Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tutorial: Vectors

  1. Apr 17, 2008 #1
    Tutorial: Vectors

    About the Tutorial
    I see a lot of questions related to vectors on these forums, so I thought it would be a good idea to post a Tutorial on how to solve some of the easier questions related to Vectors. This Guide is quite long but should cover the basics. :bugeye:

    How will this thread work?
    I will post a short tutorial on a particular sub-topic on vectors, this will then be followed by a few question and answers. This thread has been made as an introduction to people who are new to vectors and to people who want to touch up on them.

    What you will need
    • A Basic Calculator.
    • Some knowledge of both Pythagoras' Theorem and Trigonometry.
    • Patience!
    NOTE: With vectors there is sometimes the option of a scale drawing; this tutorial does that cover that. I believe that by using the method I am about to show if probably the most flexible in terms of solving a problem, though in some cases a scale drawing may be necessary.

    _Mayday_


    Please contact me via PM if you feel there is anything that may enhance the quality, or make it more user friendly.
     
    Last edited by a moderator: Jun 5, 2008
  2. jcsd
  3. Apr 17, 2008 #2
    UNIT 1


    Part 1: Vector Quantities

    Vectors are quantities that have both magnitude and direction. Examples of vectors include Force, Acceleration and Velocity. When adding vectors you need to take their direction into account. When adding vectors that are on the same plain it is just a question of addition and subtraction as shown below. If two vectors are acting in the same direction, then they can be added together, and if they are acting in opposite direction one can be subtracted from the other.

    Acting in the same direction
    [​IMG]

    Acting in opposite directions
    [​IMG]

    Adding Forces at right angles

    When two vectors are acting at right angles to each other, you can apply Pythagoras' theorem. Below I will take you through an example, please not this will only apply to forces arranged at right angles.

    [​IMG]

    In the first image you see two forces being applied on an object. It is important to include arrows which show in which direction the force is being acted in. I will take you through this and then leave you a few questions in later posts, note that the answers will be under the question in white to see it simply highlight it. :smile:

    1. Firstly you need to arrange the arrows in order, in other words in vectors you can arrange them so that all the forces lead on from on another. So in this case you can move the 4.0 N to the right so that it follows on from the 3.0 N, not that the angle between them will still be a right angle, and this helps you determine the direction of the resultant force. The resultant force is from the point at which you start the vector to the point at which it ends in a straight line, as long as the other vectors follow on. Like this:

    [​IMG]

    As you can see (a) and (b) follow on from each other, and (c) the resultant goes from he beginning to the end in a straight line.

    2. Now you can apply your knowledge of Pythagoras' theorem. So, this is how I would do it:

    [itex]a^2 + b^2 = c^2[/itex]

    [itex]4^2 + 3^2 = c^2[/itex]

    [itex]16 + 9 = c^2 (25)[/itex]

    Remember now that you have [itex]c^2[/itex] you need to root it to get the resultant, do not forget this bit!

    [itex]\sqrt{25} = 5.0 N[/itex]

    Now if you have made the forces follow on from each other then the hypotenuse is acting in the same direction as the resultant force.

    Understand? If so then why not try some of the questions posted below?

    Part 2: Using Trigonometry

    If you know your trig, this shouldn't be too difficult. If you are unsure I have provided a link to the level of trg you need.

    Sometimes it is necessary to split vectors into Horizontal and Vertical components. This is important for example when you only have one angle and one force like the example below:

    [​IMG]

    At first, to some this may look impossible. I will take you through this one, and then set a few questions that you can try yourself.

    1. Write down what you have in terms of information. You have an angle, and you have the hypoteneus. Now you need to think which of sin, cos or tan fit this. In this case it will be sine, as you are looking for the opposite (or verticle)

    2.Now arrange it into an equation similar to this:

    [itex]\sin(45) = \frac{vertical}{12.0 N}[/itex]

    Now you can rearrange it to:

    [itex]\sin(45) \times 12.0 N = Vertical (8.5 N)[/itex]

    3. Now you can do this next bit two ways. One way would be to use Pythagoras' theorum as you now have two sides of a right angle triangle. I would do it in a similar way to 2. as if you have made a mistake in finding the vertical component, you can still get this right.

