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Help with determining work from unit vectors

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    [tex]\sqrt{}[/tex]a2+b2 = c
    w = f s

    3. The attempt at a solution

    Pythagoreoms theorem on -8.6 m and -3.0 m to find the distance traveled = 9.11m

    Pythagoreoms theorem on 34 N and -38 N to find total force = 60 N

    w = fs
    w = (60N)(9.11m)
    = 714 J

    I'm not sure what I am doing wrong in this problem, I know the answer I got is incorrect though.
    Last edited: Jul 6, 2008
  2. jcsd
  3. Jul 6, 2008 #2


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    Hi joe_,

    The formula you have for work is not quite right. The work done is not just the magnitude of the f vector times the magnitude of the displacement vector (except in special cases); what is the full formula?
  4. Jul 6, 2008 #3
    Oh yes, w=f s cos[tex]\alpha[/tex]

    So I have to also find the direction of the displacement and take the cos of it.

    So I would have to do

    -8.6 / 9.11 (adj/hyp = cos [tex]\alpha[/tex])
    -.944 invcos

    19.26 deg.


    w= f s cos 19.26?
  5. Jul 6, 2008 #4
    I know this may also be semi off topic, but say something is thrown vertically upwards and the formula reads

    w = f s cos[tex]\alpha[/tex]

    So w will always = 0 because the cos of 90 is 0

    Or am I doing that wrong?
  6. Jul 6, 2008 #5


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    Hi joe_,

    I don't think that angle is right. The angle alpha is the angle between the force vector and the displacment vector. If I'm reading your work right, you found the angle the displacement vector makes with one of the coordinate axes.

    But actually I was thinking of a different formula that is very useful when the force and displacement vector are given in unit vector notation. The work is the dot product of the force and displacement vectors:

    W = \vec F \cdot \vec d

    Are you familiar with the dot product of two vectors? Using the dot product form of the work equation makes this problem very straightforward.

    (On your other post, to find the work done by gravity during an upwards motion, the f is downwards, and during the motion upwards the displacment is upwards, so alpha would be 180 degrees, since that is the angle between force and displacement.)
  7. Jul 6, 2008 #6
    I'm not familiar with the dot product method at all? Also, I think between my last question and my mess up concerning taking the angle from the coordinate axis I found my errors. I'm ALWAYS taking stuff from the axis, and not between the displacement and force. Figuring that out is helping a bit.

    So, doing this over I already know the angle from the -x axis going toward the -y axis so to speak (19.26 deg).Also, I know that the displacement of 60N. To find the direction:

    Cos x = adj/hyp
    34/60 = .5667
    inv cos .5667
    55.48 (in the fourth quadrant, going from the +x axis)

    so the difference is 180+19.26+55.48 = 254.74 deg.

    So now that I know the difference between displacement and force direction I just

    w = (60N) (9.11m) (cos 254.74)

    I'm thinking that will work? Although I do think the purpose of this problem may to use dot product.
  8. Jul 6, 2008 #7


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    I don't think you have the correct value for the total force magnitude. If you use 34 and 38 in the Pythagorean theorem, you don't get 60.

    Once you fix that, I think you are also not finding the angle between the vectors correctly. It's best to draw them out on a diagram. Since the force vector is in the fourth quadrant, and the displacment vector is in the third, you should be able to get the angle between them (and you should be able to find an angle that's less than 180 degrees). What do you get?

    I think it would be important to be able to find the correct angle here. However, as long as your class has covered dot products, I think that's the best way to solve this problem. Here you have two vector, F and s in unit vector form. One definition of the dot product has the magnitudes and angle in it (that's the one you have above); the other definition of the dot product has the components in it, and that's the one to use for this problem. Can you find that definition of the dot product?
  9. Jul 6, 2008 #8
    Er, math error on the theorem. Should be 50.99 = 51. I must have just looked at it quick.

    And when I am taking the angle between the vectors I was going the "long way" through the first and second quadrants to get there. Should I always go the shortest way?

    If I go the shortest way:

    55.48-19.26 = 36.22 deg difference.

    So my final equation would look like:

    w = f s cos[tex]\alpha[/tex]

    w = (51N)(9.11m)(cos 36.22)?
  10. Jul 6, 2008 #9


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    Well, you calculated the angle 55.48 degrees based on the total force being 60 N, which was not correct; it should be 48 degrees or so (redo it to get the accurate value)

    So let's organize the values:

    magnitude of displacement: 9 m
    direction of displacement: 19 degrees below the -x axis
    magnitude of force: 51 N
    direction of force: 48 degrees below the +x axis

    (I'll let you keep track of the more accurate values to several decimal places.) Now, you have two directions: 19 degrees below the -x axis, and 48 degrees below the +x axis. What is the angle between those two directions? Draw it out on paper and make sure it makes sense.
  11. Jul 6, 2008 #10
    Got it. My mix ups were not taking the distance correctly, a math error, and then finally not re-looking at what was affected by that math error.

    Thanks much, solved! -181.8J
  12. Jul 6, 2008 #11


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    Sounds good! To do this problem with the dot product, you would use

    W=\vec F\cdot\vec s = F_x s_x + F_y s_y

    So by multiplying and adding just the four numbers in the problem (being careful with signs) you can get the answer without having to worry about finding the angles.
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