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Sign issue in problem on work

  1. Apr 9, 2014 #1
    I am trying to find the work done on a particle when a varying force acts on it. I am familiar with W = ∫F dot ds but am running into a sign question.

    Suppose a particle is moving in the -x direction because a force is acting on in that direction. Suppose it moves from x = 4 to x = 0, and suppose the force is given by F = -5 i. Then from W = Fs cos θ, the work would be (5)(4) cos 0 = 20 J. It makes sense that the work is positive since the force and the displacement are both in the same direction.

    Now I know an integral would not be needed for this problem (but another problem I am working on does, hence my question). When I used ∫F dot ds (with unit vectors) to do this, I got a negative answer for the work done.

    I used F = -5i (since the force is in the -x direction)

    ds = -dx i (since the path is from 4 to 0, the length element points towards -x)

    and when I integrated, I did so from x = 4 to x = 0 since that was the path.

    Then ∫F dot ds = ∫(-5i) dot (-dx i) = 5∫dx = 5x, but when I plug in the limits of the interval and take 5(0) - 5(4) = -20 J.

    Where am I putting in a negative sign that does not belong?
     
  2. jcsd
  3. Apr 9, 2014 #2
    The displacement is simply dx ,irrespective of whether you move in the positive x direction or negative 'x' direction .No need to put an additional negative sign as it is taken care by the limits.In this problem since we are moving from a larger value of x to a smaller value , dx is inherently negative .
     
  4. Apr 9, 2014 #3

    BiGyElLoWhAt

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    well, here's the issue, when you integrate, you don't want to integrate from 4 to 0, you want to integrate from 0 to 4, and the integral from 0 to 4 = -integral from 4 to 0.
    *edit
    also as tanya said, use dx, not -dx
    so ∫F dot dx from 4 to 0 = ∫-5i dot dx from 4to 0 = -∫-5i dot dx from 0 to 4 = ∫5i dot dx from 0 to 4
     
    Last edited: Apr 9, 2014
  5. Apr 9, 2014 #4
    Why should the limits be 0 to 4 ?

    The particle is moving from x=4 to x=0 not the other way round . The problem is not with limits but due to an additional negative sign which is not required.
     
  6. Apr 9, 2014 #5

    BiGyElLoWhAt

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    why integrate from 4 to 0? why not just flip the coordinates and define positive x to be to the left and the origin to be at the particle, then you are moving from 0 to 4 and the force is no longer -5i, as it's in the positive x direction, it's 5i

    same thing i did with the -integral trick.
    it's not 'the' issue, but i really don't see a reason for all the - signs. I didn't really read the whole question right away, but if you look back, I edited my post almost immediately.
     
  7. Apr 9, 2014 #6
    Because the displacement of the particle is from x=4 to x=0.
     
  8. Apr 9, 2014 #7

    BiGyElLoWhAt

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    not if you define your coordinates differently: cock them 45degrees and it's from (sqrt2,sqrt2) to 0.
    you can define your coordinates however you want to. whatever makes the problem easier, as long as you either a) use a right handed coordinate system, or b)adopt left handed cross products (and your prof knows what your doing, you'll get a -component somewhere if you do this)
     
  9. Apr 9, 2014 #8
    Tanya is hitting the nail I'm driving at on the head here. I can't see why the work doesn't turn out positive like
    Fs cos (theta) says it should. F is a vector in the -x direction; it ought to be -5i, not 5i. The same with dx; it's a vector, and if I am going along a path from x=4 to x=0 then dx also points in the negative direction and should be -dx i not dx i. I am integrating along the path from x=4 to x=0, not the other way. Yet that gives a negative sign that Fs cos (theta) does not.

    And Tanya, I'm fuzzy on what you're saying here.
    The displacement is simply dx ,irrespective of whether you move in the positive x direction or negative 'x' direction .No need to put an additional negative sign as it is taken care by the limits.In this problem since we are moving from a larger value of x to a smaller value , dx is inherently negative .
    Since ds is a vector, don't we have to put in the negative direction? Especially because in the larger problem I'm working on this is one side of a closed rectangular path.


    I hate to be such a stickler, but I have a reason. I'm teaching my high school advanced physics classes about line integrals because we are about to do Ampere's law. I'm going to be showing them a problem where this same issue will come up, and I know that one or two of them will ask about this. :)

    Thanks!
     
    Last edited: Apr 9, 2014
  10. Apr 9, 2014 #9

    BiGyElLoWhAt

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    Well as far as what tanya's saying, the 'negative' sign in the dx is accounted for by the upper limit being lower than the lower limit (4 to 0, rather than 0 to 4).

    I was just trying to employ a few simple manipulations to get you the answer you thought you should get, and in turn show you that they are equal.

    To elaborate on the 'dx' dilemma, the only time you really have negative differentials is when you employ substitutions.

    start from scratch:
    [itex]W=∫F (dot) ds = ∫[/itex]mag( [itex] F [/itex] )mag([itex]ds[/itex])[itex] cos(\theta)[/itex] (still no negative signs)

    Think about this mathematically: Take a function, say [itex]f(x)=x^2[/itex]
    Now say we want to integrate it from 3 to 0 (for some reason... I still prefer to flip my coordinates or use the - integral trick) so you have [itex]∫x^2 dx[/itex] from 3 to 0. so my question for you is this. why does defining dx as a vector add a negative sign? [itex]∫x^2dx[/itex]
    from a to b lets say... that's = to [itex]\frac{b^3-a^3}{3}[/itex] so why does that change just because a is larger than b?
     
