Proving tension in rope of equilibrium system

[toforfiltum]

Third, treat the rod+rope as the 'system'. This allows you to treat T as being applied at the bottom of the rope instead of where it is tied to the rod. This makes finding the moments due to the components of T trivial.

I don't see how treating the rod+rope as a system allows me to treat T as being applied only at the bottom of the rope instead of where it is tied to the rod. In class, my teacher taught me to either resolve the components into their separate bodies, where the reaction forces among the two bodies in contact cannot be canceled out and must be considered. However, she taught me, that if treated as a whole system, the reaction forces cancel out. Though I know it is not exactly the same case as in this question, I don't see how tension only applies to the bottom of the rope, I really don't.

 

[toforfiltum]

No , at a point ⊥ to OA and ⊥ to OB ./QUOTE]

Isn't that at O?

 

[toforfiltum]

By the way, though I understand andrewkirk's explanation, I wonder if moment can be calculated around an imaginary point X?

 

[toforfiltum]

Haruspex has provided a good method . From the point he speaks about , only tension and gravity exerts a torque . The only problem in this solution is finding the proper value of tension's torque .

So Haruspex's method is different than yours, is it not? Your method takes moments about O right?

 

[Qwertywerty]

So Haruspex's method is different than yours, is it not? Your method takes moments about O right?

Yes - his does not need you to take into account the normal forces . However you need to be careful while balancing the torques .

Isn't that at O?

O isn't perpendicular to the lines it lies on ( OA and OB ) .

 

[toforfiltum]

O isn't perpendicular to the lines it lies on ( OA and OB ) .

If so, the point can lie anywhere between the diagonal through OA and OB right? Hence, it will be perpendicular to both OA and OB.

 

[Qwertywerty]

My mistake there - Consider a rectangle with three of the vertices being O , A and B . The fourth vertice would be the said point .

By the way, though I understand andrewkirk's explanation, I wonder if moment can be calculated around an imaginary point X?

What do you mean by imaginary point ? If you mean any random point in an inertial reference frame , yes , though calculating the torques might not be numerically possible .

 

[Qwertywerty]

I don't see how treating the rod+rope as a system allows me to treat T as being applied only at the bottom of the rope instead of where it is tied to the rod. In class, my teacher taught me to either resolve the components into their separate bodies, where the reaction forces among the two bodies in contact cannot be canceled out and must be considered. However, she taught me, that if treated as a whole system, the reaction forces cancel out. Though I know it is not exactly the same case as in this question, I don't see how tension only applies to the bottom of the rope, I really don't.

Tension from the rope acts on both the rod and the wall .
If you consider the rod + rope as a system , then at the point at which the rope is attached , the rope pulls at the rod with the same force as the rod pulls the rope .

Internal conservative forces always cancel . Thus , the net forces acting on a system as a whole are only the external ( in this case ) , gravity and the force due to the rope on the wall .

 

[toforfiltum]

My mistake there - Consider a rectangle with three of the vertices being O , A and B . The fourth vertice would be the said point .
What do you mean by imaginary point ? If you mean any random point in an inertial reference frame , yes , though calculating the torques might not be numerically possible .

Because according to andrewkirk, moments are taken along the fourth corner of a rectangle whose other three corners are A, O and B, which he calls X, and X doesn't exist in the picture, not on the wall nor the ground.

 

[Qwertywerty]

Because according to andrewkirk, moments are taken along the fourth corner of a rectangle whose other three corners are A, O and B, which he calls X, and X doesn't exist in the picture, not on the wall nor the ground.

Aren't all points thus imaginary ? He takes it as X simply to refer to the place from where you should balance torques .

 

[toforfiltum]

Aren't all points thus imaginary ? He takes it as X simply to refer to the place from where you should balance torques .

Ahh...this is so confusing. Firstly, I want to know if X is in fact imaginary, and if so, can you balance the moment at X, because you said that calculating moments about imaginary points is numerically impossible, and yet you said that X is the place that you should be balancing the torques.

Secondly, I think I'm confused with Haruspex's solution. In the first tip, he said that finding moments due to tension is tricky, then proceeds to say in his third tip that finding moments due to components of T is trivial because he can treat T as being applied to to the bottom of the rope. And yet you say, internal conservative forces always cancel. Then if so, why does he even need to treat T as being applied to the bottom of the rope?

 

[Qwertywerty]

Ahh...this is so confusing. Firstly, I want to know if X is in fact imaginary, and if so, can you balance the moment at X, because you said that calculating moments about imaginary points is numerically impossible, and yet you said that X is the place that you should be balancing the torques.

I said balancing from any random point might be numerically impossible - I was not aware that you were referring to andrewkirk's post . The X to which he referred is in fact a special point - the point which has already been discussed .

 

[Qwertywerty]

Secondly, I think I'm confused with Haruspex's solution. In the first tip, he said that finding moments due to tension is tricky, then proceeds to say in his third tip that finding moments due to components of T is trivial because he can treat T as being applied to to the bottom of the rope. And yet you say, internal conservative forces always cancel. Then if so, why does he even need to treat T as being applied to the bottom of the rope?

Yes , internal conservative forces cancel . But I had mentioned that the rope exerts a force on both the wall , and the rod .

The force on the rod is internal , and that on the wall is external .

 

[toforfiltum]

The force on the rod is internal , and that on the rope is external .

You meant the wall right, instead of the rope? Because, if so , it makes more sense to me.

 

[Qwertywerty]

You meant the wall right, instead of the rope? Because, if so , it makes more sense to me.

