Two bodies hanging from pulley

[rudransh verma]

Homework Statement

Two bodies of masses m1=40 kg and m2=60 kg are attached to the end of the string of negligible mass and suspended from massless pulley. The acceleration of the bodies(g=10) is-

Relevant Equations

$F=ma$

$(T- m1g)-(T-m2g)= ( m)a$. I don’t know what m?

 

[BvU]

what m

what moves ?

 

[rudransh verma]

what moves ?

m1 and m2

 

[BvU]

And that is what has to be accelerated !

$\$

 

[Chestermiller]

Homework Statement:: Two bodies of masses m1=40 kg and m2=60 kg are attached to the end of the string of negligible mass and suspended from massless pulley. The acceleration of the bodies(g=10) is-
Relevant Equations:: $F=ma$

$(T- m1g)-(T-m2g)= ( m)a$. I don’t know what m?

Please explain your rationale for obtaining this force balance. What is your force balance on each of the masses individually?

 

[BvU]

There is even a name for this setup: it's called an Atwood machine

Be careful with the signs
Ps $\LaTeX$ tip :
try $$(T- m_1g)-(T-m_2g)= ma$$
instead of (T- m1g)-(T-m2g)= ( m)a
to get $$(T- m_1g)-(T-m_2g)= ma$$
$\$

 

[rudransh verma]

Please explain your rationale for obtaining this force balance.

$(T-m_1g)$ because T is upward force and mg is downward force. Direction of the resultant force will be downward. Now $T-m_2g$. Same reason. Resultant will be that the T > m2g , so upward movement. And final resultant will be difference of these two resultants . So that’s how the required above eqn.

 

[Chestermiller]

$(T-m_1g)$ because T is upward force and mg is downward force. Direction of the resultant force will be downward. Now $T-m_2g$. Same reason. Resultant will be that the T > m2g , so upward movement. And final resultant will be difference of these two resultants . So that’s how the required above eqn.

I get $$m_2g-T=m_2a$$where a is the downward acceleration of m2. And I get $$T-m_1g=m_1a$$ where a is the upward acceleration of m1. Because of the difference in masses, m1 is accelerating upward and m2 is accelerating downward; but the magnitudes of their accelerations are the same.

 

[rudransh verma]

get m2g−T=m2awhere a is the downward acceleration of m2.

How can you take + m2g and -T. It should be other way around. T is the upward force and gravity is downward force.
Your first eqn should be $T-m_2g=m_2(-a)$
Second should be $T-m_1g=m_1(a)$
But it looks the same.
I have other query. Here I am putting the sign before a in first eqn. But in OP I am not. Is it because we know that the individual acceleration direction is known so I can put. But in OP eqn we don’t know the acceleration a of combined bodies?

 

[Chestermiller]

How can you take + m2g and -T. It should be other way around. T is the upward force and gravity is downward force.
Your first eqn should be $T-m_2g=m_2(-a)$
Second should be $T-m_1g=m_1(a)$
But it looks the same.
I have other query. Here I am putting the sign before a in first eqn. But in OP I am not. Is it because we know that the individual acceleration direction is known so I can put. But in OP eqn we don’t know the acceleration a of combined bodies?

All you need to know is that the length of rope is constant, and this means that the accelerations of the two masses are equal in magnitude and opposite in direction.

 

[rudransh verma]

@Steve4Physics Can you resolve my doubt in post#9 . I face this too often. Where should I put -ve sign. Is it always to a known quantity or can it be semi known. Things like that.

 

[Chestermiller]

@Steve4Physics Can you resolve my doubt in post#9 . I face this too often. Where should I put -ve sign. Is it always to a known quantity or can it be semi known. Things like that.

You can never go wrong if you include unit vectors in all the forces in you force balance equations.

 

[Steve4Physics]

@Steve4Physics Can you resolve my doubt in post#9 . I face this too often. Where should I put -ve sign. Is it always to a known quantity or can it be semi known. Things like that.