    4. Again, find out which of sin, cos or tan you will need to use. Now you are looking for the adjacent, so we will use cos.

    5. This is how I would do it:

    [itex]\cos(45) = \frac{Adjacent}{12.0 N}[/itex]

    Rearrange to:

    [itex]\cos(45) \times 12.0 N = Adjacent (8.5 N)[/itex]
     
    Last edited by a moderator: Apr 18, 2008
  4. Apr 17, 2008 #3
    Questions: Set 1

    Q1. Calculate the magnitude of the resultant force of these vectors

    a)
    [​IMG]

    b)
    [​IMG]

    c)
    [​IMG]


    Q2. A speed boat is directed due north at [itex]240ms^{-1}[/itex]. A side current is acting on the boat due west at [itex]53ms^{-1}[/itex]. Find the resultant force and the direction in which the force is directed?

    Q3. A football is rolled on a flat table due north at [itex]5ms^{-1}[/itex]. A side wind acts on the football due east at [itex]2ms^{-1}[/itex]. Calculate the resultant force and the direction in which the force is directed?
     
    Last edited by a moderator: Apr 18, 2008
  5. Apr 17, 2008 #4
    Answers: Set 1

    Q1. These are just a case of using [itex]a^2+b^2=c^2[/itex]

    a) [itex]5^2 + 10^2 = 125[/itex]

    [itex]25 + 100 = 125[/itex]

    [itex]\sqrt{125} = 11.2 N[/itex]

    NOTE: The answer of 125 is [itex]c^2[/itex] so you need to root your answer to get the resultant.



    b) [itex]6^2 + 17^2 = 325[/itex]

    [itex]36 + 289 = 325[/itex]

    [itex]\sqrt{325} = 18.0 N[/itex]



    c) [itex]110^2 + 250^2 = 74600[/itex]

    [itex]12100 + 62500 = 74600[/itex]

    [itex]\sqrt{74600} = 273 N[/itex]



    Q2. For the next 2 questions, as I advised a small sketch similar to ones int he previous question can help!

    a) [itex]240^2 + 53^2 = 60409[/itex]

    [itex]57600 + 2809 = 60409[/itex]

    [itex]\sqrt{60409} = 245.7 N[/itex]

    [itex]12.6 Degrees[/itex]



    Q3.

    a) [itex]5^2 + 2^2 = 29[/itex]

    [itex]25 + 4 = 29[/itex]

    [itex]\sqrt{29} = 5.4 N[/itex]

    [itex]21.8 Degrees[/itex]
     
    Last edited by a moderator: Nov 29, 2010
  6. May 30, 2009 #5
    Thanks mayday, i've been looking for vector equations like this.
     
  7. May 30, 2009 #6
    As with textbooks, It might be helpfull to use a different color to represent forces, and another color to represent the resultant force.
     
  8. Jul 29, 2009 #7
    This is a nice tutorial. I've got a lot to learn, I know the basics, but I want to be more in depth, if you have more info ready, or time to do more then please do so. Its nice to be able to reference it rather then knowing it off the top of your head. Although eventually would be nice to just know!
     
  9. Aug 12, 2009 #8
    This is useful, Mayday. Thank you. ^_^

    Could you please check the numbers in the answer section, though? It looks like the separate problems got mushed together, and it would be useful to have the correct numbers when checking the math.
     
  10. Oct 30, 2009 #9
    Thank you very much I appreciate any opportunity to get better at this. You are awesome, and the tutorial helped to clear up some issues I was having, now I can finally move on to something else.
     
  11. Jun 18, 2010 #10
    That's great, thanks.
     
  12. Jun 19, 2010 #11
    Whoah! I've got a question. So one force pushes thing A north, and another pushes thing A west. That means that thing A is actually going to have more force, and travel faster, if there are two forces acting on it rather than one? Holy s****!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Tutorial: Vectors
  1. Laser tutorials (Replies: 2)

  2. Color Tutorial (Replies: 5)

Loading...