DEFENCE AUTHORITY SHAIKH KHALIFA BIN ZAID COLLEGE…. SLIDES EXHIBITED TO PHYSICS.

Pic lgani hai

collegeapne ki

PRESENTATION

for physics:-

Topic :-

Tension in

string.

Class:-

E-1

Prepared By:-

SOHAIL

IBRAHIM.

Case to be discussed:-

‘’Motion of body connected by String’’.

Case # 1

When both bodies move vertically.

Case # 2

When one body moves vertically and the other body moves on smooth horizontal surface.

TENSION:-

The force applied on a body through a string called “Tension”.

Tension is a kind of force which is applied usually on string structures which are suspended by fixed supports and acted towards the support. Tension always directed opposite to the direction of weight where weight is another kind of force which is always directed towards the center of the earth.

M

Tension.

Weight.

PRACTICAl

Tension in the string:-

When a body of weight “W” is kept suspended by a string, the weight of the body pulls the string downward while the string pulls the body upwards with an equal force.This force is called “Tension in the string” (T).

Introduction

PRACTICAl

Conditions

1) If the body is at rest or moves with uniform velocity then;

T = W2) If the body accelerates

upward then, T > W3) If the body accelerates

downward then, W > T

CASE

#

1

WHEN BOTH THE BODIES

MOVES VERTICALLY…..

EXPALANATION

Procedure

Consider two bodies A and B of

masses m₁ and m₂ connected by

an in extensible a string which

passes over a friction less pulley .

The the body A will

acceleratedown with acceleration “a”,

and the body B will move on a

smooth horizontal surface with

the sameacceleration . Let the tension In the string be “T” .

EXPALANATION

Diagram.

EXPALANATION

Consider the

downwardmotion of body :-

“A”

EXPALANATION

Forces.

Forces acting on body “A”

Two forces are acting on the body

“A”;

1) Force of gravity m1g acting in the downward direction.

2) Tension “T” in the string in upward direction.

EXPALANATION

Forces.

Forces acting on body “A”

Since body “A” is movingDownward;

Then m1g > TNet force acting on body “A”. F1 = m1g – TBut a/c to Newton’s second

law ofmotion; F1 = m1aTherefore m1a = m1g - T

EXPALANATION

Consider the

downwardmotion of body :-

“B”

EXPALANATION

Forces.

Forces acting on body “B”

Two forces are acting on the body

“B”;

1) Force of gravity m2g acting in the downward direction.

2) Tension “T” in the string in upward direction.

EXPALANATION

Forces.

Forces acting on body “B”

Since there is no motion of body b

in the vertical direction are equal

and opposite.So, along y-axis ∑ Fy = 0. R – W2 = 0 R = W2

R = m2g.If we neglect the frictional

forcethen; F2 = T.Where; F2 = m2 a m2 a = T.

EXPALANATION

Equating.

Equating the net forcesFor ACCELERATION “a”;

For acceleration adding the net

forces of body “A” and body “B”.

m2a = T – m2

m1a = m1g - T

m1a + m2a = m1g – m2ga(m1 + m2) = (m1 - m2 )g

EXPALANATION

acceleration.

Formula for ACCELERATION “a”;

a = (m1 - m2 )g

(m1 + m2)

EXPALANATION

Equating.

Equating the net forcesFor TENSION “T”;

For Tension dividing the netforces of body “A” and body

“B”.

m2a = T – m2

m1a = m1g - T

m2 = T – m2

m1 = m1g - T

EXPALANATION

Equating.

Equating the net forcesFor TENSION “T”;

m1 (T – m2)= m2 (m1g – T)

m1 T – m1m2= m2 m1g – m2 T

m1 T + m2T = m2 m1g + m2

m1g(m1 + m2)T = 2 m2 m1g

EXPALANATION

Tension.

Formula for TENSION “T”:-

T = 2 m2 m1g

(m1 + m2)

CASE

#

2

WHEN one body moves

VERTICALLY and the other moves

on smooth HORIZONTAL

surface…..

EXPALANATION

Consider two bodies A and B of

masses m₁ and m₂ connected by

an in extensible a string which

passes over a friction less pulley .

If

m₁ > m₂Then the body A will

acceleratedown with acceleration “a”,

and the body B will move up with

thesame acceleration . Let the

tension In the string be “T” .

EXPALANATION

Diagram.

EXPALANATION

Consider the

downwardmotion of body :-

“A”

EXPALANATION

Forces.

Forces acting on body “A”

Two bodies are acting on the body

“A”;

1) Force of gravity m1g acting in the downward direction.

2) Tension “T” in the string in upward direction.

EXPALANATION

Forces.

Forces acting on body “A”

Since body “A” is movingDownward;

Then m1g > TNet force acting on body “A”. F1 = m1g – TBut a/c to Newton’s second

law ofmotion; F1 = m1aTherefore m1a = m1g - T

EXPALANATION

Consider the

downwardmotion of body :-

“B”

EXPALANATION

Forces.

Forces acting on body “B”

Three forces are acting on the

Body “B”;

1) Force of gravity m2g acting in the downward direction.

2) Tension “T” in the string which is acting horizontally towards the pulley.

3) The normal reaction “R” of the surface on the body which acts vertically upward .

EXPALANATION

Equating.

Equating the net forcesFor ACCELERATION “a”;

For acceleration adding the net

forces of body “A” and body “B”.

m2a = T m1a = m1g - T

m1a + m2a = m1g – T + Ta(m1 + m2) = m1g

EXPALANATION

acceleration.

Formula for ACCELERATION “a”;

a = m1 g

(m1 + m2)

EXPALANATION

Tension.

Formula for TENSION “T”:-

Putting the value of “a” in;

T = m2aWhere ;

a = m1 g

(m1 + m2)

Therefore ;

T = m2 m1g

(m1 + m2)