Out of curiosity how is the contact force forumla derived?

Click For Summary
SUMMARY

The contact force formula, Fc = F(m2/(m1+m2)), is derived from Newton's laws of motion applied to two blocks with masses m1 and m2 on a frictionless surface. When a horizontal force F is applied to m1, both blocks accelerate together as a single object with a combined mass, leading to the equation F = (m1+m2)*a. By applying Newton's 3rd law, the force exerted by block 1 on block 2 (F_21) can be expressed as F_21 = m2*a, which, when substituted back into the equations, confirms the derivation of the contact force formula.

PREREQUISITES
  • Understanding of Newton's 2nd law of motion
  • Familiarity with Newton's 3rd law of motion
  • Basic knowledge of force and mass concepts
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of Newton's laws of motion in detail
  • Explore applications of contact forces in real-world scenarios
  • Learn about frictionless surfaces and their implications in physics
  • Investigate the effects of varying mass on acceleration in multi-body systems
USEFUL FOR

Students of physics, educators explaining mechanics, and anyone interested in the foundational principles of force and motion.

DavidAp
Messages
44
Reaction score
0
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!
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 51 ·
2
Replies
51
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K