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Unit 1
Mechanics

Vectors
Topics Covered:

A vector is a directed line segment - a line with an arrow head at the end indicating a direction

When writing a vector we place an arrow on top of the symbol to distinguish it from its scalar counterpart.

The symbol for displacement is

Adding Vectors

A. Collinear Vectors:

These are vectors that are found along the same line of action: East - West or North - South.
When adding collinear vectors one must ensure that they are all pointing in the same direction by reversing the direction of one or more and changing the sign in front of the magnitude of the vector.

Example: Mr. Forget drives 5.0 km East from home to a gas station and then back West to the local bakery for another 2.0 km.  What is Mr.. Forgets' total displacement?

Solution:
(Here I have omitted the arrows on top of d1 and d2 for ease of editing but keep in mind that they are vectors. I will use "bold" characters instead of arrows to represent vectors.)

Given:
d1 = 5.0 km [E]
d2 = 2.0 km [E] = -2.0 km [W]

Find: dt    

Solution:   dt  = d1 + d2 = 5.0 km [E] + (-2.0 km [W] ) = 3.0  km [E]
Please note that the total distance is still 7.0 km.

B. Non-Collinear Vectors

When vectors are in the same plane (but not along the same line of action) they can be added using three different methods:

The general rule for adding vectors regardless of the method is: "add vectors from tail to head". 
 This means that the resultant VT  in the following illustration is the sum of the vectors V1 and V2 and is drawn from the tail of the first vector (V1) to the head of the last vector (V2). 
Standard Vector Direction

Note that VT is in a direction other that the standard North-South-East-West directions. To report this type of vector direction we make use of the nautical directional system (the compass).  The diagram below demonstrates how to report a vector direction properly using standard format.  The vector has direction 600 East of North (or 300 North of East) and a magnitude of 10 m.  In other words, to draw this vector using a protractor and a ruler, we started measuring 600 from the North position and we moved towards the East position.  In standard notation we report vector d as: =  10 m [N600E].
Note: we can also report this vector as = 10 m [E300N].


Method 1. Graphical Technique.

This technique requires very good accuracy when drawing vectors and  proper scaling to get precise results..

Step by step approach:

  1. Choose a scale according to the given information
  2. Start with a set of reference axis and draw the first vector using a ruler to draw its magnitude to sale and a protractor to report its direction properly as illustrated above.
  3. At the head of the vector draw a small cross.  This is a new reference axis to guide you in drawing successive vectors.
  4. Draw the second vector (and the third if necessary) using steps 1 to 3.
  5. Draw the resultant vector from the tail of the first vector to the head of the last vector.
  6. Use a ruler and protractor to report the magnitude and direction of the resultant vector properly.

Example:

A pigeon flies 5 km [N300E], then 2.5 km [E400S]. Calculate its final displacement.

Solution:

Step 1 - set a scale -- scale = 1cm = 1 km
therefore: a = 5 cm  [N300E], and b =  2.5 cm [E400S].
Step 2 - draw vectors a and b to scale
Step 3 - join the tail of vector a to the head of vector  b to find the resultant vector c.
Step 4 - measure resultant vector b using a ruler and convert to real-time units using your scale.
Step 5 - use a protractor to measure and report the direction of resultant vector c.
Step 6 - Conclusion - The final displacement  of the pigeon is c = 5.2 km [E300N]

Figure 1 - Steps 1 to 4

Figure 2 - Steps 5 & 6

Method 2. Mathematical

A. Sine-Cosine Law
i) Review

A) Pythagoras's Theorem & Trigonometry

C2 = A2  + B2

 This formula can only be used in right angle triangles.  In other words, it can be used only when vector A is perpendicular to vector B
The three basic trigonometric identities that can be derived from a right angle triangle whose included angle is q, are:

B) Sine and Cosine Laws

 This triangle is a generalized isosceles triangle where none of the sides are perpendicular to one another. Note that we use lower case letters to represent the sides of the triangle and upper case letters to represent the included angles.

For triangles that are NOT right-angle triangles we can use modified trigonometric laws:

The sine Law

The Cosine Law

Example: What was the total displacement of a balloon that flew 7 km [NE] and then encountered a wind that displaced it 3 km [S]. Use the sine/cosine laws to solve this problem.

Solution:

We start with a quick sketch

  1. Apply the cosine Law to find dtotal


where: a = 3 km [S]; c = 7 km [NE]; b = dtotal
and B = 450
therefore: b2 = (3)2 + ( 7)2 - 2(3)(7)(cos(450))
and b2 = 28.3  and b = 5.32 km
this is the magnitude of dtotal

2. Apply the Sine Law to find the value of the angle q


cross multiply and solve for sinq

= 0.398
q = invsin(0.298) = 23.500
 the vector  dtotal has direction (450 + 23.50)
3. final statement
\ dtotal = 5.32 km [N68.50]

B. Components
(click here to go to detailed notes using vector components)