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Analytical Toolbox

Learn how to explain and create graphical representations of vector quantities, resolve vectors, perform vector addition, and solve systems of vectors using math software.

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Analytical Toolbox

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

  2. Learning objectives • After the session youwill 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. Scalar Quantities • Definition • A scalar quantity is described by a single # alone (i.e. magnitude); examples include: • Length • Volume • Mass • Time

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

  5. F or more generally a force F @ an angle   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: 10N 300

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

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

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

  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 D c a A C Here we say:

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

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

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

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

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

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

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

  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. y P r b  x O a

  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. 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@5o and 8N@40o and 10N@80o Here the solution is to resolve the vectors into components and add them to give the overall result

  22. Example #2: analytic solution • Letting the forces equal F1, F2and F3 respectively • F1= i rcos+ j rsin = i6cos5o+ j6sin5o = 5.977i+0.526j (3dp) • F2= ircos+ jrsin = i8cos40o+j8sin40o = 6.128i+5.142j (3dp) • F3=i rcos+ j rsin=i10cos70o+j10sin70o = 3.420i+9.397j (3dp) • Therefore adding the individual components: • r=15.525i+15.065j (3dp)

  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: y P r b=15.065  x O a=15.525

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

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

  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: C 60o B A 300N

  27. Example # 3: Solution • i(FAB)-i(FBCCos60o)+j(FBCSin60o)=j300 i: FAB-FBCCos60o=0 j: (FBCSin60o)=300 j C i 60o B A 300N

  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: B 60o 500N A C

  29. Example # 4: Solution • Apply vector eqn, thus: • i(FABcos30o)+ j(FABsin30o)- i(FBCcos30o)+ j(FBCSin30o)=j500 B 60o 500N A C

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

  31. Examination type questions (Homework) • 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. 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] Examination Type Question 1m 1m 250 250

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

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

  36. Summary: • Have we met our learning objectives specifically are younowable 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

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