Chapter 4

Forces and Newton’s Laws of Motion

4.1 The Concepts of Force and Mass

A force is a push or a pull.

Contact forces arise from physicalcontact .

Action-at-a-distance forces do notrequire contact and include gravity and electrical forces.

4.2 Newton’s First Law of Motion

An object continues in a state of restor 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 allthe forces acting on an object.

Newton’s First Law

4.2 Newton’s First Law of Motion

Inertia is the natural tendency of anobject to remain at rest or in motion ata constant velocity along a straight line.

4.3 Newton’s Second Law of Motion

F

Mathematically, the net force is written as

where the Greek letter sigma denotes the vector sum.

4.3 Newton’s Second Law of Motion

Newton’s Second Law

When a net external force acts on an objectof mass m, the acceleration that results is directly proportional to the net force and hasa magnitude that is inversely proportional tothe mass. The direction of the acceleration isthe same as the direction of the net force.

mF

a

aF

m

4.4 The Vector Nature of Newton’s Second Law

xx maFyy maF

The direction of force and acceleration vectorscan be taken into account by using x and ycomponents.

aFm

is equivalent to

4.5 Newton’s Third Law of Motion

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.

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

4.6 Types of Forces: An Overview

Examples of nonfundamental forces:

friction

tension in a rope

normal or support forces

4.7 The Gravitational Force

Newton’s Law of Universal Gravitation

Every particle in the universe exerts an attractive force on everyother 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 linejoining the particles.

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

221

r

mmGF

2211 kgmN10673.6 G

4.8 The Normal Force

Definition of the Normal Force

The normal force is one component of the force that a surfaceexerts on an object with which it is in contact – namely, thecomponent that is perpendicular to the surface.

4.9 Static and Kinetic Frictional Forces

When an object is in contact with a surface there is a forceacting on that object. The component of this force that is parallel to the surface is called the frictional force.

4.9 Static and Kinetic Frictional Forces

When the two surfaces are not sliding across one anotherthe friction is called static friction.

4.9 Static and Kinetic Frictional Forces

The magnitude of the static frictional force can have any valuefrom zero up to a maximum value.

MAXss ff

NsMAXs Ff

10 s is called the coefficient of static friction.

4.9 Static and Kinetic Frictional Forces

Static friction opposes the impending relative motion betweentwo objects.

Kinetic friction opposes the relative sliding motion thatactually does occur.

Nkk Ff

10 k is called the coefficient of kinetic friction.

4.9 Static and Kinetic Frictional Forces

4.11 Equilibrium Application of Newton’s Laws of Motion

Definition of EquilibriumAn object is in equilibrium when it has zero acceleration.

0xF

0yF

4.12 Nonequilibrium Application of Newton’s Laws of Motion

xx maF

yy maF

When an object is accelerating, it is not in equilibrium.

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?

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

= 1515.54N

From (ii): m = [FN+ Fsin30⁰]/g

= [1515.54+700×0.5]/9.8 = 190.36kg

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?

(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

• 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 0.60. Ans: 207.89N