Equilibrium with moments of forces

[LuigiAM]

Hi everyone,

This is not really a homework question per se since it's not an assignment. Basically, our physics professor gave us a 100+ page syllabus filled with example questions and solutions and I am practicing with them by doing them one after the other. This one question I'm having trouble with. Here it is:

The first thing that pops into my mind is that the rod is 1 meter long, and according to the professor's solution point P must be 1.25 meters from the end of the rod. However, in the drawing, the point P is actually on the rod. It seems a bit unintuitive to me that the force would be acting outside of the rod?

For the position of the resultant, my calculations are like this if x is the distance between points P and O:

(10 kg)(x) - (15 kg)(0.5 - x) - (5 kg)(1 - x) = 0
(10 kg)(x) - 7.5 kg + (15 kg)(x) - (5 kg) + (5 kg)(x) = 0
x (10 kg + 15 kg + 5 kg) = 7.5 kg + 5 kg
x (30 kg) = 12.5 kg
x = 12.5 kg / 30 kg = 0.42

I'm not sure what I'm doing wrong? I think the solution has all the forces as a negative when he's adding them up, but shouldn't the 10 kg force be a different sign than the 15 kg and the 5 kg one?

Thanks for any help

(Note: I ticked the box saying that I used the template, but I don't think it applies for this question since it's not really an assignment, it's just a question with the solution and I need help understanding the solution. I hope it's ok)

 

[haruspex]

in the drawing, the point P is actually on the rod

It had to be drawn somewhere, and it might as well be within the rod. If it turns out to be beyond the rod that should be evident in the answer (negative x).

unintuitive to me that the force would be acting outside of the rod

It was immediately apparent to me. As linear forces, the three applied forces have some cancellation, but as torques around the centre of the rod they all act clockwise. Thus the answer must be a relatively small force with a relatively large torque, making it further from the centre of the rod than the applied forces.

(10 kg)(x) - (15 kg)(0.5 - x) - (5 kg)(1 - x)

Watch the signs. All of these have clockwise torque about P.

 

[LuigiAM]

Oh yes I understand why they all have the same sign now it's because they're pushing the rotation in the same direction. Hah I feel dumb I should've realized this.

But the problem still seems strange to me. How can a force that is applied outside the rod have an effect on the rod? The solution talks about an equilibrium, which means that the force R will cancel the others and prevent the rotation. It just seems impossible for this to happen unless there is actual contact with the rod, no?

 

[haruspex]

How can a force that is applied outside the rod have an effect on the rod?

The "equivalent force" is a theoretical construct. How it could be applied in practice is unimportant, but e.g. you could extend the rod using a rigid attachment, or you could balance it by applying another combination of forces with equal and opposite equivalent force.

 

[LuigiAM]

I see... so, if I understand correctly then the rod in the example actually is rotating? And R is just theoretical?

I think I got confused because of the word "equilibrium" in the solution, I thought it meant that the forces actually acting on the rod were keeping it in equilibrium, so I just couldn't wrap my mind around how it could be outside the rod

 

[haruspex]

The wording is confusing.
If you look carefully at the diagram you will see that the force up from P is labelled -R. That is, they anticipated that the resultant force will be downward so added an upward force of the same magnitude to achieve equilibrium.
This was unnecessary. They could simply have said that since the resultant had no moment about P the sum of the applied torques about P is zero.