Chapter 3Forces & Motion

Learning outcome:By the end of the lesson student be able : List the types of forces. State Newton’s First Law of Motion. State Newton’s Second Law of Motion. State Newton’s Third Law of Motion. Calculate the tension.

ForceForce is a push or pull

exerted by one object on another.

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Force is a vector quantity since it has both magnitude and direction.

Net ForceNet force at a point is the total of all forces

that acts onto/towards the point.

321 FFFF

F1

F2

F3

c

Through vector, the net force acts on point C:

For each situation, determine the net force acting upon the object.

TRY THISTRY THIS

Analyze each situation individually and determine the magnitude of the unknown forces

Effective ForceEffective force is a force that gives effect to the motion of a body whereas the direction of F is the same (parallel) as the direction of

the body.

maFFnet

m F

Normal ForceNormal force, N is a force acted on a body

by a surface and always normal (perpendicular) to that surface.

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mgNFF ynet

Forces that acted onto a body is from y-axis.

Case 1: Horizontal surface• An object lies at rest on a flat horizontal surface as shown in figure a.

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0mgNFy

mgN ThereforegmW

Figure aFigure a

Action: weight of an object is exerted on the horizontal surface

Reaction: surface is exerted a force, N on the object .

N

Case 2 : Inclined plane• An object lies at rest on a rough inclined plane as shown in figure b.

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0yy WNF

Component of the weight :

Therefore cosmgN

Figure bFigure b

yW

gmW

N

xW

y

Action:Action: y-component of the y-component of the object’s weight is exerted object’s weight is exerted on the inclined surface.on the inclined surface.

Reaction: Reaction: surface is exerted a surface is exerted a force, force, NN on the object. on the object.

os sin cmgW

mgW

y

x

Case 3 : Motion of a lift• Consider a person standing inside a lift as shown in figures i, ii and ii.

a. Lift moving upward at a uniform velocity

N

Since the lift moving at a uniform velocity, thus

Therefore

Figure i

0ya

mgNmgN

Fy

0

0

gmW

b. Lift moving upwards at a constant acceleration, a

gmW

N

By applying the newton’s 2nd law of motion, thus

Figure ii

)( gamNmamgN

maF yy

c. Lift moving downwards at a constant acceleration, a

gmW

N

By applying the newton’s 2nd law of motion, thus

Figure iii )( agmNmaNmg

maF yy

Complete the following table showing the relationship between mass and weight.

Object Mass (kg) Weight (N)Melon 1 kgApple 0.98 NPat

Eatladee 25 kg

Fred 980 N

Normal ForceNormal Force

Radial Force

RmvmaFnet

2

vR a

Radial force is the force which directs inward toward the center of a circle that moves constantly.

R = radius

Newton’s First LawLaw of Inertia

“A body that is in motion continues in motion with the same velocity (at constant speed and in a

straight line) and a body at rest continues at rest unless unbalanced (outside) force acts upon it”

Newton’s First Law“Inertia” is the property of a body that causes it to

remain at rest if it is at rest or to continue moving with a constant velocity

unless an unbalanced force acts upon it.

The more mass which an object has, the more inertia it has - the more tendency it has to resist changes in its state of motion.

Newton’s First LawNewton’s First Law

Newton’s Second Law“The net force vector is equal to the mass of the body

times the acceleration of the body”.

Unit of force is the newton (N).

maF Total

force (N)Mas

s (kg)

Acceleration (m/s2)

Newton’s Third Law“Whenever two bodies interact, the two forces that

they exert on each other are always equal in magnitude and opposite in direction”.

BonAAonB FF

In every interaction, there is a pair of forces acting on the two interacting objects.

A flexible rope, cord, or wire pulling on an object is said to be under tension and exerts a force FT.

Before we begin, we make a simple assumption that any such device is massless.

As a consequence the rope transmits force undiminished from one end to the other.

This is apparent from ∑F = ma. Since the mass is zero the net force on the cord is zero, so the force on the two ends must sum to zero.

Where does this other force come, from the object pulled!

FT-FT

FT

Two Boxes and a Cord Two boxes resting on a

frictionless surface are connected by a massless cord.

The boxes have masses of 12.0 and 10.0 kg.

A horizontal force of FP=40.0 N is applied by pulling on the lighter box.

What is the acceleration of each box and the tension in the cord?

This problem adds a force and 2 dimensions!

11amFFF TPx

Box 1 freebody diagram has 4 forces: From person pulling: FP

Tension from cord: FT

Weight: W1 = m1g Normal Force: FN1

From the 2nd Law in the x direction

22amFF Tx

Box 2 freebody diagram has three forces: Tension from cord: FT

Weight: W2 = m2g Normal Force: FN2

From the 2nd Law in the x direction

NsmkgamF

smkgN

mmFa

ammFamamFFF

amamFamamFF

T

P

P

TTP

TTP

8.21)/82.1)(0.12(

/82.10.220.40

)(

:equations two theAdding

2Box 1Box

22

2

21

21

21

222111

The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch-up which is counter to our experience.

Frictional force, • is defined as a force that resists the motion of one surface relative to

another with which it is in contact.• is independent of the area of contact between the two surfaces..• is directly proportional to the reaction force.

OR

• Coefficient of friction, – is defined as the ratio between frictional force to reaction force.

OR

– is dimensionless and depends on the nature of the surfaces.

Nf

Nf force frictional:f

friction oft coefficien : μforcereaction : N

where

f

Nf

Case 1 : Horizontal surface• Consider a box of mass m is pulled along a horizontal surface by a

horizontal force, F as shown in figures below.

– x-component :

– y-component :

maFF nettx

F

gm

N

f

mafF

0yFmgN

Case 2 : Inclined plane• Consider a box of mass m is pulled along an inclined plane by a force, F as

shown in figures below

– x-component (parallel to the inclined plane) :

y-component (perpendicular to the inclined plane:

aN

gmW

y

yWf

0yF0 yWNθmgN cos

maFx

mafWF x fθmgmaF sin