Falling Body Equation: Derivation & Meaning

[leehufford]

Hello,

I don't think I really understand the derivation of an equation of motion for a falling body under the force of Earth's gravity... I'm a little ashamed by this. I haven't been able to find someone asking quite the same question as I am about to ask.

I completely understand that the acceleration of an object is the second derivative of its position, and than an accelerated body is not in equilibrium, so ∑F ≠ 0.

I've been able to get by but it's time I truly understand what is going on. From the perspective of differential equations the derivation starts with an expression:

md²s/dt² = -mg. Then masses cancel to get d²s/dt² = -g.

What are we actually saying by equating two forces? I just see ma = -ma. But what does this statement physically mean?

I feel safe asking such a basic question on this forum and I trust that people will only give positive feedback in this situation. If I failed to find an online resource that explains this on the level I am seeking it would not be insulting to simply point me there. Thanks in advance,

Lee

 

[Doc Al]

From the perspective of differential equations the derivation starts with an expression:

md²s/dt² = -mg. Then masses cancel to get d²s/dt² = -g.

What are we actually saying by equating two forces?

Why do you think you are equating two forces? The only force is that of gravity, which equals -mg in your equation. The other side of the equation is mass times acceleration, from Newton's 2nd law.

 

[Nugatory]

md²s/dt² = -mg. Then masses cancel to get d²s/dt² = -g.

What are we actually saying by equating two forces? I just see ma = -ma. But what does this statement physically mean?

In that equation, $g$ is a constant while $s$ is a function of $t$, the time. The equation is telling you that the second derivative of $s(t)$ is equal to $-g$ and asking you to find the function $s(t)$ for which that is true.

That's what makes it a differential equation, it's an equation for an unknown function instead of just an unknown variable like you find in regular algebra.

The solution happens to be $s(t)=\frac{-1}{2}gt^2+At+B$ where $A$ and $B$ are arbitrary constants. It's a worthwhile exercise to plug this into the differential equation to verify that is in fact a solution... and then to figure out what the physical significance of the constants $A$ and $B$ is.