Can the positive direction be different for each particle in a system?

[nav888]

Homework Statement

Two particles of mass 3kg and 5kg are connected via a light inextensible string which slides over a massless frictionless pulley. Calculate the tension in the rope the instant the system is released from rest and the instantaneous acceleration of each particle.

Relevant Equations

F=ma

I'm not really struggling with the question but the coordinate systems involved more so. So due to the modelling assumptions we know that the tension will be equal throughout the rope so we can use f = ma on each particle respectively and solve the resulting equation (as acceleration will be equal for both particles). If we take the upwards direction as positive for both particles we get the following:
$T - 3g = 3a$ and $T - 5g = -5a$
Which is perfectly fine. However my textbook does something else which is quite weird, in the textbook they take the positive direction to be different for each particle, they take the positive direction to be the direction of each particles acceleration. So they get;
$T - 3g = 3a$ and $5g - T = 5a$
Now I've noticed that regardless of which way you define positive for each particle, the equations generated are always correct.
Here comes my question.
1. Is it perfectly fine to define the positive direction different for each part of a system?
2. If not, why does it work in this situation?
Thanks for your help and time

 

[PeroK]

1. Is it perfectly fine to define the positive direction different for each part of a system?

Yes, as long as you are clear about what you are doing. In this case, you could take $a$ to be the magnitude of the acceleration. I would have a diagram with $a$ and an upwards arrow on the lesser mass, and an $a$ and a downward arrow on the larger mass.

PS and I'd have the same with $T$ being a magnitude and an arrow in each case indicating the direction of the tension force.

 

[topsquark]

Here comes my question.
1. Is it perfectly fine to define the positive direction different for each part of a system?
2. If not, why does it work in this situation?
Thanks for your help and time

What PeroK said. This is why I always told my students to indicate the + directions on a FBD. (Any diagram, really.) It helps to inform the reader (and the problem solver!) which components should be labelled as positive. The Physics does not care which directions you choose as positive.

Generally I take the positive direction to be the direction of acceleration. (If I don't know that ahead of time, I'll guess, but then I take the acceleration of the second body to be in the corresponding direction. If I'm wrong I just get a negative sign.) Here's why.

If we choose + upward on the 5 kg mass, when we write Newton's 2nd Law do we write
$\sum F = T - w = ma$

or
$\sum F = T - w = -ma$?

Clearly it has to be the second equation because the acceleration of the 5 kg mass is downward. Vectorally we still have $\sum \textbf{F} = m \textbf{a}$. But in components we've just said that $\sum F = -ma$! It can easily get confusing.

-Dan

 

[kuruman]

Note that in your equations you have
$T - 3g = 3a$ and $T - 5g = -5a$

The book's equations are
$T - 3g = 3a$ and $5g - T = 5a$

If you multiply both sides of your second equation by $-1$, which you can always do and not change anything, you will get the book's set of equations.

 

[Lnewqban]

Here comes my question.
1. Is it perfectly fine to define the positive direction different for each part of a system?
2. If not, why does it work in this situation?
Thanks for your help and time

The way I visualize these problems involving simple pulleys:
Two opposite horizontal forces (mg’s) acting at the ends of a horizontal string, and only one acceleration (if any).

The only function of the pulley in this problem is to change the direction in which the string tension is transferred from one weight to another, only because the gravitational field inducing those forces is pulling from a unique direction.

 

[nav888]

Yes, as long as you are clear about what you are doing. In this case, you could take $a$ to be the magnitude of the acceleration. I would have a diagram with $a$ and an upwards arrow on the lesser mass, and an $a$ and a downward arrow on the larger mass.

PS and I'd have the same with $T$ being a magnitude and an arrow in each case indicating the direction of the tension force.

Thanks. I just find it easier to define the positive direction differently for each mass as the maths is easier

 

[kuruman]

The way I visualize these problems involving simple pulleys:
Two opposite horizontal forces (mg’s) acting at the ends of a horizontal string, and only one acceleration (if any).

The only function of the pulley in this problem is to change the direction in which the string tension is transferred from one weight to another, only because the gravitational field inducing those forces is pulling from a unique direction.

View attachment 317601

Yes, and then you can "straighten" the string and have the one-dimensional, two-mass FBD shown below. You can immediately write Newton's second $$m_2g-m_1g=(m_2+m_1)a$$ and solve for $a$. This is equivalent to adding the two equations in order to eliminate the tension.

 

[nav888]

What PeroK said. This is why I always told my students to indicate the + directions on a FBD. (Any diagram, really.) It helps to inform the reader (and the problem solver!) which components should be labelled as positive. The Physics does not care which directions you choose as positive.

Generally I take the positive direction to be the direction of acceleration. (If I don't know that ahead of time, I'll guess, but then I take the acceleration of the second body to be in the corresponding direction. If I'm wrong I just get a negative sign.) Here's why.

