Defence Authority Shaikh Khalifa Bin Zaid College…. Slides Exhibited to PHYSICS .

1. Pic lganihaicollegeapneki Defence Authority Shaikh Khalifa Bin Zaid College….Slides Exhibited to PHYSICS.

2. PRESENTATIONfor physics:- Topic :- Tension in string. Class:- E-1 Prepared By:- SOHAIL IBRAHIM.

3. Motion of body connected by String

4. 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.

5. 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. TENSION:- Tension. M Weight.

6. P • R • A • C • T • I • C • A • l 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

7. P • R • A • C • T • I • C • A • l If the body is at rest or moves with uniform velocity then; T = W 2) If the body accelerates upward then, T > W 3) If the body accelerates downward then, W > T Conditions

8. C • A • S • E • # • 1 WHEN BOTH THE BODIES MOVES VERTICALLY…..

9. E • X • P • A • L • A • N • A • T • I • O • N 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 accelerate down with acceleration “a”, and the body B will move on a smooth horizontal surface with the same acceleration . Let the tension In the string be “T” . Procedure

10. E • X • P • A • L • A • N • A • T • I • O • N Diagram.

11. E • X • P • A • L • A • N • A • T • I • O • N Consider the downward motion of body :- “A”

12. E • X • P • A • L • A • N • A • T • I • O • N Forces acting on body “A” Two forces are acting on the body “A”; Force of gravity m1g acting in the downward direction. Tension “T” in the string in upward direction. Forces.

13. Forces acting on body “A” Since body “A” is moving Downward; Then m1g > T Net force acting on body “A”. F1 = m1g – T But a/c to Newton’s second law of motion; F1 = m1a Therefore m1a = m1g - T • E • X • P • A • L • A • N • A • T • I • O • N Forces.

14. E • X • P • A • L • A • N • A • T • I • O • N Consider the downward motion of body :- “B”

15. E • X • P • A • L • A • N • A • T • I • O • N Forces acting on body “B” Two forces are acting on the body “B”; Force of gravity m2g acting in the downward direction. Tension “T” in the string in upward direction. Forces.

16. 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 force then; F2 = T. Where; F2 = m2 a m2 a = T. • E • X • P • A • L • A • N • A • T • I • O • N Forces.

17. Equating the net forces For ACCELERATION “a”; For acceleration adding the net forces of body “A” and body “B”. m2a = T – m2 m1a = m1g - T m1a + m2a = m1g – m2g a(m1 + m2) = (m1 - m2 )g • E • X • P • A • L • A • N • A • T • I • O • N Equating.

18. E • X • P • A • L • A • N • A • T • I • O • N Formula for ACCELERATION “a”; a = (m1 - m2 )g (m1 + m2) acceleration.

19. Equating the net forces For TENSION “T”; For Tension dividing the net forces of body “A” and body “B”. m2a = T – m2 m1a = m1g - T m2 = T – m2 m1 = m1g - T • E • X • P • A • L • A • N • A • T • I • O • N Equating.

20. E • X • P • A • L • A • N • A • T • I • O • N Equating the net forces For 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 Equating.

21. E • X • P • A • L • A • N • A • T • I • O • N Formula for TENSION “T”:- T = 2 m2 m1g (m1 + m2) Tension.

22. WHEN one body moves VERTICALLY and the other moves on smooth HORIZONTAL surface….. • C • A • S • E • # • 2

23. E • X • P • A • L • A • N • A • T • I • O • N 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 > m₂ Then the body A will accelerate down with acceleration “a”, and the body B will move up with the same acceleration . Let the tension In the string be “T” .

24. E • X • P • A • L • A • N • A • T • I • O • N Diagram.

25. E • X • P • A • L • A • N • A • T • I • O • N Consider the downward motion of body :- “A”

26. E • X • P • A • L • A • N • A • T • I • O • N Forces acting on body “A” Two bodies are acting on the body “A”; Force of gravity m1g acting in the downward direction. Tension “T” in the string in upward direction. Forces.

27. Forces acting on body “A” Since body “A” is moving Downward; Then m1g > T Net force acting on body “A”. F1 = m1g – T But a/c to Newton’s second law of motion; F1 = m1a Therefore m1a = m1g - T • E • X • P • A • L • A • N • A • T • I • O • N Forces.

28. E • X • P • A • L • A • N • A • T • I • O • N Consider the downward motion of body :- “B”

29. Forces acting on body “B” Three forces are acting on the Body “B”; Force of gravity m2g acting in the downward direction. Tension “T” in the string which is acting horizontally towards the pulley. The normal reaction “R” of the surface on the body which acts vertically upward . • E • X • P • A • L • A • N • A • T • I • O • N Forces.

30. Equating the net forces For ACCELERATION “a”; For acceleration adding the net forces of body “A” and body “B”. m2a = T m1a = m1g - T m1a + m2a = m1g – T + T a(m1 + m2) = m1g • E • X • P • A • L • A • N • A • T • I • O • N Equating.

31. E • X • P • A • L • A • N • A • T • I • O • N Formula for ACCELERATION “a”; a = m1 g (m1 + m2) acceleration.

32. Formula for TENSION “T”:- Putting the value of “a” in; T = m2a Where ; a = m1 g (m1 + m2) Therefore ; • E • X • P • A • L • A • N • A • T • I • O • N Tension. T = m2 m1g (m1 + m2)