I remember being initially confused by signs in Atwood machine calculations at school many (and I mean many) years ago. But then I sussed it out. Not sure if this will help or not but...

Here are 3 approaches you can use for problems like this. They all boil down to the same thing of course.

Method 1. Consistently use ‘upwards is positive’.
For the left mass: $T – m_1g = m_1a_1$
For the right mass: $T – m_2g = m_2a_2$
and then remember
$a_2=-a_1$ (as already noted by @Chestermiller)

Method 2. Take the direction of acceleration as locally positive (so each mass has the same positive acceleration).
For the rising left mass: $T – m_1g = m_1a$ (upwards is positive)
For the descending right mass: $m_2g – T = m_2a$ (downwards is positive)

Method 3 (for acceleration).
Realise the magnitude of acceleration is the same as that of a total mass of $M = m_1+m_2$ accelerated by a force, F, which is the difference in weights ($F = (m_2 – m_1)g$). Then use $F=Ma$. That gives the acceleration very simply; then you can easily find tension if required.

Method 1 is the most general/rigorous and preferred for more complex problems. But Methods 2 and 3 are useful if you understand them well enough to use them appropriately.

[Minor edits only]

 

[rudransh verma]

Method 3 (for acceleration).
Realise the magnitude of acceleration is the same as that of a total mass of M=m1+m2 accelerated by a force, F, which is the difference in weights (F=(m2–m1)g). Then use F=Ma. That gives the acceleration very simply; then you can easily find tension if required.

I took all the forces and added them up by superposition principle. That will be my net force which will act on my system of bodies. Now what is a? That can be found out. Right?
Or taking the approach of @Chestermiller
Your first eqn should be $T−m_2g=m_2(−a)$
Second should be $T−m_1g=m_1(a)$
Both has same magnitude a but opposite signs. According to you these should be local accelerations.

 

[Chestermiller]

Let $L_1$ be the length of the portion of the rope supporting m1 and $L_2$ be the length of the portion of the rope supporting m2. Since the rope is of constant length we have: $$L_1+L_2=C$$where C is constant. Differentiating this equation twice with respect to time, we have $$\frac{d^2L_1}{dt^2}+\frac{d^2L_2}{dt^2}=0. The downward acceleration of m1 is $\frac{d^2L_1}{dt^2}$ and the downward acceleration of m2 is $\frac{d^2L_2}{dt^2}$ .

 

[Steve4Physics]

I took all the forces and added them up by superposition principle. That will be my net force which will act on my system of bodies. Now what is a?

If you add all the forces vectorially, you will get the resultant force.

Using ‘upwards is positive’, resultant force on $m_1$ and $m_2$ is:
$F = (T – m_1g) + (T – m_2g)$

Total mass of $m_1$ and $m_2$ is $M = m_1 +m_2$.

Applying $F=Ma$ gives

$a = \frac F M = \frac {(T – m_1g) + (T – m_2g)}{ m_1 +m_2}$

where $a$ is the acceleration of the system’s centre of mass. That’s not the acceleration you are interested in for this problem,

 

[kuruman]

I remember being initially confused by signs in Atwood machine calculations at school many (and I mean many) years ago. But then I sussed it out. Not sure if this will help or not but...

Here are 3 approaches you can use for problems like this. They all boil down to the same thing of course.

Method 1. Consistently use ‘upwards is positive’.
For the left mass: $T – m_1g = m_1a_1$
For the right mass: $T – m_2g = m_2a_2$
and then remember
$a_2=-a_1$ (as already noted by @Chestermiller)

Method 2. Take the direction of acceleration as locally positive (so each mass has the same positive acceleration).
For the rising left mass: $T – m_1g = m_1a$ (upwards is positive)
For the descending right mass: $m_2g – T = m_2a$ (downwards is positive)

Method 3 (for acceleration).
Realise the magnitude of acceleration is the same as that of a total mass of $M = m_1+m_2$ accelerated by a force, F, which is the difference in weights ($F = (m_2 – m_1)g$). Then use $F=Ma$. That gives the acceleration very simply; then you can easily find tension if required.

