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Simple Simple Physics problem I need help with

  1. Jan 21, 2007 #1

    I am in a Physics 2 course and havnt taken physics in two years. I know this is an extremley easy problem but could some one lay out the steps for it. I figuerd I'd try to use this site as a sort of review to get me back in the swing of things. I appreciate it!
  2. jcsd
  3. Jan 21, 2007 #2
    Welcome to the forums. You'll find the people here very willing to help out, but one of the rules in the homework help section is that you first have to show what you've tried. Generally, the default format for homework help is
    1. statement of problem, all variables and given/known data
    2. relevant equations
    3. attempt at a solution

    For starters though, you might want to glance at a section of your text on breaking vectors into components; and/or perhaps, a quick simple trig review.
  4. Jan 21, 2007 #3
    Ok I see. IWell for starts is R hat a vector that starts at d and goes to c? I know how to use Trig to kind the components of the right triangles. Basically. How do I get started?
  5. Jan 21, 2007 #4
    Okay, finding those components of the right triangles would be a great start. R hat isn't a vector that starts at d and goes to c; it's the sum of those two vectors. In other words, vector C means 20 meters up and to the right (or more specifically, at an angle of 25 degrees to the x-axis. Vector D means 30 meters up and to the left (at that angle). R, standing for "Resultant" means to do both.

    incidentally, if you really know your trig, there's a quick formula that does it all in a step or two when you have two components (some particular law.) However, since you're reviewing, I think it may be more productive toward understanding if you break each separate vector into components. It doesn't matter what the units are on the vectors; you add them the same way. These problems might make a bit more intuitive sense if the displacement were replaced by force. Thus, you would have two forces acting on an object at the origin. The resultant is "these two individual forces would have the same effect as what individual force on the object."
    Last edited: Jan 21, 2007
  6. Jan 21, 2007 #5
    ok I assume by components you mean break the C and D vextors down in to Cx Cy, Dx Dy and then you can add Cx and Dx together and Cy and Dy together. I still don't inderstand how that gets the resultant vector. Here's what I have so far.

    Cy = 20*sin(25) = 8.45 *isn't x i hat, and y j hat if I remember right?
    Cx = 20*cos(25) = 18.13

    Dy = 30*sin(20) = 10.26
    Dx = 30*cos(20) = 29.19

    This doesnt look right to me.
  7. Jan 21, 2007 #6
    Good so far; you have a typo on the Dx. And, as you said, add the Cx and Dx together and the Cy and Dy together. However, note that Cx is to the right and Dx is to the left. Most people would define the right as the positive direction and the left as the negative direction.

    Once you have those two individual components of R, you'll need the pythagorean theorem to find the magnitude, and some more simple trig to find the angle. (draw a right triangle with those components.)
  8. Jan 22, 2007 #7
    Ok Heres what I have so far.

    C hat = (18.13)ihat +(8.45)jhat

    D hat = -(29.19)ihat+(10.26)jhat

    So for the first question in the picture...

    R hat = -(11.06)ihat + (18.71)jhat... Correct?

    Then using a right triangle and the components from R hat from above as the x and y components of that triangle I simply use Pythag's therom to find the hypotenous so...

    11.06^2 + 18.71^2 = c^2

    So the magnitude is 472.38??? seems large? Then how do I find the direction?
  9. Jan 22, 2007 #8
    that equals c^2; you forgot the square root.
    Use your components and tangent to find the angle. Starting at the origin, your triangle goes 11.06 to the left and 18.71 up.
  10. Jan 22, 2007 #9
    Ok so the answeres I got for this problem are...

    R hat = -(11.06)ihat + (18.71)jhat

    Magnitude = 21.73
    Direction = 1.69 degrees north of west....

    Look okay?
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