1. Forces and Newton’s Laws of Motion
Chapter 4
Forces and Newton’s Laws of Motion

2. 4.1 The Concepts of Force and Mass
A force is a push or a pull.
Contact forces arise from physical
contact .
Action-at-a-distance forces do not
require contact and include gravity and electrical forces.

3. 4.2 Newton’s First Law of Motion
An object continues in a state of rest
or in a state of motion at a constant
velocity along a straight line, unless
compelled to change that state by a
net force.
The net force is the vector sum of all
the forces acting on an object.

4. 4.2 Newton’s First Law of Motion
Inertia is the natural tendency of an
object to remain at rest or in motion at
a constant velocity along a straight
line.

5. 4.3 Newton’s Second Law of Motion
Mathematically, the net force is
written as
where the Greek letter sigma denotes the vector sum.

6. 4.3 Newton’s Second Law of Motion
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.

7. 4.4 The Vector Nature of Newton’s Second Law
The direction of force and acceleration vectors
can be taken into account by using x and y
components.
is equivalent to

8. 4.5 Newton’s Third Law of Motion
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal magnitude on the first body.

9. 4.6 Types of Forces: An Overview
In nature there are two general types of forces,
fundamental and nonfundamental.
Fundamental Forces
1. Gravitational force
2. Strong Nuclear force
3. Electromagnetic force
4. Weak Nuclear force

10. 4.6 Types of Forces: An Overview
Examples of nonfundamental forces:
friction
tension in a rope
normal or support forces

11. 4.7 The Gravitational Force
Newton’s Law of Universal Gravitation
Every particle in the universe exerts an attractive force on every
other particle.
A particle is a piece of matter, small enough in size to be
regarded as a mathematical point.
The force that each exerts on the other is directed along the line
joining the particles.

12. 4.7 The Gravitational Force
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by

13. Definition of the Normal Force
The normal force is one component of the force that a surface
exerts on an object with which it is in contact – namely, the
component that is perpendicular
to the surface.

14. 4.9 Static and Kinetic Frictional Forces
When an object is in contact with a surface there is a force
acting on that object. The component of this force that is
parallel to the surface is called the
frictional force.

15. 4.9 Static and Kinetic Frictional Forces
When the two surfaces are
not sliding across one another
the friction is called
static friction.

16. 4.9 Static and Kinetic Frictional Forces
The magnitude of the static frictional force can have any value
from zero up to a maximum value.
is called the coefficient of static friction.

17. 4.9 Static and Kinetic Frictional Forces
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motion that
actually does occur.
is called the coefficient of kinetic friction.

18. 4.9 Static and Kinetic Frictional Forces

19. 4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of Equilibrium
An object is in equilibrium when it has zero acceleration.

20. 4.12 Nonequilibrium Application of Newton’s Laws of Motion
When an object is accelerating, it is not in equilibrium.

21. Example
A man can exert a force of 700N on a rope attached to a sled. The rope is at an angle of 30⁰ with the horizontal. If the coefficient of kinetic friction between the sled and ground is 0.4, what is the maximum load on the sled that the man can pull at constant speed?

22. Solution ƩFx = Fcos30⁰ - fk = 0 …………(i)
ƩFy = FN + Fsin30⁰ - mg = 0 ………..(ii)
From (i): μkFN = Fcos30⁰
Therefore FN = [Fcos30⁰]/μk
= [700×0.866]/0.4
= N

23. From (ii): m = [FN+ Fsin30⁰]/g
= [ ×0.5]/9.8
= kg

24. Problems to be solved
4.34, 4.40, 4.58, 4.61, 4.91, 4.93, 4.107
B4.1 A subway train has three cars, each weighing 1.2×105N. The frictional force on each car is 103N, and the first car, acting as an engine, exerts a horizontal force of 4.8×104N on the rails. (a) What is the acceleration of the train? (b) What is the tension in the coupling between the first and the second cars?

25. (c) What is the tension between the second and third cars. Ans: (a) 1
(c) What is the tension between the second and third cars? Ans: (a) 1.3m/s2 (b) 32000N (c) 16000N
B4.2 Block 1 of mass m1= 8.0kg is moving on a frictionless 30.0⁰ incline. This block is connected to block 2 of mass m2= 22.0kg by a massless cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord. Ans: 5.88m/s2
86.24N

26. B4.3 What minimum force is required to drag a carton of books across the floor if the force is applied at an angle of 45⁰ to the horizontal. Take the mass of carton as 40kg and the coefficient of friction as Ans: N