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Finding Force Given Potential Energy

  1. Oct 13, 2012 #1
    Now let me start off by saying that I, along with over 100 students in my class are having issues with this concept, thanks to our "amazing" professor, while our midterm exam is a few days away. I don't want to sound like I am complaining, but I am frustrated with the fact that the professor refuses to go over an example with us completely. The following is a practice exam question and for some reason, he refuses to provide answers to the questions (forget providing an answer key). Any help would be greatly appreciated.

    V(r; C6, C12) = C12/r12 - C6/r6 with C12 = 1.0 and C6 = 2.0

    2. Relevant equations
    None, all equations you need for the question are in the question itself.

    3. The attempt at a solution
    So I know that F(vector) = -∇V, where F is the force and V is the potential energy. Basically this means that I need to take the partial derivative of V(r) and multiply that by r(vector)/r and so for part (a), I get: (12C12r-13-6C6r-7)(r(vector)/r)
    Have I done this correct? And if so, how do I proceed to part (b)?
  2. jcsd
  3. Oct 13, 2012 #2
    You are correct in part a, but you can simplify the answer a little.
    What is the relation between r,r(vector) and x,y,z,x^,y^,z^?
  4. Oct 13, 2012 #3
    By simplifying the answer, do you mean just substitute the values of C6 and C12 with their respective values (given in the question)?

    In addition, I know that r(vector)/r is just a normalized vector. In other words, it gives me the direction of the vector force. In addition, r = √x2+y2(+z2). However, I still do not know how to solve part (b). I can't seem to put my ideas together.
  5. Oct 13, 2012 #4
    By simplify, I just meant multiply r^-13 and r^-12 by the 1/r you had grouped with r(vector).

    Ok, you have r^2=x^2+y^2+z^2, what about r(vector)?
  6. Oct 13, 2012 #5
    By multiplying the derived equation by 1/r, I get:

    And since r = √x2+y2(+z2), I get:

    Finally r(vector) gives me the exact location using the x,y,z co-ordinates. So, now, do I just substitute the values of x,y,z given by r(vector) into the equation above?
    Last edited: Oct 13, 2012
  7. Oct 13, 2012 #6
    quick note: Sqrt(...)^-14 and Sqrt(...)^-8 simplify
  8. Oct 13, 2012 #7
    Hm, by simplifying, do you mean that they become:
  9. Oct 13, 2012 #8
    As for plugging in, remember that [itex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/itex]
  10. Oct 13, 2012 #9
    I am not sure what you mean with that equation above... what does the hat above the variables mean?

    After substituting the numbers, I got r = √35 and plugging that into the original derived equation, I got an answer of -7.997 [itex]\times[/itex]10-6 (which I am assuming is an answer in Newtons).
  11. Oct 14, 2012 #10
    [itex]\hat{x}, \hat{y}, \hat{z}[/itex] are the unit vectors in the x,y, and z directions. If you haven't seen this notation yet, you are overdue. It's just a way of writing out a vector as a sum of terms instead of as a matrix, which has the benefit of being very useful for multi-variable calculus. I don't know how you're doing gradients without having seen unit vectors yet.