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I have a problem with variable forces

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A force F has components F sub x = axy-by^2, F sub y= -axy+bx^2 where a= 2N/m^2 & b=4N/m^2 Calculate the work done on an object of mass 4kg if it is moved on a closed path from (x,y) values of (0,1) to (4,1), to (4,3) to (0,3) and back to (0,1). all coordinates are in metres. The path between points is always a straight line.

    2. Relevant equations
    deltaK=F.dr

    3. The attempt at a solution
    Looking at the forces components, the force can be expressed as F=F sub x + F sub y=(axy-by^2)+(-axy+bx^2)=bx^2-by^2

    Considering 4 separate motions, we can treat them as 2 dimensional problems, so delta K = F.dr=F.dx
    where delta K sub 1+ delta K sub 2+ delta K sub 3 + delta K sub 4 = delta K sub total. In each motion, since 1 component (x or y) will remain constant, leaving me with 4 line integrals of a quadratic with the constant term determined on which motion I am dealing with. IE: A horizontal motion from (0,1) to (4,1) has no change in y, so y=1 for the duration of the motion. Conversely, when dealing with a vertical motion, the x value remains constant, again leaving me with a line integral of a quadratic where say in the motion (4,1) to (4,3), x remains constant where x=4.

    Is this the proper approach? Am I able to treat the 2 components as a single force?

    Further, I notice here that the mass is not included in my calculations. I can justify not needing this since the change in kinetic energy describes the work done, I do not have a component of time, and do not require any information about velocity or acceleration, thus although it is very nice that the object is 4kg, there is no need to use this in any computations.

    Without showing all of my work, I determined the work performed to be equal to 256J.

    I have handed this problem in for last weeks problem set, and have a similar problem on my new problem set this week. If truly needed I can reproduce my calculations, but as mentioned they have been handed in already, and I am simply looking for reassurance I have managed the problem appropriately.

    Thank you for your help in advance.
     
  2. jcsd
  3. Nov 11, 2014 #2

    collinsmark

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    Hello blair chiasson,

    Welcome to PF!

    By the way, for future reference, you can use the [itex] x_2 [/itex] button on the menu for subscript and the [itex] x^2 [/itex] button for superscript. That way, you can write out things such as
    Fx = axy - by2, Fy = -axy + bx2

    And if you want to get really fancy, you use [itex] \LaTeX [/itex] described here.
    https://www.physicsforums.com/threads/latex-faq.617570/#post-3596252

    That way, you could write
    [tex] \vec F = \left( axy - by^2 \right) \hat x + \left(-axy + bx^2 \right) \hat y [/tex]

    Yes, that's the right approach.

    And yes, you can treat them as a single force. Just make sure you account for the direction of the vectors within the dot product of [itex] \vec F \cdot \vec{ds} [/itex]. Because the path (i.e., [itex] \vec{ds} [/itex]) moves in only one x- or y-direction (on any given leg), the non-zero part of the dot product will only involve one of the force components on any given leg.

    For what it's worth, I arrived at a different answer. :(

     
    Last edited: Nov 11, 2014
  4. Nov 11, 2014 #3
    Thank you for your response!

    It is possible I performed a line integral in the wrong direction. I am just learning to work with integration this semester. I am happy that the basis of my answer is on the right track, I'll surely grab partial marks for it, it was said to be the tricky one on the problem set. You mention that only one of the force components are involved on any given leg. Am I understanding this that I should not sum up Fx+Fy, and simply use Fx for my horizontal motions? If this is the case, I'd like to try calculating it again and comparing my new result with the one which you have obtained.

    Thanks again!

    PS Thanks for the tip typing out my messy maths.
     
  5. Nov 11, 2014 #4

    collinsmark

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    Yes, that's right.

