Defence Authority Shaikh Khalifa Bin Zaid College….
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Transcript of Defence Authority Shaikh Khalifa Bin Zaid College….
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DEFENCE AUTHORITY SHAIKH KHALIFA BIN ZAID COLLEGE….
Pic lgani hai
collegeapne ki
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PRESENTATION
for physics:-
Topic :-
Tension in
string.
Class:-
E-1
Prepared By:-
SOHAIL
IBRAHIM.
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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.
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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.
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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
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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
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CASE
#
1
WHEN BOTH THE BODIES
MOVES VERTICALLY…..
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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” .
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EXPALANATION
Diagram.
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EXPALANATION
Consider the
downwardmotion of body :-
“A”
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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.
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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
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EXPALANATION
Consider the
downwardmotion of body :-
“B”
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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.
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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.
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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
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EXPALANATION
acceleration.
Formula for ACCELERATION “a”;
a = (m1 - m2 )g
(m1 + m2)
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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
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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
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EXPALANATION
Tension.
Formula for TENSION “T”:-
T = 2 m2 m1g
(m1 + m2)
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CASE
#
2
WHEN one body moves
VERTICALLY and the other moves
on smooth HORIZONTAL
surface…..
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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” .
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EXPALANATION
Diagram.
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EXPALANATION
Consider the
downwardmotion of body :-
“A”
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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.
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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
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EXPALANATION
Consider the
downwardmotion of body :-
“B”
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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 .
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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
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EXPALANATION
acceleration.
Formula for ACCELERATION “a”;
a = m1 g
(m1 + m2)
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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)
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CASE 1
Tension and acceleration on a string when two bodies of unequal in masses are attached passing over frictionless pulley hanging vertically downward.
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DATAm1 Kg m2 Kg a={(m1 - m2) / (m1 +
m2) } x g m/sec 2
T= {2(m1 m2) / (m1 +
m2) } x g Newton
5 4.8 0.2 48 6 5.4 0.52 55.71 7 6.4 0.44 65.53 8 7.3 0.45 74.81 9 8.8 0.11 87.21
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MULTIPLE BAR
10
1
2
3
4
5
6
7
8
9
m1#REF!#REF!
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BUBBLES WITH LINES
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.50
10
20
30
40
50
60
70
80
90
100
m1m2aT
Axis Title
Axis Title
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DOUGHNUT CHART
12345
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CASE II When one body in vertical and other is
lying on the horizontal plane.
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DATAm1 Kg m2 Kg a={ m1 / (m1 + m2) }
x g m/sec 2T= {(m1 m2) / (m1 +
m2) } x g Newton
5 4.8 5 24 6 5.4 5.157 27.85 7 6.4 5.11 32.76 8 7.3 5.12 35.86 9 8.8 4.95 43.6
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3-D BAR
m1 Kg
5
6
7
8
9
0 5 10 15 20 25 30 35 40 45
Series3Series2Series1
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3-D AREA
1 2 3 4 50
10
20
30
40
50
60
70
T= {(m1 m2) / (m1 + m2) } x g Newton a={ m1 / (m1 + m2) } x g m/sec 2 m2 Kg m1 Kg
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CONE CHARTS
1 2 3 4 5
0
5
10
15
20
25
30
35
40
45
m1 Kg a={ m1 / (m1 + m2) } x g m/sec 2
m1 Kg m2 Kg a={ m1 / (m1 + m2) } x g m/sec 2T= {(m1 m2) / (m1 + m2) } x g Newton