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How to determine the direction on tension

I'm doing M1 atm and I come across so many tension questions everywhere everyday, but sometimes I determine the direction of tension wrong, is there like a thought process you go through to determine which way tension is pointing? Thanks this is one my key issues with the dynamics chapter

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Original post by Robbie242
I'm doing M1 atm and I come across so many tension questions everywhere everyday, but sometimes I determine the direction of tension wrong, is there like a thought process you go through to determine which way tension is pointing? Thanks this is one my key issues with the dynamics chapter


Tension is the "stretching" force (i.e it creates tension), so is always in the direction that would stretch a string or something.

You must have heard of the saying "you could cut the tension with a knife" :biggrin:
(edited 11 years ago)
Reply 2
Original post by Indeterminate
Tension is the "stretching" force (i.e it creates tension), so is always in the direction that would stretch a string or something.
So say on this question Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Inititally they are at rest on a rough hortizontal plane with the string taut.

First of all would there be two tensions?
and secondly would the tension from p be -> inbetween particles P and Q whilst for Q it would also be -> I don't know, I'm really bad at this
Original post by Robbie242
So say on this question Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Inititally they are at rest on a rough hortizontal plane with the string taut.

First of all would there be two tensions?
and secondly would the tension from p be -> inbetween particles P and Q whilst for Q it would also be -> I don't know, I'm really bad at this



There are two tension forces.

Hint:

To make a string taut ("stiff") in your hands, you'd pull it out in both directions as far as possible, right?
Reply 4
Original post by Indeterminate
There are two tension forces.

Hint:

To make a string taut ("stiff") in your hands, you'd pull it out in both directions as far as possible, right?
Oh that really helps, so would it be say Particle P-<------>---Q is what your saying right?
if it is, I hope I don't get in trouble for doing that in the exam haha i.e moving my hands apart in mid-air :P
(edited 11 years ago)
Original post by Robbie242
Oh that really helps, so would it be say Particle P-<------>---Q is what your saying right?


they act in the direction that's inwards (i.e towards the centre of the string) :smile:
(edited 11 years ago)
Reply 6
Original post by Indeterminate
Yes, they act in the direction that's outwards from the centre of the string :smile:
are there special cases where the concept of tension is slightly different?, i.e. what if its string holding an object below the pulley, would 1 tension force be acting upwards, as the weight would be stretching it downwards?
(edited 11 years ago)
Reply 7
Original post by Robbie242
are there special cases where the concept of tension is slightly different?, i.e. what if its string holding an object below the pulley, would 1 tension force be acting upwards, as the weight would be stretching it downwards?


The pulley pulls on the particle as you assume that the pulley is fixed, that's how i like to think of it makes sense in a way
Reply 8
Original post by Indeterminate
You would still have 2 tensions, but resolving (in equilibrium) would give

Tmg=0T=mgT - mg = 0 \Rightarrow T=mg

so they would cancel each other out.

Note also that, if a string is slack ("loose"), then there is no tension.

What is the special name given when the arrows are facing out instead of in? I can't remember the name
Original post by Robbie242
are there special cases where the concept of tension is slightly different?, i.e. what if its string holding an object below the pulley, would 1 tension force be acting upwards, as the weight would be stretching it downwards?


resolving (in equilibrium) would give



so they would cancel each other out.

Note also that, if a string is slack ("loose"), then there is no tension.
(edited 11 years ago)
Reply 10
and would it be correct saying one is acting up and the other one down? by saying they cancel out do you mean the same effect, so you count it as one, because surely mg cannot equal 0 since a block must have some sort of degree of mass

Original post by Indeterminate
You would still have 2 tensions, but resolving (in equilibrium) would give



so they would cancel each other out.

Note also that, if a string is slack ("loose"), then there is no tension.


