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Finding Magnitude

  1. Sep 6, 2004 #1
    I have a problem here. That I believe, I have half way figured out.

    Here is the problem shortened.

    Force A has a magnitude of 465 newtons and is directed due west.
    Force B has a magnitude of 350 newtons and is directed due north.

    I have to find the magnitude and direction of the resultant force A + B.

    So what I first did was draw a triangle that looks like the one below.

    I used the formula R = (The sguare root of) A^2 + B^2. To find the resultant. Was that correct? If so my answer was R = - 582 newtons. If not, then I'm back to the freaking drawing board.

    I then needed to find the degree of the angle. Pointed out in the other attachment. For that I used the formula: theta = sin^-1 (ho/h).

    So my answer for that was 37 degrees south of west. Did I do the right thing here? Did I even find the right angle?


    One of my biggest problems is figuring out what formual to use to find the angle I need. And how to figure out what angle it is I need to find.

    I would appreciate any help that would be provided.

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Sep 6, 2004 #2
    Ok well it's nice to know I'm at least going in the right direction.

    I applied the negative sign, to indicate it's negative direction which was south of west. I wasn't sure whether I needed that or not.

    But I am confused as to why the the angle I would be finding would be outside the triangle?

    Also my other big problem, is knowing when to use cos, sin, or tan to find the angle I need. Is there something, that could help me to know when I need to use which ever function is the right one for the angle I need?

    One last thing, just to check if I'm going in the right direction is. The above problem has a second part to it. Instead of B pointing North, B points south.

    So again I drew another triangle and figured out that the resultant = 306.1 N and according to my drawing that I attached, would the angle that I'm pointing to be the correct angle to find? And would I be correct in using sin again to find that angle, or would I use cos, or tan?

    Thanks again for the help, it is much appreciated
     

    Attached Files:

  4. Sep 6, 2004 #3
    You don't need to put the negative sign in front of R. The negative sign in this case doen't have any meaning in term of direction and i think it is wrong. To indicate the direction of vector R, the angle you have specified is more than enough.

    Sorry to realize now that the diagram that you and I have drawn is not right. the direction of the resultant force is wrong. Omit the diagram. This is the new one.

    if you want to give the direction in term of [tex]\theta[/tex];
    then
    [tex]tan\theta=\frac{465}{350}; \theta=53.0^0[/tex]
    then the direction is 53.0 degree west of north.

    if you want to give the direction in term of [tex]\alpha[/tex];
    then
    [tex]\alpha=90^0-53.0^0=37.0^0[/tex]

    if you want to give the direction in term of the horizontal, you can say that the resultant force makes a (90+53)=143 degree from the horizontal in anti-clockwise direction or makes a (360-143)=217 degree from the horizontal in clockwise direction etc.

    you can use the formula for sin, cos or tangen since you have got the R as well. but for safety reason, i suggest that you use the information provided to get the angle in case you have found your R wrong.
     

    Attached Files:

    Last edited: Sep 6, 2004
  5. Sep 6, 2004 #4
    Thank you. So in my second part, would I be correct in using sin again to find the angle needed?
     
  6. Sep 6, 2004 #5
    your diagram is wrong; should look like this one.

    since you prefer to use sin; then
    [tex]sin\theta=\frac{350}{582}; \theta=37.0^0[/tex] ; the resultant force is 37.0 degree south of west or is (90-37.0)=53.0 degree west of south.

    Rule for finding the direction of the resultant force :
    align the two vectors so that the head of one meets the tail of the other. the resultant force will point from the tail of the first vector to the head of the second one. this is how we get the diagram.do not change the original direction of the vectors.
     

    Attached Files:

    Last edited: Sep 6, 2004
  7. Sep 7, 2004 #6
    Ok now, I'm really confused. Why would B,when drawn, be placed above force A instead of below as I had originally drawn it? Oh wait, I think I know.....please let me know if I am correct... It would be drawn as you drew it because of the tail-to-head method. Am I correct in thinking that?

    If so, then my first drawing was wrong as well then as you have stated it.

    Ok, ok, well I think I'm getting somewhere now.
     
  8. Sep 7, 2004 #7
    Oh I see that perhaps I wasn't specific enough in the description of the second part.

    Force B is applied as a -B. And I'm to find the direction and magnitude of the resultant force A - B.

    So for this part I used the earlier square root function only subtracting instead of adding. And came up with a resultant of 306.1 N.

    To find theta, I thought I would use cos, but my calculator is giving me an error, so using cos apparently is not correct here. So I tried it using sin and again got an error. I'm guessing this is due to my resultant being 306.1 N but our class uses webassign and our answers require 3 significant digits.

    So in this case I am stumped again.

    Thank you for all your help you have given.
     