  11. Apr 9, 2014 #10
    The vector ds is given by ds = (-dx)(-i)

    Chet
     
  12. Apr 9, 2014 #11
    In cartesian coordinates ## \vec{ds} = dx \hat{i}+dy \hat{j}+dz \hat{k}## .Since the particle is moving along the 'x' axis ## \vec{ds} = dx \hat{i}##

    What is dx ?

    'dx' represents change in the displacement of the particle along the 'x' axis .If it moves from higher to lower value of x (as in this case) ,it is a negative quantity. No need to put an extra negative sign.OTOH if the particle moves from lower to higher value ,'dx' is positive.

    ## Work = \int \vec{F} \cdot \vec{ds} ##

    ## Work = \int -5 \hat{i} \cdot dx \hat{i} ##

    ## Work = \int -5dx ##

    This is the same problem many students face while deriving potential at a point due to a point charge.

    Don't be surprised if there are more than a couple of students raising their hands :smile:
     
    Last edited: Apr 9, 2014
  13. Apr 9, 2014 #12
    Thanks, Tanya, that does clarify things for me. I appreciate your time, and the time of all those that commented.
     
  14. Apr 9, 2014 #13
    Did my post #10 make sense to anyone?

    Chet
     
  15. Apr 9, 2014 #14
    Hi Chet

    I beg to differ :smile: . Did post#11 make sense to you ?
     
  16. Apr 9, 2014 #15
    Hi Tanya,

    Hu?? My final result in #10 is the same as your final result in #11.

    Chet
     
  17. Apr 9, 2014 #16
    I apologize for my inability to understand the reasoning behind your expression of displacement. But I know you surely have a sound logic behind it.
     
  18. Apr 9, 2014 #17
    Hi Tanya,

    My rationale was, if the displacement is in the negative x direction, the magnitude of the displacement vector is -dx, and the direction of the displacement vector is -i, so that the displacement vector is ds=idx. I was just trying to show where the OP went wrong in his original analysis, and where, if he had done it correctly, he would have arrived at your expression.

    Chet
     
  19. Apr 9, 2014 #18
    Thanks for the input .

    Could you show me how you would derive expression for the potential at a point due to a point charge by calculating the work done in bringing a unit positive charge from infinity to a point .

    It is quite confusing for the students as it involves quite a few negative signs . I am interested in knowing your approach .
     
  20. Apr 10, 2014 #19

    ehild

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    The definite integral is the area under the graph of the function, therefore it is additive: [tex]\int_a^b{f(x) dx}+\int_b^c{f(x) dx}=\int_a^c{f(x) dx}[/tex]

    If c=a, the whole integral is zero, as the area is zero :
    [tex]\int_a^b{f(x) dx}+\int_b^a{f(x) dx}=\int_a^a{f(x) dx}=0[/tex]

    that is

    [tex]\int_b^a{f(x) dx}=-\int_a^b{f(x) dx}[/tex],

    The backward integral is opposite to the forward one. dx is just a symbol. Do not change its sign when the integration goes from a higher value to a lower one. [tex]\int_b^a{f(x) dx}\neq\int_b^a{f(x) (-dx)}[/tex]

    What would you do if you have the integration limits symbolically? As, for example, in

    [tex]\int_y^{y^2}{\cos(x) dx}[/tex], when the upper limit can be both greater and less than the lower limit?

    In case of work in one-dimensional motion, with F parallel to the x axis, ##dW=\vec F (\vec x)d\vec x## . During the motion, the elementary displacement can be both positive and negative. Think of a body projected vertically up. For example, ## F=-mg\vec i = -4 \vec i## and ##\vec x=(v_0 t-g/2 t^2)\vec i=(10 t-5t^2)\vec i## . First it moves upward, then downward. You want to get the work done in the first 1.5 s of flight, the net displacement is 3.75 m, while the body raises up to 5 m, and then falls down 1.25 m. The work done by gravity is ##\int _0 ^{3.75}{Fdx}=-15 J## equal to the upward work -20 J plus the downward work 5J.


    ehild
     
  21. Apr 10, 2014 #20
    Hi Tanya. No problem.

    Use spherical coordinates with the point charge Q at the origin, and the test charge q located at r. The force that you have to apply to the test charge to keep it from moving is:

    [tex]\vec{F}=-k\frac{Qq}{r^2}\vec{i}_r[/tex]

    (This is equal and opposite to the force that the charge Q is exerting on q). If you move the test charge from r to r+dr, the amount of work you do is:

    [tex]dW = \vec{F}\centerdot (\vec{i}_rdr)=-k\frac{Qq}{r^2}dr[/tex]
    (If dr is negative, then the amount of work you do is positive, since Q is repelling q).

    The previous equation can be rewritten as:
    [tex]dW = kQqd(1/r)[/tex]
    To get the work in moving the test charge from ∞ to radial location r, we integrate from 1/r=0 to 1/r:
    [tex]W=\frac{kQq}{r}[/tex]
    Per unit test charge, this is:
    [tex]\frac{W}{q}=\frac{kQ}{r}[/tex]

    Chet
     
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