resultant of the forces acting on a particle is
equal to the rate of change of linear momentum of
the particle .
• The sum of the moments about “O” of the forces
acting on a particle is equal to the rate of change of
angular momentum of the particle .
• When a particle of mass m is acted upon by a force
the acceleration of the particle must satisfy,
F=ma

9. • If force acting on particle is zero, particle will
not accelerate, i.e., it will remain stationary or
continue on a straight line at constant velocity.

10. Linear Momentum Conservation Principle:
• If the resultant force on a particle is zero, the linear
momentum of the particle remains constant in both
magnitude and direction .
• Replacing the acceleration by the derivative of the
velocity yields
ΣF = m(dv/dt)
= d/dt (mv)
=dL/dt
Where ,
L = linear momentumof the particle

11. SOLUTION:
Q#1
•
•
•
•
•
•
A 200-lb block rests on a
horizontal
plane. Find the magnitude of
the force
P required to give the block
an acceleration
or 10 ft/s2 to the right. The
coefficient
of kinetic friction between
the
block and plane is u^k = 0.25
• Resolve the equation of motion for
the block into two rectangular
component equations.
• Unknowns consist of the applied
force P and the normal reaction N
from the plane. The two equations
may be solved for these unknowns.

12. SOLUTION:
• Resolve the equation of motion for the
block into two rectangular component
equations.
ΣFx = ma :
Pcos30 - 0.25N = (6.21lb.s2/ft)(10ft/s2)= 6.21lb
ΣFy = 0 :
N − Psin 30° − 200lb = 0
• Unknowns consist of the applied force P
and the normal reaction N from the plane.
The two equations may be solved for these
unknowns.
N = P sin 30° + 200 lb
P cos 30° - 0.25(P Sin 30° + 200 lb) = 62.1 lb
P = 151 lb

13. • PRESENTATION TOPICS
• Relation between force , mass and
acceleration .
• Applications of Newton second law of motion.
• Equations of motions .
• Introduction to kinetics of particles .

14. Applications of Newton second law of motion :
• When a force is applied
on a body of mass(m),
there is a change in
velocity w.r.t time , and
change in velocity
means acceleration is
produced in a body .

16. We know that ,
Acceleration(a) is directly
proportional to Force(F)
• In the above figure, if the
person applies lesser
force on the vehicle . The
less acceleration is
produced in a body .
• On the other hand , if the
person applies greater
force on the vehicle . The
more acceleration is
produced in a body .

22. • PRESENTATION TOPICS
• Relation between force , mass and
acceleration .
• Applications of Newton second law of motion.
• Equations of motions .
• Introduction to kinetics of particles .

23. Equations of motion
PROOF OF EQUATION OF MOTION.
1st Equation of motion.
Vf = vi + at
Proof.
Consider a body moving with initial velocity vi .if a force F act on the
body and produces an acceleration a in it, then after time t seconds. Its
final velocity becomes vf. Then change in velocity of the body is vf - vi, and
the rate of change of velocity of the body is equal to the acceleration
acting on the body,i.e,
a = vf – vi/t
vf – vi = at
vf = vi + at

24. 2nd Equation of Motion.
S = vit + at2/2
Proof.
According to the first equation of motion, we have
Vf = vi + at
eq no (1)
If the body has constant acceleration, then its average velocity is given by
Vav = vi + vf/2
eq no (2)
Distance covered by the body in time t is giver by
S = vav * t
eq no (3)
Putting the mauve of Vav from eq (2) in eq (3) we get
S = t * (vi + vf)/2
Putting the value of vf from eq,(1) in eq,(4), we get
S = t (vi + vi + a t)/2 = 2vit + at2/2
S = vi t + at2/2

25. Third equation of motion.
2as = vf2 – vi2
Proof.
We know that
S = vav . t
eq(1)
Vav = vi + vf/2
eq(2)
From 1st equation of motion
Vf = vi + at
t= vf – vi/a this is
eq (3)
Putting the value of vav and t from eq,(2) and (3), in eq,(1),
S = (vi + vf)/2* (vf – vi)/a
2as = (vf +vf). (vf – vi)
2as = vf2 – vi2

26. Some Rules to be used in the above three
equation.
• If the body starts its motion from rest then use vi=0
• If a moving body stops its motion then vf=0
• If the velocity of the body is incressing then its
acceleration is positive and its directed is the same as
that of velocity.
• If the velocity of the body is decreasing then its
acceleration is negative and its directed is opposite to
that of the velocity .

