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Angle of a F*r calculation

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Force F= (-6.93 N) + (7.85 N) acts on a particle with position vector r = (4.66 m) + (1.36 m). What are (a) the magnitude of the torque on the particle about the origin and (b) the angle between the directions of and ?

    2. Relevant equations



    3. The attempt at a solution

    So I find (a) to be 46.0058 N*m which is correct and for (b) I use t/f*r

    So 46.0058/sqrt(6.93^2+7.85^2)*sqrt(4.66^2+1.36^2) = .9050610046

    sin-1(of that) = 64.8315006 = angle. But this is incorrect?
     
  2. jcsd
  3. Mar 16, 2008 #2

    G01

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    Remember this definition of the dot product?

    [tex]\vec{A}\cdot\vec{B}=|A||B|\cos\theta[/tex]

    Try using this to solve for the angle and see if it gives the same value. (I think the cross product formula should work, but for some reason I'm getting a different value for the angle in each one...)
     
  4. Mar 16, 2008 #3
    Does that work with a vector? I've been using t = rFsin(Theta)
     
  5. Mar 16, 2008 #4
    Only thing I could think of is maybe its in the wrong quadrant but that doesn't seem to be the problem.
     
  6. Mar 16, 2008 #5

    G01

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    Yes. It's the formula for the dot product between two vectors. Theta is the angle between the vectors.


    Also, you have to know that:

    [tex]\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y[/tex]

    in order to use the dot product formula to find the angle between the vectors.

    Let A=F and B=r and see what angle the dot product formula gives you.

    (Sorry if you haven't learned what I'm talking about here. I also got the same angle you did with the formula your using. So I'm trying to check that answer with another equation.)
     
  7. Mar 16, 2008 #6
    Really appreciate the help, not sure what its looking for :(
     
  8. Mar 16, 2008 #7

    G01

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    Well. I think the method your using should be correct. Here is what I suggest.

    If your answer for the torque is correct, and the angle you get from the torque formula is wrong, I suggest the following alternative method:

    Find the angle of the F vector from the x axis. (You should be able to do this using the components.)

    Do the same for the r vector.

    Now you should be able to subtract the two angles to find the angle between them.

    If you get the same answer as you got before. Chances are that your probably right, especially if your answer to part a) is correct. I suggest asking your instructor about this problem. It is possible that the answer in the back of the book is wrong.

    Also, this may seem simple after all of this, but we should make sure just in case: Are you sure the answer in the back of the book is in degrees and not radians?
     
  9. Mar 16, 2008 #8
    Its not in degrees, its actually online homework that changes the variables for each person. I know (a) is right because it told me so and it also said the units for (b) were degrees :(

    I tried it the way you suggested and again still wrong. I'll ask the instructor tomorrow but its really not a big deal I guess.
     
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