# Out of curiosity how is the contact force forumla derived?

1. Sep 26, 2011

### DavidAp

Hello!

I know that the contact force formula is
Fc = F(m2/(m1+m2))

but I was wondering, how did this equation come into being? How is it derived? No this is not a homework question, I was just curious to know.

Thanks!

2. Sep 26, 2011

### Hechima

Consider two blocks with masses m1 and m2 placed side by side on a frictionless surface.

_-F->__|m1||m2|____
If there's a horizontal force F applied to m1, and m1 and m2 remain in contact, then the blocks will be accelerating together, as if they were one object with a combined mass. (Newton's 2nd law)

(equation 1) F = (m1+m2)*a

Meanwhile, block 2 is being pushed on by block 1 (Newton's 3rd law), let's call this force F_21. Since block 2 is accelerating, and F_21 is the only force acting on it, we have (Newton's 2nd law, again)

(equation 2) F_21 = m2*a

Take equation 1, solve for a, then substitute into equation 2, and you get

F_21 = $\frac{m2}{m1+m2}$F

Also, block 2 is pushing back on block 1 with a force F_12=-F_21 (Newton's 3rd law)
However, this is not the total force acting on block 1 because we still have F from the left.

So F1(total) = F - F_21 = F - $\frac{m2}{m1+m2}$F = m1*a, as expected.