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Using force vector to integrate work in xy-plane

  1. Mar 26, 2015 #1
    Hello, I picked up a challenging problem (at least to me) and I'm having difficulties.

    1. The problem statement, all variables and given/known data

    An object moves in xy-plane from point O = (0; 0) to point A = (1 m; 0) and from there to point B = (1 m; 2 m). All this time when the object moves a force [itex]\vec F[/itex] = ax2[itex]\vec i[/itex] + by[itex]\vec j[/itex] affects the object, where a = 3,0 N/m2 and b = 1,5 N/m2. Calculate the work done by this force in this path.

    2. Relevant equations

    [itex]W = \int_{x_0}^x F_x(x) \, dx[/itex]

    3. The attempt at a solution

    After trying with different methods and failing, I thought the solution should be obtained by using a vector from O to B.

    So I created vectors:
    [itex]\vec {OA}[/itex] = a(1 m - 0 m)2[itex]\vec i[/itex] + b(0 m - 0 m)2[itex]\vec j[/itex] = a * 1 m2[itex]\vec i[/itex]
    [itex]\vec {AB}[/itex] = a(1 m - 1 m)2[itex]\vec i[/itex] + b(2 m - 0 m)2[itex]\vec j[/itex] = b * 2 m[itex]\vec j[/itex]
    [itex]\vec {OB}[/itex] = [itex]\vec {OA}[/itex] + [itex]\vec {AB}[/itex] = a * 1 m2[itex]\vec i[/itex] + b * 2 m[itex]\vec j[/itex] = [itex]\vec s[/itex]

    After this I'm stuck. I think I need the dot product of [itex]\vec F[/itex] * [itex]\vec s[/itex] ([itex]\vec i[/itex] and [itex]\vec j[/itex] would disappear then) which could be integrated, but by what? dx? Or maybe somehow by dx and dy..? I found some similiar problems of line integrals and vector fields, but I don't know apply those ideas to my problem if it is even possible.

    There is also one other difficulty. By using the equation [itex]W = \int_{x_0}^x F_x(x) \, dx[/itex], I need to know x0 and x. x0 = 0 m, but do I use the length of |[itex]\vec s[/itex]| as ending point?

    Am I completely off track or what? Any hints? Thanks for reading
     
  2. jcsd
  3. Mar 26, 2015 #2

    RUber

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    It seems like you have all the pieces you need for this problem.
    ## W = \int_{x_0}^{x_1} F_x(x) dx = \int_{x_0}^{x_1} ax^2 +by dx ##
    This equation is for the work done by moving in the x direction.

    Note that the problem says the path is 1 unit in x first, then 2 units in y. So integrate the first path in x and then integrate the second path in y.

    The total distance travelled is 3m, so drawing the vector OB probably won't help much.
     
  4. Mar 26, 2015 #3
    This really cleared things up

    So from O to A, x1 = 1m y = 0

    [itex]W_1 = \int_{x_0}^{x_1} F_x(x) \, dx [/itex] = [itex]\int_{x_0}^{x_1} ax^2 + by \, dx [/itex] = [itex]\int_{x_0}^{x_1} ax^2 \, dx [/itex] = [itex]\left. \frac {1}{3}ax^3\right|_{x_0}^{x_1}[/itex] = 1 Nm = 1 J

    AB, x = 0 y1 = 2m

    [itex]W_2 = \int_{y_0}^{y_1} F_y(y) \, dy [/itex] = [itex]\int_{y_0}^{y_1} ax^2 + by \, dy [/itex] = [itex]\int_{y_0}^{y_1} by \, dy [/itex] = [itex]\left. \frac {1}{2}by^2\right|_{y_0}^{y_1}[/itex] = 3 Nm = 3 J

    [itex]W_{total} = W_1 + W_2 = 4 J[/itex]

    That is the correct answer, thanks!

    One more thing, how did [itex]\vec i[/itex] and [itex]\vec j[/itex] disappear? I thought it would require a dot product
     
  5. Mar 26, 2015 #4

    RUber

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    Essentially, you did the dot product intuitively.
    When you took the first path, you ended up with just the x component of F, which is exactly what you would have gotten with F dot s with s being the first path.
    Similarly if you dot F with the second path, you come to the same conclusion, keeping only the y component of F.
    When the paths are purely in the direction of your basis vector, the dot product does exactly what you think it should--keeps only the relevant parts.
     
  6. Mar 31, 2015 #5
    Can I mark it like this? [itex]\vec i[/itex] being the first path and [itex]\vec j[/itex] being the second path.

    [itex]\vec F_x[/itex] = [itex]\vec F[/itex] * [itex]\vec i[/itex] = (ax2[itex]\vec i[/itex] + by[itex]\vec j[/itex]) * [itex]\vec i[/itex] = ax2
    [itex]\vec F_y[/itex] = [itex]\vec F[/itex] * [itex]\vec j[/itex] = (ax2[itex]\vec i[/itex] + by[itex]\vec j[/itex]) * [itex]\vec j[/itex] = by

    Or is this incorrect way to think about it? Since if I create vectors for those two paths, I get [itex]\vec {AB}[/itex] = [itex]\vec i[/itex] and [itex]\vec {AB}[/itex] = [itex]\vec 2j[/itex]. But [itex]\vec {AB}[/itex] * [itex]\vec F[/itex] gives me a wrong answer, 2by that is
     
    Last edited: Mar 31, 2015
  7. Mar 31, 2015 #6

    BvU

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    Not good. It doesn't match dimensionally either (##\vec i\;## has the dimension of length too).


    With (for OA) ##\ \vec ds = \hat\imath dx\ ## you can write $$
    W_{OA} = \int_O^A \vec F \cdot d\vec s = \int_0^1 \left ( ax^2 \; \hat \imath + by \; \hat \jmath \right ) \cdot \hat \imath \;dx = \\ \qquad \int_0^1 \left ( ax^2 * \hat \imath\cdot \hat \imath + by * \hat \jmath\cdot \hat \imath \right ) dx = \int_0^1 \left ( ax^2 * 1 + by * 0 \right ) dx
    $$
     
  8. Apr 7, 2015 #7
    I see, so can AB be written the same way too? ##\ d\vec s = \hat\jmath dy\ ##

    Does ##\hat\imath dx\ ## and ##\hat\jmath dy\ ## just symbolize the path taken and thus the actual length is taken in to account only in the integrals?
     
  9. Apr 7, 2015 #8

    RUber

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    Right. Since ##\vec{ds}## refers to a small distance in the direction of the path. The vector component is ##\hat i## or ##\hat j## and the magnitude is the tiny step dx or dy. This gets much more complicated when your paths are not the straight lines you have in this example.
     
  10. Apr 7, 2015 #9

    BvU

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    The path goes from (1,0) to (1,2) and it is subdivided in small steps ##d\vec s## so that we can take the inner product ##\ \vec F\cdot\vec ds\ ## for each of those little steps and add up all the contributions.

    The vector ##\Delta \vec s## that takes you from ##\vec A## to ##\vec B## is ##\vec B - \vec A##, because ##\vec B = \vec A\ + \ (\vec B - \vec A)##, so
    ##\Delta \vec s = (0,2)## and if we subdivide that in little steps we get ##\ d\vec s = \hat\jmath dy\ ## where y runs from 0 to 2.

    It's a representation, a symbolization if you want to call it that. Not of the path, but of an infinitesimal little piece of the path. And yes, where the path begins and where it ends are the bounds of the integral.

    [edit]crossed RUber's reply - but I think we all agree.
     
    Last edited: Apr 7, 2015
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