# Mechanics A level maths

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#1
I had a couple of questions and was wondering if anyone could help me out with them xx
1. my teacher mentioned that in pulleys, the tension is in two directions on the string, which means that the string doesn't break. But I'm just casting my mind back to some physics scenarios where tension on the rope only acted in one direction? (and obvs the string didn't break) so is what she said wrong?
2. what does -ve time mean? i know we just ignore it, but im just wondering, for my own understanding, what does a negative time mean?
3. In Pulleys, for example, what is the significance of a light inextensible string?

thank you very much edit: i just watched a vid and it said that the significance of a light inextensible string si that the tension is the same on both sides, but why would that mean the tension is the same on both sides? if you imaging a pulley and a very heavy object on one side and a lighter one on the other side, the rope on the side of the heavy object is more likely to snap?
Last edited by username3477548; 1 year ago
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1 year ago
#2
(Original post by vix.xvi)
I had a couple of questions and was wondering if anyone could help me out with them xx
1. my teacher mentioned that in pulleys, the tension is in two directions on the string, which means that the string doesn't break. But I'm just casting my mind back to some physics scenarios where tension on the rope only acted in one direction? (and obvs the string didn't break) so is what she said wrong?
2. what does -ve time mean? i know we just ignore it, but im just wondering, for my own understanding, what does a negative time mean?
3. In Pulleys, for example, what is the significance of a light inextensible string?

thank you very much edit: i just watched a vid and it said that the significance of a light inextensible string si that the tension is the same on both sides, but why would that mean the tension is the same on both sides? if you imaging a pulley and a very heavy object on one side and a lighter one on the other side, the rope on the side of the heavy object is more likely to snap?
The last one is no. If the light mass is zero, there is no tension in the string. It's good to have intuition, but consider limiting cases and write down the equations.

Editing ...
3. Light inextensible, means precisely that. No mass to affect motion and no (Hooke-type) extension to consider.

2. Negative time is simply the time before the arbitrary time 0. I posted this at t=0. You posted at t=-15min.

1. Strings can break.
Last edited by mqb2766; 1 year ago
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1 year ago
#3
1. maybe it has something to do with equilibrium, or the starting position of an experiment.

2. negative time is like before you start timing an experiment, you might give the person managing the stopwatch a countdown, before you actually do the experiment, so that the experiment starts at the same time as the stopwatch.

3. a light inextensible string means that the string itself has no mass, so if there's a weight on the end of it, the string doesn't contribute to the mass, and inextensible means it cannot be stretched, so it's pulled taut.
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#4
(Original post by nm12345)
1. maybe it has something to do with equilibrium, or the starting position of an experiment.

2. negative time is like before you start timing an experiment, you might give the person managing the stopwatch a countdown, before you actually do the experiment, so that the experiment starts at the same time as the stopwatch.

3. a light inextensible string means that the string itself has no mass, so if there's a weight on the end of it, the string doesn't contribute to the mass, and inextensible means it cannot be stretched, so it's pulled taut.
(Original post by mqb2766)
The last one is no. If the light mass is zero, there is no tension in the string. It's good to have intuition, but consider limiting cases and write down the equations.

Editing ...
3. Light inextensible, means precisely that. No mass to affect motion and no (Hooke-type) extension to consider.

2. Negative time is simply the time before the arbitrary time 0. I posted this at t=0. You posted at t=-15min.

1. Strings can break.
i see, thanks!
For 2. if i calculate the time taken for an object to fall and get two values of time, one being -ve, what does the -ve one mean though?
also for 1., so tension has to be in two directions (= n opp) to ensure a string doesn't break?
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1 year ago
#5
(Original post by vix.xvi)
i see, thanks!
For 2. if i calculate the time taken for an object to fall and get two values of time, one being -ve, what does the -ve one mean though?
also for 1., so tension has to be in two directions (= n opp) to ensure a string doesn't break?
the negative value for time is just a result of the quadratic equation being solved, so it's just ignored
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1 year ago
#6
For 2) I guess you're solving a quadratic? Sketch it. There will be "a solution" running the equation backwards from the initial condition at t=0.

1. Can you upload a question/diagram.
Last edited by mqb2766; 1 year ago
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1 year ago
#7
(Original post by vix.xvi)
I had a couple of questions and was wondering if anyone could help me out with them xx
1. my teacher mentioned that in pulleys, the tension is in two directions on the string, which means that the string doesn't break. But I'm just casting my mind back to some physics scenarios where tension on the rope only acted in one direction? (and obvs the string didn't break) so is what she said wrong?

edit: i just watched a vid and it said that the significance of a light inextensible string si that the tension is the same on both sides, but why would that mean the tension is the same on both sides? if you imaging a pulley and a very heavy object on one side and a lighter one on the other side, the rope on the side of the heavy object is more likely to snap?
The key word here is really "light" (the tension would be the same even if it was extensible, as long as it was light). By F=ma, if m = 0 ("light" ), then the net force on the string must be 0. Note that this also applies to small subsections of the string - you can imagine it as a set of infinitesimally small "links" - the tension on both sides of the link has to be equal.