Cheers btw
(edited 11 years ago)
Reply 11
Original post by Robbie242
and would it be correct saying one is acting up and the other one down? by saying they cancel out do you mean the same effect, so you count it as one, because surely mg cannot equal 0 since a block must have some sort of degree of mass



Cheers btw

The resultant force is zero, T is the same force as the weight (mg) acting in opposite direction
(edited 11 years ago)
Original post by Robbie242
and would it be correct saying one is acting up and the other one down? by saying they cancel out do you mean the same effect, so you count it as one, because surely mg cannot equal 0 since a block must have some sort of degree of mass



Cheers btw


Sorry, I should've been more clear. You needn't think of it as two tensions. Instead, think of it as T cancelling out mg (in equilibrium)
Reply 13
Original post by Indeterminate
Sorry, I should've been more clear. You needn't think of it as two tensions. Instead, think of it as T cancelling out mg (in equilibrium)
Cheers, yeah so when an objects hanging down and I resolve upwards, tension will be positive and downward forces negative, ty
Original post by Indeterminate
Yes, they act in the direction that's outwards from the centre of the string :smile:


I think this is all wrong!

Tension acts along the string, but in M1 we don't consider the string! We only consider the free body diagram of the particles and not the string, i.e. the forces that act on the particle and not the string. In this context the tension will be the force by which the string pulls the particles (when it is taut) and that will be along the line of the string. Therefore P and Q both will both experience a tension force exerted by the string that tries to draw both towards each other.
If a taut string is involved with a pulley or beads etc the tension force along it will be the same in magnitude if the pulley etc is smooth.

As tension is the magnitude of a force, it is measured in newtons (or sometimes pounds-force) and is always measured parallel to the string on which it applies

Tension forces can be modeled using ideal strings which are massless, frictionless, unbreakable, and unstretchable. They can be combined with ideal pulleys which allow ideal strings to switch physical direction. Ideal strings transmit tension forces instantaneously in action-reaction pairs so that if two objects are connected by an ideal string, any force directed along the string by the first object is accompanied by a force directed along the string in the opposite direction by the second object.
(edited 11 years ago)
Original post by StUdEnTIGCSE
I think this is all wrong!

Tension acts along the string, but in M1 we don't consider the string! We only consider the free body diagram of the particles and not the string, i.e. the forces that act on the particle and not the string. In this context the tension will be the force by which the string pulls the particles (when it is taut) and that will be along the line of the string. Therefore P and Q both will both experience a tension force exerted by the string that tries to draw both towards each other.
If a taut string is involved with a pulley or beads etc the tension force along it will be the same in magnitude if the pulley etc is smooth.


No, the tension would only pull in in towards the centre of the string if the two particles are on an inclined plane. This stops the particle that's lowest from falling away.

On a horizontal surface, the particles "hold on" to the string to keep it taut and stop it from "relaxing", so tension pulls away from the centre.
Original post by Indeterminate


On a horizontal surface, the particles "hold on" to the string to keep it taut and stop it from "relaxing", so tension pulls away from the center.


Yes but we don't consider the string! We consider the particles.
We only consider the Newton Third Law pair of that, the tension force exerted by the string on the particles!

eg; Particle P pulls the string away from its center of mass with a tension force of x magnitude
So the string pulls P towards its center with a tension force of x magnitude.
(edited 11 years ago)
Original post by StUdEnTIGCSE
Yes but we don't consider the string! We consider the particles.
We only consider the Newton Third Law pair of that, the tension force exerted by the string on the particles!

eg; Particle P pulls the string away from its center of mass with a tension force of x magnitude
So the string pulls P towards its center with a tension force of x magnitude.


Nonsense.

By the way, a light string doesn't have any mass!
Reply 18
Original post by Indeterminate
Nonsense.

By the way, a light string doesn't have any mass!
Glad we got that cleared up lol, I was starting to get confused
Original post by Indeterminate
Nonsense.

By the way, a light string doesn't have any mass!


Original post by Robbie242
Glad we got that cleared up lol, I was starting to get confused


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