Method 1 is the most general/rigorous and preferred for more complex problems. But Methods 2 and 3 are useful if you understand them well enough to use them appropriately.

[Minor edits only]

I concur that method 1 is the most general/rigorous. For this particular problem, however, there is method 4 which I add for completeness.

Since the ideal pulley changes the direction of the tension but not its magnitude, one can rotate the right string together with $m_2$ and the force of gravity counterclockwise by 90°, do the same but clockwise for $m_1$ and end up with the equivalent situation of two horizontal masses connected with a string. $F_1=m_1g$ is pulling to the left and $F_2=m_2g$ is pulling to the right. What is the net force? What is the acceleration of the system of two masses?

 

[Steve4Physics]

Or taking the approach of @Chestermiller
Your first eqn should be $T−m_2g=m_2(−a)$
Second should be $T−m_1g=m_1(a)$
Both has same magnitude a but opposite signs. According to you these should be local accelerations.

No. Check again. @Chestermiller's approach and my (Post 13) Method 1 are the same. They both use the upwards-is-positive sign convention, not a local convention.

 

[rudransh verma]

They both use the upwards-is-positive sign convention, not a local convention.

I too have used the upward as +ve convention.

 

[BvU]

I too have used the upward as +ve convention.

the forces and added them up

So why the minus sign ?

Thanks @Chestermiller, and @Steve4Physics and @kuruman -- I should have jumped on the 'be careful with the signs' and not spent any time on the $\TeX$ !

$\$

 

[Steve4Physics]

I too have used the upward as +ve convention.

Are you referring to your equation in Post #1:
$(T- m_1g)-(T-m_2g)= (m)a$?

This does not correctly use the ‘upwards-is-positive’ convention and also the right-hand side is wrong. If you replace $'m'$ in your right hand side by $'m_1+m_2'$ your equation becomes:
$(T- m_1g)-(T-m_2g)= (m_1 + m_2)a$
which simplifies to
$m_2g - m_1g= (m_1 + m_2)a$
This is the same as using Method 3 in Post #13. [Edit - or equivalently, @kuruman's Method 4 in Post #17.]

To correctly use the ‘upwards-is-positive convention you need to:
a) use two separate equations, one for each mass;
b) use two different accelerations, $a_1$ and $a_2$ where $a_2=-a_1$. If you prefer you can use symbols $a$ and $(-a)$ instead of $a_1$ and $a_2$.

 

[kuruman]

I too have used the upward as +ve convention.

You did, but you messed up the net force. Consider what happens when you use unit vectors. The net force on each mass is
$\vec F_{net,1}=T~\hat y-m_1g~\hat y$
$\vec F_{net,2}=T~\hat y-m_2g~\hat y.$

You did that right, but you did not pay attention to the other side of each equation which is $m_1\vec a_1$ for the first equation and $m_2\vec a_2$ for the second.

Since the rope does not stretch or shrink, $\vec a_1=-\vec a_2=\vec a$. Newton's 2nd law for each mass separately gives
$T~\hat y-m_1g~\hat y=m_1\vec a$
$T~\hat y-m_2g~\hat y=-m_2\vec a$.

Then, since everything is one dimension, the unit vectors and vector signs can be dropped to get
$T-m_1g=m_1a$
$T-m_2g=-m_2a$
and solve a system of two equations and two unknowns. Introducing unit vectors, sometimes helps sorting out negative signs in one dimensional cases.

Addendum on edit:
Note that in the last two equations $T$ and $g$ are magnitudes of vectors while $a$ is the vertical component of the acceleration of mass $m_1$. Having chosen "up" as positive, it could be positive or negative depending on whether $m_1$ is, respectively, less than or greater than $m_2$.

 

[haruspex]

because the combined mass is going downwards

One is going up, one down. How are you defining combined mass here? Answering that should tell you how to interpret the sign of the acceleration.