If we choose + upward on the 5 kg mass, when we write Newton's 2nd Law do we write
$\sum F = T - w = ma$

or
$\sum F = T - w = -ma$?

Clearly it has to be the second equation because the acceleration of the 5 kg mass is downward. Vectorally we still have $\sum \textbf{F} = m \textbf{a}$. But in components we've just said that $\sum F = -ma$! It can easily get confusing.

-Dan

Could you define the positive direction different for each part of a system no matter how complicated the system? This is just a theoretical question I know I'm not crazy enough to do that

 

[PeroK]

Thanks. I just find it easier to define the positive direction differently for each mass as the maths is easier

Didn't you say the opposite in your original post? That you chose up as positive for both particles and your textbook did something "weird" by choosing up as positive of one mass and down as positive for the other?

If we take the upwards direction as positive for both particles we get the following:
$T - 3g = 3a$ and $T - 5g = -5a$
Which is perfectly fine.

However my textbook does something else which is quite weird, in the textbook they take the positive direction to be different for each particle, they take the positive direction to be the direction of each particles acceleration.

 

[Lnewqban]

... Imaginarily removing the gravitational field, I forgot to mention.
Exactly as shown in @kuruman ’s excellent diagram of post #7.

I even use this method for more complicated pulley systems, also visualizing each pulley as an instantaneous lever.

 

[nav888]

Didn't you say the opposite in your original post? That you chose up as positive for both particles and your textbook did something "weird" by choosing up as positive of one mass and down as positive for the other?

Well now I prefer the other method because the maths is simpler yes, purely because I know that it's a valid method. I only thought it was weird because I wasn't sure if it was correct physics or not

 

[nav888]

Is it purely by convention then that we tend to stick to one positive direction for the duration of the problem? (During introductory physics courses)

 

[kuruman]

Could you define the positive direction different for each part of a system no matter how complicated the system? This is just a theoretical question I know I'm not crazy enough to do that

You can define the positive direction any way you please. However, you must make sure that the signs in front of each quantity are consistent with your choice. Say you choose the direction of the acceleration of each mass as positive. Then you write

T-m₁g = m₁a
T-m₂g = m₂a

Certainly the tension and the weights have opposite signs in each equation which is as it should be. However, the equations are inconsistent. One mass is accelerating up and the other is accelerating down. This means that in one equation (it doesn't matter which) the right-hand side term must have the same sign as the weight in the left-hand side whilst in the other equation the corresponding signs must be opposite. This inconsistency implies that the connecting string is being stretched as the masses accelerate.

 

[nav888]

You can define the positive direction any way you please. However, you must make sure that the signs in front of each quantity are consistent with your choice. Say you choose the direction of the acceleration of each mass as positive. Then you write

T-m₁g = m₁a
T-m₂g = m₂a

Certainly the tension and the weights have opposite signs in each equation which is as it should be. However, the equations are inconsistent. One mass is accelerating up and the other is accelerating down. This means that in one equation (it doesn't matter which) the right-hand side term must have the same sign as the weight in the left-hand side whilst in the other equation the corresponding signs must be opposite. This inconsistency implies that the connecting string is being stretched as the masses accelerate.

Yup ok thanks for the clarification.

 

[Lnewqban]

Yup ok thanks for the clarification.

Welcome, @nav888 !

 

[Scottpm]

Homework Statement:: Two particles of mass 3kg and 5kg are connected via a light inextensible string which slides over a massless frictionless pulley. Calculate the tension in the rope the instant the system is released from rest and the instantaneous acceleration of each particle.
Relevant Equations:: F=ma

I'm not really struggling with the question but the coordinate systems involved more so. So due to the modelling assumptions we know that the tension will be equal throughout the rope so we can use f = ma on each particle respectively and solve the resulting equation (as acceleration will be equal for both particles). If we take the upwards direction as positive for both particles we get the following:
$T - 3g = 3a$ and $T - 5g = -5a$
Which is perfectly fine. However my textbook does something else which is quite weird, in the textbook they take the positive direction to be different for each particle, they take the positive direction to be the direction of each particles acceleration. So they get;
$T - 3g = 3a$ and $5g - T = 5a$
Now I've noticed that regardless of which way you define positive for each particle, the equations generated are always correct.
Here comes my question.
1. Is it perfectly fine to define the positive direction different for each part of a system?
2. If not, why does it work in this situation?
Thanks for your help and time

While others have answered this, I thought I'd throw my 2 cents in (or is it 25 cents now with inflation?).

It is perfectly fine to have different coordinate systems for two (or more parts) of a problem.

I like to think of it this way: I hate it when the two ends of the string (even massless) have different coordinate systems. So, if the positive direction for the 5kg mass is down that iss also the positive direction for the end of the string attached to this mass. Now, the other end of the string will go up, so I prefer that also be the positive direction. So, the for the 3kg mass positive is up.

I find that this usually makes the math easier. And that's how I choose the coordinate system(s).