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Mechanics 2. Center of masses of bodies. A levels

So, I am studying mechanics for my A levels and came across an interesting exercise, which got me thinking, is it possible for a body to be in equilibrium if the vector sum of all forces is not equal to 0? The book doesn't seem to mention anything like that.
No, it's not possible.
Reply 2
I attached a screenshot of the exercise. How come the exercise says that the body is in equilibrium even tho the horizontal sum of all the vectors are not 0? What am I missing?
Original post by DFranklin
No, it's not possible.
Original post by MrCarving
I attached a screenshot of the exercise. How come the exercise says that the body is in equilibrium even tho the horizontal sum of all the vectors are not 0? What am I missing?

There may be something else in the question, but just from the diagram, the body would not be in equilibrium as the resolved T2 would move the body to the right as there is no force to stop/balance it.
(edited 1 year ago)
Reply 4
Thank you for you reply. If you don't mind, here is the full question. Do you have ideas why it is in equilibrium?
Can only presume its wrong. Do you have the solutions / which book is it from?
(edited 1 year ago)
Reply 6
Original post by MrCarving
Thank you for you reply. If you don't mind, here is the full question. Do you have ideas why it is in equilibrium?

I don't see anything wrong with the question? In the original model you aren't told it's in equilibrium, just that it's hanging in that position at that particular moment.
Frictional forces could also be large enough to resist motion provided by the horizontal component of T2, this isn't mentioned anywhere in the question but could be an assumption made.

In part b it moves to equilibrium but that offending string has snapped so isn't a problem any more.
(edited 1 year ago)
Original post by mqb2766
Can only presume its wrong. Do you have the solutions / which book is it from?

The question is from Edexcel (2017 Spec) FM2 Ex 2G. The model working for part (a) ignores any consideration of equilibrium, dealing instead in terms of locating the CoM then calculating moments about two points on the framework.
My suggestion to the OP would be to locate the CoM and then move on to part (b), ignoring the nonsense about tensions in part (a).
Agreed. Id found the question and it looks like a cut and paste from the preivous question where they had two vertical strings and it makes sense. Id guess (was 1/2 way through working) they equated vertical and 1 moment to get the two "solutions" 7W/18 and 11W/9, but its nonsense (horizontally).
Original post by Skiwi
I don't see anything wrong with the question? In the original model you aren't told it's in equilibrium, just that it's hanging in that position at that particular moment.
Frictional forces could also be large enough to resist motion provided by the horizontal component of T2, this isn't mentioned anywhere in the question but could be an assumption made.

In part b it moves to equilibrium but that offending string has snapped so isn't a problem any more.


There is no force acting towards "the plane producing friction" and even if there was, friction would also act vertically so the two tensions would not sum (resolved) to W (which they do), so there is no friction. The question and the previous one seem to assume equilibrium in the first part, even though its not explicitly stated (obviously not good).
Reply 10
Original post by mqb2766
Can only presume its wrong. Do you have the solutions / which book is it from?

It is from Edexcel Mechanics 2. (not physics, my bad)
(edited 1 year ago)
Reply 11
Original post by old_engineer
The question is from Edexcel (2017 Spec) FM2 Ex 2G. The model working for part (a) ignores any consideration of equilibrium, dealing instead in terms of locating the CoM then calculating moments about two points on the framework.
My suggestion to the OP would be to locate the CoM and then move on to part (b), ignoring the nonsense about tensions in part (a).


Wait a second. If the body is not in equilibrium then I can't say that T1 plus the vertical component of T2 equals to the weight of the object and therefore I can't find T2. I can find T1 by taking moments around D (assuming i found the center of mass), but how do i find T2 after that without using Newton's second law?
Ive got it in Further Maths/Mech 2, Pearson, but the examples and other questions in the section (called "Frameworks in Equilibrium") are generally based on vertical strings so I can only assume they've simply messed up with ;setting T2 to be diagonal and have not thought about resolving horizontally, so its wrong.
(edited 1 year ago)
Reply 13
Original post by mqb2766
There is no force acting towards "the plane producing friction" and even if there was, friction would also act vertically so the two tensions would not sum (resolved) to W (which they do), so there is no friction. The question and the previous one seem to assume equilibrium in the first part, even though its not explicitly stated (obviously not good).

Looking at the mark scheme for the question it does seem they have just disregarded horizontal component completely.

A lot of questions in the fm2 textbook are almost copy and paste from previous ones. The dynamics and kinematics chapters are especially bad for this and led to a couple of questions with bizzare scenarios that didn't really make sense. I found it was best to just solve it how I was going to have to in the exam and to try to ignore anything else.


I don't really understand why you say friction must be upwards, if you tried to move the framework through a horizontal force wouldn't you also have a horizontal frictional force due to air resistance acting against it? ( I haven't worked through the question so that may be why I'm missing something)
Original post by Skiwi
Looking at the mark scheme for the question it does seem they have just disregarded horizontal component completely.

A lot of questions in the fm2 textbook are almost copy and paste from previous ones. The dynamics and kinematics chapters are especially bad for this and led to a couple of questions with bizzare scenarios that didn't really make sense. I found it was best to just solve it how I was going to have to in the exam and to try to ignore anything else.


I don't really understand why you say friction must be upwards, if you tried to move the framework through a horizontal force wouldn't you also have a horizontal frictional force due to air resistance acting against it? ( I haven't worked through the question so that may be why I'm missing something)

Im assuming that youre saying there could be a vertical plane "behind" the framework which is producing friction? If so, there would also have to be a force acting into it so there would be friction produced between the body and the plane and there is no reason to assume the resulting friction would only be horizontal, rather a component would likely act upwards (against weight) and the tension in the strings reduced. But Ive not crunched the numbers to work it through (neither am I going to).

But there is no plane and no force acting into it so there is no friction, diagonal, horizontal or whatever.
(edited 1 year ago)
Reply 15
Original post by mqb2766
Im assuming that youre saying there could be a vertical plane "behind" the framework which is producing friction? If so, there would also have to be a force acting into it so there would be friction produced between the body and the plane and there is no reason to assume the resulting friction would only be horizontal, rather a component would likely act upwards (against weight) and the tension in the strings reduced. But Ive not crunched the numbers to work it through (neither am I going to).

But there is no plane and no force acting into it so there is no friction, diagonal, horizontal or whatever.

One of the modelling assumptions for a rod is that it has no thickness, completely forgot about that. So you're correct that there would be no friction.

Thank you
Original post by MrCarving
Wait a second. If the body is not in equilibrium then I can't say that T1 plus the vertical component of T2 equals to the weight of the object and therefore I can't find T2. I can find T1 by taking moments around D (assuming i found the center of mass), but how do i find T2 after that without using Newton's second law?

That's why my suggestion was to ignore the nonsense about tensions and move on to part (b).

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