Ah yes , post edited .

 

[haruspex]

O isn't perpendicular to the lines it lies on ( OA and OB ) .

As I think you now realize, that was rather confusing. A point cannot be perpendicular to anything.

Toforfiltum,
The normal forces are perpendicular to their respective surfaces, but a force also has a line of action. These pass through points A and B. The point X (good to have a name for it) is where those two lines intersect.

 

[haruspex]

Ahh...this is so confusing. Firstly, I want to know if X is in fact imaginary,

Secondly, I think I'm confused with Haruspex's solution. In the first tip, he said that finding moments due to tension is tricky, then proceeds to say in his third tip that finding moments due to components of T is trivial because he can treat T as being applied to to the bottom of the rope. And yet you say, internal conservative forces always cancel. Then if so, why does he even need to treat T as being applied to the bottom of the rope?

As Qwertywerty notes, all points are in a sense imaginary. It is just a point in space that you choose to single out for a purpose. It makes no difference whether it is in one of the objects or in the air. What does matter is that it is a fixed point in the inertial frame, not a point fixed to an accelerating object.

I meant that finding the moment looks difficult, involving a little geometry, until you apply the two tricks I then described, which makes it trivial.
I understand that taking the point of application of T as being the bottom end of the rope is the most conceptually challenging. In general, you can take any assemblage of the bodies in the problem as being a system and consider the forces acting on it from outside. The laws of physics apply to that system just as much as to any individual body within it. As Qwertywerty wrote, the force the rope exerts on the rod is internal to the rod+rope system, whereas the force exerted on the rope at O by ground/wall is external to it.

 

[toforfiltum]

As Qwertywerty notes, all points are in a sense imaginary. It is just a point in space that you choose to single out for a purpose. It makes no difference whether it is in one of the objects or in the air. What does matter is that it is a fixed point in the inertial frame, not a point fixed to an accelerating object.

I meant that finding the moment looks difficult, involving a little geometry, until you apply the two tricks I then described, which makes it trivial.
I understand that taking the point of application of T as being the bottom end of the rope is the most conceptually challenging. In general, you can take any assemblage of the bodies in the problem as being a system and consider the forces acting on it from outside. The laws of physics apply to that system just as much as to any individual body within it. As Qwertywerty wrote, the force the rope exerts on the rod is internal to the rod+rope system, whereas the force exerted on the rope at O by ground/wall is external to it.

To try and understand your solution, I tried to use your proposed method of solution. However, while resolving the components of T into its horizontal and vertical components, I do in fact, as you say, encounter difficulty in finding their perpendicular distances from X. So , you said that by treating the T as being applied to the bottom of the rope, finding moments due to T will be trivial. To be honest, I don't get what you say, and I have been trying hard but coming up short. If finding moments due to T is trivial, what other moments are there to balance moments from W? Can you please just show your working, so I can maybe understand what you say more clearly?

And also, one last question about the internal external forces. You said that the force the rope exerts on the rod is internal because you are treating the rod and the rope as a system. So the internal conservative forces are the force that the rod exerts on the rope and the force that the rope exerts on the rod, in opposite directions, right?
Now, if I were to treat the rope and wall as a system instead, the internal force will be the force of the rope on the wall, and for the external force, the rope on the rod, is it not?

 

[Qwertywerty]

So the internal conservative forces are the force that the rod exerts on the rope and the force that the rope exerts on the rod, in opposite directions, right?

Yes , following Newton's third law .

Now, if I were to treat the rope and wall as a system instead, the internal force will be the force of the rope on the wall, and for the external force, the rope on the rod, is it not?

Yes , these would be respectively , the internal and external force that the rope would be exerting .

 

[toforfiltum]

Yes , following Newton's third law .
Yes , these would be respectively , the internal and external force the rope would be exerting .

Thanks Qwertywerty, for all your explanations in this thread. Really appreciate it.

 

[Qwertywerty]

Haruspex , I was wondering - the fact that balancing torques about point X gives us a single equation is all well and good - but does it necessarily give you a faster solution than say , about point O ?

 

[SammyS]

For this problem, I like finding the torque about point O. Using point X is also handy. In this problem, the unknown distances OC and BC make any method using them fairly complicated, whereas the normal forces (especially N_A) are not too difficult to work with.

If the student understands how to use the line of action of a force, then using point X does give a nice result. But using the "line of action" also makes point O and even point B good points for evaluating the torque.

SammyS

 

[andrewkirk]

If one uses torques about O, T does not come into the equation because it does not provide any torque about O. So one would need to use additional equations to solve for T.

 

[SammyS]

If one uses torques about O, T does not come into the equation because it does not provide any torque about O. So one would need to use additional equations to solve for T.

Sure, but T comes back in by considering the forces equation(s).

 

[Qwertywerty]

Sure, but T comes back in by considering the forces equation(s).

Yes , I think he has mentioned that you would need additional equations .

Haruspex , I was wondering - the fact that balancing torques about point X gives us a single equation is all well and good - but does it necessarily give you a faster solution than say , about point O ?

What I was trying to say here was that although you would get more equations , time would not be much of a factor , as the three equations required to be solved while using O could be , easily and fast .

 

[haruspex]

What I was trying to say here was that although you would get more equations , time would not be much of a factor , as the three equations required to be solved while using O could be , easily and fast .

If the other two tricks do not come immediately to mind, I agree there would be no time saved in the problem in this thread. However, it is a useful option to look for in general.