If the string isn't light, the tension doesn't have to be the same at different parts, and typically won't be. The classic example is a heavy rope hanging downwards - the top part of the rope is bearing the full weight of the rope, the bottom of the rope is bearing no tension.

A more "fun" example is to hold a slinky so it hangs vertically, and then let go of the top. The "bottom" of the slinky seems to float initially, as it remains held up by the differential tension in the spring.

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#8
(Original post by mqb2766)
For 2) I guess you're solving a quadratic? Sketch it. There will be "a solution" running the equation backwards from the initial condition at t=0.

1. Can you upload a question/diagram.
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1 year ago
#9
(Original post by vix.xvi)
The tension in the string is really irrelevant. It's how it acts on the masses.
So the second diagram is fine as it shows tension opposing weight for the masses on each side. The string is light, it means the same tension is applied either side of the pulley.
Last edited by mqb2766; 1 year ago
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1 year ago
#10
(Original post by vix.xvi)
Again: if the tension at the two ends of the string aren't equal, the net force will cause the string to have infinite acceleration, which is impossible (certainly in this case, as we know the things it's attached to won't have infinite acceleration).

(If you consider what is happens to infinitesimal parts of the string, you're going to find they want to all move at different accelerations, which means the string would fall apart, but I think that's unnecessarily complicating things here).
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#11
(Original post by DFranklin)
The key word here is really "light" (the tension would be the same even if it was extensible, as long as it was light). By F=ma, if m = 0 ("light" ), then the net force on the string must be 0. Note that this also applies to small subsections of the string - you can imagine it as a set of infinitesimally small "links" - the tension on both sides of the link has to be equal.

If the string isn't light, the tension doesn't have to be the same at different parts, and typically won't be. The classic example is a heavy rope hanging downwards - the top part of the rope is bearing the full weight of the rope, the bottom of the rope is bearing no tension.

A more "fun" example is to hold a slinky so it hangs vertically, and then let go of the top. The "bottom" of the slinky seems to float initially, as it remains held up by the differential tension in the spring.

thank you very much the vid was very interesting

sorry last question, so the tension in the string is the same everywhere (for a light inextensible string) because in effect the tension force is spread all over the string? sorry im just getting confused alot aha
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#12
(Original post by DFranklin)
Again: if the tension at the two ends of the string aren't equal, the net force will cause the string to have infinite acceleration, which is impossible (certainly in this case, as we know the things it's attached to won't have infinite acceleration).

(If you consider what is happens to infinitesimal parts of the string, you're going to find they want to all move at different accelerations, which means the string would fall apart, but I think that's unnecessarily complicating things here).
(Original post by mqb2766)
The tension in the string is really irrelevant. It's how it acts on the masses.
So the second diagram is fine as it shows tension opposing weight for the masses on each side. The string is light, it means the same tension is applied either side of the pulley.
ok thank you for all your help!
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1 year ago
#13
(Original post by vix.xvi)
thank you very much the vid was very interesting

sorry last question, so the tension in the string is the same everywhere (for a light inextensible string) because in effect the tension force is spread all over the string? sorry im just getting confused alot aha
It's not really about it being spread over the string - the tension in the slinky is spread over the string as well, but it's not the same everywhere in that case. (A similar argument shows that the tension in the slinky has to be *continuous* (differentiable, even) - at any point in the slinky there are two equal+opposite tensions pulling in opposite directions, but the size of those tensions can still vary over time).

The problem is that to analyse this properly, you really need to divide the string into infinitesimal paces and do a bunch of post-A-level calculus).

But I think in this case you don't really need to do that.

You can see that for a piece of (light, inextensible) string, you can see the net force must be zero, so the tension at the two ends must be equal and opposite. But this must be true for any point P in the string, because you can simply imagine the string as being two shorter strings joined at P and do the same analysis.
Last edited by DFranklin; 1 year ago
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#14
(Original post by DFranklin)
It's not really about it being spread over the string - the tension in the slinky is spread over the string as well, but it's not the same everywhere in that case. (A similar argument shows that the tension in the slinky has to be *continuous* (differentiable, even) - at any point in the slinky there are two equal+opposite tensions pulling in opposite directions, but the size of those tensions can still vary over time).

The problem is that to analyse this properly, you really need to divide the string into infinitesimal paces and do a bunch of post-A-level calculus).

But I think in this case you don't really need to do that.

You can see that for a piece of (light, inextensible) string, you can see the net force must be zero, so the tension at the two ends must be equal and opposite. But this must be true for any point P in the string, because you can simply imagine the string as being two shorter strings joined at P and do the same analysis.
thank you very much!!
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