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Find Missing Vector

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data

    When adding Vectors A & B; Vector A = 26a 35 deg Resultant = 50a 12 deg

    2. Relevant equations

    None, no idea how to start

    3. The attempt at a solution

    Identify Vector B

    Yes, first class on Physics and I am stumbling big time
     

    Attached Files:

    Last edited: Aug 29, 2011
  2. jcsd
  3. Aug 28, 2011 #2

    PeterO

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    Do you understand what "Vector A = 26a 35 deg" means?
     
  4. Aug 28, 2011 #3
    Well, Vector A is a measure of 26a, with A being a variable which could be pounds, millisecond, etc and it is at 35 degrees.

    Am I on track?

    It has been 40 years since I finished college, and now taking a crash course in Physics since my son has his first class. He has given up out of frustration and gone to bed. I want to give him a hand, but it is a lot of work trying to remember Trig along with understanding Physics.
     
  5. Aug 29, 2011 #4

    PeterO

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    Getting there.

    at 35o to what? ie. making an angle of 35o with what?
     
  6. Aug 29, 2011 #5
    Well, since I am adding 2 vectors, I am guessing they are going in the same direction, but at different degrees.
     
  7. Aug 29, 2011 #6

    PeterO

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    Best way here is to draw the two vectors with a common start point - it would look like the hands on a clock.

    The 26a 35 deg could be an hour hand pointing at approx 2'oclock
    The 35a 12 deg could be a minute hand pointing at about 13mins past

    what vector is needed to go from the end of the first to the end of the second?
    You could use the cosine rule, or do x- and y- components of them to work it out.
    The angle will be negative, that is for sure.
     
  8. Aug 29, 2011 #7
    PeterO, no question you are trying to help. Problem is I am just now re-learning the concepts and terminology.

    I read the "why aren't I getting anwers" post, and thankfully you were willing to provide asssistance. Clearly, I am not ready to be posting on this forum since I am trying to solve a problem I do not understand.

    My hopes were for a quick fix that someone would show me the solution steps, then I would discuss with my son.

    Clearly this is not the intent of this forum and for this reason, I appreciate your willingness to provide assistance.
     
  9. Aug 29, 2011 #8

    ehild

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    Anester,

    Look at the books and notes of your son. They had to learn "vector addition". What method did they use? Is it geometry, drawing a vector triangle or parallelogram or using x and y components of the vectors?

    ehild
     
  10. Aug 29, 2011 #9

    PeterO

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    One system of defining vector - especially in 2 dimensions - is like polar co-ordinates.
    You give the amplitude, and the angle above the positive x-axis of the Cartesian system.

    I wonder whether 26a 35 deg was really meaning 26 (amplitude) 35 deg
    As I said the hour hand on a clock at 2:00 pm does a pretty good impression of [26, 35o] , especially of it is 26 cm long. [the angle is slightly wrong] note Cartesian co-ordinates (x,y) are written in round brackets. Polar Co-ordinates eg [25,34o] are written in square brackets.

    In some courses you learn how to work with polar co-ordinates as polar co-ordinates. Other wise it is common to convert them to a Cartesian equivalent, add them then convert back.
     
  11. Aug 29, 2011 #10
    ehild,

    This is an introductory Physics class at his Community College. First class was last week and he has not picked up the book yet. He "lost" his notes. Hit me at 8pm tonight with the problem, now it is 1am. Class at 9:30am so I wanted to get him some confidence before he turned in his first assignment.

    Fortunately, I figured out the Resultant steps; have a few hours left to get the Equilibrant. Will have to give up on the missing Vector question which are 2 of the 9 questions. Not terrible for his first assignment. Obviously, we are going to talk about getting an earlier start on these assignments.
     
    Last edited: Aug 29, 2011
  12. Aug 29, 2011 #11
    PeterO,

    I won't begin to attempt to use words to explain my ignorance, so this is the problem directly from the worksheet:

    PROCEDURE
    1. Use appropriate math processes to provide missing information in the table and requested information in the problems following.

    DATA -- Be sure to include appropriate units of measure.

    VECTOR A VECTOR B CALCULATED CALCULATED
    RESULTANT EQUILIBRANT
    6 26 a 35*________ 50 a 12* ________
     
  13. Aug 29, 2011 #12

    rl.bhat

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    You should have given the full text of the problem.
    Is it given in the problem that the angle between vector A and vector B is 35 degrees and the angle between the vector R and vector B is 12 degrees?
     
  14. Aug 29, 2011 #13
    rl.bhat,

    Actually figured out how to attach the assignment - Vectors.doc.
     
  15. Aug 29, 2011 #14

    ehild

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    Equilibrant is the opposite of the original vector. The magnitude is the same and it makes an angle of 180° with the original. So its angle is 180°+ the angle of the original vector.
    As for vector addition, the simplest way is using the coordinates. They are the projections of the vector to the x and y axis (ax and ay ). See figure attached.
    If you have vector with length a and angle [itex]\alpha[/itex] with respect to the positive x axis, ax=acos([itex]\alpha[/itex])
    and ay=asin([itex]\alpha[/itex])

    You get the components of the sum of two vectors a and b by adding the components :c=a+b ---> cx=ax+bx, cy=ay+by.

    If you have the components of a vector, the length is √(ax2+ay2)
    and tan([itex]\alpha[/itex])=ay/ax.

    Can I give a piece of advice: no reason to help if your son does not listen to the lectures. He must make notes. Do not try to help without seeing what they learn and how. You might explain on a different way than his teacher does, and you will confuse the boy.

    ehild
     

    Attached Files:

  16. Aug 29, 2011 #15
    ehild,

    Your advice is well taken - guessing you may be a Professor with first hand knowledge.

    I won't be able to be a long term solution, just wanted to kick start his first weeks. You are correct, I cannot replace his Prof.

    I appreciate your detail and will be able to work with it. One question - do you know of a website that has an Equilibrant calculator, then I can validate my steps are correct?
     
  17. Aug 29, 2011 #16

    ehild

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    You can add two numbers without calculator, do you not? Add or subtract 180° from the angle of the resultant vector (so as the result is less than 360°) to get the angle of the equilibrant. The magnitudes are the same.

    ehild
     

    Attached Files:

  18. Aug 29, 2011 #17
    Got it, and my highest gratitude for the Physics Forums team assistance.

    ehild, loved the jpg; will actually get a few hours extra sleep with the assistance from the forum.

    Thanks to all
     
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