# A case of normal force working in the “wrong direction” [closed]

Consider a painter on a scaffold pulling themself up by two ropes connected to the ends of the scaffold via a set of pulleys. The painter has mass $$M$$, the scaffold has mass $$m$$, and the painter pulls themself up with a force $$F$$ on each rope. Say the painter rises with uniform acceleration $$a$$.

Modeling this situation, one arrives at the system of linear equations $$2F - N - mg = ma$$ $$2F + N - Mg = Ma$$ (for example in this video), where $$N$$ is the normal force of the scaffold on the painter. Solving this system of equations, we find: $$N = \frac{2F(M-m)}{M+m}$$ But this seems very strange to me: this means that $$N$$ is negative if the scaffold has more mass than the painter. If I’m interpreting this correctly, that would mean the normal force of the scaffold on the painter is working downward, i.e., the scaffold is somehow pulling the painter down. Intuitively, that doesn’t seem possible. Am I interpreting this correctly? Or is the appropriate response to treat the term $$M-m$$ as $$\max\{0, M-m\}$$ instead (i.e., no normal force if the scaffold is too massive)?

By assuming that the acceleration of the painter and the scaffold are both equal to $$a$$, you are implicitly assuming that the painter and the scaffold are connected together e.g. the painter has tied or glued or nailed themselves to the scaffold. In this case $$N$$ can be negative, and indeed $$N$$ will be negative if

$$M - m < 0 \\ \Rightarrow M < m$$

i.e. if the scaffold has a greater mass than the painter.

However, if the painter and the scaffold are not connected together then you are right, $$N$$ cannot be negative. In this case, if $$M < m$$ then we have to drop the assumption that the accelerations of the painter and the scaffold are the same. So if $$M < m$$ we set $$N$$ to $$0$$, and the accelerations of the painter and the scaffold to $$A$$ and $$a$$, and we have

$$2F -mg = ma \\ 2F - Mg = MA$$

which we can rearrange to find

$$\displaystyle a = \frac {2F} m - g \\ \displaystyle A = \frac {2F} M - g$$

and since $$M < m$$ (and we assume $$F >0$$) then we find that $$A > a$$. In other words, if the scaffold has a greater mass than the painter, then by pulling on the rope the painter pulls themselves up off the scaffold.