2. Addition and resolution of forces

3. 1Which of the following schoolbags would you feel heavier to carry if the same number of books are put in them? AB

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

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

6. aGraphical method Adding forces F2F2 F1F1 F2F2 + F1F1 F2F2 F1F1 'Tip-to-tail' method resultant force

7. aGraphical method Adding forces F2F2 F1F1 F2F2 + F1F1 Parallelogram of forces method resultant force parallel

8. bAlgebraic method (1-dimension) 1Adding forces F2F2 F1F1 =F1F1 F2F2 +F F2F2 F1F1 =F1F1 F2F2 +F Adding forces in the same direction & along the same line. The magnitude of the resultant force is the algebraic sum of the forces. direction is the same as the forces Adding forces in the opposite direction & along the same line. direction is the same as the larger force

9. cAlgebraic method (2-dimensions) 1Adding forces Forces in 2 dimensions can also be added algebraically. See this example:

10. 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. Example 1

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

12. = 5 N tan  = 3 4  = 36.9  R  = 36.9  3 N 4 N 5 N Example 2 2 forces, 3 N and 4 N, act at right angles to each other. Find the magnitude & direction of the resultant force. Resultant force is 5 N with 36.9  to the 4-N force.

13. If Janice is pushing with a force of 60 N & Maximum force Minimum force A 100 N 80 N B 120 N60 N C 140 N20 N D 160 N 40 N Q1If Janice is pushing with... what are the maximum and the minimum magnitudes of their combined force? Tommy is pushing with 80 N,

14. Find the resultant of the forces below with graphical method. The resultant force is ________ N making an angle __________ with the 3-N force. Q2Find the resultant of the... 8.9 43  3 N 7 N 60° (1 cm represents 1.5 N)

15. 2 ropes are used to pull a tree as shown. Q32 ropes are used to pull... Find the magnitude of the resultant force acting on the tree. 120 N Magnitude of resultant = ____________ (Pythagoras’ theorem) = ________ N 22 120  170

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

17. 2Resolving 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. aGraphical method Resolving forces into components  C FxFx FyFy y x component O Suppose a force F is represented by OC : F Magnitudes of F x & F y can be measured directly.

19. bAlgebraic method Resolving forces into components Magnitudes of F x & F y can be also be found by  C FxFx FyFy y x O F F 2 = F x 2 + F y 2 (Pythagoras’ theorem) F x = F cos  F y = F sin  tan  = FxFx FyFy algebraic method:

20. 180 N 200 N 12 10 A B C 15 Example 3 170 N Find the magnitude & direction of the resultant force the skiers exert on the speedboat?

21. Example 3 12  Resolve tension along each rope into 2 components  along line of travel  normal to line of travel 180 cos 12  N 180 N 180 sin 12  N Components in each direction are added

22. 180 N 200 N 12  10  A B C 15  Example 3 170 N Total force along line of travel: 180 cos 12  + 170 cos 10  + 200 cos (10  +15  ) = 525 N

23. 180 N 200 N 12 10 A B C 15 Example 3 170 N Total force normal to line of travel: 180 sin 12  170 cos 10  200 cos (10  +15  ) =  76.6 N +ve

24. tan  = 76.6 525  = 8.30 N76.6 2 525 2 R += = 531 N Resultant force exerted by the skiers is 531 N at an angle of 8.30° to the line of travel. Example 3 76.6 N 525 N  R

25. A 1-kg block rests on a wedge. Q1A 1-kg block on rest on a... What is the friction acting on the block? Use graphical method to find the weight components of the block along & normal to the wedge. A B C O 30 

26. Q1A 1-kg block on rest on a... Use a scale of 1 cm to represent _____ N. Length of OB = _____ cm  Weight component along the wedge = ______ N 5 1.75 8.75 A B C O Length of OC = _____ cm  Weight component normal to the wedge = ______ N 1 5 Since the block is at rest, friction = weight component ____________ (along/normal to) the wedge = ________ N. along 5 at rest 30 

27. Weight component along the wedge = 10 × _______ = ______ N sin 30  cos 30  5 8.66 Q2Find the weight components... Find the weight components of the block by algebraic method. A B C O at rest 30  Weight component normal to the wedge = 10 × _______ = _______ N

28. A 1-kg trolley runs down a friction- compensated runway at a constant speed. Hence find friction between trolley & runway. Find the component of weight of the trolley along runway,  = 20  Example 4 mass = 1 kg

29. Weight (W ) = mg Component of weight along runway = 10 N = 10 sin 20  = 3.42 N W 20   = 20  mass = 1 kg W Component of weight along runway: Example 4

30. Friction between trolley & runway: Example 4 constant speed   net force on trolley = 0  friction = 3.42 N (up the runway)  = 20  constant v friction component of weight along runway = 3.42 N

31. AThe 3-kg trolley moves downwards. BBoth trolleys remain at rest. CThe 2-kg trolley moves downwards. Q3What happens to the trolleys... What happens to the trolleys when they are released ? smooth pulley 20  30  3 kg 2 kg

32. 2 spring balances are used to support an 1-kg mass. 30  1-kg mass F1F1F1F1 F2F2F2F2 Example 5 What are the readings of the 2 balances?

33. 30  F2F2 F1F1 Example 5 F2F2 F1F1 30  F1F1 10 N = sin 30  10 F1F1  F 1 = 10 sin 30  = 20 N = tan 30  10 F2F2  F 2 = 10 tan 30  = 17.3 N The mass is in equilibrium.  resultant force of F 2 & weight is equal & opposite to F 1

34. F2F2 F1F1 30  The mass is in equilibrium. horizontal resultant = 0 vertical resultant = 0 horizontal: F 1 cos 30  = F 2 vertical: F 1 sin 30  = 10  F 1 = 20 N; F 2 = 17.3 N  net force = 0 10 N  Example 5 Alternative method by resolution of forces