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Another Work problem

  1. Oct 18, 2004 #1
    A force F = (11.5i + 9.62j + 13.2k)kN acts on a small object of mass 109g. If the displacement of the object is d = (5.63i + 3.81j) m, calculate the work done by the force.

    I know W=F*d*cos theta so W=101.4 cos theta
    How do I find theta? Do I use the inverse tangant of the F and the d?
  2. jcsd
  3. Oct 18, 2004 #2
    The work done is defined by the dot product (or inner product) between the force vector and the displacement vector. In the cartesian coordinate system, we have

    W = F_x*d_x + F_y*d_y + F_z*d_z

    where F_x is the x-component of the force, d_x is the x-component of the displacement and similarly for others.

    Best regards,
  4. Oct 18, 2004 #3
    Right, got that part, I think.
    W=11.5*5.63 + 9.62*3.81 + 13.2*0 = 101.4 cos theta
    My question is how to find theta. would theta be the inverse tangant of 36.65/64.75?
  5. Oct 18, 2004 #4

    Doc Al

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    Staff: Mentor

    W=11.5*5.63 + 9.62*3.81 + 13.2*0 = 101.4, not 101.4 cos (theta)!

    There are (at least) two ways to calculate work, depending upon what you are given:

    (1) W = F*d*cos(theta), is good if you are given the magnitude of the force and displacement and the angle between them.

    (2) W = F_x*d_x + F_y*d_y + F_z*d_z, is good if you are given the components.

    When you use method #2, theta is not needed.
  6. Oct 18, 2004 #5
    It says that the answer is not 101.4

    And the next questions says What is the angle between F and d?
  7. Oct 18, 2004 #6

    Doc Al

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    Staff: Mentor

    Check the units. The force was given in kN.

    If you know W, F, and d, then you can find theta using W = Fd cos(theta). F and d are the magnitudes of the vectors.
  8. Oct 18, 2004 #7
    Ahh that makes sense. I was trying to figure it out without first finding the work. Thanks
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