AS Physics Unit 7 Forces in Equilibrium

1. AS Physics Unit 7 Forces in Equilibrium Ks5 AS Physics AQA 2450 Mr D Powell

2. Forces in Equilibrium 7

3. Chapter Map

4. 7.1 Vectors and Scalars What is a vector quantity? How do we represent vectors? How do we add and resolve vectors? Representing a vector Addition of vectors using a scale diagram Addition of two perpendicular vectors using a calculator Resolving a vector into two perpendicular components Spec 3.2.1

5. M A Definition.... TASK: Make a note of each quantity, then look at the units and give a reason why they have been sorted as such..

6. Displacement and velocity A runner completes one lap of an athletics track. What distance has she run? 400m What is her final displacement? If she ends up exactly where she started, her displacement from her starting position is zero. What is her average velocity for the lap, and how does it compare to her average speed?

7. What did you decide? Despite the trip moving at various speeds, because it ended up at the starting point, the average velocity was zero. This will always be true when the final displacement is zero.

8. Vector equations An equation is a statement of complete equality. The left hand side must match the right hand side in both quantity and units. In a vector equation, the vectors on both sides of the equation must have equal magnitudes anddirections. Take Newton’s second law, for example: force = mass × acceleration Force and acceleration are both vectors, so their directions will be equal. Mass is a scalar: it scales the right-hand side of the equation so that both quantities are equal. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in ms-2. The units on both sides must be equal, so 1N = 1kgms-2.

9. Displacement vectors Harry and Sally are exploring the desert. They need to reach an oasis, but choose to take different routes. • Harry travels due north, then due east. N • Sally simply travels in a straight line to the oasis. When Harry met Sally at the oasis, they had travelled different distances. However, because they both reached the same destination from the same starting point, their overall displacements were the same.

10. Vector notation A scalar quantity is often represented by a lower case letter, e.g. speed, v. A vector quantity can also be represented by a lower case letter, but it is written or printed in one of the following formats to differentiate it from the scalar equivalent: y The value of a vector can be written in magnitude and direction form: 6 e.g. v = (v, θ) a Or as a pair of values called components: x 8 8 e.g. a = (8, 6) or 6

11. Vector addition Displacement vectors can always be added ‘nose to tail’ to find a total or resultantvector. y b This can be done approximately by scale drawing: a 10 a + b x 7 It can also be done by calculation, breaking each vector down into perpendicular components first and then adding these together to find the components of the resultant: 2 -2 0 c + d = + = 3 2 5

12. Calculating a resultant When adding two perpendicular vectors, it is often necessary to calculate the exact magnitude and direction of the resultant vector. This requires the use of Pythagoras’ theorem, and trigonometry. For example, what is the resultant vector of a vertical displacement of 3km and a horizontal displacement of 4km? magnitude: direction: R2 = 32 + 42 tan θ = 4/3 R θ = tan-1(4/3) 4km R = √ 32 + 42 = 53° = √ 25 θ = 5km 3km

13. Vector components vertical component horizontal component Just as it is possible to add two vectors together to get a resultant vector, it is very often useful to break a ‘diagonal’ vector into its perpendicular components. This makes it easier to describe the motion of an object, and to do any relevant calculations.

14. Calculating components A vector can be separated into perpendicular components given only its magnitude and its angle from one of the component axes. This requires the use of trigonometry. For example, what are the horizontal and vertical components of a vector with a magnitude of 6ms-1 and a direction of 60° from the horizontal? cos60° = x / 6 x = 6 × cos60 sin60° = y / 6 y = 6 × sin60 6ms-1 y = 3ms-1 = 5.2ms-1 60° x

15. M Pythagoras & Adding Vectors TASK: Now try out the same techniques for two more triangles; opp = 2m, adj = 8m opp = 1.5m, adj = 4m Answers  = 14 , R =8.25m  = 20.6 , R =4.27m

16. M Applications... Answers  = 67.83 , R =58.3m  =39.4 , R =65.6ms-1 TASK: Now try out the same techniques for two more triangles; What is R &  if opp = 54N, adj = 22N What is adj &  if opp = 54ms-1, R = 85ms-1

17. T Splitting a vector into components... TASK: Put on your maths brain and try and explain using trig why Vx is related to cos and Vy is related to sin

18. Resolving in a context.... Answers Vy =8.8ms-1 Vy =70.7N Vx = 70.7N TASK: A bird is flying at a speed of 15ms-1 at an angle of 36 to the horizontal. What is its vertical speed. A javelin is thrown at a resultant force of 100N at a 45  angle. What are the horizontal and vertical forces on it.

19. P Questions to try on your own.... Answers .....

20. More Trig Practice • How many radians in a circle? • How many radians in quarter of a circle? • Work out  in degrees for a right-angled triangle if hyp = 5, opp = 4, adj = 3? • Work out opp if  = 450 and the adj = 7cm? • Work out tan if opp = 53.0m, adj = 42.0m? • Work out opp if  = /2 and the adj = 3cm? • Work out  in radians for a right-angled triangle if hyp = 5, opp = 4, adj = 3? • Rearrange sin=opp/hyp to make  the subject. • Work out sin (/2) =, sin (2) = , cos () = , cos(cos-1(/4)) = , tan(/4) = • Use formulae & numbers from your calculator to prove that;

21. Vector Problems 1 Question 1 • Both sides contribute as vectors so double it to If using only pythag do simply 2 x 15N Sin20 = 10.26 = 10.3N • label upper part of triangle as a. • Hence a2 = (152+152-2x152Cos140)0.5 = 28.19N, then half for triangle. To 14.1N. Then pythag (152-14.12)0.5 = 5.12N • (x2) = 10.23N a c b