    When moving in the x-direction, the path vector is [itex] \vec {ds} = (dx) \hat x + 0 \hat y [/itex]

    So the dot product in that case is

    [tex] W = \int_a^b \vec F \cdot \vec{ds} = \int_a^b \left( F_x dx + F_y (0) \right) = \int_a^b F_x dx[/tex]

    Similarly, when moving only along the y-direction, the x-component of the force gets multiplied by 0 when taking the dot product.
     
  6. Nov 11, 2014 #5
    AHA! That explains my poor result. I will recalculate this and share my solution with you as soon as I have the kids in bed. (Learning physics is hard, but can be more difficult with kids running about!)

    Thanks again!
     
  7. Nov 11, 2014 #6

    collinsmark

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    Oh, and by the way, there is a more general (and better) way to look at the vector calculus (better than what I wrote earlier).

    Generally, [itex] \vec {ds} = (dx) \hat x + (dy) \hat y [/itex]

    Then we have,

    [tex] W = \int_a^b \vec F \cdot \vec{ds} \\ = \int_{a_x}^{b_x} F_x dx + \int_{a_y}^{b_y} F_y dy [/tex]

    However, notice what happens when we move only in the x-direction: [itex] a_y = b_y [/itex] and with that the second integral reduces to zero.
     
  8. Nov 11, 2014 #7
    That makes it much clearer why one of the components zeros out. I redid my calculations, please mind that I am new at integration and I likely did make an error along the way. My new result is 134 J. Is this what I should be looking to obtain?

    Thanks again for all the help :)
     
  9. Nov 11, 2014 #8

    collinsmark

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    Unfortunately, my answer is different.

    If you post your work we might be able to figure out where the discrepancy is coming from.
     
  10. Nov 11, 2014 #9
    While typing my work, I found errors computing the integral. I will recompute and type it out.
     
  11. Nov 11, 2014 #10
    I did it all again, I got this time -96 J. I'm so certain I don't know what I am doing at this point that I cant really feel good about taking up your time. I will go see my prof tomorrow during his office hours. I've never had a single class on the topic of integration, and do not take any applied math classes either. It leaves me in a bit of a pinch when I bump into things like this but it is the nature of the program I guess. Overall, I have to move on from this problem and complete the rest of my assigned work with the time that I do have.

    Thank you again for all the help, it is much appreciated.
     
  12. Nov 11, 2014 #11

    collinsmark

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    That's still different than the answer I obtained.

    That sounds like quite a challenge. Vector calculus can can be quite tricky in its own right, even if one is very familiar with normal calculus. I can imagine that learning this material without prior knowledge of integration could be a little daunting.

    On a positive note, if you are able to tackle this material, the regular calculus should be a breeze. ;)
     
  13. Nov 12, 2014 #12
    Hi again. I spent today going over integration with a friend of mine. My issue seems that when I was calculating the work done in the negative x direction, I considered the work a negative value. Also, I was having difficulties with choosing which x/y terms to subtract from the others. After a little tap on my shoulder I recognized that while looking for area under the curve, we are limiting the spread of values in the same way we do for calculating the change in temperature, or anything else. So the final position subtracts from the initial position. Which is also what was leading to my negative values in the reverse direction. Although we didn't come to a solid solution, since we were doing a review of integration in general, I do feel as if I have a better hold on it now, and am starting to make good sense of it. I have a question from this weeks problem set that I am going to ask about tonight, if you have time, I'd love to hear your feedback on this.

    Thanks for all your help, although it was a struggle, that seems to be when you obtain the most knowledge.

    Cheers!
     
  14. Nov 13, 2014 #13

    Are you still using this in your calculations?
    You cannot cancel the terms in the x any y components. The components add as vectors not scalars.
    If you dropped, never mind.
     
  15. Nov 13, 2014 #14
    No, I have gotten past that. With some help from collinsmark and a classmate I got through the problem. It had been handed in already before coming here for direction, so all the points for it were already lost. The good news is that I have learned a lot about integration in the past couple of days, and being faced with a similar problem on a test I believe that I will do just fine. Thanks for all the help guys.
     
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