University Physics I - Classical Mechanics, 2019

118 CHAPTER 6. INTERACTIONS, PART 2: FORCES
I claim that in that case you can again get the force on object 1, F 21 , by taking the derivative
of U(x 2 − x 1 ) with respect to x 1 (leaving x 2 alone), and reciprocally, you get F 12 by taking the
derivative of U(x 2 − x 1 ) with respect to x 2 . Here is how it works, again using the chain rule:
F 21 = − d
dx 1
U(x 12 )=− dU
dx 12
F 12 = − d
dx 2
U(x 12 )=− dU
dx 12
d
(x 2 − x 1 )= dU
dx 1 dx 12
d
(x 2 − x 1 )=− dU
(6.18)
dx 2 dx 12
and you can see that this automatically ensures that F 21 = −F 12 . In fact, it was in order to ensure
this that I required that U should depend only on the difference of x 1 and x 2 , rather than on each
one separately. Since we got the condition F 21 = −F 12 originally from conservation of momentum,
you can see now how the two things are related 1 .
The only example we have seen so far of this kind of potential energy function was in last chapter’s
Section 5.1.1, for two carts interacting through an “ideal” spring. I told you there that the potential
energy of the system could be written as 1 2 k(x 2 − x 1 − x 0 ) 2 ,wherek was the “spring constant” and
x 0 the relaxed length of the spring. If you apply Eqs. (6.18) to this function, you will find that the
force exerted (through the spring) by cart 2 on cart 1 is
F 21 = k(x 2 − x 1 − x 0 ) (6.19)
Note that this force will be negative under the assumptions we made last chapter, namely, that cart
2 is on the right, cart 1 on the left, and the spring is compressed, so that x 2 − x 1 x 0 > 0 (spring stretched, pulling force) and positive if x