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Addition and resolution of forces

Addition and resolution of forces. 1 Which of the following schoolbags would you feel heavier to carry if the same number of books are put in them?. A . B. 2 A light rope is stretched tightly between two poles. . A T-shirt of weight 8 N is hung at the midpoint of the rope.

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Addition and resolution of forces

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  1. Addition and resolution of forces

  2. 1 Which of the following schoolbagswould you feel heavier to carry if thesame number of books are put inthem? A B

  3. 2 A light rope is stretched tightlybetween two poles. A T-shirt of weight8 N is hung at the midpoint of therope. What is the tension T in the string? AT < 4 N BT = 4 N CT > 4 N 8 N

  4. Adding forces a Graphical method Force is a vector quantity. It has both magnitude and direction. Like displacements, forces can also be added graphically using the 'tip-to-tail' method. Lines representing forces should be drawn to scale.

  5. F2 F1 F1 + F2 F2 F1 Adding forces a Graphical method 'Tip-to-tail' method resultant force

  6. F2 F1 F1 + F2 Adding forces a Graphical method Parallelogram of forces method parallel parallel resultant force

  7. Adding forces a Graphical method Video Simulation

  8. F1 F2 Addition of forces Stretch a rubber band with a spring balance. F B O A Stretch the rubber band by thesame amount using 2 springbalances. Note the magnitude & direction of the F1 & F2. Check if resultant = F(in magnitude & direction) by –using 'tip-to-tail' method –using parallelogram of forces method

  9. Addition of forces Video

  10. F1 F1 = = F1 F1 + + F2 F2 F2 F2 1 Adding forces b Algebraic method (1-dimension) The magnitude of the resultant force isthe algebraic sum of the forces. Adding forces in the samedirection & along the same line. Adding forces in the oppositedirection & along the same line. F F direction is the same as the forces direction is the same as the larger force

  11. 1 Adding forces c Algebraic method (2-demensions) Forces in 2 dimensions can also be added algebraically. See this example:

  12. Example 8 C Use 1 cm to represent 25 000 N Length of AC = 6.9 cm resultant Resultant D B = 6.9  25 000 N = 173 000 N 100 000 N 100 000 N (with an  of 30 to either force) 60 A oil rig

  13. 3 N 3 4 4 N Example 9 2 forces, 3 N and 4 N, act at right angles to each other. Find the magnitude & direction of the resultant force. R = 5 N tan  = 5 N  =36.9 = 36.9  Resultant force is 5 N with 36.9 to the 4-N force.

  14. Q1 If Janice is pushing with... If Janice is pushing with a force of 60 N & Tommy ispushing with 80 N, what are the maximum and the minimum magnitudes of their combined force? Maximum force Minimum force A 100 N 80 N B120 N60 N C 140 N20 N D 160 N 40 N

  15. 3 N 60° 7 N Q2 Find the resultant of the... Find the resultant of the forces below with graphical method. (1 cm represents 1.5 N) The resultant force is ________ Nmaking an angle__________with the 3-N force. 8.9 43

  16. 2 2 + 120 120 Q3 2 ropes are used to pull... 2 ropes are used to pull a tree as shown. Find themagnitude of the resultant force acting on the tree. 120 N Magnitude of resultant = ____________ (Pythagoras’ theorem) = ________ N 170 120 N

  17. 2 Resolving forces into components We 've learnt how to combine 2 forces into 1. Let's see how to spilt (resolve) 1 force into 2.

  18. Fy Fx Resolving forces into components a Graphical method Suppose a force F is represented by OC : y component C Magnitudes of Fx & Fy can be measured directly. F component  x O

  19. Fy Fx Resolving forces into components b Algebraic method Magnitudes of Fx & Fy can be also be found by algebraic method: y C F 2= Fx2 + Fy2 (Pythagoras’ theorem) Fx = F cos  F Fy = F sin  Fy tan =  Fx O x

  20. Resolving forces into components b Algebraic method Simulation

  21. Example 8 2 tug boats pull an oil rig, each exerting a force of 100 000 N, they are at 60 to each other. Find the resultant force.

  22. Example 10 A 1-kg trolley runs down a friction-compensated runway at a constant speed. Find the component of weight of the trolley along runway, Hence find friction between trolley & runway. mass = 1 kg  = 20

  23. mass = 1 kg 20  = 20 W W Example 10 Component of weight along runway: Weight (W ) = mg = 10 N Component of weight along runway = 10 sin 20 = 3.42 N

  24. Example 10 Friction between trolley & runway: constant v friction component of weight along runway = 3.42 N  = 20 constant speed   net force on trolley = 0 friction = 3.42 N (up the runway)

  25. Example 11 180 sin 12 N 180 N 12 180 cos 12 N Resolve tension along each rope into 2 components  along line of travel  normal to line of travel Components in each direction are added

  26. Example 11 Total force along line of travel: 180 cos 12 + 170 cos 10 + 200 cos (10+15) = 525 N A 180 N 12 10 B 15 170 N 200 N C

  27. Example 11 Total force normal to line of travel: 180 sin 12  170 cos 10  200 cos (10+15) = 76.6N +ve A 180 N 12 10 B 15 170 N 200 N C

  28. = + R 5252 76.62 N Example 11 525 N  R 76.6 N 76.6 tan  = 525 = 531N  =8.30 Resultant force exerted by the skiers is 531 N at an angle of 8.30°to the line of travel.

  29. Forces in equilibrium Forces acting on 1-kg mass: weight 2 forces from 2 springs But it is at rest, 1-kg mass  net force = 0 Implication: in equilibrium Resultant of any 2 forces  equal & opposite to the 3rd force

  30. Q1 A 1-kg block on rest on a... A 1-kg block rests on a wedge. Use graphicalmethod to find the weight components of the block along & normal to the wedge. O C B 30 A What is the frictionacting on the block?

  31. Q1 A 1-kg block on rest on a... 5 Use a scale of 1 cm to represent _____ N. at rest O Length of OB = _____ cm 1.75 C  Weight componentalong the wedge = ______ N 8.75 B Length of OC = _____ cm 1 30 A  Weight componentnormal to the wedge = ______ N 5 Since the block is at rest, friction = weight component ____________ (along/normal to) the wedge = ________ N. along 5

  32. Q2 Find the weight components... Find the weight components of the block by algebraic method. at rest O C Weight componentalong the wedge = 10×_______ = ______ N B 30 sin 30 A 5 Weight componentnormal to the wedge = 10×_______ = _______ N cos 30 8.66

  33. Q3 What happens to the trolleys... What happens to the trolleys when they are released ? smooth pulley 3 kg 2 kg 20 30 A The 3-kg trolley moves downwards. B Both trolleys remain at rest. C The 2-kg trolley moves downwards.

  34. Example 11 Find the magnitude & direction of the resultant force the skiers exert on the speedboat? 180 N A 12 10 170 N 15 B 200 N C

  35. Example 12 2 spring balances are used to support an 1-kg mass. What are the readings of the 2 balances? F1 F2 30 1-kg mass

  36. F1 F1 F1 F2 F2 Example 12 30 30 30 10 N The mass is in equilibrium.  resultant force of F2 & weight is equal & opposite to F1 10 10  F1 = = 20 N = sin 30 sin 30 F1 10 10  F2 = = 17.3 N = tan 30 tan 30 F2

  37. F1 F2 Example 12 Alternative method by resolution of forces 30 The mass is in equilibrium.  net force = 0 10 N horizontal resultant = 0  vertical resultant = 0 horizontal: F1 cos 30 = F2  F1= 20 N; vertical: F1 sin 30 = 10 F2= 17.3 N

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