  9. Sep 7, 2004 #8

    Doc Al

    User Avatar

    Staff: Mentor

    A few comments. To add vectors graphically, draw a proper diagram. Start with the first force (A): the vector will be an arrow pointing west with a length proportional to 465 N. Now, starting at the tip (head) of the A vector, start drawing the vector for B: the tail of vector B starts at the head of A and points north. The resultant force is a vector that starts at the tail of A and ends up at the head of B. (If you are drawing it like most would, then A points left, and B points up. So the resultant points in a direction north of west. Draw this!)

    Once you've drawn this, you'll see you have a right triangle. Find the resultant like this: [itex]R^2 = A^2 + B^2[/itex].

    You can find the angle the resultant makes using some trig. Find the tangent of the angle.

    Of course you can also add the vectors using components:
    [itex]A_x = - 465[/itex]; [itex]A_y = 0[/itex]
    [itex]B_x = 0[/itex]; [itex]B_y = + 350[/itex]

    You take it from here.
     
  10. Sep 7, 2004 #9
    Thanks, I realized with the help of Leong that I wasn't drawing my diagrams properly. I wasn't using the tail-to-head method. So I straightened that out. And I have the first part completed.

    It's the second part, with which I have gotten stuck. I redrew my diagram. Which looks like the one below.

    My equation looks like

    R = (the square root of) 465^2 - 350^2
    R = 306.1 N

    Then however when I need to find theta I get errors on my calculator. My original thought was to use cos to find theta. However like I mentioned I get an error on my calculator.

    I know the error lies somewhere with my resultant, but I cannot for the life of me figure out why?
     

    Attached Files:

  11. Sep 7, 2004 #10

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand Leong's or your diagram. Why is B, a force pointing north, drawn as an arrow pointing down? (Did you change the definition of the forces from what you stated in your first post?) Redraw it.

    Use the equation I gave you. You are essentially finding the hypotenuse of a right triangle. Here it is again: [itex]R^2 = A^2 + B^2[/itex].

    Does an answer of 306 N even make sense? Think about it. Can the hypotenuse of a right triangle be smaller than the other sides?


    First get the resultant correct. Then worry about the angle.
     
  12. Sep 7, 2004 #11
    This is because R = SQRT(A^2 + B^2)

    For some reason you subtracted??

    I think your diagrams are confusing, if using North South East West coords, North is generally taken as upwards and west as to the left (just like a map).

    I have attached the diagram I used to solve this problem.

    I got R = 582 N, In the direction 37 degrees North of West
     

    Attached Files:

  13. Sep 7, 2004 #12
    I am drawing it south because the second part of the above problem states. Suppose a force of -B instead of B is applied. What then are the magnitude and direction of the resultant force A - B applied?

    I don't understand this: [itex]R^2 = A^2 + B^2[/itex] Please keep in mind it's been about 5 years since my last trig class, and my physics I class is only in it's second week. This is all greek to me.

    So I have no idea what does and doesn't make sense. When it's explained I can then start to grasp it but until then all of this is non understandable on my part.
     
  14. Sep 7, 2004 #13
    Please people stay with me here. Forget about force B pointing north!

    There is a second part to the problem which asks what is the magnitude and direction of the resultant force A - B if a force - B was applied instead of just B.
     
  15. Sep 7, 2004 #14
    It would have been clearer if you had stated this earlier.

    The only thing that would change in this case is the direction of the resultant force and therefore you would get 37 degrees South of West.

    R^2 = A^2 + B^2 is Pythagoras' theorem if you don't remember........... the square of the hypotenuse of a right triangle is the sum of the squares of the two adjacent sides.
     
  16. Sep 7, 2004 #15
    referring to post #9;
    do you know what the phytagorean theorem is all about ?
    you diagram in post #9 is correct.
    find the angle and magnitude using the geometry method. forget about the vector notation; i think it is confusing you.
     
  17. Sep 7, 2004 #16
    Well I did mention it, but that doesn't really matter. I'm not here to get nit picky with the help I'm getting. I appreciate any and all help I can get.

    However, why then would you not subtract A - B as the question asks?
     
  18. Sep 7, 2004 #17
    Yes I know what the pythagorean theorm is. And yes negative vectors are definitely confusing me. Because here I'm told to ignore it and then with another problem I am not supposed to ignore it. Let me just say I hate physics. If I didn't have to take this stupid class I never would have.
     
  19. Sep 7, 2004 #18
    You can think of -B as being a vector with the same magnitude as B, but in the opposite direction, then you can simply add the vectors as before

    That is A-B = A+(-B)

    See my diagram
     

    Attached Files:

  20. Sep 7, 2004 #19
    Ok. Thank you.
     
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