Two bodies hanging from pulley

[jbriggs444]

Why do you say that?

Because it is true. The combined mass is not a compact body. It does not have a well-defined position. Which leaves the first and second derivatives of position likewise undefined.

 

[rudransh verma]

Because it is true. The combined mass is not a compact body. It does not have a well-defined position. Which leaves the first and second derivatives of position likewise undefined.

Ok. Then yes center of mass

 

[jbriggs444]

Ok. Then yes center of mass

OK. Suppose that you have this $a_\text{cm}$ defined.

Can you write an equation for $a_1$ and $a_2$ in terms of $a_\text{cm}$? And do you really want to go down this road to solve the problem?

 

[rudransh verma]

OK. Suppose that you have this $a_\text{cm}$ defined.

Can you write an equation for $a_1$ and $a_2$ in terms of $a_\text{cm}$? And do you really want to go down this road to solve the problem?

No! I don’t. But I think if I get the resultant force and combined mass then I can find the acceleration a of mass 1 , mass2, center of mass, etc .All will be same.

 

[jbriggs444]

No! I don’t. But I think if I get the resultant force and combined mass then I can find the acceleration a of mass 1 , mass2, center of mass, etc .All will be same.

I believe that the problem can be attacked in this manner. It is not a wise approach. But it is possible.

You will have to account for the force from the pulley. This is currently an unknown that does not appear in your equations. More unknowns. More equations. More stuff to solve. Not ideal.

We can write one equation down. $\sum F = ma$ $$-gm_1 + -gm_2 + F_\text{pr} = (m_1+m_2)a_\text{cm}$$Here $F_\text{pr}$ denotes the net force of the pulley on the rope.

 

[Steve4Physics]

No! I don’t. But I think if I get the resultant force and combined mass then I can find the acceleration a of mass 1 , mass2, center of mass, etc .All will be same.

For information, I get:
$a_1 = +2m/s^2$
$a_2 = -2m/s^2$
$a_{cm} = -0.4m/s^2$
Three different values!

(If you wamt to find $a_{cm}$ for yourself, here’s a hint: the pulley is not accelerating so the upwards force on it equals 2T where T is the tension. The total downwards force on the system is $m_1g + m_2g$.)

 

[rudransh verma]

For information, I get:
a1=+2m/s2
a2=−2m/s2
acm=−0.4m/s2
Three different values!

I am currently on topic laws of motion. So let’s not introduce center of anything.

 

[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.

To be clear since the tension is same and opposite(even though it doesn't look in the diagram) the net force is simply the difference of the weights and we can say the system will move in the direction of heavier weight.
But this is not always true as in the case of incline with two masses one hanging from the pulley and other kept on the incline.
https://www.physicsforums.com/threads/two-blocks-a-pulley-and-an-inclined-plane.1011992/

 

[Steve4Physics]

To be clear since the tension is same and opposite(even though it doesn't look in the diagram) the net force is simply the difference of the weights and we can say the system will move in the direction of heavier weight.
But this is not always true as in the case of incline with two masses one hanging from the pulley and other kept on the incline.
https://www.physicsforums.com/threads/two-blocks-a-pulley-and-an-inclined-plane.1011992/

'Method 3' (in Post #13) is for solving Atwood Machine problems with both masses hanging vertically.

However if one or both masses are on an incline, Method 3 can easily be modified.

If $m_1$ and $m_2$ are on inclines of $θ_1$ and $θ_2$ (to the horizontal) respectively, we must consider the downhill component of each weight: $m_1gsinθ_1$ and $m_2gsinθ_2$.

For example if $m_1gsinθ_1 > m_2gsinθ_2$ then $m_1$ accelerates downhill.

(To get the direction, you don’t actually need to include 'g' as it cancels; you can just compare $m_1sin θ_1$ and $m_2sinθ_2$.)

The net accelerating force is $(m_1sin θ_1 – m_2sinθ_2)g$.

Note:
If $θ_1=θ_2=90º$ we have the Atwood machine.
If $θ_1=0$ and $θ_2=90º$ we have $m_1$ on a horizontal surface and $m_2$ hanging vertically.

 

[rudransh verma]

'Method 3' (in Post #13) is for solving Atwood Machine problems with both masses hanging vertically.

I am not talking about this method. I am asking you to verify what I said.

 

[jbriggs444]

I am not talking about this method. I am asking you to verify what I said.

So you are asking for a verification of of the correctness of:

To be clear since the tension is same and opposite(even though it doesn't look in the diagram) the net force is simply the difference of the weights and we can say the system will move in the direction of heavier weight.

Yes, correct for a single pulley from which two weights are suspended vertically.

Strictly speaking, the system will "accelerate" in this direction and will "move" in that direction if it begins at rest.

And you want a verification of:

But this is not always true as in the case of incline with two masses one hanging from the pulley and other kept on the incline.

Yes. If one or both of the masses slides down a slope (with or without friction), things are more complicated. The heavier weight may not always accelerate down-slope. Or, if friction is high enough, may not move at all.

 

[rudransh verma]

Yes, correct for a single pulley from which two weights are suspended vertically

Thanks

The net accelerating force is (m1sinθ1–m2sinθ2)g.

yes! I got you

 

[rudransh verma]

@jbriggs444 I have another question. We know acceleration due to gravity is the acceleration which every body follows no matter what the mass. Both falls at same rate and touch the ground at same time. So why now its changed? Why heavier mass or the incline angle changes this phenomenon? ( and let's assume the friction on incline is zero)

 

[jbriggs444]

@jbriggs444 I have another question. We know acceleration due to gravity is the acceleration which every body follows no matter what the mass. Both falls at same rate and touch the ground at same time. So why now its changed? Why heavier mass or the incline angle changes this phenomenon? ( and let's assume the friction on incline is zero)

Newton's second law.

$\sum F = ma$ If you add up all the forces and divide by the mass, you can solve for the acceleration.

If the only force is gravity then $\sum F = mg$ and $\sum F = ma$ and $a = g$.

If there are other forces involved such as from strings and inclined planes then acceleration can vary. That is why a mass sitting on a table does not fall through the table at 9.8 meters per second per second.

 

[rudransh verma]

If there are other forces involved such as from strings and inclined planes than acceleration can vary. That is why a mass sitting on a table does not fall through the table at 9.8 meters per second per second.

I felt something else. Sometimes you know the answer but even then you overthink and ask silly question.