- #36
toforfiltum
- 341
- 4
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 said: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 said:O isn't perpendicular to the lines it lies on ( OA and OB ) .
toforfiltum said:By the way, though I understand andrewkirk's explanation, I wonder if moment can be calculated around an imaginary point X?
toforfiltum said: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.
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 said: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 .
toforfiltum said: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.
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.Qwertywerty said:Aren't all points thus imaginary ? He takes it as X simply to refer to the place from where you should balance torques .
toforfiltum said: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.
toforfiltum said: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?
You meant the wall right, instead of the rope? Because, if so , it makes more sense to me.Qwertywerty said:The force on the rod is internal , and that on the rope is external .
toforfiltum said:You meant the wall right, instead of the rope? Because, if so , it makes more sense to me.
As I think you now realize, that was rather confusing. A point cannot be perpendicular to anything.Qwertywerty said:O isn't perpendicular to the lines it lies on ( OA and OB ) .
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.toforfiltum said: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?
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?haruspex said: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 said: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?
toforfiltum said: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?
Thanks Qwertywerty, for all your explanations in this thread. Really appreciate it.Qwertywerty said:Yes , following Newton's third law .
Yes , these would be respectively , the internal and external force the rope would be exerting .
Sure, but T comes back in by considering the forces equation(s).andrewkirk said: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 said:Sure, but T comes back in by considering the forces equation(s).
Qwertywerty said: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 ?
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.Qwertywerty said: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 .