1. Analytical Toolbox Vectors and Applications By Dr J.P.M. Whitty

2. 2 Learning objectives After the session you will be able to: Explain two types of physical quantities Create graphical representation of vector quantities Resolve vectors Perform vector addition Use math software (or otherwise) to solve systems of vectors in order to answer examination type questions

3. 3 Scalar Quantities Definition A scalar quantity is described by a single # alone (i.e. magnitude); examples include: Length Volume Mass Time

4. 4 Vector Quantities Definition: A vector quantity is described by a magnitude AND a direction Force Velocity Acceleration Displacement

5. 5 Vector Quantities Cont… A vector quantity (such as force) can be depicted as an arrow at an angle to the horizontal, e.g. 10 Newtons acting at 30 degrees: 30 0 10N  F or more generally a force F @ an angle 

6. 6 Example 1. Which of the following are vector and which are scalar quantities: a) Temperature at 373K b) An acceleration downwards of 9.8ms -2 c) A weight of mass 7kg d) £500 e) A north-westerly in of 20 knots Scalar Vector Scalar Vector

7. 7 Vector Representations A (position) vector may join two points in space (A and B say), then, we may say: B A a Bold face They are usually written as: With the magnitude written as:

8. 8 Equal vectors Two vectors are equal if they have the same magnitude and direction B A a Here we say: D C c

9. 9 Equal and opposite vectors Two vectors are equal and opposite then they have the same magnitude but act in opposite directions (sometimes referred to as negative vectors) B A a Here we say: D C c

10. 10 Addition of vectors Any quantities can be added using the a parallelogram of triangular rules: Parallelogram rule: Vectors drawn from a single point Triangular rule: Vectors placed end to end Resultant vector

11. 11 Example #1: Find the resultant force of two 5N@10 o and 8N@70 o 5Units @10 o 8Units @70 o Measure R to give: 11.4N@42 o R 8Units @70 o 5Units @10 o

12. 12 Example #2: Find the resultant force of three 6N@5 o and 8N@40 o and 10N@80 o 6Units@5 o 10Units@70 o 8Units@40 o Solution 1: Apply the Parallelogram rule twice Measure R to give: 21.6@44 o R

13. 13 Example #2: Triangular rule Here we simply place the vectors end to end, thus: 6Units@5 o 10Units@70 o 8Units@40 o Measure R to give: 21.6@44 o R

14. 14 Sum of a number of vectors In general the triangular rule takes less construction and it is also easier extended to account for a number of vectors, thus: Let a be a vector from A to B, b be from B to C and so on… Then a+b+c+d, can be evaluated by formation of vector chain

15. 15 Vector chains The sum a+b+c+d, is constructed thus: A B C D E ab c d r Here we can say: Notice the pattern

16. 16 Example: Given that P,Q,R and S are point in three dimensional space, find the vector sum of Solution: This has the same pattern as previously, i.e. a connected path thus: Note: No need to draw the diagram the outside letters render the result so long as they are connected

17. 17 The null vector Suppose we consider another case where the resultant vector r=-e, we have: A B C D E ab c d e Here we now have: i.e. the same position: i.e. 0, the null vector, has no length & hence direction

18. 18 Class Examples Time Find the sum of the position vectors: Solution:

19. 19 Vector components The vector OP is defined by its magnitude r and its direction . It can also be defined in terms of the components a and b in the directions OX and OY, respectively. O  r b a P x y

20. 20 Unit vectors Hence OP=a (along OX) + b (along OY). If a unit vectors i,j (i.e. vectors of length unity) are introduced along OX and OY respectively then: r=a i + b j = i a + j b r= i rcos  + j rsin  Where a and b are the lengths along OX and OY, equal to the magnitudes of the original vectors

21. 21 Vector addition (analytic solution) The use of unit vector allows the calculation of vector addition analytically. Returning to the previous example #2, viz: Find the resultant force of three 6N@5 o and 8N@40 o and 10N@80 o Here the solution is to resolve the vectors into components and add them to give the overall result

22. 22 Example #2: analytic solution Letting the forces equal F 1, F 2 and F 3 respectively 1) F 1 = i rcos  + j rsin  = i6cos5 o + j6sin5 o = 5.977i+0.526j (3dp) 2) F 2 = ircos  + jrsin  = i8cos40 o +j8sin40 o = 6.128i+5.142j (3dp) 3) F 3 =i rcos  + j rsin  =i10cos70 o +j10sin70 o = 3.420i+9.397j (3dp) 4) Therefore adding the individual components: 5) r=15.525i+15.065j (3dp)

23. 23 Example #2: analytic solution The result is in Cartesian form, however we have been asked for the magnitude and direction of the resultant vector. To do this we must resort back to elementary trigonometry and Pythagoras, thus: O  r 15.065 b=15.065 15.525 a=15.525 P x y

24. 24 Example #2: Alternative notation The previous example can be evaluated using column or row vectors as follows:

25. 25 Example #2: MatLab This notation allows to solve such problems using math software such as MatLab

26. 26 Example # 3 Find the forces in the members of the structure; and evaluate the stress in each given that each bar is 50mm in diameter: 300N 60 o A B C

27. 27 Example # 3: Solution i(F AB )-i(F BC Cos60 o )+j(F BC Sin60 o )=j300 300N 60 o A B C j i i: F AB -F BC Cos60 o =0 j: (F BC Sin60 o )=300

28. 28 Example # 4 Find the forces in the members of the structure; and evaluate the stress in each given that each bar is 25mm in diameter: 500N 60 o A B C

29. 29 Example # 4: Solution Apply vector eqn, thus: i(F AB cos30 o )+ j(F AB sin30 o )- i(F BC cos30 o )+ j(F BC Sin30 o )=j500 500N 60 o A B C

30. 30 Examination type questions (Homework) 1. Find the value of the resultant force given that the following act on a specific point in a roof truss.

31. 31 Examination type questions (Homework) 2. Given that the following three forces act on a 12mm diameter bar: Find the resultant force on the bar [3], and evaluate the maximum stress that bar can experience.

32. 32 Examination Type Question Exploit symmetry conditions and find the stresses in each of red members the 20mm dia, steel members (E=200GPa). Hence or otherwise evaluate the resulting strains. [20 marks] 250 1m

33. 33 Examination Type: Solution Exploit symmetry thus: 250 A B C Now @ A:

34. 34 Examination Type Question: Strain value solutions These can be evaluated from the elasticity definitions as well! Note: the units here are of utmost importance

35. 35 Examination Type Question: Stress value solutions These can be evaluated from the elasticity definitions

36. 36 Summary: Have we met our learning objectives specifically are you now able to: Explain two types of physical quantities Create graphical representation of vector quantities Resolve vectors Perform vector addition Use math software (or otherwise) to solve systems of vectors in order to answer examination type questions