Finding Force Given Potential Energy

[CallMeShady]

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.

Homework Statement

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

Homework Equations

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

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)?

 

[frogjg2003]

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^?

 

[CallMeShady]

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^?

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.

 

[frogjg2003]

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)?

 

[CallMeShady]

By multiplying the derived equation by 1/r, I get:
(12C₁₂r^-14-6C₆r^-8)(r(vector))

And since r = √x²+y²(+z²), I get:
(12C₁₂(√x²+y²+z²)^-14-6C₆(√x²+y²+z²)^-8)(r(vector))

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?

 

[frogjg2003]

Yes
quick note: Sqrt(...)^-14 and Sqrt(...)^-8 simplify

 

[CallMeShady]

Hm, by simplifying, do you mean that they become:
(x²+y²+z²)^-7
&
(x²+y²+z²)^-4
?

 

[frogjg2003]

Yes.
As for plugging in, remember that [itex]\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}[/itex]

 

[CallMeShady]

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).

 

[frogjg2003]

[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.