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Quick force question

  1. Jun 23, 2005 #1
    Points A and B have position vectors (3i + j)m and (5i + 4j + 2k)m respectively. A particle moves from rest at the point A to the point B under the action of a constant force F newtons only. Given that the work done by the force in moving the bead from A to B is 34 Nm, find F.

    I've found that [itex]|\textbf{F}| = \sqrt{68}[/itex], which is correct, but I'm supposed to find it in vector form: [itex]\textbf{F} = 4\textbf{i} + 6\textbf{j} + 4\textbf{k}[/itex]. How do I do that?
  2. jcsd
  3. Jun 23, 2005 #2
    Draw a vector from point A to point B, this is vector F. Now break it down into its directional components
  4. Jun 23, 2005 #3
    Isn't that the position vector of B relative to A?
  5. Jun 23, 2005 #4
    Yes, thats where the force acted on the particle.
  6. Jun 23, 2005 #5
    Oh! I see what you're getting at:
    \textbf{F} = \sqrt{\frac{68}{17}} (2\textbf{i} + 3\textbf{j} + 2\textbf{k})

    And that gives the required answer. Thanks! :biggrin:
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