Sign of force for assumed $mgy$ gravity [closed]

If gravity force of earth is $$mg$$:

1. if positive y is pointing upwards, then: $$m\vec a = -mg\hat y$$ and $$\ddot y = -g$$
2. if positive y is pointing down, then: $$m\vec a = mg\hat y$$ and $$\ddot y = g$$

If you assume force is $$mgy$$:

1. if positive y is pointing upwards, then: $$m\ddot y = -mgy$$ and $$\ddot y = -gy$$

2. if positive y is pointing down, then: $$m\ddot y = mgy$$ and $$\ddot y = gy$$

Solving the (3) and (4) differential eqs end up having completely different trajectories((3) with $$c_1cos(\sqrt{2g}t)$$ while (4) with $$Ae^{\sqrt{g}t} + Be^{-\sqrt{g}t}$$ which shouldn't be the case. There seems to be a flaw in (3) or (4), but no idea what.

• "If you assume force is mgy..." The force is not $mgy$. The quantity $mgy$ is the gravitational potential energy (or negative of the gravitational potential energy depending on your conventions). The force is $mg$ in magnitude and its direction is towards the earth, regardless of your conventions for which direction of y is positive.
– hft
Commented Nov 12, 2023 at 20:48
• @user1079505 correct and i know potential is $mgy$, but i am assuming that force is $mgy$. Commented Nov 12, 2023 at 21:08
• In all 4 equations units on the left and right sides disagree. Commented Nov 12, 2023 at 21:22
• Otherwise, there are systems where the force grows linearly with the distance: harmonic oscillators. You may want to learn about them. Commented Nov 12, 2023 at 21:25
• i know potential is $mgy$, but i am assuming that force is $mgy$ That makes no sense at all. Force and potential don’t have the same dimensions, so mgy can’t be a force. It’s like saying “distance is $vt$ but I am assuming mass is $vt$.” You can’t make arbitrary assumptions. Physics has rules, and dimensional consistency is one of them. Commented Nov 12, 2023 at 23:20

If you assume force is $$mgy$$:

After reading OP's comments, it seems that OP just wants to understand how to solve a problem where the force is proportional to the displacement.

The main problem the commenters (myself included) seem to have is one of units. In particular, OP already said that $$m$$ and $$g$$ are the usual mass and gravitational constant. Therefore, $$mgy$$ can not have units of force.

So, we can introduce another parameter with units of length $$\ell$$ and write, more appropriately: $$F = \frac{mg}{\ell}y\;.$$

In fact, to save us some writing, let's define: $$k\equiv -\frac{mg}{\ell}$$ and thus we can write: $$F = -ky\;.$$

Such a force has been studied since at least the 17th century. In particular, by Robert Hooke, who famously quipped "ut tensio sic vis." Such a force is often called a spring force.

Certainly one can solve the equation: $$m\frac{d^2 y}{dt^2} = -k y\;,$$ (given appropriate boundary conditions).

The solutions for position $$y(t)$$ are well-known to be sinusoidal. For example $$y(t) = A\sin(\sqrt{\frac{k}{m}} t)\;$$ is one such solution.