Out of curiosity how is the contact force forumla derived?

AI Thread Summary
The contact force formula Fc = F(m2/(m1+m2)) is derived from Newton's laws of motion applied to two blocks, m1 and m2, on a frictionless surface. When a horizontal force F is applied to m1, both blocks accelerate together as a single system, leading to the equation F = (m1+m2)*a. The force exerted by block 1 on block 2, denoted as F_21, is determined using Newton's second law, resulting in F_21 = m2*a. By substituting the expression for acceleration into this equation, the contact force formula is obtained. This derivation illustrates the interaction between the two blocks under the influence of an external force.
DavidAp
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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!
 
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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.
 
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