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Oblogog
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#1
Why, if you are given a rod in equilibrium with forces acting on it, are you allowed to resolve vertically/ horizontally (forces up = forces down / forces left = forces right)? Can you think of some way to derive it? It doesn't make sense to me.
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ghostwalker
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#2
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#2
(Original post by Oblogog)
Why, if you are given a rod in equilibrium with forces acting on it, are you allowed to resolve vertically/ horizontally (forces up = forces down / forces left = forces right)? Can you think of some way to derive it? It doesn't make sense to me.
Why, if you are given a rod in equilibrium with forces acting on it, are you allowed to resolve vertically/ horizontally (forces up = forces down / forces left = forces right)? Can you think of some way to derive it? It doesn't make sense to me.
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Oblogog
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#3
(Original post by ghostwalker)
If something is in equilibrium, i.e. its acceleration is zero, then the vector sum of forces acting on it must be zero. If it's zero, then the sum of the forces in any direction is zero. So, you can resolve in any direction you like. Horizontal and vertical just tend to be the most useful.
If something is in equilibrium, i.e. its acceleration is zero, then the vector sum of forces acting on it must be zero. If it's zero, then the sum of the forces in any direction is zero. So, you can resolve in any direction you like. Horizontal and vertical just tend to be the most useful.
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JN17
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#4
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#4
(Original post by Oblogog)
But the forces are acting on different places on the rod? If you change where you are applying the force but keep the magnitude and direction constant then according to this nothing would change but in reality things would change wouldn't they?
But the forces are acting on different places on the rod? If you change where you are applying the force but keep the magnitude and direction constant then according to this nothing would change but in reality things would change wouldn't they?
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ghostwalker
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#5
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#5
(Original post by Oblogog)
But the forces are acting on different places on the rod? If you change where you are applying the force but keep the magnitude and direction constant then according to this nothing would change but in reality things would change wouldn't they?
But the forces are acting on different places on the rod? If you change where you are applying the force but keep the magnitude and direction constant then according to this nothing would change but in reality things would change wouldn't they?
Your question was IF it is in eqilibrium, why can you resolve.... And that is what I was attempting to answer.
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Oblogog
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#6
I just can't get my head around it. I can see why you can resolve forces on the same axis when the forces are acting at the same point but in the case of the rod they're all acting at different points?
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Zacken
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#7
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#7
(Original post by Oblogog)
I just can't get my head around it. I can see why you can resolve forces on the same axis when the forces are acting at the same point but in the case of the rod they're all acting at different points?
I just can't get my head around it. I can see why you can resolve forces on the same axis when the forces are acting at the same point but in the case of the rod they're all acting at different points?
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Oblogog
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#8
(Original post by Zacken)
According to you, what does being in equilibrium mean?
According to you, what does being in equilibrium mean?
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Zacken
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#9
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#9
(Original post by Oblogog)
The rod isn't moving
The rod isn't moving
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Oblogog
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#10
(Original post by Zacken)
And why do you think that? Why can't it be moving at a constant velocity with no acceleration? Have you heard of Newtons first two laws?
And why do you think that? Why can't it be moving at a constant velocity with no acceleration? Have you heard of Newtons first two laws?
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Student403
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#11
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#11
(Original post by Oblogog)
I was going to add (/constant linear velocity) but it seemed pointless to say
I was going to add (/constant linear velocity) but it seemed pointless to say
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Oblogog
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#12
(Original post by Student403)
You should have added it because they are two very different things
You should have added it because they are two very different things
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Student403
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#13
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#13
(Original post by Oblogog)
I thought if you move with it it will be at rest relative to you and the laws of physics are the same in all inertial reference frames. Maybe should have added it anyway.
I thought if you move with it it will be at rest relative to you and the laws of physics are the same in all inertial reference frames. Maybe should have added it anyway.
What about in the case you are a stationary observer? (As in most physics questions)
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Oblogog
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#14
(Original post by Student403)
Well yes relative to you IF you move with it
What about in the case you are a stationary observer? (As in most physics questions)
Well yes relative to you IF you move with it
What about in the case you are a stationary observer? (As in most physics questions)
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Student403
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#15
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#15
(Original post by Oblogog)
It's easier to think about the original problem I asked if you don't have to picture it moving away from you.
It's easier to think about the original problem I asked if you don't have to picture it moving away from you.
Equilibrium means the resultant force on the object is 0. Newton's second law tells us this will result in no acceleration. For an object moving at a velocity V, it will continue to move in that direction with velocity V. If it is at rest, it will stay at rest. (N1L). That's all there is to equilibrium
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Oblogog
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#16
(Original post by Student403)
I think you are overcomplicating this by talking about relativity.
Equilibrium means the resultant force on the object is 0. Newton's second law tells us this will result in no acceleration. For an object moving at a velocity V, it will continue to move in that direction with velocity V. If it is at rest, it will stay at rest. (N1L). That's all there is to equilibrium
I think you are overcomplicating this by talking about relativity.
Equilibrium means the resultant force on the object is 0. Newton's second law tells us this will result in no acceleration. For an object moving at a velocity V, it will continue to move in that direction with velocity V. If it is at rest, it will stay at rest. (N1L). That's all there is to equilibrium
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samb1234
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#17
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#17
(Original post by Oblogog)
Why, if you are given a rod in equilibrium with forces acting on it, are you allowed to resolve vertically/ horizontally (forces up = forces down / forces left = forces right)? Can you think of some way to derive it? It doesn't make sense to me.
Why, if you are given a rod in equilibrium with forces acting on it, are you allowed to resolve vertically/ horizontally (forces up = forces down / forces left = forces right)? Can you think of some way to derive it? It doesn't make sense to me.
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Oblogog
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#18
I don't think I've asked my question very clearly. If a rod is in equilibrium why can you set the sum of the forces in any direction equal to 0 even though they are acting at different points on the rod. I'm familiar with this happening if the forces are acting at the same point but it confuses me why you can do it in this case.
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Oblogog
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#19
I should have said that to start with. Someone answer my question please 
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atsruser
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#20
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#20
(Original post by Oblogog)
I don't think I've asked my question very clearly. If a rod is in equilibrium why can you set the sum of the forces in any direction equal to 0 even though they are acting at different points on the rod. I'm familiar with this happening if the forces are acting at the same point but it confuses me why you can do it in this case.
I don't think I've asked my question very clearly. If a rod is in equilibrium why can you set the sum of the forces in any direction equal to 0 even though they are acting at different points on the rod. I'm familiar with this happening if the forces are acting at the same point but it confuses me why you can do it in this case.
For example, consider a metal rod with a forces applied along the length of the rod, by pads at both ends. You can probably accept that the centre 1 cm of a 1 m rod feels the same accelerating force as the end 1 cms of the rod, even though the external force is about 0.5 m away. That's because each part of the rod pushes on the bit next to it.
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