Is this mechanics problem underdetermined or am I just missing the final piece?

Image two people carrying some rod up a flight of stairs, at an angle $\alpha$. I'm trying to find the ratio $$\frac{F_1}{F_2}$$ as a function of $\alpha$, where $F_1$ is the total force applied by the person carrying the lower end and $F_2$ is the same, but for the upper end, as seen in the diagram below.

When breaking the vectors $\overline{F_1}$ and $\overline{F_2}$ into components along and perpendicular to the rod, I get

\begin{align} \sum \overline{F}= \overline{0} \Rightarrow F_{1,x}-F_{2,x}&= F_g\sin\alpha\\ F_{1,y}+F_{2,y}&= F_g\cos\alpha \end{align}

and from considering the torques I get \begin{align} \sum \overline{\tau}= \overline{0} \Rightarrow F_{1,y}=F_{2,y} \end{align}

From this I get

\begin{align} \overline{F_1}&= \begin{pmatrix}F_{2,x}+F_g\sin\alpha \\ \frac{F_g}{2}\cos \alpha \end{pmatrix}\\ \overline{F_2}&= \;\;\;\;\;\begin{pmatrix}F_{2,x} \\ \frac{F_g}{2}\cos \alpha \end{pmatrix} \end{align}

However, I end up with one unknown (here $F_{2,x}$) too much.

Any ideas?

OK I'll take another go at this. I think it make sense that there is an extra $F_{2x}$ in your equations. It accounts for the fact that if one force pushes harder along the rod then the other force could still balance by pushing harder itself.

For example if the rod had $\alpha=0$ each side would only need to apply a force of $F_g/2$. But it is also possible to apply $F_g/2$ in the vertical while pushing arbitrarily hard in the horizontal so long as the other side applies the same force.

You should reword the question to ask, "what is the minimum force needed to keep the object up." We can do that by setting $F_{2x}=0$.

Edit

I would like to add that $F_{2x}$ could be negative (a pull) which could result in $F_{1x}=0$. We probably need a better constraint than what I gave. One may be "have both lifters perform the same work and assume the rob only moves in the X direction." In this case the X components of the force must be equal in magnitude.

• That makes sense, thanks! The ratio then becomes $\frac{F_1}{F_2}=2\sqrt{\tan^2\alpha+\frac{1}{4}}$. Aug 11 '16 at 23:00
• Are you sure? If $F_{2x} = 0$, doesn't $F_{1x} = 0$, and $F_{1y} = F_{2y} = F_g/2$? Aug 11 '16 at 23:07
• @alephzero : I think you are overlooking the fact that x and y are along and perpendicular to the rod, not horizontal and vertical as is more common. Aug 12 '16 at 13:23
• @sammygerbil I made that mistake at first. Aug 12 '16 at 13:38

If you need to find four quantities (e.g. the $x$ and $y$ components of $F_1$ and $F_2$) you need four equations. Resolving forces and moments will only give you three equations.

If you solve this graphically by drawing a triangle of forces, it is clear that if you extend the line $F_g$ upwards and take any arbitrary point $P$ on the line, there is a solution where the lines of action of vectors $F_1$ and $F_2$ both pass through point $P$.

So, either the question can't be answered, or there is another piece of information that you haven't told us.

In your diagram, it looks as if $F_2$ is perpendicular to the slope of the rod. Is that just a coincidence, or is it the missing piece of information?