Sure thing!! Jake and I's Physics background is virtually identical, so I don't think I'll add much
The approach to the first question is (I'm quite sure) definitely correct, I saw something similar in Physics last year. I do think there would be other methods, perhaps examining the components of force on the beam, but I like Jake's approach!
Less sure about the next bit, but I like Jake's approach and the reasoning is sound. However, I think the fact that the sign is not at the centre of the beam will change the answer, because the distance from the pivot point makes a massive difference to the force on the beam.
In general, that sign is what makes this question tricky. I'm not quite sure how to compensate for it. One possibility would be to move the sign to the end of the beam, so that it can be compensated for more easily, then find the lesser, equivalent mass that makes this valid. For example, if we find that moving it to the end of the beam doubles the torque, we could then adjust the mass to return it to its initial value. This would allow us to compensate for the vertical force of the sign in our vector diagrams on the right of the beam, which not changing the overall effect.
Sorry if that last thing was not explained quite nicely, but on the whole, I like Jake's approach! Just that one finer point I'd disagree with, but this is a bit beyond my capabilities
Having known really very little about classical mechanics, my reasoning for the final parts goes something like this. The example questions I found (linked above) bears a striking resemblance to our current predicament, except it doesn't include a sign. I think that the horizontal force approach is likely correct (it does not require taking into account any mass, just the Tension force and an angle). However, like Jamon said, the vertical force takes into account the mass of the beam, therefore perhaps the sign changes the formula (ie. it won't be just mass one + mass two * g).
However, that formula also doesn't require use of the
length of the bar. Length appears to be independent of the force: all that is required is a Tension force, an angle and a 'mass'.
I'm not sure whether this is true, but this was my line of reasoning. If you add the mass of the sign at a point, this will change the center of mass. The new center (which will likely be somewhere *on* the point of contact between the sign and the rod, a % way along the infinitesimal connection) *should* have half of the mass on its right, and half of its mass on the left. Therefore, you could create a new rod (with a new length) that is the same mass as the original rod (including the sign), with a center of mass at the same point.
This new rod will have a mass of (mass one + mass two), probably be longer, and probably be denser. (Note: If you're worried about the length change affecting tension, just increase the density of the metal. No harm done) However, the formula doesn't care about the length of the rod (or density); only its mass. In all respects, this new rod is the same as the old, so I figure that the formula will still apply.
This could be completely off, I'm just explaining my reasoning from a currently-purely-quantum-mechanical-and-definitely-not-classical perspective.
I don't know if I explained myself well: hopefully it made sense!
Jake