27. Some Numerical Problems.
Q # 2: What force is required to move a mass of 5kg through 150 meters in 10
sec. if the mass is originally at rest.
Given data
m=5kg
s=150m
t=10sec
For this we have to find the acceleration acting on the body and according to the
gives data we apply second equation of motion.
F=?
S=vit+1/2at2
150=0*10*a*100/2
a=3m/sec2
F= ma
F=5*3=15N

28. Equation of motion for a freely falling body.
If a body is moving freely upward of falling downward then its acceleration a
will be the acceleration due to gravity.
i.e g=9.8m/sec2 and its distance S will show the height h of the body. Then the
equation of motion will take the form.
Downward motion
Vf=vi+at
S=vit+1/2at2
2as=vf2-vi2
vf=vi+gt
h=vit+1/2gt2
2gh=vf2-vi2
Upward motion
Vf=vi+at
vf=vi-gt
S=vit+at2/2
s=vit-gt2/2
2as=vf2-vi2
-2gt=vf2-vi2

29. Q # 3:. A 50kg shell of conical shape is
dropped from a height of 1000m. It falls freely
under gravity and sinks in into the ground .
Find
(a) Its velocity of the moment it touches the
ground.
(b). the time it takes to reach the groun.
(c). the deceleration of the shell as it
penetrates into the ground.
Given data is
m=50kg
vi=0
s=1m
using 2nd equation of motion find the vf.
2gh=vf2-vi2
2*(9.8)*1000 = vf2-0
Vf = 140ms-1

30. Now find the t?
Using equation of motion.
From 1st equation on motion.
Vf = vi+gt
140 = 0+9.8*t
t = 140/9.8= 14.3sec
t = 14.3sec
(c) for penetrating into ground.
For ground the velocity of 140ms-1 is intial
velocity. In ground it comes to rest after
covering a distance of 1m.
Given data is
Vi’ = 140ms-1
Vf ’ = 0
S = 1m
a= ?
Using equation motion
From 3rd equation motion
2as = vf2-vi2
2*a*1 = 0(140)2
a= -19600/2
a = -9800ms-1

31. • PRESENTATION TOPICS
•
•
•
•
Relation between force , mass and acceleration .
Applications of Newton second law of motion.
Equations of motions .
Introduction to kinetics of particles .

32. Introduction to kinetics of particles
Difference between kinematics and kinetics :
Kinematics:- kinematics is the study of motion (not focusing upon forces , just
the motion) .
Kinetics :- kinetics is the study of motion in relation to forces .
Introduction to kinetics of particles
According to Newton second law, a particle will accelerate when it is subjected
to unbalanced forces . kinetics is the study of the relations between unbalanced
forces and resulting changes in motion .
Methods of solution of problems of particle kinetics:There are three basic methods of solution of problems of particle kinetics .
These three methods are summarized as follows :

33. (1)DIRECT APPLICATION OF NEWTON SECOND LAW:First we applied the Newton second law F=ma to determine the
instantaneous relation between force and acceleration they produce .
Then solve a problem using x-y coordinates for plane motion problems
and x-y-z coordinates for space motion problems .
(2)WORK-ENERGY EQUATIONS:Next, we apply F=ma with respect to displacement and derived the
equations for work and energy . These equations enable us to relate the
initial and final velocities to the work done during an interval of forces .
(3)IMPULSE-MOMENTUM
EQUATIONS:-
Then we write Newton second law in terms of force equals time rate of
change of linear momentum and moment equals time rate of change of
angular momentum . Then solve these relations with respect to time and
derived the impulse and momentum equations .

34. Q#4:
Determine the rated speed of a
highway curve of radius ρ = 400 ft
banked through an angle θ = 18o.
The
rated speed of a banked highway
curve
is the speed at which a car should
travel if no lateral friction force is
to
be exerted at its wheels.
SOLUTION:
 The car travels in a horizontal
circular path with a normal
component of acceleration directed
toward the center of the path . The
forces acting on the car are its
weight and a normal reaction from
the road surface.
 Resolve the equation of motion for
the car into vertical and normal
components.
 Solve for the vehicle speed.

35. • Resolve the equation of motion for
the car into vertical and normal
components.
ΣFy = 0 :
R Cos θ – W = 0
R = W/Cos θ
ΣFn = man :
R sin θ = (W/g) *an
(W/Cos θ)*Sin θ =(W/g)* (v ^2/ ρ )
SOLUTION:
• The car travels in a horizontal
circular
path with a normal component
of
acceleration directed toward
the center
of the path.The forces acting on
the
car are its weight and a normal
reaction from the road surface.
• Solve for the vehicle speed.
v^2 = g ρ Tan θ
=(32.2ft/s^2)(400 ft)Tan 18°
v = 64.7 ft/s = 44.1mi/h

36. Thank you
Any Question please !!!!!