SnoppDoge said:
There is no way anyone on the Forum can tell for sure, we can only guess.
F
c is sometimes used for centripetal force. If so, I can try to make sense of the rest of the equation, which you originally posted as:
SnoppDoge said:
Fc = Mr(theta - Oo) ^ 2 / (t - tn) ^ 2
You later put that r as a subscript, but I suspect that was wrong. It's probably a radius.
The Oo could be θ
0, an initial angle. The tn maybe t
0, an initial time. That gives us
F
c=Mr(θ-θ
0)
2/(t-t
0)
2.
That makes sense if a (small) object mass M is rotating at distance r from an axis at constant speed and moves from an angular position θ
0 at time t
0 to θ at time t. The formula would give the centripetal force required.
SnoppDoge said:
what is m in number 2? is it mass?
Almost surely.
You originally posted this as
SnoppDoge said:
M = Fk / Mk(m)(g)cos(theta)
When you put in subscripts you changed it to
SnoppDoge said:
m = Fk / Mk(m)(g)cos(theta)
which resulted in the variable m appearing on both sides. I now assume you mean either ##M = \frac{F_k}{M_kmg\cos(\theta)}## or ##M = \frac{F_k}{M_k}mg\cos(\theta)##. Neither makes obvious sense. F
k smells like a force, maybe of kinetic friction. mg cos(θ) would also be a force. Not sure what M and M
k are supposed to be, but one would guess they are of the same type.
If ##M = \frac{F_k}{M_kmg\cos(\theta)}## then ##M M_k= \frac{F_k}{mg\cos(\theta)}##, making the M's dimensionless.
If ##M = \frac{F_k}{M_k}mg\cos(\theta)## then ##M M_k= F_kmg\cos(\theta)##, making the Ms also forces.