 

[rudransh verma]

@Steve4Physics I got you. One approach is to take individual mass. So $T-{m_1}g={m_1}{a_1}$ and $T-{m_2}g=-{m_2}{a_1}$.
Other one is to take combined masses. Net force is downwards $(T-{m_1}g)-(T-{m_2}g)= ({m_1}+{m_2})a$.
All accelerations ${a_1}=-{a_2}=-a$.

I have one last doubt. When finding something It’s very often the case that we know the direction of that. Like here we know what a’s direction should be , downwards. So should we include that in our eqns or not?

 

[haruspex]

When finding something It’s very often the case that we know the direction of that. Like here we know what a’s direction should be , downwards. So should we include that in our eqns or not?

You should not have to do anything special to include it in the equations. The equations are the same whether you can predict that or not (except perhaps when a nonlinear force is involved, like friction). But certainly use it as a sanity check of the answer.

 

[rudransh verma]

You should not have to do anything special to include it in the equations. The equations are the same whether you can predict that or not (except perhaps when a nonlinear force is involved, like friction). But certainly use it as a sanity check of the answer.

downwards (T−m1g)−(T−m2g)=(m1+m2)a.

So can I write -a in this eqn instead of a?
If I take $a$ I get $2m/s^2$. But if I take $-a$ I get $a=-2m/s^2$
Which one is more correct. Obviously second one!

 

[haruspex]

$(T-{m_1}g)-(T-{m_2}g)= ({m_1}+{m_2})a$.

Not sure why you write T- in both terms on the left. You have a force $m_1g$ pulling down on the left and a force $m_2g$ pulling down on the right. So $m_2g-m_1g$ represents the net "m2 down, m1 up" force. It follows that the equation defines a as positive if $m_2$ descends and negative if it rises.

 

[Steve4Physics]

So can I write -a in this eqn instead of a?
If I take $a$ I get $2m/s^2$. But if I take $-a$ I get $a=-2m/s^2$
Which one is more correct. Obviously second one!

Take a step back.

The original question (Post #1) asks for the “acceleration of the bodies”.
But the bodies have different accelerations. Using ‘upwards is positive’:
$m_1$’s acceleration is +2m/s²
$m_2$’s acceleration is -2m/s²

The accelerations have different values. Neither value is ‘more correct’ than the other. $m_2$ is no more important than $m_1$.

The question is (I assume) asking for the magnitude of the accelerations - since that is the same for both masses.

 

[rudransh verma]

The accelerations have different values. Neither value is ‘more correct’ than the other. m2 is no more important than m1.

You took individual masses and so accelerations will be opposite. I am taking combined mass and net downward force. Obviously the acceleration will be downwards. Now I am simply asking while finding this acceleration should I include -ve sign or not in the final eqn and then solve ?
I am assuming here we need to find $vector a$.

 

[Steve4Physics]

You took individual masses and so accelerations will be opposite. I am taking combined mass and net downward force. Obviously the acceleration will be downwards.

There is a net downwards force on the combined masses. This means acceleration of the centre of mass of the combined masses will be downwards - but this is of no interest in this particular question.

It looks like you are specifically thinking about the acceleration of $m_2$ and the string on the right side of the pulley. I agree this acceleration is downwards (-2m/s²).

 

[rudransh verma]

This means acceleration of the centre of mass of the combined masses will be downwards

It looks like you are specifically thinking about the acceleration of m2 and the string on the right side of the pulley

Both are same.

 

[jbriggs444]

Obviously the acceleration will be downwards.

The acceleration of what, exactly?

 

[rudransh verma]

The acceleration of what, exactly?

Of combined mass

 

[jbriggs444]

Of combined mass

The combined mass does not have a position or a velocity or an acceleration. Perhaps you mean the center of mass.

With this acceleration defined, one could indeed write down an equation for the individual accelerations of the individual masses $a_1$ and $a_2$ in terms of this $a_\text{cm}$

 

[rudransh verma]

The combined mass does not have a position or a velocity or an acceleration.

Why do you say that?