22. Vector Problems 1 Question 2

23. Vector Problems 1 Question 3 3) A mass of 20.0kg is hung from the midpoint P of a wire. Calculate the tension in each suspending wire in Newton’s. Assume g = 10ms-2. (Hint resolve…..) • Use the idea of point of equilibrium. The forces must balance vertical and horizontal. • Weight balances the tension so for each wire Use the 2T Cos70 = 200 T Cos70 = 100 T = 100 / Cos70 = 292.38N

24. Vector Problems 1 Question 5/6 Magnitude = sqrt(35002+ 2792) = 3511.1m = 3510m Direction = tan-1 (279/3500) = 4.557° = 4.6° south from vertical 6) Vertical motion: 21 sin (43°) = 14.32 ms-1 Horizontal motion: 21 cos (43°) =15.358 ms-1 S = 3 x 15.358 ms-1 = 46.1 ms-1 3500km 0.279km

25. Vector Problems 2 Question 2 Use the parallelogram method to resolve the forces acting on this object placed. (Hint employ both cosine & sine rule). (4 marks)

26. Vector Problems 2 Question 2 Use the parallelogram method to find the resultant acting on this object and angle. (Hint employ both cosine & sine rule). (4 marks)

27. Vector Problems 2 Question 3 3) Two forces of magnitude 10.0N and F Newton’s produce a resultant of magnitude 30.0N in the direction OA. Find the direction and magnitude of F. (2 marks) (Hint use Pythagoras)

28. Vector Problems 2 Question 4 The graph shows a part completed vector diagram. You task is to find out the vector R by mathematical analysis. The vectors A & B are shown in both coordinate and bearing formats. Show working for both a mathematical method (2 marks) and drawn out scaled method. (2 marks) Hint: this is not as hard as it might initially look!

29. Vector Problems 2 Question 4 - stage 1 Finding the components of vectors for vector addition involves forming a right triangle from each vector and using the standard triangle trigonometry. The vector sum can be found by combining these components and converting to polar form.

30. Vector Problems 2 Question 4 – stage 2 After finding the components for the vectors A and B, and combining them to find the components of the resultant vector R, the result can be put in polar form by

31. Vector Problems 2 - Question 5

32. 5a N

33. 5b N N

34. 6) Forces of 60.0N and F Newton’s act at point O. Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks)

35. Vector Problems 2 - Question 7 Forces of 60.0N and F Newton’s act at point O. Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks) c2 = a2 + b2 - 2abCosC

36. Using the Sine or Cosine Rule 30N In this problem we can solve it with other methods of trigonometry. Make a triangle of vectors where we are trying to find length c and angle A; b = 30N 120 A a = 60N c = F Sine Rule Use reciprocal version of above SinC / c = SinA / a (60N x Sin 120) / 79.4N) = SinA Sin-1 (60N x Sin 120) / 79.4N) =A A = 40.876 = 41  Cosine Rule c2 = a2 + b2 - 2abCosC c2 = 60N2 + 30N2 – 2x60Nx30N xCos120 c2 = 4500N2 +1800N2 c2 = 6300N c = 79.4N c = F = 79.4N

37. 7) Simultaneous Equations (A*+) If this problem is to be solved mathematically we must establish two formulae. One in the vertical and one in the horizontal; 30N The key is knowing that the resultant force is 30N at 0º. Hence; Resolving Horizontally; 30N = FCos + 60NCos(360-120) 30N = FCos + 60N x -0.5 30N = FCos - 30N 60N = FCos  Eq 1 Resolving Vertically; 0 = FSin + 60NSin(360-120) 0 = FSin + 60NSin(240) 0 = FSin + 60N x 3/2 0 = FSin -51.96N 51.96N = FSin Eq 2 So by dividing Eq 2 by Eq 1 FSin / FCos  = 51.95N/ 60N tan  = 0.866  = 40.89 F = 60N / cos  = 79.4N Or F = 51.96N / sin  = 78.96N

38. Wires Example - Correct So by turning it on its side we can use the same formulae (which is confusing but correct) right side Left side

39. WRONG!

40. Alternative Can do by using (90- ) as angle; If we take the outside angle instead of the inner angle we can do this and use the triangle; (90-)

41. 7.2 Balanced Forces Why do we have to consider the direction in which a force acts? When do two (or more) forces have no overall effect on a point object? What is the parallelogram of forces? Equilibrium of a point object Testing three forces in equilibrium Practical - Coplanar forces

42. S Why are the objects not moving...

43. M Equilibrium Point of an Object

44. M Example 1

45. M Example 1

46. M Example 2

47. M Practical... • Assemble the equipment shown below. Use some weights of different types and measure the angles shown. • The basic system will adjust in angle to whatever masses you add. • The tension in each string must equal the force exerted by W1 or W2. This enables the problem to be solved by resolving • Try it practically so get a range of angles & weights then prove the equilibrium.

48. M Example 3

49. M Example 1... If you think that F1 & F2 must resolve and their components in H & V directions must balance with the weights to make the point P be in equilibrium. Define the problem as shown below with two new angles and triangles. Make sure you use the correct angles! Using the following figures the problem is very simple... W1 = 5N W2 = 3N W3 = 6N 1 = 150 2 = 124 F1 F2 F2sin F1sin   F2cos  F1cos 

50. M Example 1... • Using the following figures the problem is very simple... • W1 = 5N W2 = 3N W3 = 6N • 1 = 150 2 = 124 • =60  = 34  F1 F2 F2sin F1sin   F2cos  F1cos 